446 free, high-quality past-paper-style questions on Electricity and magnetism for Cambridge IGCSE Physics, each with a full mark scheme. Covers: Electromagnetic induction, motors and transformers; Charge, current, potential difference, resistance and power; Electrical safety; and more. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,610 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IGPC-038-04-01Electromagnetic induction, motors and transformersPaper CEasy1 mark
A coil is connected to a sensitive ammeter. In which situation is an e.m.f. induced in the coil?
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An e.m.f. is induced only when the magnetic field linking the coil changes, which happens when there is relative motion between the magnet and the coil. Pulling the magnet out changes the field through the coil. A stationary magnet gives no change. Moving the magnet and coil together gives no relative motion. A wire held still produces no change in the field through the coil.
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Answer C
Answer: C
Common error: A magnet does not need to be strong or close for induction to occur. What matters is that the field through the coil is changing. A strong magnet held still induces nothing.
2IGPC-038-04-02Electromagnetic induction, motors and transformersPaper CEasy1 mark
A transformer has 200 turns on its primary coil and 1000 turns on its secondary coil. An alternating voltage of 12 V is connected to the primary coil. What is the output voltage?
Diagram not to scale
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Vs/Vp = Ns/Np, so Vs = 12 × 1000/200 = 60 V. The secondary has more turns, so this is a step-up transformer. 2.4 V comes from inverting the turns ratio. 12 V assumes a transformer does not change the voltage. 240 V comes from multiplying by the number of primary turns.
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Answer C
Answer: C
Exam tip: Check your answer makes sense. More turns on the secondary means a bigger output voltage.
3IGPC-038-04-06Electromagnetic induction, motors and transformersPaper CEasy1 mark
Which material is most suitable for the core of a transformer?
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Soft iron is easily magnetised and demagnetised. It therefore follows the rapidly changing magnetic field produced by the alternating current. It also links the field from the primary coil to the secondary coil. Steel is magnetic but keeps its magnetism, so it cannot follow the changes well. Copper and aluminium are not magnetic; they are chosen for the coils because they are good conductors.
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Answer B
Answer: B
Common error: Do not confuse the coil material (copper, a good conductor) with the core material (soft iron, easily magnetised).
4IGPC-038-04-08Electromagnetic induction, motors and transformersPaper CEasy1 mark
A coil carrying a current sits between the poles of a magnet, and it turns. Which change would not increase the turning effect on the coil?
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The turning effect increases with the current, the number of turns and the strength of the magnetic field. Reversing the connections reverses the current. This reverses the direction of the turning effect but does not change its size.
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Answer D
Answer: D
Common error: Reversing the current or the field changes the direction of rotation, not how strong the turning effect is.
5IGPC-038-04-10Electromagnetic induction, motors and transformersPaper CEasy1 mark
The primary coil of a transformer is connected to a 12 V battery. The secondary coil has twice as many turns as the primary coil. What is the steady output voltage?
The primary coil is connected to a battery. Diagram not to scale
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A battery supplies a steady direct current. This produces a magnetic field in the core that does not change, so no e.m.f. is induced in the secondary coil. The steady output is 0 V; there is only a brief pulse when the battery is first connected. 24 V would be correct only for an alternating supply. 6 V inverts the turns ratio. 12 V assumes the output equals the input.
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Answer A
Answer: A
Exam tip: Transformers need a changing magnetic field, so they work only with a.c. Check whether the supply is a.c. or d.c. before using the turns-ratio equation.
6IGPC-038-04-11Electromagnetic induction, motors and transformersPaper CEasy5 marks
A student connects a coil of wire to a centre-zero galvanometer. She then moves a bar magnet in and out of the coil.
(a)
State what happens to the galvanometer needle as she pushes the north pole of the magnet into the coil.
(1)
(b)
She stops the magnet so that it rests inside the coil. State the reading on the galvanometer.
(1)
(c)
She now pulls the magnet out of the coil. Describe how the movement of the needle differs from that in (a).
(1)
(d)
Suggest two changes she could make to obtain a larger deflection when the magnet is pushed in.
(2)
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Part (a)
The needle deflects / moves to one side
✓ 1
Part (b)
Zero / the needle returns to the centre
✓ 1
Part (c)
The needle deflects in the opposite direction
✓ 1
Part (d)
Any two of: move the magnet faster ; use a stronger magnet ; use a coil with more turns
✓ 3
Exam tip: In (d), give two clearly different changes. 'Move it faster' and 'move it more quickly' count as only one.
7IGPC-038-01-01Charge, current, potential difference, resistance and powerPaper CEasy1 mark
Which unit is the unit of electric charge?
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Charge is measured in coulombs (C). The ampere is the unit of current, which is a rate of flow of charge, not the charge itself. The joule is the unit of energy. The volt is the unit of potential difference, which is energy per unit charge.
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Answer B
Answer: B
Common error: Do not confuse the coulomb (amount of charge) with the ampere (charge flowing per second).
8IGPC-038-01-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A student wants to measure the current in a lamp and the potential difference across it. How should the ammeter and the voltmeter be connected?
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The current through the lamp must also pass through the ammeter, so the ammeter goes in series. Potential difference is measured between the two ends of the lamp, so the voltmeter goes in parallel. The other options either swap the meters or put both in the same arrangement.
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Answer C
Answer: C
Exam tip: Remember: ammeter in series, voltmeter in parallel, across the component being tested.
9IGPC-038-01-03Charge, current, potential difference, resistance and powerPaper CEasy1 mark
The potential difference across a resistor is 6.0 V and the current in it is 0.30 A. What is the resistance of the resistor?
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R = V / I = 6.0 / 0.30 = 20 Ω. The value 1.8 Ω comes from multiplying V by I, and 0.050 Ω from dividing I by V. 180 Ω is a power-of-ten slip.
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Answer C
Answer: C
Common error: Resistance is pd divided by current, not current divided by pd. Check your answer is reasonable: a small current at several volts means a fairly large resistance.
10IGPC-038-01-09Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A 2.0 kW heater is switched on for 3.0 hours. Electrical energy costs 25 cents per kW h. What is the cost of using the heater?
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Energy = power × time = 2.0 kW × 3.0 h = 6.0 kW h. Cost = 6.0 × 25 = 150 cents. 6 cents gives the energy value but not the cost. 50 cents uses only the power, and 75 cents uses only the time.
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Answer D
Answer: D
Exam tip: For kW h calculations keep the power in kW and the time in hours. Do not convert to W or s.
11IGPC-038-01-11Charge, current, potential difference, resistance and powerPaper CEasy5 marks
A student investigates a resistor connected to a variable power supply. The student measures the current in the resistor and the potential difference across it.
(a)
State the name of the meter used to measure the current, and state how it is connected in the circuit.
(2)
(b)
When the potential difference is 4.5 V, the current is 0.15 A. Calculate the resistance of the resistor.
(2)
(c)
The potential difference is increased to 9.0 V. The resistance stays constant. Calculate the new current.
(1)
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Part (a)
Ammeter . Connected in series with the resistor .
✓ 2
Part (b)
R = V / I . R = 4.5 / 0.15 = 30 Ω .
✓ 2
Part (c)
I = V / R = 9.0 / 30 = 0.30 A .
✓ 1
Exam tip: If the resistance is constant, doubling the pd doubles the current. Use this as a quick check on your answer.
12IGPC-038-01-12Charge, current, potential difference, resistance and powerPaper CEasy5 marks
A television has a power rating of 120 W. It is connected to a 230 V mains supply.
(a)
Calculate the current in the television.
(2)
(b)
The television is used for 5.0 hours each day for 30 days. Calculate the energy transferred, in kW h.
(2)
(c)
Electrical energy costs 20 cents per kW h. Calculate the cost of using the television for the 30 days.
(1)
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Part (a)
I = P / V = 120 / 230 . I = 0.52 A .
✓ 2
Part (b)
Total time = 5.0 × 30 = 150 h . Energy = 0.120 kW × 150 h = 18 kW h .
✓ 2
Part (c)
Cost = 18 × 20 = 360 cents ($3.60) .
✓ 1
Common error: Using 120 instead of 0.120 for the power gives an answer 1000 times too large. Convert watts to kilowatts before calculating energy in kW h.
13IGPC-038-03-01Electrical safetyPaper CEasy1 mark
In a mains plug, which wire must the fuse be connected in?
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The fuse goes in the live wire. The live wire is at a high voltage, so when the fuse melts the appliance is cut off from that voltage. A fuse in the neutral wire would break the circuit, but the appliance would still be connected to the live supply and could still give a shock. The earth wire normally carries no current, and a fuse joining live and neutral would melt as soon as the plug was switched on.
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Answer B
Answer: B
Exam tip: Fuses, switches and circuit breakers all go in the live wire. That way, when they open, the appliance is disconnected from the high voltage.
14IGPC-038-03-02Electrical safetyPaper CEasy1 mark
What is the purpose of a fuse in an electrical circuit?
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A fuse is a thin wire that heats up when current passes through it. If the current goes above the fuse rating, the wire melts and the circuit is broken. This stops the cables overheating and so prevents fires. A fuse does not change the voltage or store charge. Connecting the case to the ground is the job of the earth wire.
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Answer C
Answer: C
Common error: A fuse does not reduce the current to a safe value. It switches the circuit off completely by melting.
15IGPC-038-03-03Electrical safetyPaper CEasy1 mark
Which situation is most likely to make the cables in a household circuit overheat?
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Plugging several high-power appliances into one socket overloads it. The total current becomes very large, and a large current makes the cables heat up, which can melt the insulation or start a fire. Thick, low-resistance cables heat up less, not more. A correctly rated fuse or a circuit breaker protects against overheating rather than causing it.
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Answer B
Answer: B
Exam tip: The three standard hazards are damaged insulation, overheating cables (from too much current) and damp conditions. Be ready to name each one and give its cause.
16IGPC-038-03-08Electrical safetyPaper CEasy1 mark
Which of these is the greatest electrical hazard?
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Impure water conducts electricity, and wet skin has a much lower resistance than dry skin. If water reaches live parts, a current can pass through the person and give a dangerous shock. Each of the other options describes safe practice.
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Answer B
Answer: B
Exam tip: When you explain why damp conditions are dangerous, say that water conducts electricity, so current can pass through the person.
17IGPC-038-03-12Electrical safetyPaper CEasy5 marks
A table lamp is rated 60 W, 230 V. Its plug can be fitted with a 1 A, 3 A or 13 A fuse.
(a)
State the equation that links power, current and voltage.
(1)
(b)
Calculate the current in the lamp when it is working normally.
(2)
(c)
State which fuse should be fitted. Explain your choice.
(2)
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Part (a)
P = IV (power = current × voltage)
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Part (b)
I = P/V = 60/230 ; I = 0.26 A
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Part (c)
1 A fuse ; it is the smallest rating that is above the normal current, so it lets the lamp work but melts quickly if a fault causes a larger current
✓ 2
Common error: A 13 A fuse would let the lamp work, but it is a poor choice. A fault current could be far larger than the lamp's normal current without melting the fuse, so the flex could overheat.
18IGPC-038-00-01Magnets and magnetic fieldsPaper CEasy1 mark
Which metal is a magnetic material?
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Nickel is one of the magnetic materials (with iron, cobalt and steel). A magnet attracts it, and it can be magnetised. Aluminium, copper and brass are metals, but they are non-magnetic, so a magnet does not noticeably attract them.
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Answer C
Answer: C
Common error: Not all metals are magnetic. Learn the short list: iron, steel, nickel and cobalt.
19IGPC-038-00-02Magnets and magnetic fieldsPaper CEasy1 mark
The north pole of a bar magnet is brought close to one end of an unmagnetised iron rod. What happens to the rod?
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The magnet induces magnetism in the iron. The near end of the rod becomes a south pole. Unlike poles attract, so the rod is attracted. This is why a magnet attracts unmagnetised magnetic materials whichever pole is used.
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Answer B
Answer: B
Exam tip: Attraction can happen between a magnet and any magnetic material. Repulsion only happens between two magnets.
20IGPC-038-00-03Magnets and magnetic fieldsPaper CEasy1 mark
The direction of a magnetic field at a point is defined as
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By definition, the field direction is the direction of the force on a north pole placed at that point. So field lines outside a magnet run from N to S. Option C gives the reverse of this. Option A describes a south pole, which feels a force in the opposite direction.
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Answer B
Answer: B
Common error: Outside a magnet, field lines go from north to south, not south to north.
21IGPC-038-00-11Magnets and magnetic fieldsPaper CEasy5 marks
A strong bar magnet holds a chain of four small iron nails. The top nail touches the magnet and each of the others hangs from the one above.
(a)
State two materials, other than iron, that are attracted by a magnet.
(2)
(b)
Explain why each nail holds up the nail below it, even though the nails were not magnets to begin with.
(2)
(c)
The magnet is lifted away. The nails fall apart. State what this shows about the magnetism of iron.
(1)
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Part (a)
any two from: steel / nickel / cobalt
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Part (b)
the magnet induces magnetism in the nails / the nails become temporary magnets ; each nail has an induced pole that attracts the opposite (unlike) pole induced in the next nail
✓ 2
Part (c)
iron is easily demagnetised / its magnetism is temporary
✓ 1
Common error: Aluminium and copper are metals but are not magnetic, so they do not score here.
22IGPC-038-02-01Series and parallel circuitsPaper CEasy1 mark
Three resistors of resistance 2.0 Ω, 3.0 Ω and 5.0 Ω are connected in series. What is their combined resistance?
Diagram not to scale
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For resistors in series the combined resistance is the sum of the individual resistances: 2.0 + 3.0 + 5.0 = 10 Ω. 0.97 Ω is the result of using the parallel formula. 3.3 Ω is the average of the three values. 30 Ω comes from multiplying the resistances.
✓ 1
Answer C
Answer: C
Exam tip: A series combination always has a larger resistance than its largest single resistor. Use this to check your answer quickly.
23IGPC-038-02-02Series and parallel circuitsPaper CEasy1 mark
A cell, a switch, a lamp and a resistor are connected in series. An ammeter placed between the cell and the lamp reads 0.40 A. The ammeter is then moved to a point between the lamp and the resistor. What does it now read?
Diagram not to scale
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A series circuit has only one path, so the same charge flows past every point each second. The current is 0.40 A everywhere in the circuit. Components transfer energy, not charge, so current is not 'used up'. The current is also not shared out between components in series.
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Answer C
Answer: C
Common error: Components use up energy, not current. The current returning to the cell equals the current leaving it.
24IGPC-038-02-04Series and parallel circuitsPaper CEasy1 mark
A 4.0 Ω resistor and a 12 Ω resistor are connected in parallel. Which value could be their combined resistance?
Diagram not to scale
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The combined resistance of two resistors in parallel is less than either resistor alone. Only 3.0 Ω is less than 4.0 Ω. Using 1/R = 1/4.0 + 1/12 = 4/12 gives R = 3.0 Ω. 16 Ω is the series total. 8.0 Ω is the average. 12 Ω is just the larger resistor.
✓ 1
Answer A
Answer: A
Common error: Adding a resistor in parallel provides an extra path for charge, so the combined resistance goes down, not up.
25IGPC-038-02-11Series and parallel circuitsPaper CEasy7 marks
A 6.0 V battery, a lamp, an 8.0 Ω resistor and an ammeter are connected in series. The ammeter reads 0.50 A. The battery has negligible internal resistance.
Diagram not to scale
(a)
State the current in the 8.0 Ω resistor.
(1)
(b)
Calculate the potential difference across the resistor.
(2)
(c)
Calculate the potential difference across the lamp.
(2)
(d)
Calculate the total resistance of the circuit.
(2)
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Part (a)
0.50 A (the current is the same at all points in a series circuit)
✓ 1
Part (b)
V = IR = 0.50 × 8.0
✓ 1
= 4.0 V
✓ 1
Part (c)
The pds across series components add up to the battery e.m.f.: 6.0 − 4.0
✓ 1
= 2.0 V
✓ 1
Part (d)
R = V/I = 6.0/0.50
✓ 1
= 12 Ω
✓ 1
Exam tip: Always write the equation you are using before substituting. A correct method can earn credit even if the final arithmetic slips.
26IGPC-038-00-12Magnets and magnetic fieldsPaper CEasy5 marks
A student places a bar magnet on a sheet of paper. She wants to plot its magnetic field using a small plotting compass.
(a)
Describe how she uses the plotting compass to draw one field line.
(3)
(b)
State the direction of the field lines outside the magnet.
(1)
(c)
State where the magnetic field of the bar magnet is strongest.
(1)
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Part (a)
place the compass on the paper near one pole of the magnet and mark a dot at the position of the north (N) end of the needle (accept: mark dots at both ends of the needle) ; move the compass so that the S end of the needle is over the last dot, mark a new dot at the N end, and repeat many times ; join the dots with a smooth curve and add an arrow in the direction the N end of the needle points
✓ 3
Part (b)
from the north pole to the south pole
✓ 1
Part (c)
at / near the poles (ends) of the magnet
✓ 1
Exam tip: Field lines never cross. Outside the magnet they always carry an arrow pointing from N to S, and a field line drawn without an arrow is incomplete.
27IGPC-038-00-13Magnets and magnetic fieldsPaper CEasy6 marks
A crane in a scrapyard uses a large electromagnet to pick up old cars and drop them into a crusher.
(a)
State one advantage of using an electromagnet rather than a permanent magnet for this job.
(1)
(b)
Explain why the core of the electromagnet is made of soft iron and not steel.
(2)
(c)
Suggest one device that uses a permanent magnet.
(1)
(d)
The coil of the electromagnet has a resistance of 12 Ω and is connected to a 230 V supply. Calculate the current in the coil.
(2)
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Part (a)
it can be switched off to release the car / its strength can be changed by changing the current
✓ 1
Part (b)
soft iron is easily magnetised, so it makes a strong magnet when there is a current ; soft iron loses its magnetism when the current is switched off, so the car is released, whereas steel would stay magnetised
✓ 2
Part (c)
any suitable example, e.g. fridge-door seal / compass / loudspeaker / microphone
✓ 1
Part (d)
I = V / R = 230 / 12 ; I = 19 A (19.2 A)
✓ 2
Exam tip: When explaining a choice of core material, say what happens both when the current is on and when it is off.
28IGPC-038-02-12Series and parallel circuitsPaper CEasy7 marks
The ceiling lamps in a kitchen are connected in parallel to the 230 V mains supply. Each lamp has its own switch.
(a)
State the potential difference across each lamp when it is switched on.
(1)
(b)
Give two advantages of connecting the lamps in parallel rather than in series.
(2)
(c)
Another lamp is switched on. State and explain what happens to the current from the supply.
(2)
(d)
The current in one lamp is 0.26 A. Calculate the resistance of this lamp when it is lit.
(2)
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Part (a)
230 V (each branch is connected across the full supply)
✓ 1
Part (b)
First advantage, e.g. each lamp can be switched on and off independently
✓ 1
Second advantage, e.g. if one lamp fails, the others stay lit
✓ 1
Also accept for either mark: each lamp receives the full supply voltage, so it lights at full brightness.
—
Part (c)
The current from the supply increases
✓ 1
An extra branch / path is added, so the combined resistance decreases
✓ 1
Also accept for the second mark: the new lamp draws its own current, which adds to the currents in the other branches.
—
Part (d)
R = V/I = 230/0.26
✓ 1
= 885 Ω (accept 880 Ω)
✓ 1
Common error: Do not halve the supply voltage for lamps in parallel. The p.d. is shared only between components in series.
29IGPC-038-03-11Electrical safetyPaper CEasy4 marks
A metal desk lamp is connected to the mains by a three-core cable containing live, neutral and earth wires.
(a)
State which wire is connected to the metal case of the lamp.
(1)
(b)
The switch on the lamp is connected in the live wire. Explain why.
(2)
(c)
The insulation of the cable has become cracked, so that the wires inside can be seen. State one hazard this causes.
(1)
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Part (a)
the earth wire
✓ 1
Part (b)
When the switch is off, the lamp is disconnected from the live wire / the high voltage ; so the lamp cannot give a shock when it is switched off, for example while the bulb is being changed
✓ 2
Part (c)
Any one of: a person may touch the exposed live wire and get an electric shock; or the wires may touch each other, causing a short circuit that could start a fire
✓ 1
Exam tip: In an "explain" question about the switch position, say what happens when the switch is open: the appliance is no longer connected to the live voltage.
30IGPC3-4-04c-01Magnets and magnetic fieldsPaper CEasy1 mark
A crane in a scrap yard uses an electromagnet to pick up steel and then release it. Why is soft iron, rather than steel, used for the core?
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Soft iron magnetises and demagnetises easily, so the load is released when the current stops (C). A: steel is magnetic. B describes steel, which stays magnetised. D: the core does not carry the coil current, and conductivity is not the reason.
✓ 1
Answer C
Answer: C
Common error: Soft magnetic materials are chosen for electromagnets because they lose their magnetism; steel is chosen for permanent magnets.
31IGPC3-4-04c-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A phone charger supplies a steady current of 0.50 A for 2.0 minutes. How much charge flows through the phone in this time?
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Q = It = 0.50 × 120 s = 60 C (B). A uses 2.0 minutes without converting to seconds. C divides time by current (120 / 0.50). D divides current by time (0.50 / 120).
✓ 1
Answer B
Answer: B
Common error: Convert minutes to seconds before using Q = It.
32IGPC3-4-04c-08Magnets and magnetic fieldsPaper CEasy1 mark
A plastic rod becomes positively charged when it is rubbed with a dry cloth. What has happened?
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Only electrons transfer when objects are rubbed. The rod loses negative charge and becomes positive, so electrons moved from the rod to the cloth (D). A and C involve protons, which are fixed in nuclei. B would make the rod negative.
✓ 1
Answer D
Answer: D
Common error: Charging by rubbing moves electrons only; a positive object has lost electrons.
33IGPC3-4-04c-09Charge, current, potential difference, resistance and powerPaper CEasy6 marks
A room heater is labelled 2.3 kW, 230 V. It is switched on for 4.0 hours.
(a)
Calculate the current in the heater.
(2)
(b)
Calculate the energy transferred in kW h.
(2)
(c)
Electricity costs 15 cents per kW h. Calculate the cost in dollars.
(2)
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Part (a)
I = P / V = 2300 / 230
✓ 1
I = 10 A
✓ 1
Part (b)
E = P t = 2.3 × 4.0
✓ 1
E = 9.2 kW h
✓ 1
Part (c)
9.2 × 15 = 138 cents, ECF from (b)
✓ 1
cost = 1.38 dollars
✓ 1
Common error: Use power in kilowatts and time in hours to get energy in kW h, not joules.
34IGPC3-4-04c-10Magnets and magnetic fieldsPaper CEasy5 marks
A student investigates the magnetic field of a bar magnet and then builds an electromagnet.
(a)
State what is meant by a magnetic field.
(1)
(b)
Describe how a plotting compass can be used to find the shape of the field around the bar magnet.
(2)
(c)
State two ways to increase the strength of the electromagnet.
(2)
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Part (a)
a region where a magnet or magnetic material experiences a force
✓ 1
Part (b)
place the compass near the magnet and mark the positions of both ends of the needle
✓ 1
move the compass so that it starts at the last mark, repeat, then join the marks to draw a field line
✓ 1
Part (c)
increase the current
✓ 1
increase the number of turns of the coil (or add a soft iron core)
✓ 1
Exam tip: Field lines are drawn with arrows pointing from the north pole to the south pole outside the magnet.
35IGPC3-4-04c-11Series and parallel circuitsPaper CEasy6 marks
A 9.0 V battery, a 12 Ω fixed resistor and a thermistor are connected in series. The current in the circuit is 0.30 A.
Diagram not to scale
(a)
Calculate the potential difference across the fixed resistor.
(2)
(b)
Calculate the potential difference across the thermistor.
(1)
(c)
Calculate the resistance of the thermistor at this temperature.
(2)
(d)
The temperature of the thermistor rises and its resistance decreases. State what happens to the current in the circuit.
(1)
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Part (a)
V = I R = 0.30 × 12
✓ 1
V = 3.6 V
✓ 1
Part (b)
9.0 − 3.6 = 5.4 V, ECF from (a)
✓ 1
Part (c)
R = V / I = 5.4 / 0.30
✓ 1
R = 18 Ω
✓ 1
Part (d)
the current increases
✓ 1
Exam tip: In a series circuit the potential differences add up to the battery voltage, so subtract to find the missing one.
36IGPC3-4-06r6-01Magnets and magnetic fieldsPaper CEasy1 mark
An electric guitar pickup contains a magnet that magnetises the metal string above it. Which of these metals is suitable for the string?
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C is correct: iron is a magnetic material, so it is attracted to and magnetised by the pickup magnet. Aluminium, copper and zinc are metals but are not magnetic, so A, B and D are wrong.
✓ 1
Answer C
Answer: C
Common error: Not all metals are magnetic. Only iron, steel, cobalt and nickel are attracted by a magnet.
37IGPC3-4-06r6-02Series and parallel circuitsPaper CEasy1 mark
A string of three identical lamps is connected in series to a battery to light a greenhouse. An ammeter placed before the first lamp reads 0.30 A. What does an identical ammeter placed after the third lamp read?
Three identical lamps in series, with an ammeter before the first lamp and an identical ammeter after the third.
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B is correct: in a series circuit the current is the same at every point, because charge is not used up. A divides the current between the lamps as if they were in parallel. C adds the currents of the lamps. D assumes the lamps use up all the current.
✓ 1
Answer B
Answer: B
Common error: Lamps transfer energy, not charge. Current is not used up as it passes round a series circuit.
38IGPC3-4-06r6-11Series and parallel circuitsPaper CEasy6 marks
The lighting circuit in an aircraft cockpit contains a 10 Ω resistor and a 20 Ω resistor connected in series to a 12 V battery.
Diagram not to scale
(a)
Calculate the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
State what happens to the total resistance when a third resistor is added in series.
(1)
(d)
Calculate the potential difference across the 20 Ω resistor.
(2)
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Part (a)
10 + 20 = 30 Ω
✓ 1
Part (b)
I = V / R = 12 / 30 (ECF from (a))
✓ 1
0.40 A
✓ 1
Part (c)
the total resistance increases
✓ 1
Part (d)
V = IR = 0.40 × 20 (ECF from (b))
✓ 1
8.0 V
✓ 1
Exam tip: In a series circuit the same current passes through each resistor, so use V = IR with that current for each one.
39IGPC3-4-07r6-01Magnets and magnetic fieldsPaper CEasy1 mark
A small plotting compass is placed very close to the north pole of a bar magnet that is being used to test a rail-signal sensor. The magnet's field is much stronger than the Earth's field at this position. Which statement describes the compass needle?
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C is correct: the north end of a compass points in the direction of the field, and field lines leave a north pole. A is wrong because it assumes like poles attract (they repel). B is wrong because the magnet's field dominates here. D is wrong because a steady field gives a fixed needle direction.
✓ 1
Answer C
Answer: C
Common error: Field lines leave a north pole and enter a south pole; a compass north end follows the field direction, not the nearest pole.
40IGPC3-4-07r6-02Series and parallel circuitsPaper CEasy1 mark
Three identical lamps are connected in series with a battery on a test rig for a train carriage. The filament of one lamp breaks. What happens to the other two lamps?
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Notes
C is correct: a series circuit has one path, so a break stops the current in every component. A is wrong because no extra current can flow. B is wrong because that describes parallel lamps. D is wrong because there is no current at all.
✓ 1
Answer C
Answer: C
Common error: In a series circuit one break stops the current everywhere; in parallel only the broken branch is affected.
41IGPC3-4-07r6-09Electrical safetyPaper CEasy6 marks
A rowing machine in a gym is connected to the 230 V mains. Its motor is rated 460 W and the machine has a metal frame.
(a)
Explain why the metal frame is connected to the earth wire.
(2)
(b)
Calculate the current in the motor when it works normally.
(2)
(c)
Fuses rated 1 A, 3 A, 5 A and 13 A are available. State which fuse should be fitted and give a reason.
(2)
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Notes
Part (a)
If a fault makes the live wire touch the frame, a large current flows to earth through the earth wire
✓ 1
this large current melts the fuse and cuts off the supply, so the user does not get an electric shock
✓ 1
Part (b)
I = P / V
✓ 1
460 / 230 = 2.0 A
✓ 1
Part (c)
3 A
✓ 1
it is just above the normal current of 2.0 A, so it blows if the current rises only a little (ECF from (b))
✓ 1
Exam tip: The fuse and the switch go in the live wire so that a fault leaves nothing at a dangerous voltage.
42IGPC3-4-07r6-16Series and parallel circuitsPaper CEasy6 marks
The lamps in a train carriage are connected in parallel to a 24 V supply. Each lamp is rated 24 V, 12 W.
(a)
State two advantages of connecting the lamps in parallel rather than in series.
(2)
(b)
Calculate the current in one lamp when it works normally.
(2)
(c)
The carriage has 6 such lamps, all lit. Calculate the current from the supply.
(2)
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Part (a)
if one lamp fails the others stay lit
✓ 1
each lamp has the full supply pd so shines at normal brightness
✓ 1
Part (b)
I = P / V
✓ 1
12 / 24 = 0.50 A
✓ 1
Part (c)
branch currents add: 6 × 0.50
✓ 1
3.0 A (ECF from (b))
✓ 1
Exam tip: Write the formula first, then substitute; it secures the method mark even if the arithmetic slips.
43IGPC3-4-05r7-01Series and parallel circuitsPaper CEasy1 mark
In a hospital ward, two identical lamps are connected in parallel to a mains supply. One lamp fails because its filament breaks. What happens to the other lamp and to the total current drawn from the supply?
Two identical lamps in parallel.
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Notes
B is correct: each branch of a parallel circuit has the full supply p.d., so the working lamp is unchanged; with one branch removed the total current halves. A is wrong because the circuit is not broken in the other branch (that is series behaviour). C is wrong because the p.d. across the working lamp does not change, so it cannot get brighter. D is wrong because the working lamp's current is unchanged, so it does not dim.
✓ 1
Answer B
Answer: B
Common error: Do not apply series reasoning to a parallel circuit: a break in one branch leaves the other branches working normally.
44IGPC3-4-05r7-02Magnets and magnetic fieldsPaper CEasy1 mark
A hospital door is held shut by an electromagnet. Which material is the best choice for the core of the electromagnet, so that the door releases at once when the current is switched off?
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Notes
B is correct: soft iron is magnetised easily by the coil's field and loses almost all its magnetism when the current stops. A is wrong because steel is magnetically hard and would stay magnetised, so the door would not release. C is wrong because copper is not a magnetic material, so the core would not strengthen the field. D is wrong because plastic is not magnetic at all.
✓ 1
Answer B
Answer: B
Exam tip: For a switch-off electromagnet choose soft iron; for a permanent magnet choose steel.
45IGPC3-4-05r7-04Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A small heater inside a beehive carries a steady current of 0.50 A for 2.0 minutes. How much charge flows through the heater in this time?
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C is correct: Q = It = 0.50 × (2.0 × 60) = 60 C. A is wrong because it divides current by the time in minutes. B is wrong because the time was left in minutes (0.50 × 2.0). D is wrong because it uses the time in seconds (120) as the charge, ignoring the current.
✓ 1
Answer C
Answer: C
Common error: Convert minutes to seconds before using Q = It.
46IGPC3-4-05r7-09Charge, current, potential difference, resistance and powerPaper CEasy6 marks
A beekeeper uses a 12 V heater of resistance 6.0 Ω to keep a hive warm.
(a)
Calculate the current in the heater.
(2)
(b)
Calculate the power of the heater.
(2)
(c)
Calculate the energy transferred by the heater in 30 minutes. Give your answer in joules.
(2)
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Part (a)
I = V / R
✓ 1
I = 12 / 6.0 = 2.0 A
✓ 1
Part (b)
P = IV
✓ 1
P = 2.0 × 12 = 24 W (ECF from (a))
✓ 1
Part (c)
E = Pt with t = 30 × 60 = 1800 s
✓ 1
E = 24 × 1800 = 43200 J (ECF from (b))
✓ 1
Exam tip: Change minutes to seconds before using E = Pt.
47IGPC3-4-05r7-10Series and parallel circuitsPaper CEasy6 marks
In a Formula 1 garage, two heating elements for tyre warmers, of resistance 12 Ω and 6.0 Ω, are connected in parallel to a 12 V supply.
Diagram not to scale
(a)
State the potential difference across each element.
(1)
(b)
Calculate the current in each element.
(2)
(c)
State the current in the supply.
(1)
(d)
Calculate the combined resistance of the two elements.
(2)
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Notes
Part (a)
12 V across each element
✓ 1
Part (b)
12 Ω element: I = 12 / 12 = 1.0 A
✓ 1
6.0 Ω element: I = 12 / 6.0 = 2.0 A
✓ 1
Part (c)
3.0 A (ECF from (b))
✓ 1
Part (d)
R = V / I = 12 / 3.0 (ECF from (c))
✓ 1
R = 4.0 Ω
✓ 1
Common error: The combined resistance of resistors in parallel is always smaller than the smallest individual resistance.
48IGPC3-4-06r9-01Magnets and magnetic fieldsPaper CEasy1 mark
A mountain rescue team builds an electromagnet to lift steel equipment boxes and release them by switching off the current. Which material should be used for the core, and why?
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Correct: C. Soft iron magnetises strongly when current flows and demagnetises when it stops, so the boxes are released. A is wrong because steel keeps its magnetism and the boxes would not drop. B is wrong because copper is not magnetic, so conduction is irrelevant. D is wrong because aluminium is not magnetic.
✓ 1
Answer C
Answer: C
Common error: Soft iron is used for electromagnet cores and steel for permanent magnets; the key property is whether the material keeps its magnetism.
49IGPC3-4-06r9-05Electromagnetic induction, motors and transformersPaper CEasy1 mark
A mountaineer's wind-up torch contains a small generator in which a magnet spins next to a coil. Which change would increase the e.m.f. induced in the coil?
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Correct: D. A faster change in the magnetic field through the coil induces a larger e.m.f. A, B and C each reduce the rate of cutting of field or the number of turns cutting it, so the e.m.f. would be smaller.
✓ 1
Answer D
Answer: D
Common error: The induced e.m.f. increases with faster motion, a stronger magnet and more turns; reducing any of these lowers it.
50IGPC3-4-06r9-06Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A heating element in a food dryer passes a charge of 450 C in 3.0 minutes. What is the current in the element?
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Correct: C. Time = 3.0 × 60 = 180 s, and I = Q ÷ t = 450 ÷ 180 = 2.5 A. A divides by 3.0 without converting minutes to seconds. B inverts the formula (180 ÷ 450). D multiplies charge by time.
✓ 1
Answer C
Answer: C
Common error: Always convert minutes to seconds before using I = Q/t.
51IGPC3-4-06r9-09Magnets and magnetic fieldsPaper CEasy5 marks
A mountaineer carries a compass and a bar magnet.
(a)
The north pole of the bar magnet is brought close to the north pole of a second bar magnet. State what happens.
(1)
(b)
Describe how a plotting compass can be used to draw one magnetic field line of the bar magnet.
(3)
(c)
State one way to demagnetise a steel needle.
(1)
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Notes
Part (a)
the magnets repel
✓ 1
Part (b)
place the compass near the north pole and mark the positions of both ends of the needle
✓ 1
move the compass so that one end of the needle is at the previous mark, mark the other end, and repeat
✓ 1
join the marks and add an arrow pointing from the north pole to the south pole
✓ 1
Part (c)
heat it strongly / hammer it / place it in a coil carrying alternating current and slowly withdraw it
✓ 1
Exam tip: Field lines always point away from a north pole and towards a south pole, and they never cross.
52IGPC3-4-06r9-10Charge, current, potential difference, resistance and powerPaper CEasy6 marks
An incubator in a food-science laboratory has a heating element that takes a current of 5.0 A from the 230 V mains supply. It runs for 2.0 hours. Electricity costs £0.30 per kWh.
(a)
Calculate the power of the heating element.
(2)
(b)
Calculate the energy transferred in 2.0 hours, in kWh.
(2)
(c)
Calculate the cost of running the heater for 2.0 hours.
(2)
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Part (a)
power = voltage × current
✓ 1
power = 230 × 5.0 = 1150 W
✓ 1
Part (b)
power in kW = 1.15 kW
✓ 1
energy = 1.15 × 2.0 = 2.3 kWh
✓ 1
Part (c)
cost = energy × price = 2.3 × 0.30
✓ 1
£0.69
✓ 1
Common error: For kWh, the power must be in kilowatts and the time in hours; do not convert the time to seconds.
53IGPC3-4-07r10-01Electrical safetyPaper CEasy1 mark
A metal-cased electric kettle in a mountain hut is connected to the mains by a three-pin plug. What is the main purpose of the earth wire?
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Notes
B is correct: if the live wire touches the case, a large current flows through the earth wire, the fuse melts and the case is no longer live. A is wrong because normal current flows in the live and neutral wires only. C is wrong because the earth wire has no effect on the power used. D is wrong because overheating of the live wire is the job of the fuse, not the earth wire.
✓ 1
Answer B
Answer: B
Common error: The earth wire carries no current in normal use; it only matters when there is a fault.
54IGPC3-4-07r10-06Series and parallel circuitsPaper CEasy1 mark
Two identical LEDs in a headtorch are connected in parallel to a battery. One LED fails and becomes an open circuit. What happens to the other LED?
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Notes
Each branch of a parallel circuit is connected directly across the battery, so the working LED still has the same p.d. and the same current: B is correct. A treats the LEDs as if they were in series. C and D wrongly suggest that the current in the working branch changes; only the total current from the battery falls.
✓ 1
Answer B
Answer: B
Exam tip: In a parallel circuit, each branch works independently of the others.
55IGPC3-4-07r10-08Magnets and magnetic fieldsPaper CEasy1 mark
A quarry crane uses an electromagnet to lift steel plates and must release them when the current is switched off. Which material should be used for the core?
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Notes
Soft iron is magnetised by the coil current and demagnetises when the current is off, so the plates are released: A is correct. Steel would stay magnetised and keep holding the plates. Copper and aluminium are not magnetic materials, so they would not strengthen the field, whatever their conductivity or density.
✓ 1
Answer A
Answer: A
Common error: A good electrical conductor is not necessarily a good magnetic material.
56IGPC3-4-07r10-09Electrical safetyPaper CEasy6 marks
A mountain hut is supplied with 230 V mains electricity. A kettle in the hut has a metal case and is rated at 1.84 kW.
(a)
Describe what happens inside a fuse when the current through it is too large.
(2)
(b)
Calculate the normal current in the kettle.
(2)
(c)
The plug can be fitted with a 3 A, 5 A, 10 A or 13 A fuse. State which fuse should be used and give a reason.
(2)
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Part (a)
the fuse wire heats up
✓ 1
it melts, breaking the circuit so the current stops
✓ 1
Part (b)
I = P / V
✓ 1
= 1840 / 230 = 8.0 A
✓ 1
Part (c)
10 A
✓ 1
it is the lowest rating above the normal current of 8.0 A, so it melts soon after a fault current starts
✓ 1
Exam tip: Give the fuse rating and link it to the normal current in the same answer.
57IGPC3-4-07r10-12Series and parallel circuitsPaper CEasy6 marks
Two lamps, P and Q, light a tent. They are connected in parallel to a 12 V battery. An ammeter in the branch containing lamp P reads 0.50 A. Lamp Q has a resistance of 48 Ω.
The ammeter is in the branch containing lamp P.
(a)
State the potential difference across lamp Q.
(1)
(b)
Calculate the current in lamp Q.
(2)
(c)
Calculate the current supplied by the battery.
(1)
(d)
Lamp Q fails and becomes an open circuit. State and explain the effect on the ammeter reading in the branch containing lamp P.
(2)
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Notes
Part (a)
12 V
✓ 1
Part (b)
I = V / R
✓ 1
= 12 / 48 = 0.25 A
✓ 1
Part (c)
0.50 + 0.25 = 0.75 A (ECF from (b))
✓ 1
Part (d)
the reading is unchanged (0.50 A)
✓ 1
lamp P is still connected directly across the 12 V battery (branches work independently)
✓ 1
Exam tip: Branch currents in a parallel circuit add up to the current from the supply.
58IGPC3-4-05r11-01Magnets and magnetic fieldsPaper CEasy1 mark
A mountaineer places a small plotting compass on the ground in line with the axis of a bar magnet, just beyond its north pole. In which direction does the north-seeking end of the compass needle point?
The compass lies on the axis of the magnet, just beyond its north pole. Diagram not to scale
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Notes
Magnetic field lines leave a north pole, so at this position the field points away from the magnet, and the north end of a compass lies along the field direction. A is wrong because it treats the north-seeking end as attracted to a north pole, but like poles repel. C is wrong because the field is perpendicular to the axis only beside the middle of the magnet, not in line with it. D is wrong because the needle settles in the field direction rather than spinning.
✓ 1
Answer B
Answer: B
Common error: Remember that a compass needle shows the direction of the field, so its north end is repelled by a north pole.
59IGPC3-4-05r11-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A rechargeable battery at a remote telecoms mast is charged with a steady current of 4.0 A for 2.5 minutes. How much charge flows through the battery?
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Notes
Q = It with t = 2.5 × 60 = 150 s, so Q = 4.0 × 150 = 600 C. A multiplies by 2.5 without converting minutes to seconds. B divides the time by the current (150 / 4.0 = 37.5 C). D divides current by time in minutes (4.0 / 2.5).
✓ 1
Answer C
Answer: C
Common error: Convert minutes to seconds before using Q = It.
60IGPC3-4-05r11-09Series and parallel circuitsPaper CEasy5 marks
A mountaineer's headlamp is powered by three 1.5 V cells connected in series. When the lamp is switched on there is a current of 0.20 A in it.
Three 1.5 V cells in series with the lamp. The long line of each cell symbol is its positive terminal.
(a)
State the e.m.f. of the battery of cells.
(1)
(b)
Calculate the charge that passes through the lamp in 5.0 minutes.
(2)
(c)
Calculate the power of the lamp. Assume the potential difference across the lamp equals the e.m.f. of the battery.
(2)
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Notes
Part (a)
4.5 V
✓ 1
Part (b)
t = 5.0 × 60 = 300 s
✓ 1
Q = It = 0.20 × 300 = 60 C
✓ 1
Part (c)
P = IV = 0.20 × 4.5
✓ 1
= 0.90 W
✓ 1
Common error: Always change minutes into seconds when using charge = current × time.
61IGPC3-4-05r11-10Magnets and magnetic fieldsPaper CEasy5 marks
A flour mill uses an electromagnet above a conveyor belt to remove small steel fragments from the flour.
(a)
Name one metal element, other than iron, that is attracted by a magnet.
(1)
(b)
The core of the electromagnet is made of soft iron rather than steel. Explain why soft iron is used.
(2)
(c)
State two ways to increase the strength of the electromagnet.
(2)
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Notes
Part (a)
nickel or cobalt
✓ 1
Part (b)
soft iron is magnetised easily when the current is switched on
✓ 1
and it loses its magnetism quickly when the current is switched off, so the fragments drop into a bin
✓ 1
Part (c)
increase the current in the coil
✓ 1
increase the number of turns of wire in the coil
✓ 1
Exam tip: Soft iron is used for temporary magnets; steel is used for permanent magnets.
62IGPC3-4-05r11-11Series and parallel circuitsPaper CEasy5 marks
In a telecoms test box, two resistors of resistance 20 Ω and 30 Ω are connected in parallel to a 6.0 V supply of negligible internal resistance.
Diagram not to scale
(a)
State the potential difference across the 30 Ω resistor.
(1)
(b)
Calculate the current in the 30 Ω resistor.
(2)
(c)
The current in the 20 Ω resistor is 0.30 A. Calculate the current in the supply.
(2)
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Notes
Part (a)
6.0 V
✓ 1
Part (b)
I = V / R = 6.0 / 30
✓ 1
= 0.20 A
✓ 1
Part (c)
currents in parallel branches add together
✓ 1
0.30 + 0.20 = 0.50 A, ECF from (b)
✓ 1
Exam tip: In a parallel circuit the potential difference is the same across each branch and the branch currents add.
63IGPC3-4-07r11-01Magnets and magnetic fieldsPaper CEasy1 mark
A coffee roasting plant uses an electromagnet to remove iron fragments from green beans on a conveyor. The core of the electromagnet is soft iron rather than steel. Which statement explains this choice?
Show mark scheme
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Notes
C is correct: soft iron is a magnetically soft material, so it is magnetised by the coil and demagnetises when the current stops, letting the fragments be released. A is wrong because the current in the coil is set by the coil and supply, not the core. B describes steel and would stop the magnet being switched off. D is wrong because soft iron is a conductor.
✓ 1
Answer C
Answer: C
Common error: Do not mix up the two materials: steel is used where permanent magnetism is wanted, soft iron where magnetism must be temporary.
64IGPC3-4-07r11-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A charge of 4.5 C passes through a sensor on a space probe in a time of 30 s. What is the average current in the sensor?
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Notes
I = Q / t = 4.5 / 30 = 0.15 A, so A is correct. B comes from dividing the wrong way round (t / Q). C comes from multiplying charge by time. D comes from converting the 30 s into minutes and then dividing again as if it were seconds.
✓ 1
Answer A
Answer: A
Exam tip: Write the equation I = Q / t first and check the units: coulombs per second gives amperes.
65IGPC3-4-07r11-07Magnets and magnetic fieldsPaper CEasy1 mark
Which statement about the magnetic field around a bar magnet is correct?
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Notes
A is correct. B reverses the direction. C is wrong because field lines never cross: the field has one direction at each point. D is wrong because field lines form continuous paths and do not stop.
✓ 1
Answer A
Answer: A
Exam tip: When drawing field lines add arrowheads pointing away from the north pole, and never let lines cross.
66IGPC3-4-07r11-10Series and parallel circuitsPaper CEasy6 marks
Two lighting circuits on a footbridge are connected in parallel to a 230 V supply. Circuit X has a resistance of 46 Ω and circuit Y has a resistance of 115 Ω.
Diagram not to scale
(a)
Calculate the current in circuit X and in circuit Y.
(2)
(b)
State the current in the supply wire.
(1)
(c)
Circuit Y develops a fault and stops conducting. State the current in circuit X now.
(1)
(d)
Calculate the combined resistance of the two circuits when both are working. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
I(X) = 230 / 46 = 5.0 A
✓ 1
I(Y) = 230 / 115 = 2.0 A
✓ 1
Part (b)
7.0 A (ECF from (a))
✓ 1
Part (c)
unchanged, 5.0 A
✓ 1
Part (d)
R = V / I = 230 / 7.0
✓ 1
= 32.9 Ω (ECF from (b))
✓ 1
Exam tip: The combined resistance of resistors in parallel is always less than the smallest single resistance, which is a quick check on your answer.
67IGPC3-4-05r12-01Magnets and magnetic fieldsPaper CEasy1 mark
A technician places a small plotting compass near a bar magnet in a laboratory. Which statement about the magnetic field outside the bar magnet is correct?
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Notes
Outside a magnet the field direction is the direction a free north pole would move, so lines run from north to south: option C is correct. A reverses the direction. B is wrong because closely spaced lines show a stronger field. D is wrong because the field has one direction at each point, so lines never cross.
✓ 1
Answer C
Answer: C
Common error: Field lines are drawn from north to south outside the magnet; the arrow shows the direction of the force on a north pole.
68IGPC3-4-05r12-02Electrical safetyPaper CEasy1 mark
A heater on a hospital trolley is rated 230 V, 1.84 kW and is connected through a plug containing a fuse. The available fuses are 3 A, 5 A, 10 A and 13 A. Which fuse is the most suitable?
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Notes
Normal current I = P/V = 1840/230 = 8.0 A. The fuse must be above this so it does not blow in normal use, but as close as possible so that it protects the wiring: 10 A. 3 A and 5 A are below 8.0 A and would blow straight away. 13 A is too high: a fault current between 10 A and 13 A would not melt it.
✓ 1
Answer C
Answer: C
Exam tip: Calculate the normal current first, then choose the smallest fuse rating that is larger than it.
69IGPC3-4-05r12-05Series and parallel circuitsPaper CEasy1 mark
Three different lamps are connected in parallel across a supply in a shop display. An ammeter in the main wire reads 0.90 A. Ammeters in the branches of lamp 1 and lamp 2 read 0.30 A and 0.25 A. What is the current in the branch containing lamp 3?
Ammeters in the main wire and in each branch. Diagram not to scale
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Notes
In a parallel circuit the currents in the branches add up to the current in the main wire: 0.90 − 0.30 − 0.25 = 0.35 A. 0.55 A is the sum of the two known branch currents. 0.90 A assumes every branch carries the main current. 1.45 A adds all three known readings.
✓ 1
Answer A
Answer: A
Common error: Current splits between parallel branches; it is not the same in every branch.
70IGPC3-4-05r12-07Electrical safetyPaper CEasy1 mark
Why is the fuse in a mains appliance placed in the live wire rather than in the neutral wire?
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Mark
Notes
With the fuse in the live wire, a melted fuse isolates the appliance from the high potential of the live supply, so touching its internals is safe: D is correct. A is wrong because live and neutral carry the same current. B is wrong because the neutral is close to 0 V. C is wrong because a fuse does not control the current in normal use.
✓ 1
Answer D
Answer: D
Common error: Live and neutral carry the same current; the difference is that the live wire is at a dangerous potential.
71IGPC3-4-05r12-09Series and parallel circuitsPaper CEasy5 marks
A shop uses a 12 V supply, with negligible internal resistance, for an illuminated sign. Two resistors of resistance 8.0 Ω and 22 Ω are connected in series with the supply.
Diagram not to scale
(a)
Calculate the total resistance of the two resistors.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
Calculate the potential difference across the 22 Ω resistor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R = 8.0 + 22 = 30 Ω
✓ 1
Part (b)
I = V / R
✓ 1
I = 12 / 30 = 0.40 A (ECF from (a))
✓ 1
Part (c)
V = IR
✓ 1
V = 0.40 × 22 = 8.8 V (ECF from (b))
✓ 1
Exam tip: In a series circuit the same current passes through every component, so use the circuit current with each resistor's own resistance.
72IGPC3-4-06r13-01Magnets and magnetic fieldsPaper CEasy1 mark
Which statement about the magnetic field around a bar magnet is correct?
Show mark scheme
Marking point
Mark
Notes
Correct: by convention field lines run from N to S outside the magnet. The reverse direction is wrong; field lines never cross; they are closest together near the poles where the field is strongest.
✓ 1
Answer C
Answer: C
Common error: Field lines never cross, and the closer together they are, the stronger the field.
73IGPC3-4-06r13-02Series and parallel circuitsPaper CEasy1 mark
Two identical lamps are connected in series with a battery in a telescope dome. An ammeter placed before the first lamp reads 0.40 A. What does a second ammeter read when placed between the two lamps?
Two identical lamps in series; the second ammeter is between the lamps.
Show mark scheme
Marking point
Mark
Notes
Correct: in a series circuit the current is the same at every point, so 0.40 A. 0.20 A assumes the current is shared between lamps; 0.80 A assumes the current adds; 0 A assumes the first lamp uses up the current.
✓ 1
Answer A
Answer: A
Common error: Current is not used up by components; the same charge per second passes every point in a series circuit.
74IGPC3-4-06r13-08Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A sports-science sensor draws a current of 0.50 A for 2.0 minutes. How much charge flows through the sensor?
Show mark scheme
Marking point
Mark
Notes
Correct: Q = I t = 0.50 × 120 s = 60 C. 1.0 C does not convert minutes to seconds; 0.0042 C divides current by time; 240 C divides time by current.
✓ 1
Answer D
Answer: D
Exam tip: Convert minutes to seconds before using Q = It.
75IGPC3-4-06r13-09Magnets and magnetic fieldsPaper CEasy6 marks
A sports scientist investigates magnets.
(a)
State what is meant by a magnetic field.
(1)
(b)
Describe how a plotting compass can be used to find the shape of the magnetic field around a bar magnet.
(3)
(c)
A crane in a railway yard uses an electromagnet with a soft iron core. Explain why soft iron is used rather than steel.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a region in which a magnetic pole or magnetic material experiences a force
✓ 1
Part (b)
place the magnet on paper and mark the positions of the compass needle's ends near one pole
✓ 1
move the compass so one end is at the last mark and mark the other end; repeat to build a line of dots and join them
✓ 1
repeat from different start points; direction is the way the needle's north pole points
✓ 1
Part (c)
soft iron is easily magnetised
✓ 1
soft iron loses its magnetism when the current is switched off so the load is released
✓ 1
Exam tip: In a plotting-compass method, mention marking both ends of the needle and joining the dots.
76IGPC3-4-06r13-11Electrical safetyPaper CEasy5 marks
A railway maintenance depot has an electric drill with a metal case, connected to the mains by a three-core cable.
(a)
State the purpose of a fuse in the plug.
(1)
(b)
Explain how the earth wire protects the user if the live wire touches the metal case.
(3)
(c)
Explain why the switch must be placed in the live wire.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
breaks the circuit / melts if the current is too large
✓ 1
Part (b)
a large current flows through the earth wire to earth
✓ 1
the fuse melts and disconnects the live supply
✓ 1
so the case does not stay at a dangerous voltage
✓ 1
Part (c)
when switched off the appliance is disconnected from the live (high) voltage
✓ 1
Common error: The fuse goes in the live wire; the earth wire does not carry current in normal use.
77IGPC3-4-06r15-01Magnets and magnetic fieldsPaper CEasy1 mark
Archaeologists use a lifting electromagnet on a site crane to move iron fittings, and the magnet must be switched on and off repeatedly. Which material is most suitable for the core of the electromagnet?
Show mark scheme
Marking point
Mark
Notes
Soft iron is easily magnetised and loses its magnetism when the current stops, so C is correct. Steel keeps its magnetism, so the load would not be released. Copper and aluminium are not magnetic materials.
✓ 1
Answer C
Answer: C
Common error: Steel is used for permanent magnets; soft iron is used where magnetism must be switched on and off.
78IGPC3-4-06r15-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A cyclist's dynamo lamp carries a current of 0.25 A for 2.0 minutes. How much charge passes through the lamp?
Show mark scheme
Marking point
Mark
Notes
Q = It = 0.25 × 120 s = 30 C, so B is correct. A uses 2.0 minutes without converting to seconds. C is t ÷ I. D is I ÷ t.
✓ 1
Answer B
Answer: B
Exam tip: Convert time to seconds before using Q = It.
79IGPC3-4-06r15-07Magnets and magnetic fieldsPaper CEasy1 mark
Which statement about the magnetic field around a bar magnet is correct?
Show mark scheme
Marking point
Mark
Notes
By convention field lines outside the magnet run from north to south, so A is correct. B reverses the direction. C is wrong because field lines never cross. D is wrong: lines are closest where the field is strongest.
✓ 1
Answer A
Answer: A
Common error: Field lines are closest together where the field is strongest, near the poles.
80IGPC3-4-06r15-09Charge, current, potential difference, resistance and powerPaper CEasy6 marks
A bicycle lamp is connected to a 6.0 V battery and the current in the lamp is 0.50 A.
(a)
Calculate the charge that flows through the lamp in 10 minutes.
(2)
(b)
Calculate the resistance of the lamp.
(2)
(c)
Calculate the power of the lamp.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Q = It = 0.50 × 600
✓ 1
300 C
✓ 1
Part (b)
R = V/I = 6.0 ÷ 0.50
✓ 1
12 Ω
✓ 1
Part (c)
P = IV = 0.50 × 6.0
✓ 1
3.0 W
✓ 1
Exam tip: Check units: charge in coulombs needs time in seconds.
81IGPC3-4-06r15-12Magnets and magnetic fieldsPaper CEasy6 marks
Archaeologists study the magnetic field of a buried iron object using a plotting compass.
(a)
Describe how a plotting compass can be used to show the magnetic field pattern around a bar magnet.
(3)
(b)
State what the direction of a magnetic field line represents.
(1)
(c)
State two differences between iron and steel as magnetic materials.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
place the magnet on paper and mark the position of the compass needle ends near one pole
✓ 1
move the compass so one end is at the previous mark, mark again, and repeat
✓ 1
join the marks to make a field line; repeat from other starting points
✓ 1
Part (b)
direction of the force on a north pole placed at that point
✓ 1
Part (c)
iron is magnetised and demagnetised easily (magnetically soft)
✓ 1
steel keeps its magnetism (magnetically hard) and is used for permanent magnets
✓ 1
Common error: A field line points in the direction a free north pole would move, away from a north pole.
82IGPC3-4-07r16-01Magnets and magnetic fieldsPaper CEasy1 mark
A robot arm on a fish-processing line lifts steel sheets using an electromagnet. The sheet must drop as soon as the current is switched off. Which core material is best for the electromagnet?
Show mark scheme
Marking point
Mark
Notes
Soft iron is magnetised easily by the coil's field and loses almost all its magnetism when the current is switched off, so the sheet is released. Copper is not magnetic. Steel keeps its magnetism, so the sheet would stay held. Aluminium is not a magnetic material, so its low mass does not help.
✓ 1
Answer C
Answer: C
Common error: Steel is used for permanent magnets, not for the core of an electromagnet that has to switch off completely.
83IGPC3-4-07r16-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
An aerator pump in a fish farm carries a steady current of 0.80 A for 15 minutes. How much charge passes through the pump in this time?
Show mark scheme
Marking point
Mark
Notes
Q = It with t = 15 × 60 = 900 s, so Q = 0.80 × 900 = 720 C. 12 C comes from using 15 minutes without converting to seconds. 1125 C comes from dividing time by current (900 / 0.80). 0.00089 C comes from dividing current by time (0.80 / 900).
✓ 1
Answer B
Answer: B
Common error: Always convert minutes to seconds before using Q = It.
84IGPC3-4-07r16-07Magnets and magnetic fieldsPaper CEasy1 mark
A small plotting compass is placed close to the north pole of a bar magnet used in a robot's position sensor. Which statement about the compass needle is correct?
Show mark scheme
Marking point
Mark
Notes
The field outside a magnet points away from its north pole, and the north-seeking end of the compass lines up with the field direction, so it points away from the magnet's north pole. Pointing towards it assumes like poles attract. The south-seeking end therefore points towards that pole, not away. A compass settles; it does not spin.
✓ 1
Answer C
Answer: C
Common error: Like poles repel: the north end of a compass is repelled by a north pole.
85IGPC3-4-07r16-09Magnets and magnetic fieldsPaper CEasy6 marks
A robot on a recycling line uses magnets to sort metal objects.
(a)
State two metals that are attracted to a magnet.
(2)
(b)
Describe how you could show that a steel bar is a magnet, and not just an unmagnetised piece of steel.
(2)
(c)
State the direction of the magnetic field lines outside a bar magnet and state where the field is strongest.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 any one of iron, steel, nickel or cobalt
✓ 1
B1 a second, different metal from that list
✓ 1
Part (b)
B1 bring one end of the bar close to a known pole of a magnet (or hang the bar on a thread and bring a known pole near)
✓ 1
B1 repulsion is seen at one end; only a magnet is repelled, unmagnetised steel is always attracted
✓ 1
Part (c)
B1 field lines run from the north pole to the south pole
✓ 1
B1 field is strongest at the poles, where the lines are closest together
✓ 1
Common error: Attraction alone does not show that a bar is a magnet, because magnets attract unmagnetised iron and steel too; only repulsion is a reliable test.
86IGPC3-4-07r16-15Charge, current, potential difference, resistance and powerPaper CEasy6 marks
A sensor in a robot draws a steady current of 0.25 A for 4.0 minutes from a 7.5 V battery.
(a)
Calculate the charge that flows through the sensor in this time.
(2)
(b)
Calculate the energy transferred by the battery in this time. Use 60 C for the charge.
(2)
(c)
Calculate the resistance of the sensor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Q = It with t = 4.0 × 60 = 240 s
✓ 1
A1 60 C
✓ 1
Part (b)
M1 E = QV = 60 × 7.5
✓ 1
A1 450 J
✓ 1
Part (c)
M1 R = V / I = 7.5 / 0.25
✓ 1
A1 30 Ω
✓ 1
Common error: Convert minutes to seconds before using Q = It.
87IGPC3-4-08r21-01Magnets and magnetic fieldsPaper CEasy1 mark
An orienteering coach places a plotting compass close to the north pole of a bar magnet. The magnet's field here is much stronger than the Earth's field. In which direction does the north end of the compass needle point?
Show mark scheme
Marking point
Mark
Notes
The north end of a compass points in the direction of the magnetic field, and field lines leave a north pole, so the needle points away from it; B is correct. A is wrong because like poles repel. C is wrong because the field is not perpendicular to that line near a pole. D ignores the fact that the magnet's field dominates here.
✓ 1
Answer B
Answer: B
Exam tip: Magnetic field lines point away from a north pole and towards a south pole.
88IGPC3-4-08r21-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A cooling unit on a telescope camera runs for 40 s, during which a charge of 120 C passes through it. What is the average current?
Show mark scheme
Marking point
Mark
Notes
Current = charge ÷ time = 120 ÷ 40 = 3.0 A, so C is correct. A inverts the equation (40 ÷ 120). B adds charge and time. D multiplies them.
✓ 1
Answer C
Answer: C
Exam tip: Current is charge per unit time: I = Q ÷ t.
89IGPC3-4-08r21-09Magnets and magnetic fieldsPaper CEasy5 marks
A student investigates magnetic materials in a sports-science laboratory.
(a)
Explain why the core of an electromagnet is made of iron rather than steel.
(2)
(b)
Describe how a plotting compass can be used to find the shape of the magnetic field around a bar magnet.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
iron is magnetically soft, so it is easily magnetised
✓ 1
and it loses its magnetism when the current is switched off, unlike steel
✓ 1
Part (b)
place the magnet on paper with the compass near one pole and mark the positions of both ends of the needle
✓ 1
move the compass so that one end of the needle is at the previous mark, and repeat
✓ 1
join the marks with a smooth line and show the direction with an arrow from north to south
✓ 1
Exam tip: In a plotting-compass method, state how each new position is chosen from the previous mark.
90IGPC3-4-08r21-10Charge, current, potential difference, resistance and powerPaper CEasy6 marks
A heater in a train driver's cab is rated 230 V, 2.3 kW.
(a)
Calculate the current in the heater.
(2)
(b)
Calculate the energy transferred by the heater in 3.0 hours. Give your answer in kWh.
(2)
(c)
Electrical energy costs £0.30 per kWh. Use 6.9 kWh to calculate the cost of running the heater for 3.0 hours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
I = P ÷ V = 2300 ÷ 230
✓ 1
10 A
✓ 1
Part (b)
E = P × t = 2.3 × 3.0
✓ 1
6.9 kWh
✓ 1
Part (c)
cost = 6.9 × 0.30
✓ 1
£2.07
✓ 1
Exam tip: Keep power in kW and time in hours when the answer is needed in kWh.
91IGPC3-4-07r23-01Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A motor in a fish-sorting robot is switched on. A charge of 150 C passes through the motor in 50 s. What is the current in the motor?
Show mark scheme
Marking point
Mark
Notes
Current = charge ÷ time = 150 ÷ 50 = 3.0 A. 0.33 A comes from dividing time by charge. 200 A comes from adding the two numbers. 7500 A comes from multiplying charge by time.
✓ 1
Answer B
Answer: B
Exam tip: Write the equation I = Q / t first and check that the units (coulomb, second) fit it.
92IGPC3-4-07r23-02Magnets and magnetic fieldsPaper CEasy1 mark
A tailor uses a bar magnet to pick up steel pins. Which statement about the magnetic field around a bar magnet is correct?
Show mark scheme
Marking point
Mark
Notes
The spacing of the field lines shows the strength of the field: close lines mean a strong field, which is near the poles. Outside a magnet the lines point from north to south, so the first alternative has the direction reversed. Field lines never cross, because the field has only one direction at each point. The lines are not evenly spaced because the field gets weaker with distance from the magnet.
✓ 1
Answer C
Answer: C
Common error: Field lines are drawn from north to south outside the magnet; students often reverse the arrows.
93IGPC3-4-07r23-08Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A freezer on a fishing boat has a power of 0.40 kW. It runs for 6.0 hours each day for 5 days. Electricity costs £0.30 per kWh. What is the total cost?
Show mark scheme
Marking point
Mark
Notes
Energy = 0.40 kW × 6.0 h × 5 = 12 kWh, and the cost = 12 × £0.30 = £3.60. The value £0.72 is the cost for one day only. The value £0.60 leaves out the hours. The value £9.00 leaves out the power.
✓ 1
Answer B
Answer: B
Exam tip: For energy in kWh, use power in kW and time in hours, then multiply by the cost per kWh.
94IGPC3-4-07r23-09Charge, current, potential difference, resistance and powerPaper CEasy6 marks
The gripper motor in a packing robot works from a 6.0 V supply. The current in the motor is 0.80 A and it runs for 25 s.
(a)
Calculate the charge that passes through the motor in 25 s.
(2)
(b)
Calculate the electrical energy transferred in the motor in 25 s.
(2)
(c)
Calculate the power of the motor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Q = I × t
✓ 1
= 0.80 × 25 = 20 C
✓ 1
Part (b)
E = I × V × t, or E = 6.0 × 20
✓ 1
= 120 J
✓ 1
Part (c)
P = I × V, or P = E / t
✓ 1
= 0.80 × 6.0 = 4.8 W
✓ 1
Exam tip: Write the equation, then substitute with units; you earn the method mark even if the arithmetic slips.
95IGPC3-4-07r23-12Magnets and magnetic fieldsPaper CEasy6 marks
A robot arm has a steel hand and a small permanent magnet in a catch.
(a)
Name one other magnetic material, apart from steel.
(1)
(b)
Describe how a small plotting compass can be used to show the magnetic field around a bar magnet.
(3)
(c)
State one difference between the behaviour of a permanent magnet and a temporary magnet.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Iron or nickel or cobalt
✓ 1
Part (b)
Place the compass near the magnet and mark the positions of the ends of the needle
✓ 1
Move the compass so that one end of the needle is at the previous mark, and repeat many times
✓ 1
Join the marks to make a field line, with the direction shown as the way the north pole of the needle points
✓ 1
Part (c)
A permanent magnet keeps its magnetism after the magnetising field is removed
✓ 1
A temporary magnet loses (most of) its magnetism when removed from the field
✓ 1
Common error: A compass needle points along the field line in the direction a north pole would move, which is away from a north pole of the magnet.
96IGPC3-4-06r24-01Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A beekeeper's electric fence energiser draws a constant current of 0.25 A for 2.0 minutes. How much charge flows through the energiser in this time?
Show mark scheme
Marking point
Mark
Notes
Charge = current × time = 0.25 A × 120 s = 30 C, so 30 C is correct. 0.0021 C comes from dividing current by time. 0.50 C uses the time in minutes instead of seconds. 480 C comes from dividing time by current.
✓ 1
Answer B
Answer: B
Common error: Convert minutes to seconds before using charge = current × time.
97IGPC3-4-06r24-04Magnets and magnetic fieldsPaper CEasy1 mark
Which statement about the magnetic field lines outside a bar magnet is correct?
Show mark scheme
Marking point
Mark
Notes
By convention the field direction is the direction a free north pole would be pushed, so outside the magnet the lines run from north to south. The reverse direction is wrong. Lines do not run from both poles to the middle, and they are not circles around one pole only.
✓ 1
Answer D
Answer: D
Exam tip: Draw an arrow on every field line, pointing away from the north pole.
98IGPC3-4-06r24-06Series and parallel circuitsPaper CEasy1 mark
Three identical lamps are connected in parallel to a 12 V supply. An ammeter in the supply wire reads 1.5 A. One lamp is removed. What does the ammeter now read?
Circuit diagram
Show mark scheme
Marking point
Mark
Notes
Each lamp carries 1.5 A ÷ 3 = 0.50 A and the lamps work independently. With two lamps left the supply current is 2 × 0.50 A = 1.0 A. 0.50 A is only the current in one lamp. 0.75 A halves the current. 1.5 A assumes the supply current stays the same.
✓ 1
Answer D
Answer: D
Common error: In a parallel circuit, removing one branch lowers the supply current but leaves the current in the other branches unchanged.
99IGPC3-4-07r24-01Magnets and magnetic fieldsPaper CEasy1 mark
A beekeeper sweeps a bar magnet over a hive frame to pick up stray steel staples. Which statement about the magnetic field lines of the bar magnet is correct?
Show mark scheme
Marking point
Mark
Notes
Field lines point in the direction a free north pole would move, so outside the magnet they go from north to south, and the closer the lines the stronger the field. The reverse direction is wrong. Field lines never cross, because the field at a point has only one direction. The spacing is not uniform: the field is weaker further from the magnet, so the lines spread out.
✓ 1
Answer B
Answer: B
Common error: Field lines never cross, and the direction rule (north to south outside the magnet) is for the outside only.
100IGPC3-4-07r24-02Electrical safetyPaper CEasy1 mark
A hospital bedside lamp has a metal case and is connected to the mains through a three-pin plug. A fuse is fitted in the plug. In which wire should the fuse be placed, and why?
Show mark scheme
Marking point
Mark
Notes
A fuse in the live wire means that when it melts the appliance is cut off from the high potential of the supply, so no part of the appliance remains live. A fuse in the neutral wire would leave the lamp connected to the live wire when it blows. A fuse in the earth wire would not stop a fault current reaching the case in the first place, and the earth wire carries no current in normal use. Although the current is the same in live and neutral, the position matters for safety.
✓ 1
Answer D
Answer: D
Common error: The current being the same in the live and neutral wires does not mean the fuse can go in either: it must be in the live wire.
101IGPC3-4-07r24-03Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A motor in a hospital infusion pump carries a constant current of 0.40 A for 5.0 minutes. What charge passes through the motor in this time?
Show mark scheme
Marking point
Mark
Notes
Q = It with t converted to seconds: 5.0 minutes = 300 s, so Q = 0.40 × 300 = 120 C. 2.0 C comes from using 5.0 as the time in minutes without converting. 750 C comes from dividing the time by the current instead of multiplying. 0.0013 C comes from dividing the current by the time.
✓ 1
Answer A
Answer: A
Exam tip: Convert minutes to seconds before using Q = It.
102IGPC3-4-07r24-09Magnets and magnetic fieldsPaper CEasy5 marks
A hospital technician checks whether a steel surgical tray has become magnetised, using a plotting compass.
(a)
State what a plotting compass shows when it is placed at a point in a magnetic field.
(1)
(b)
Explain why steel is used for permanent magnets but soft iron is used for the core of an electromagnet.
(2)
(c)
State two ways to increase the strength of an electromagnet.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The direction of the magnetic field at that point
✓ 1
Part (b)
Steel is magnetically hard: it keeps its magnetism once magnetised
✓ 1
Soft iron is magnetically soft: it is easily magnetised and loses its magnetism when the current is switched off
✓ 1
Part (c)
Any two different valid ways, for example: increase the current in the coil, increase the number of turns of wire, use a soft-iron core
✓ 1
second different valid way
✓ 1
Exam tip: Say that steel retains its magnetism and soft iron does not; just writing that one is 'stronger' earns no credit.
103IGPC3-4-07r24-10Charge, current, potential difference, resistance and powerPaper CEasy6 marks
An electric fence energiser protects a beekeeper's apiary from bears. It is powered by a 12 V battery and draws a constant current of 0.25 A.
(a)
Calculate the power of the energiser.
(2)
(b)
The energiser is on for 8.0 hours each night. Calculate the charge that passes through it in this time.
(2)
(c)
Calculate the energy transferred in 8.0 hours. Give your answer in kJ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P = IV or 12 × 0.25
✓ 1
A1 3.0 W
✓ 1
Part (b)
M1 convert 8.0 h to 28 800 s and use Q = It
✓ 1
A1 7200 C
✓ 1
Part (c)
M1 E = Pt (3.0 × 28 800) or E = QV (7200 × 12)
✓ 1
A1 86.4 kJ (ECF from (a) or (b))
✓ 1
Common error: Convert hours to seconds (×3600) and joules to kilojoules (÷1000) before giving the final answer.
104IGPC3-4-07r26-01Magnets and magnetic fieldsPaper CEasy1 mark
A bar magnet lies on a bench. A small plotting compass is placed on the bench in line with the axis of the magnet, close to its north pole. Ignore the Earth's magnetic field. In which direction does the north pole of the compass needle point?
Show mark scheme
Marking point
Mark
Notes
Magnetic field lines outside a magnet leave the north pole, and the north pole of a compass needle points along the field direction, so it points directly away from the magnet's north pole. Pointing towards the north pole confuses the field direction with the force on a south pole. A right-angled direction is only correct beside the middle of the magnet, not in line with its axis. The last option is wrong because the field near the north pole is directed away from that pole, not round to the south pole.
✓ 1
Answer B
Answer: B
Common error: A compass needle's north pole shows the direction of the field, which is away from a north pole and towards a south pole.
105IGPC3-4-07r26-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A hospital ward has an ultraviolet lamp that draws a steady current of 0.50 A for 6.0 minutes. What charge passes through the lamp in this time?
Show mark scheme
Marking point
Mark
Notes
Charge = current × time, with the time in seconds: 0.50 × (6.0 × 60) = 180 C. The 3.0 C option forgets to convert minutes to seconds. The 12 C option divides time by current instead of multiplying. The 0.0014 C option divides current by the time in seconds.
✓ 1
Answer D
Answer: D
Common error: Always convert minutes to seconds before using Q = It.
106IGPC3-4-07r26-09Charge, current, potential difference, resistance and powerPaper CEasy6 marks
A hospital ward uses an ultraviolet lamp to disinfect surfaces. The lamp has a power rating of 40 W and operates from the 230 V mains supply.
(a)
Calculate the current in the lamp. Give your answer to 2 significant figures. [2]
(2)
(b)
Calculate the energy transferred by the lamp in 15 minutes. [2]
(2)
(c)
The lamp is used for 3.0 hours each day for 30 days. Electricity costs €0.30 per kWh. Calculate the cost of using the lamp for the 30 days. [2]
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 I = P / V = 40 / 230
✓ 1
A1 0.17 A
✓ 1
Part (b)
M1 E = P × t = 40 × (15 × 60)
✓ 1
A1 36 000 J
✓ 1
Part (c)
M1 energy = 0.040 kW × 3.0 h × 30 = 3.6 kWh
✓ 1
A1 cost = 3.6 × 0.30 = €1.08
✓ 1
Exam tip: Convert watts to kilowatts and minutes to hours or seconds before using the energy formulae.
107IGPC3-4-07r26-10Magnets and magnetic fieldsPaper CEasy7 marks
A scrap-metal recycling depot uses a large electromagnet on a crane to lift iron and steel items. The electromagnet is a coil of insulated wire wound on a soft iron core.
(a)
Describe how a plotting compass can be used to show the shape and direction of the magnetic field around a bar magnet. [3]
(3)
(b)
Explain why the core is made of soft iron rather than steel. [2]
(2)
(c)
State two ways in which the electromagnet could be made stronger. [2]
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Place the magnet on paper and put the plotting compass near one pole, marking the positions of both ends of the needle
✓ 1
Move the compass so that one end of the needle sits on the previous mark and repeat to build up a line of dots
✓ 1
Join the dots with a smooth line, with an arrow in the direction the north pole of the needle points (from north pole of the magnet to its south pole)
✓ 1
Part (b)
Soft iron is easily magnetised when current flows in the coil
✓ 1
It loses its magnetism easily when the current is switched off, so the load is released
✓ 1
Part (c)
Any valid first way, for example increasing the current in the coil
✓ 1
Any different valid second way, for example increasing the number of turns on the coil
✓ 1
Common error: Steel keeps its magnetism once magnetised, so it suits permanent magnets but not electromagnets that must switch off.
108IGPC3-4-09r30-02Magnets and magnetic fieldsPaper CEasy1 mark
A small plotting compass is placed beside the north pole of a bar magnet. In which direction does the north pole of the compass needle point?
Show mark scheme
Marking point
Mark
Notes
Magnetic field lines leave the north pole of a magnet, and the north pole of a compass needle points in the direction of the field, so it points away from the magnet's north pole. Pointing towards the north pole confuses attraction with field direction (like poles repel). Pointing straight to the south pole ignores that the field lines are curved. The needle always lines up along the field, never across it.
✓ 1
Answer D
Answer: D
Common error: A compass needle's north pole points along the field direction, which is away from a north pole and towards a south pole.
109IGPC3-4-09r30-06Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A sensor on a fish farm carries a steady current of 0.25 A for 2.0 minutes. How much charge flows through the sensor?
Show mark scheme
Marking point
Mark
Notes
Charge = current × time = 0.25 × 120 s = 30 C. Using 2.0 as the time in minutes gives 0.50 C. Dividing current by time gives 0.0021 C. Dividing time by current gives 480 C.
✓ 1
Answer A
Answer: A
Common error: Convert minutes to seconds before using Q = It.
110IGPC3-4-09r30-07Electromagnetic induction, motors and transformersPaper CEasy1 mark
How can the direction of rotation of a simple d.c. motor be reversed?
Show mark scheme
Marking point
Mark
Notes
The force on the coil depends on the directions of both the current and the field. Reversing just one of them reverses the force and so the rotation. Reversing both leaves the force direction unchanged. A stronger magnet or more turns changes the size of the force but not its direction.
✓ 1
Answer C
Answer: C
Common error: Changing both the current direction and the field direction cancels out, so the motor turns the same way.
111IGPC3-4-09r30-08Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A lamp on a fishing boat is marked 12 V, 36 W. What is its resistance when working normally?
Show mark scheme
Marking point
Mark
Notes
The current is I = P/V = 36/12 = 3.0 A, so R = V/I = 12/3.0 = 4.0 Ω. 3.0 Ω is the current used as if it were the resistance. 432 Ω multiplies power by voltage. 0.25 Ω is current divided by voltage (inverted).
✓ 1
Answer B
Answer: B
Exam tip: Find the current from P = IV first, then use R = V/I.
112IGPC3-4-09r30-09Series and parallel circuitsPaper CEasy5 marks
A warehouse robot has a sensor circuit. A 12 V battery is connected in series to two resistors of 40 Ω and 20 Ω.
Circuit diagram
(a)
Calculate the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
Calculate the potential difference across the 20 Ω resistor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
40 + 20 = 60 Ω
✓ 1
Part (b)
M1 I = V/R = 12/60
✓ 1
A1 0.20 A
✓ 1
Part (c)
M1 V = IR = 0.20 × 20
✓ 1
A1 4.0 V (ECF from (b))
✓ 1
Common error: In a series circuit the same current flows through each resistor, but the potential difference is shared between them.
113IGPC3-4-09r30-15Charge, current, potential difference, resistance and powerPaper CEasy6 marks
A warehouse robot's motor draws a steady current of 4.0 A from a 24 V supply for 15 minutes.
(a)
Calculate the charge that flows through the motor.
(2)
(b)
Calculate the electrical energy transferred to the motor.
(2)
(c)
Calculate the power of the motor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Q = It = 4.0 × (15 × 60)
✓ 1
A1 3600 C
✓ 1
Part (b)
M1 E = QV = 3600 × 24
✓ 1
A1 86 400 J (ECF from (a))
✓ 1
Part (c)
M1 P = IV = 4.0 × 24
✓ 1
A1 96 W
✓ 1
Common error: Time must be in seconds when using Q = It, so 15 minutes is 900 s.
114IGPC3-4-07r31-01Magnets and magnetic fieldsPaper CEasy1 mark
In a railway workshop, the north pole of a bar magnet is brought close to one end of an unmagnetised iron nail. What happens to the nail?
Show mark scheme
Marking point
Mark
Notes
Iron is a magnetic material, so the nail is magnetised by induction: the end nearest the north pole becomes a south pole and is attracted. Repulsion would need the near end to become a north pole, which is the wrong induced polarity. Attraction does not happen without magnetisation, so the idea of no induced poles is wrong. An unmagnetised nail does respond, because magnetic materials are attracted to either pole.
✓ 1
Answer A
Answer: A
Common error: Only a magnet can repel; a magnetic material such as iron can only be attracted, whichever pole is used.
115IGPC3-4-07r31-06Magnets and magnetic fieldsPaper CEasy1 mark
Which statement about the magnetic field lines around a bar magnet is correct?
Show mark scheme
Marking point
Mark
Notes
Closer lines show a stronger field, for example near the poles. Field lines never cross, because the field has only one direction at each point. Outside the magnet the direction is from north to south, not south to north. Lines further apart show a weaker field, so the idea of closer lines for a weaker field is reversed.
✓ 1
Answer B
Answer: B
Common error: Field lines show the direction a free north pole would move, so outside a magnet they run from north to south.
116IGPC3-4-07r31-09Charge, current, potential difference, resistance and powerPaper CEasy5 marks
A signal lamp beside a railway track is connected to a 12 V supply. A steady current of 0.50 A passes through the lamp for 15 minutes.
(a)
Calculate the charge that passes through the lamp in 15 minutes.
(2)
(b)
Calculate the energy transferred by the lamp in 15 minutes.
(2)
(c)
State what is meant by a potential difference of 12 V across the lamp.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 Q = It with t = 15 × 60 = 900 s
✓ 1
A1 Q = 450 C
✓ 1
Part (b)
M1 E = QV (or E = IVt) = 450 × 12
✓ 1
A1 5400 J (accept 5.4 kJ; ECF from (a))
✓ 1
Part (c)
B1 12 J of energy is transferred per coulomb of charge passing through the lamp (energy transferred per unit charge)
✓ 1
Common error: Always convert minutes to seconds before using Q = It or E = IVt.
117IGPC3-4-07r31-10Magnets and magnetic fieldsPaper CEasy7 marks
A student investigates the magnetic field around a bar magnet using a plotting compass.
(a)
State what is meant by a magnetic field.
(1)
(b)
Describe how the student uses the plotting compass to draw one field line.
(3)
(c)
State the direction of the field lines outside the magnet.
(1)
(d)
Explain why soft iron is used for the core of an electromagnet but steel is used for a permanent magnet.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 a region in which a magnet or magnetic material experiences a force
✓ 1
Part (b)
B1 place the magnet on paper and put the compass near one pole, marking the positions of both ends of the needle
✓ 1
B1 move the compass so that one end of the needle is at the previous end mark and mark the new position, repeating this many times
✓ 1
B1 join the marks with a smooth line and add an arrow showing the direction
✓ 1
Part (c)
B1 from the north pole to the south pole
✓ 1
Part (d)
B1 soft iron is magnetised easily and loses its magnetism easily when the current is switched off
✓ 1
B1 steel keeps its magnetism for a long time
✓ 1
Exam tip: In a plotting-compass question, say how you mark the needle ends and how you find the direction.
118IGPC3-4-07r31-15Electromagnetic induction, motors and transformersPaper CEasy5 marks
A cyclist in a sports-science laboratory has a bicycle fitted with a dynamo that lights a lamp.
(a)
State what is meant by an induced e.m.f.
(1)
(b)
State two changes that would increase the e.m.f. produced by the dynamo.
(2)
(c)
A student pushes a bar magnet into a coil connected to a sensitive voltmeter, then holds the magnet still inside the coil. Describe the voltmeter reading.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 an e.m.f. produced in a conductor when it cuts magnetic field lines or when the magnetic field through it changes
✓ 1
Part (b)
B1 one valid change, for example the magnet turning faster
✓ 1
B1 a second different valid change, for example a stronger magnet or a coil with more turns
✓ 1
Part (c)
B1 the voltmeter shows a reading (deflects) while the magnet is moving
✓ 1
B1 the reading is zero when the magnet is stationary in the coil
✓ 1
Common error: An e.m.f. is induced only while the magnetic field is changing, not just because a magnet is present.
119IGPC3-4-09r31-01Electrical safetyPaper CEasy1 mark
A gaming console has a metal case and a three-core mains cable. Why is the earth wire connected to the metal case?
Show mark scheme
Marking point
Mark
Notes
The earth wire gives a low-resistance path from the case to earth, so a fault current becomes large enough to melt the fuse and isolate the console. The normal current returns through the neutral wire, not the earth wire. The earth wire has very low resistance, so it does not make the cable cooler. Keeping the case at live potential would make it dangerous to touch.
✓ 1
Answer A
Answer: A
Common error: The neutral wire carries the normal return current; the earth wire carries current only when there is a fault.
120IGPC3-4-09r31-05Electromagnetic induction, motors and transformersPaper CEasy1 mark
A bar magnet is pushed into a coil connected to a sensitive voltmeter, then held still inside the coil. Which statement about the voltmeter is correct?
Show mark scheme
Marking point
Mark
Notes
An e.m.f. is induced only while the magnetic field through the coil is changing, which happens while the magnet moves. When the magnet is still there is no cutting of field lines, so the reading is zero. A steady or largest reading at rest is wrong because a strong field alone induces nothing. A zero reading throughout is wrong because the voltmeter does respond while the magnet is moving.
✓ 1
Answer D
Answer: D
Common error: A strong magnetic field does not induce an e.m.f.; the field must be changing or moving relative to the coil.
121IGPC3-4-09r31-06Charge, current, potential difference, resistance and powerPaper CEasy1 mark
The motor of an infusion pump carries a current of 0.40 A for 2.5 minutes. How much charge passes through the motor?
Show mark scheme
Marking point
Mark
Notes
Charge = current × time. The time must be in seconds: 2.5 min = 150 s, so Q = 0.40 × 150 = 60 C. 1.0 C uses 2.5 as seconds. 375 C comes from 150 ÷ 0.40. 0.0027 C comes from 0.40 ÷ 150.
✓ 1
Answer B
Answer: B
Common error: Convert minutes to seconds before using Q = It.
122IGPC3-4-09r31-09Electrical safetyPaper CEasy5 marks
A gaming PC has a power supply rated 575 W and is connected to the 230 V mains.
(a)
State what happens in a fuse when the current through it is larger than its rating.
(1)
(b)
Calculate the normal working current of the power supply.
(2)
(c)
Fuses rated 3 A, 5 A and 13 A are available. State which fuse should be used and give a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The thin wire in the fuse heats up and melts, breaking the circuit
✓ 1
Part (b)
M1 use of I = P ÷ V = 575 ÷ 230
✓ 1
A1 2.5 A
✓ 1
Part (c)
3 A
✓ 1
Its rating is just above the working current of 2.5 A, so it carries the normal current but melts if a larger fault current flows
✓ 1
Exam tip: Choose the fuse with the lowest rating that is still above the normal working current.
123IGPC3-4-09r31-10Charge, current, potential difference, resistance and powerPaper CEasy5 marks
A water-tank alarm circuit has a 9.0 V battery connected in series to a 150 Ω resistor and a 300 Ω resistor.
Circuit diagram
(a)
State the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
Calculate the potential difference across the 300 Ω resistor.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
450 Ω
✓ 1
Part (b)
M1 I = V ÷ R = 9.0 ÷ 450
✓ 1
A1 0.020 A
✓ 1
Part (c)
M1 V = IR = 0.020 × 300
✓ 1
A1 6.0 V (ECF from (b))
✓ 1
Common error: Use the total resistance to find the current, then use only the single resistor's resistance for its own p.d.
124IGPC3-4-10r33-01Magnets and magnetic fieldsPaper CEasy1 mark
On a farm, a bar magnet is used to pick up iron fragments from animal feed. A fragment is attracted when the north pole of the magnet is brought close to it. What is the pole induced at the end of the fragment nearest the magnet?
Show mark scheme
Marking point
Mark
Notes
The magnet induces magnetism in the iron, and the nearest end becomes a south pole, which is attracted to the north pole. A north pole would be repelled. The iron does become magnetised while near the magnet, and the induced pole does not depend on the fragment's size.
✓ 1
Answer C
Answer: C
Common error: A magnet attracts unmagnetised iron because it induces an opposite pole nearest to it; only repulsion proves a magnet.
125IGPC3-4-10r33-02Electrical safetyPaper CEasy1 mark
The metal casing of an electric drill used in an aircraft hangar is connected to the earth wire. What is the purpose of this connection?
Show mark scheme
Marking point
Mark
Notes
If a fault connects the live wire to the casing, the earth wire gives a low-resistance path, so a large current flows, the fuse blows and the casing is made safe. The earth wire does not affect speed, it prevents the casing staying at a dangerous voltage, and the neutral wire still completes the circuit.
✓ 1
Answer A
Answer: A
Exam tip: Link the earth wire to the fuse: the earth path lets a fault current become large enough to blow the fuse.
126IGPC3-4-10r33-07Charge, current, potential difference, resistance and powerPaper CEasy1 mark
In a milking machine sensor a steady current of 0.40 A flows for 2.5 minutes. How much charge flows through the sensor?
Show mark scheme
Marking point
Mark
Notes
Charge = current × time = 0.40 × 150 s = 60 C, with the time converted to seconds. 1.0 C uses 2.5 minutes without converting, 375 C divides 150 by 0.40, and 0.0027 C divides 0.40 by 150.
✓ 1
Answer D
Answer: D
Common error: Convert minutes to seconds before using Q = It.
127IGPC3-4-10r33-09Charge, current, potential difference, resistance and powerPaper CEasy6 marks
An aircraft cabin has a reading lamp in series with a resistor. The circuit is connected to a 12 V supply. The potential difference across the lamp is 6.0 V and the current is 0.50 A.
Circuit diagram
(a)
Calculate the resistance of the lamp.
(2)
(b)
Calculate the potential difference across the resistor and the resistance of the resistor.
(2)
(c)
Calculate the power of the lamp.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of R = V / I
✓ 1
A1 R = 12 Ω
✓ 1
Part (b)
A1 pd across resistor = 6.0 V
✓ 1
A1 R = 12 Ω (using 6.0 V ÷ 0.50 A)
✓ 1
Part (c)
M1 use of P = VI
✓ 1
A1 P = 3.0 W
✓ 1
Exam tip: In a series circuit the supply voltage is shared, so the pds add up to the supply voltage.
128IGPC3-4-10r33-14Charge, current, potential difference, resistance and powerPaper CEasy5 marks
A guitar amplifier on a stage is connected to the 230 V mains. It has a power of 460 W and is used for 2.5 hours.
(a)
Calculate the current in the amplifier.
(2)
(b)
Calculate the energy used, in kilowatt-hours.
(2)
(c)
Electricity costs €0.20 per kWh. Calculate the cost of using the amplifier for 2.5 hours. Use the exact value from (b).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 use of I = P / V
✓ 1
A1 I = 2.0 A
✓ 1
Part (b)
M1 energy = power in kW × time in hours (0.46 × 2.5)
✓ 1
A1 1.15 kWh
✓ 1
Part (c)
A1 cost = €0.23; ECF from (b)
✓ 1
Common error: For kWh, use power in kilowatts and time in hours; do not use watts and seconds.
129IGPC3-4-08r35-01Magnets and magnetic fieldsPaper CEasy1 mark
A food factory uses an electromagnet to lift steel lids off a conveyor and then release them. Which material is the best choice for the core of the electromagnet?
Show mark scheme
Marking point
Mark
Notes
Soft iron is easily magnetised when current flows and loses almost all its magnetism when the current is switched off, so the lids drop on cue. Steel stays magnetised, so the lids would not be released. Copper is not a magnetic material, so it would not strengthen the field. Plastic is not magnetic at all.
✓ 1
Answer D
Answer: D
Common error: Steel is used for permanent magnets because it keeps its magnetism; that same property makes it a poor core for an electromagnet.
130IGPC3-4-08r35-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A mountain rescue radio draws a steady current of 0.40 A for 2.0 minutes. How much charge flows through the radio in this time?
Show mark scheme
Marking point
Mark
Notes
Charge = current × time, and the time must be in seconds: 0.40 × 120 = 48 C. Using 2.0 minutes directly gives 0.80 C. Dividing time by current gives 300 C, and dividing current by time gives 0.0033 C; neither is the relationship for charge.
✓ 1
Answer B
Answer: B
Exam tip: Convert minutes to seconds before using Q = I t.
131IGPC3-4-08r37-01Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A navigation lamp on a ship's mast carries a constant current of 0.40 A for 2.5 minutes. How much charge passes through the lamp in this time?
Show mark scheme
Marking point
Mark
Notes
Time must be in seconds: 2.5 min = 150 s, so Q = It = 0.40 × 150 = 60 C. Using 2.5 as the time without converting gives 1.0 C. Dividing the time by the current gives 375 C. Dividing the current by the time gives 0.0027 C, which is not a charge.
✓ 1
Answer A
Answer: A
Common error: Leaving the time in minutes when using Q = It; the equation needs seconds.
132IGPC3-4-08r37-02Magnets and magnetic fieldsPaper CEasy1 mark
A dockside crane uses an electromagnet to lift scrap steel and must drop the scrap when the current is switched off. Which material is best for the core of the electromagnet?
Show mark scheme
Marking point
Mark
Notes
Soft iron becomes strongly magnetic when current flows and loses its magnetism when the current stops, so the scrap is released. Copper is not magnetic, so being a conductor does not help. Steel would stay magnetised and keep holding the scrap. Plastic is not magnetic at all.
✓ 1
Answer C
Answer: C
Exam tip: Link the property to the use: soft magnetic materials suit electromagnets, hard magnetic materials suit permanent magnets.
133IGPC3-4-08r37-03Series and parallel circuitsPaper CEasy1 mark
In a theatre, two identical lamps are connected in parallel to a supply that keeps a constant p.d. of 12 V. A third identical lamp is then connected in parallel with the first two. What happens?
Circuit diagram
Show mark scheme
Marking point
Mark
Notes
Each parallel branch has the full 12 V across it, so every lamp keeps the same current and brightness. A new branch adds more current, so the total current from the supply increases. The p.d. across the original lamps stays at 12 V. Adding a branch lowers the total resistance, so it does not increase it.
✓ 1
Answer B
Answer: B
Common error: Thinking that adding a parallel branch increases the total resistance; it provides another path, so the total resistance falls.
134IGPC3-4-08r37-07Electrical safetyPaper CEasy1 mark
A stage lantern has a metal casing and a three-pin plug. What is the main purpose of the earth wire connected to the casing?
Show mark scheme
Marking point
Mark
Notes
If a fault makes the live wire touch the metal casing, the earth wire gives a low-resistance path to earth. The large current melts the fuse and removes the danger. The return path in normal use is the neutral wire. The earth wire does not change the voltage and has no effect on the lamp's temperature.
✓ 1
Answer B
Answer: B
Common error: Confusing the earth wire with the neutral wire; the earth wire carries current only when there is a fault.
135IGPC3-4-08r37-09Electrical safetyPaper CEasy8 marks
A theatre spotlight is rated 800 W and is connected to a 230 V mains supply.
(a)
Calculate the current in the spotlight. Give your answer to 2 significant figures.
(2)
(b)
The spotlight is on for 2.5 hours. Calculate the energy transferred, in kWh.
(2)
(c)
Electricity costs €0.24 per kWh. Calculate the cost of using the spotlight for 2.5 hours.
(2)
(d)
The plug may contain a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 I = P/V = 800 / 230
✓ 1
A1 3.5 A
✓ 1
Part (b)
M1 E = Pt = 0.80 kW × 2.5 h
✓ 1
A1 2.0 kWh
✓ 1
Part (c)
M1 cost = 2.0 × 0.24 (ECF from (b))
✓ 1
A1 €0.48
✓ 1
Part (d)
A1 5 A fuse
✓ 1
A1 it is the lowest rating above the normal current of 3.5 A, so the fuse melts for a fault but not in normal use
✓ 1
Exam tip: Convert watts to kilowatts and minutes to hours before calculating kilowatt-hours; do not mix with joules.
136IGPC3-4-07r39-01Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A space probe's spectrometer draws a steady current of 0.25 A for 2.0 hours. How much charge passes through the spectrometer in this time?
Show mark scheme
Marking point
Mark
Notes
Charge = current × time, with time in seconds: 0.25 × (2.0 × 3600) = 1800 C. 30 C comes from using the time in minutes. 8.0 C comes from dividing time by current. 3.5×10−5 C comes from dividing current by time.
✓ 1
Answer B
Answer: B
Common error: Convert time to seconds before using Q = I t; the coulomb is defined using the second.
137IGPC3-4-07r39-02Electrical safetyPaper CEasy1 mark
In the supply to the drum motor of a coffee roaster, the on/off switch is placed in the live wire. Why is this done?
Show mark scheme
Marking point
Mark
Notes
With the switch in the live wire, opening it removes the high potential from the motor and its wiring, so touching them is safe. Disconnecting the earth wire would remove protection, not add it. The switch does not change the current in the fuse. The neutral wire stays near zero potential, not above the live wire.
✓ 1
Answer D
Answer: D
Exam tip: Switches and fuses always go in the live wire so that the appliance is not at a dangerous potential when it is off.
138IGPC3-4-07r39-08Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A floodlight on a bridge has a power of 500 W and is on for 6.0 hours each night. Electrical energy costs €0.30 per kWh. What is the cost of running the floodlight for one night?
Show mark scheme
Marking point
Mark
Notes
Energy = 0.500 kW × 6.0 h = 3.0 kWh, so the cost is 3.0 × €0.30 = €0.90. €900 comes from using 500 W instead of 0.500 kW. €0.15 comes from forgetting to multiply by the time. €3.60 comes from dividing the time by the power.
✓ 1
Answer B
Answer: B
Common error: The kilowatt-hour needs power in kW and time in hours; convert watts to kilowatts first.
139IGPC3-4-07r39-09Charge, current, potential difference, resistance and powerPaper CEasy6 marks
A heater keeps the camera of a space probe warm. The heater has a resistance of 18 Ω and is connected to a 24 V supply.
Circuit diagram
(a)
Calculate the current in the heater.
(2)
(b)
Calculate the power of the heater. Use the unrounded value of the current from (a).
(2)
(c)
The heater is on for 15 minutes. Calculate the energy transferred, in joules.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
I = V / R
✓ 1
= 24 / 18 = 1.33 A
✓ 1
Part (b)
P = I V (or V²/R)
✓ 1
= 32 W
✓ 1
Part (c)
E = P t with t = 900 s
✓ 1
= 32 × 900 = 28 800 J (accept 2.88×104 J)
✓ 1
Common error: Change minutes into seconds before using E = P t.
140IGPC3-4-07r39-11Series and parallel circuitsPaper CEasy5 marks
Two lamps, P and Q, are connected in parallel to a 6.0 V battery to make a warning indicator on a bridge. An ammeter in the lead from the battery reads 0.90 A. The current in lamp P is 0.60 A.
Circuit diagram
(a)
Calculate the current in lamp Q.
(1)
(b)
Calculate the resistance of lamp Q.
(2)
(c)
Suggest two advantages of connecting the lamps in parallel rather than in series.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.90 − 0.60 = 0.30 A
✓ 1
Part (b)
p.d. across Q is 6.0 V and R = V / I
✓ 1
= 6.0 / 0.30 = 20 Ω
✓ 1
Part (c)
Each lamp has the full battery p.d., so each is as bright as designed
✓ 1
If one lamp fails the other stays lit (accept: lamps can be switched separately)
✓ 1
Exam tip: In a parallel circuit the p.d. across each branch equals the supply p.d., and the branch currents add to the supply current.
141IGPC3-4-08r42-01Magnets and magnetic fieldsPaper CEasy1 mark
A small plotting compass is placed on the axis of a long bar magnet, very close to its north pole, where the magnet's field is much stronger than the Earth's field. In which direction does the north pole of the compass needle point?
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The north pole of a compass needle points along the field direction, and magnetic field lines leave a north pole. Close to the north pole of a strong magnet the needle therefore points directly away from it, since like poles repel. Pointing towards the north pole would happen near a south pole. Pointing at right angles is not the direction of the field on the axis. The Earth's field is negligible here.
✓ 1
Answer C
Answer: C
Common error: Field lines point away from a north pole and towards a south pole; a compass needle follows them.
142IGPC3-4-08r42-02Charge, current, potential difference, resistance and powerPaper CEasy1 mark
A pump motor in an irrigation system carries a current of 2.5 A for 40 s. How much charge passes through the motor in this time?
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Charge = current × time = 2.5 × 40 = 100 C. The value 16 C comes from dividing the time by the current. The value 0.0625 C comes from dividing the current by the time. The value 42.5 C comes from adding the two numbers.
✓ 1
Answer A
Answer: A
Exam tip: Write Q = It first; then check the units: amperes multiplied by seconds give coulombs.
143IGPC3-4-08r42-09Magnets and magnetic fieldsPaper CEasy5 marks
A farmer uses a bar magnet to collect steel nails from a barn floor.
(a)
State what is meant by a magnetic field.
(1)
(b)
Describe how a plotting compass can be used to find the direction of the magnetic field around the bar magnet.
(2)
(c)
The magnet is replaced by an electromagnet so that the nails can be released when it is switched off. Suggest which of soft iron or steel is the better material for its core, and give a reason.
(2)
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Part (a)
A region in which a magnet (or magnetic material) experiences a force
✓ 1
Part (b)
Place the compass at a point near the magnet; the direction in which its north pole points shows the field direction there
✓ 1
repeat at other positions and join the marks to build up a field line
✓ 1
Part (c)
Soft iron
✓ 1
it is easily magnetised and loses its magnetism when the current is switched off
✓ 1
Common error: Steel keeps its magnetism, so it is used for permanent magnets, not for cores of electromagnets.
144IGPC3-4-08r42-14Electromagnetic induction, motors and transformersPaper CEasy4 marks
A flight simulator uses a small d.c. motor in which a coil carrying a current lies between the poles of two magnets.
(a)
State what causes the coil to turn.
(1)
(b)
State two ways to make the motor turn faster.
(2)
(c)
State what happens to the direction of rotation if the current is reversed.
(1)
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Part (a)
A force acts on a current-carrying conductor in a magnetic field
✓ 1
Part (b)
Any one from: increase the current / use stronger magnets / increase the number of turns on the coil
✓ 1
a different second change from the same list
✓ 1
Part (c)
The direction of rotation reverses
✓ 1
Exam tip: Give two different changes when asked for two; repeating the same idea in other words earns only one mark.
145IGPC-038-04-03Electromagnetic induction, motors and transformersPaper CMedium1 mark
A student moves a bar magnet into a coil connected to a voltmeter and records the maximum reading. Which change would not give a larger maximum reading?
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The induced e.m.f. depends on how quickly the magnetic field through the coil changes. It also depends on the number of turns. More turns, faster movement and a stronger magnet all increase the e.m.f. A magnet held still produces no change in field, so no e.m.f. is induced, however long it stays there.
✓ 1
Answer D
Answer: D
Common error: Induced e.m.f. depends on the rate of change of field, not on the time the magnet spends inside the coil.
146IGPC-038-01-05Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A hair dryer has a power rating of 1.2 kW and is connected to a 230 V mains supply. What is the current in the hair dryer?
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P = IV, so I = P / V = 1200 / 230 = 5.2 A. 0.19 A results from calculating V / P. 190 A comes from dividing 230 by 1.2 without converting kW to W. 2.8 × 105 A comes from multiplying P by V.
✓ 1
Answer B
Answer: B
Common error: Convert kilowatts to watts (1 kW = 1000 W) before using P = IV.
147IGPC-038-01-07Charge, current, potential difference, resistance and powerPaper CMedium1 mark
Two different lamps, X and Y, are connected in series with a battery. Lamp X is brighter than lamp Y. The current in lamp X is 0.40 A. What is the current in lamp Y?
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In a series circuit there is only one path, so the same charge flows through every component each second. The current in Y is therefore 0.40 A. Current is not used up: energy is transferred in the lamps, but charge is not lost. Current divides only between parallel branches. Lamp X is brighter because it has a larger pd across it, not a larger current.
✓ 1
Answer C
Answer: C
Common error: Components use energy, not current. In a series circuit the current is the same everywhere, even when the lamps have different brightness.
148IGPC-038-01-10Charge, current, potential difference, resistance and powerPaper CMedium1 mark
The potential difference across a filament lamp is gradually increased from zero. Which statement describes how the resistance of the lamp changes, and why?
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As the current increases, the filament gets hotter, and the resistance of the metal increases as its temperature rises. This is why the current–pd graph for a lamp curves, with the current rising less and less for each extra volt. A constant resistance applies only to a resistor at constant temperature. Resistance that decreases as temperature rises describes a thermistor, not a filament.
✓ 1
Answer B
Answer: B
Common error: Ohm's law behaviour (current proportional to pd) applies only when the temperature stays constant. A lamp filament does not stay at constant temperature.
149IGPC-038-01-14Charge, current, potential difference, resistance and powerPaper CMedium5 marks
A 12 Ω resistor and an 18 Ω resistor are connected in series with a 6.0 V battery. The battery has negligible internal resistance.
Diagram not to scale
(a)
Calculate the combined resistance of the two resistors.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
Calculate the potential difference across the 18 Ω resistor.
(2)
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Part (a)
R = 12 + 18 = 30 Ω .
✓ 1
Part (b)
I = V / R . I = 6.0 / 30 = 0.20 A .
✓ 2
Part (c)
V = IR = 0.20 × 18 . V = 3.6 V .
✓ 2
Exam tip: Check your answer: the pd across the 12 Ω resistor is 2.4 V, and 2.4 V + 3.6 V = 6.0 V, which equals the battery pd.
150IGPC-038-03-04Electrical safetyPaper CMedium1 mark
An electric kettle is rated 2.3 kW, 230 V. Fuses rated 3 A, 5 A, 10 A and 13 A are available. Which fuse is the most suitable for the kettle?
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The normal current is I = P/V = 2300/230 = 10 A. The fuse rating should be just above the normal current, so 13 A is the right choice. A 10 A fuse is too close to the working current and would melt during normal use. The 3 A and 5 A fuses are below the working current and would melt as soon as the kettle was switched on.
✓ 1
Answer D
Answer: D
Common error: Picking the fuse rating that equals the working current is wrong. Choose the next rating above it.
151IGPC-038-03-05Electrical safetyPaper CMedium1 mark
A toaster has a metal case that is connected to the earth wire. A fault makes the live wire touch the metal case. What happens?
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The earth wire has a very low resistance, so it gives the current an easy path to the ground. A very large current flows, the fuse melts and the toaster is cut off from the live supply. The case therefore does not stay live. A fuse never lowers the current to a safe working value; it breaks the circuit completely.
✓ 1
Answer A
Answer: A
Exam tip: To explain earthing fully, give the whole chain: low-resistance path, then large current, then fuse melts, then appliance disconnected.
152IGPC-038-03-07Electrical safetyPaper CMedium1 mark
Many homes now use circuit breakers (trip switches) instead of fuses. What is an advantage of a circuit breaker compared with a fuse?
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After it trips, a circuit breaker can simply be switched back on once the fault has been fixed. A fuse has to be replaced. Like a fuse, a breaker must be in the live wire. It is chosen to trip at a suitable current, not a larger one. It does not change the supply voltage.
✓ 1
Answer A
Answer: A
Exam tip: Other advantages of circuit breakers are that they act quickly and that it is easy to see which circuit has tripped.
153IGPC-038-03-14Electrical safetyPaper CMedium6 marks
A washing machine has a metal case that is connected to the earth wire. It is rated 2.5 kW and works from the 230 V mains.
(a)
A loose live wire inside the machine touches the metal case. Explain how the earth wire and the fuse together protect a person who touches the case.
(3)
(b)
Calculate the normal working current of the washing machine.
(2)
(c)
Suggest a suitable fuse rating from 3 A, 5 A and 13 A.
(1)
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Part (a)
The earth wire provides a low-resistance path from the case to the ground ; a very large current flows through the live and earth wires ; the fuse melts, disconnecting the live supply so the case is no longer live
✓ 3
Part (b)
I = P/V = 2500/230 ; I = 11 A (10.9 A)
✓ 2
Part (c)
13 A
✓ 1
Common error: Do not write that the current "goes to earth instead of through the person" and stop there. You must also say that the fuse melts and disconnects the supply.
154IGPC-038-00-04Magnets and magnetic fieldsPaper CMedium1 mark
An engineer needs one material for a permanent magnet in a loudspeaker and another for the core of an electromagnet in a door lock. Which choice is correct?
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Steel is hard to demagnetise, so it keeps its magnetism and suits a permanent magnet. Soft iron magnetises easily and loses its magnetism as soon as the current is switched off. That makes soft iron right for an electromagnet core, because the lock must release when the current stops.
✓ 1
Answer B
Answer: B
Exam tip: Link soft iron with 'temporary' and steel with 'permanent'.
155IGPC-038-00-05Magnets and magnetic fieldsPaper CMedium1 mark
A student has a metal bar and wants to prove that it is a magnet rather than just a magnetic material. Which observation proves this?
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Repulsion only happens between two magnets, with like poles facing. An unmagnetised piece of iron would be attracted by either pole of a magnet. So attraction cannot prove the bar is a magnet, but repulsion does.
✓ 1
Answer C
Answer: C
Exam tip: 'Repulsion is the only true test for a magnet' is a statement that earns marks.
156IGPC-038-00-07Magnets and magnetic fieldsPaper CMedium1 mark
A plotting compass is placed a short distance beyond the north pole of a bar magnet, on the line through the centre of the magnet. In which direction does the north-seeking end of the compass needle point?
The compass lies on the line through the centre of the magnet. Diagram not to scale
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The north-seeking end of the needle points along the field direction. Field lines leave the north pole of the magnet. Beyond the N pole, they point away from the magnet. Like poles repel, so the needle's N end is pushed away from the magnet's N pole.
✓ 1
Answer B
Answer: B
Exam tip: A compass needle's north end always points along the field line, in the direction of the arrow.
157IGPC-038-00-08Magnets and magnetic fieldsPaper CMedium1 mark
A long straight vertical wire passes through a horizontal sheet of card. There is a steady current in the wire. Which describes the magnetic field pattern on the card?
Side view of the wire passing through the horizontal card.
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The field around a long straight current-carrying wire forms concentric circles centred on the wire. The circles lie in a plane at right angles to the wire. Radiating lines would mean the field points directly away from the wire, which is wrong.
✓ 1
Answer B
Answer: B
Common error: Do not confuse the circular field around a wire with the radial lines of an electric field around a charge.
158IGPC-038-02-05Series and parallel circuitsPaper CMedium1 mark
The lamps in a house are connected in parallel to the mains supply. What is an advantage of this arrangement?
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Each lamp is on its own branch connected directly across the supply. A broken lamp opens only its own branch, so current still flows through the others. Each lamp can also be switched on and off independently. In parallel, the combined resistance is smaller, not larger. The supply voltage is not shared, because each lamp receives the full pd. Adding a lamp increases the supply current.
✓ 1
Answer B
Answer: B
Exam tip: When asked for advantages of parallel lighting, give two separate points: independent switching, and the other lamps keep working if one fails. Each lamp also receives the full supply voltage.
159IGPC-038-02-07Series and parallel circuitsPaper CMedium1 mark
A torch contains three 1.5 V cells connected in series. One cell has been inserted the wrong way round. What is the combined e.m.f. of the cells?
The long line of each cell symbol is its positive terminal.
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Notes
The reversed cell's e.m.f. opposes the other two, so it must be subtracted: 1.5 + 1.5 − 1.5 = 1.5 V. 4.5 V ignores the reversed cell. 3.0 V simply leaves the reversed cell out. 0 V assumes the reversed cell cancels both of the others.
✓ 1
Answer B
Answer: B
Exam tip: Combined e.m.f. of cells in series: add the e.m.f.s of cells facing the same way, then subtract any that face the opposite way.
160IGPC-038-04-12Electromagnetic induction, motors and transformersPaper CMedium7 marks
A transformer in a model railway controller is connected to the 230 V a.c. mains supply. Its primary coil has 2000 turns. The output voltage is 6.9 V.
Diagram not to scale
(a)
State whether this is a step-up or a step-down transformer.
(1)
(b)
Calculate the number of turns on the secondary coil.
(2)
(c)
Explain why the transformer would not give a steady output if it were connected to a d.c. supply.
(2)
(d)
State a suitable material for the core of the transformer and give a reason for your choice.
(2)
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Part (a)
Step-down
✓ 1
Part (b)
Ns = Np × Vs/Vp or 2000 × 6.9/230
✓ 1
Ns = 60 turns
✓ 1
Part (c)
A d.c. supply gives a constant current, so the magnetic field in the core is constant / unchanging
✓ 1
An e.m.f. is induced in the secondary only while the field is changing, so there is no output except a brief pulse when the supply is switched on or off
✓ 1
Part (d)
Soft iron (accept iron or another magnetically soft material, e.g. ferrite; reject steel)
✓ 1
Any one reason: it is easily magnetised and demagnetised, so it follows the alternating field; OR it concentrates the magnetic field and links it from the primary coil to the secondary coil
✓ 1
Exam tip: Write the equation Vp/Vs = Np/Ns before substituting. This can earn a mark even if you make an arithmetic slip.
161IGPC-038-02-15Series and parallel circuitsPaper CMedium6 marks
A student has two identical lamps and a 3.0 V battery of negligible internal resistance. She connects the lamps first in series with the battery, and then in parallel with the battery.
Diagram not to scale
(a)
State the potential difference across each lamp when they are connected in series.
(1)
(b)
State the potential difference across each lamp when they are connected in parallel.
(1)
(c)
State the arrangement in which the lamps are brighter, and explain why.
(2)
(d)
Each lamp has a resistance of 5.0 Ω. Assume this stays constant. Calculate the current in the battery for the series arrangement.
(2)
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Part (a)
1.5 V (the pd is shared equally between identical lamps)
✓ 1
Part (b)
3.0 V
✓ 1
Part (c)
Parallel
✓ 1
Each lamp has a larger pd across it (3.0 V rather than 1.5 V), so there is a larger current in each lamp / more power is transferred in each lamp (either explanation)
✓ 1
Part (d)
Total resistance = 5.0 + 5.0 = 10 Ω
✓ 1
I = 3.0 / 10 = 0.30 A
✓ 1
Exam tip: When you explain brightness, link it to the current in, pd across, or power of the individual lamp, not just the circuit as a whole.
162IGPC-038-03-06Electrical safetyPaper CMedium1 mark
A hairdryer is double insulated. Its plug is connected by a cable with only two wires and no earth wire. Why is it safe to use the hairdryer without an earth wire?
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In a double-insulated appliance, the live parts are insulated and also enclosed in an insulating (usually plastic) outer case. Even if a wire comes loose inside, no part a user can touch becomes live, so no earth connection is needed. Option C is therefore correct. A is wrong because a fuse melts only when the current becomes too large. It protects the cable from overheating but does not prevent a shock. B is wrong because a hairdryer draws several amperes from the mains, which is far more than enough to be dangerous. D is wrong because the appliance still runs at the full mains voltage.
✓ 1
Answer C
Answer: C
Common error: Do not say the fuse is what makes a double-insulated appliance safe. The fuse only melts when the current is too large. Protection from shock comes from the two layers of insulation, the insulated live parts and the insulating case, which keep every touchable surface from becoming live.
163IGPC3-4-04c-03Electrical safetyPaper CMedium1 mark
An electric kettle is rated 1500 W and is used on a 230 V mains supply. The plug can be fitted with a 3 A, 5 A, 13 A or 30 A fuse. Which fuse is most suitable?
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Current = P / V = 1500 / 230 = 6.5 A. The fuse must be just above this, so 13 A (C). A and B are below the normal current and would blow immediately. D is far too high, so the cable could overheat before it melted.
✓ 1
Answer C
Answer: C
Exam tip: Calculate the normal current first, then choose the next fuse rating above it.
164IGPC3-4-04c-04Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A heating unit contains three identical 6.0 Ω elements. Two are connected in parallel with each other, and this pair is connected in series with the third. What is the total resistance of the unit?
Diagram not to scale
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The parallel pair has resistance 6.0 × 6.0 / (6.0 + 6.0) = 3.0 Ω; adding the series element gives 9.0 Ω (D). A adds only two of the elements. B treats all three as in series. C treats all three as in parallel.
✓ 1
Answer D
Answer: D
Exam tip: Reduce the parallel part to a single resistance before adding series parts.
165IGPC3-4-04c-12Series and parallel circuitsPaper CMedium8 marks
Two lamps are connected in parallel to a 6.0 V battery. Lamp A has a resistance of 12 Ω and lamp B has a resistance of 4.0 Ω.
Diagram not to scale
(a)
Calculate the current in each lamp.
(3)
(b)
State the current in the battery.
(1)
(c)
Calculate the combined resistance of the two lamps.
(2)
(d)
Lamp A fails and stops conducting. State what happens to the current in lamp B and to the current in the battery.
(2)
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Notes
Part (a)
lamp A: I = V / R = 6.0 / 12 = 0.50 A
✓ 1
lamp B: I = 6.0 / 4.0
✓ 1
I = 1.5 A
✓ 1
Part (b)
0.50 + 1.5 = 2.0 A
✓ 1
Part (c)
R = V / I = 6.0 / 2.0, ECF from (b)
✓ 1
R = 3.0 Ω
✓ 1
Part (d)
the current in lamp B is unchanged
✓ 1
the current in the battery decreases
✓ 1
Common error: The combined resistance of parallel resistors is smaller than the smallest individual resistance.
166IGPC3-4-06r6-04Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A heating element in a poultry house has a resistance of 24 Ω. The current in it is 6.0 A. What is the power of the element?
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Notes
P = I2R = 6.02 × 24 = 864 W, so C is correct. A is R/I. B is I × R, which is a potential difference of 144 V and not a power. D is I × R2, squaring the wrong quantity.
✓ 1
Answer C
Answer: C
Common error: Square the current, not the resistance, in P = I²R.
167IGPC3-4-06r6-07Series and parallel circuitsPaper CMedium1 mark
Three identical heaters in a barn are connected in parallel to a 230 V supply. One heater is switched off using its own switch. What happens to the current in the supply cable and to the potential difference across each remaining heater?
Three identical heaters in parallel, each with its own switch. Diagram not to scale
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Notes
Each parallel branch has the full 230 V supply, so the p.d. across the remaining heaters is unchanged. With one branch removed, the total current drawn is smaller, so A is correct. B wrongly says the p.d. is shared. C and D forget that the removed heater no longer draws current.
✓ 1
Answer A
Answer: A
Common error: In a parallel circuit each branch works independently. Removing one branch reduces the total current but leaves the other branches unchanged.
168IGPC3-4-06r6-09Magnets and magnetic fieldsPaper CMedium6 marks
A speaker maker tests two metal bars. One is a steel bar that may be magnetised and the other is an unmagnetised iron bar.
(a)
Describe how a known permanent magnet can be used to show that the steel bar is magnetised.
(2)
(b)
Explain why steel is used for the permanent magnets in the speakers, whereas soft iron is used in electromagnets.
(2)
(c)
Describe how a plotting compass can be used to show the shape of the magnetic field around a bar magnet.
(2)
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Part (a)
bring one pole of the known magnet close to each end of the steel bar
✓ 1
if the end of the bar is repelled, it is a magnet (attraction alone does not prove it)
✓ 1
Part (b)
steel is magnetically hard, so it keeps its magnetism
✓ 1
soft iron is easily magnetised and demagnetised, so the field can be switched off
✓ 1
Part (c)
place the compass near the magnet, mark the positions of the needle ends, then move the compass so one end sits at the previous mark and repeat
✓ 1
join the marks to give a field line, with the arrow pointing from the north pole to the south pole
✓ 1
Common error: Attraction does not prove a bar is a magnet, because an unmagnetised iron bar is also attracted. Only repulsion does.
169IGPC3-4-06r6-12Series and parallel circuitsPaper CMedium6 marks
Two heating cables for seed trays, of resistance 12 Ω and 4.0 Ω, are connected in parallel to a 6.0 V battery.
Diagram not to scale
(a)
Calculate the current in each cable.
(2)
(b)
The 12 Ω cable breaks. State what happens to the current in the 4.0 Ω cable and explain your answer.
(2)
(c)
Before the break, calculate the total current supplied by the battery.
(2)
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Notes
Part (a)
12 Ω cable: I = V / R = 6.0 / 12 = 0.50 A
✓ 1
4.0 Ω cable: I = 6.0 / 4.0 = 1.5 A
✓ 1
Part (b)
it stays the same at 1.5 A
✓ 1
the 4.0 Ω cable is still connected directly across the full 6.0 V of the battery
✓ 1
Part (c)
total current is the sum of the branch currents = 0.50 + 1.5 (ECF from (a))
✓ 1
2.0 A
✓ 1
Common error: In a parallel circuit the p.d. across each branch equals the battery p.d., and the total current is the sum of the branch currents.
170IGPC3-4-06r6-13Electrical safetyPaper CMedium5 marks
A ground crew uses a mains-powered floodlight with a metal casing on an airport apron. The plug contains a fuse.
(a)
State why the fuse is connected in the live wire.
(1)
(b)
Explain how the earth wire protects a person if the live wire touches the metal casing.
(2)
(c)
State why using the floodlight in wet conditions is more dangerous.
(1)
(d)
State one advantage of using a circuit breaker instead of a fuse.
(1)
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Notes
Part (a)
so that when the fuse melts, the appliance is disconnected from the live (high-voltage) supply
✓ 1
Part (b)
the casing is connected to earth, so the casing does not become live
✓ 1
a large current flows to earth through the earth wire, which melts the fuse and disconnects the supply
✓ 1
Part (c)
water lowers the resistance of the body (or provides a conducting path), so a larger current could pass through a person
✓ 1
Part (d)
a circuit breaker can be reset (it does not need replacing)
✓ 1
Exam tip: The fuse or circuit breaker must be in the live wire so that the appliance is isolated from the high voltage when it operates.
171IGPC3-4-07r6-03Electrical safetyPaper CMedium1 mark
A treadmill in a sports-science laboratory has a motor labelled 230 V, 1.8 kW. The plug can be fitted with a 3 A, 5 A, 13 A or 30 A fuse. Which fuse is most suitable?
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Notes
Normal current = 1800 / 230 = 7.8 A. The fuse must be rated just above this, so 13 A. 3 A and 5 A are below the normal current and would blow immediately. 30 A is far above the normal current, so it would not protect the flex from overheating before blowing.
✓ 1
Answer B
Answer: B
Common error: Choose the fuse rating just above the normal working current, not the largest available.
172IGPC3-4-07r6-04Series and parallel circuitsPaper CMedium1 mark
Two heaters in an observatory dome, of resistance 8.0 Ω and 24 Ω, are connected in parallel across a 12 V supply with negligible internal resistance. What is the total current supplied?
Diagram not to scale
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Notes
Current in 8.0 Ω = 12/8.0 = 1.5 A; in 24 Ω = 12/24 = 0.50 A; the total is 2.0 A. 1.5 A counts only one branch. 0.38 A comes from adding the resistances as if in series (12/32). 6.0 A is the combined resistance, not a current.
✓ 1
Answer B
Answer: B
Exam tip: In parallel, each branch has the full supply pd, so find each branch current and add them.
173IGPC3-4-07r6-08Magnets and magnetic fieldsPaper CMedium1 mark
A crane in a rail scrap yard uses an electromagnet to lift steel parts and release them when the current is switched off. Which core material and reason are correct?
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Notes
D is correct: soft iron magnetises and demagnetises easily, so the load drops when the current stops. A is wrong because a steel core would stay magnetised. B is wrong because copper is not magnetic. C is wrong because soft iron is not permanently magnetised.
✓ 1
Answer D
Answer: D
Common error: Soft iron is used where magnetism must be switched off; steel is used for permanent magnets.
174IGPC3-4-07r6-10Series and parallel circuitsPaper CMedium6 marks
In an observatory dome a 12 V supply with negligible internal resistance powers a heater of resistance 10 Ω and a fan of resistance 30 Ω. The two are connected in parallel, and an ammeter measures the current from the supply.
The ammeter measures the current from the supply.
(a)
Calculate the combined resistance of the heater and the fan.
(2)
(b)
Calculate the reading on the ammeter.
(2)
(c)
The fan is disconnected. Calculate the new ammeter reading.
(2)
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Part (a)
1/R = 1/10 + 1/30
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R = 7.5 Ω
✓ 1
Part (b)
I = V / R
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12 / 7.5 = 1.6 A (ECF from (a))
✓ 1
Part (c)
the pd across the heater is still 12 V because it is on its own branch of the supply
✓ 1
I = 12 / 10 = 1.2 A
✓ 1
Exam tip: Removing one parallel branch does not change the current in the other branches.
175IGPC3-4-07r6-11Magnets and magnetic fieldsPaper CMedium6 marks
A railway depot uses an electromagnet to lift steel rail clips. A technician also studies the field of a bar magnet.
(a)
State two ways to increase the strength of the electromagnet.
(2)
(b)
Describe how a plotting compass can be used to plot the magnetic field lines of the bar magnet.
(3)
(c)
State what the spacing of the field lines shows.
(1)
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Part (a)
increase the current in the coil
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increase the number of turns of the coil (accept: add a soft-iron core)
✓ 1
Part (b)
place the magnet on paper and draw round it, then put the compass near one pole and mark the positions of both ends of the needle
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move the compass so that one end of the needle is at the previous mark and mark the new position, repeating until the other pole is reached
✓ 1
join the marks to make a field line and repeat from other starting points
✓ 1
Part (c)
lines closer together where the field is stronger
✓ 1
Exam tip: Add arrows to any field line you draw; they point away from north and towards south.
176IGPC3-4-07r6-12Series and parallel circuitsPaper CMedium7 marks
A sports scientist builds a circuit to time a sprinter. A 12 V cell of negligible internal resistance is connected to a resistor X in series with two resistors in parallel, one of 6.0 Ω and one of 12 Ω. A voltmeter connected across the parallel pair reads 4.8 V.
The voltmeter is connected across the parallel pair of resistors.
(a)
Calculate the current in the 6.0 Ω resistor.
(2)
(b)
Calculate the current in the cell.
(2)
(c)
Calculate the resistance of X.
(3)
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Part (a)
I = V / R
✓ 1
4.8 / 6.0 = 0.80 A
✓ 1
Part (b)
current in the 12 Ω resistor = 4.8 / 12 = 0.40 A
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the branch currents add: 0.80 + 0.40 = 1.2 A
✓ 1
Part (c)
pd across X = 12 − 4.8 = 7.2 V
✓ 1
R = V / I
✓ 1
7.2 / 1.2 = 6.0 Ω (ECF from (b))
✓ 1
Common error: The pd across a series resistor is the supply pd minus the pd across the rest of the circuit, not the full supply pd.
177IGPC3-4-05r7-03Electrical safetyPaper CMedium1 mark
A kettle in a hospital kitchen is rated at 2.3 kW and is used on a 230 V mains supply. What current does the kettle draw?
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B is correct: I = P / V = 2300 / 230 = 10 A. A is wrong because it uses V / P, inverting the equation. C is wrong because 13 A is a typical fuse rating, not the current drawn. D is wrong because it multiplies P by V (2300 × 230).
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Answer B
Answer: B
Common error: Write P = IV with the quantities in base units first (kW to W) and rearrange before substituting.
178IGPC3-4-05r7-08Series and parallel circuitsPaper CMedium1 mark
A 6.0 V cell is connected in series to a 4.0 Ω resistor and an 8.0 Ω resistor. What is the potential difference across the 4.0 Ω resistor?
Diagram not to scale
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A is correct: I = 6.0 / 12 = 0.50 A, so V = 0.50 × 4.0 = 2.0 V. B is wrong because it shares the p.d. equally. C is wrong because 4.0 V is the p.d. across the 8.0 Ω resistor. D is wrong because it is the full cell p.d., which is shared between both resistors in series.
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Answer A
Answer: A
Exam tip: In a series circuit the larger resistor takes the larger share of the p.d.
179IGPC3-4-05r7-11Electrical safetyPaper CMedium9 marks
A bedside monitor in a hospital has a metal case. It is connected to a 240 V mains supply and has a power rating of 1.8 kW.
(a)
Explain how the earth wire connected to the metal case protects a patient if a fault makes the live wire touch the case.
(3)
(b)
State why the fuse is fitted in the live wire rather than the neutral wire.
(1)
(c)
Calculate the current in the monitor at its rated power. Give your answer in amperes.
(3)
(d)
Fuses are available with ratings of 3 A, 5 A, 10 A and 13 A. State which fuse is the most suitable and give a reason.
(2)
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Part (a)
A low-resistance path to earth lets a large current flow
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The large current makes the fuse melt (or the circuit breaker trip)
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The supply is disconnected so the case cannot stay at a dangerous potential
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Part (b)
When the fuse melts the appliance is completely disconnected from the live (high-voltage) supply
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Part (c)
I = P / V
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I = 1800 / 240
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I = 7.5 A
✓ 1
Part (d)
10 A fuse
✓ 1
It is the lowest rating above the normal current of 7.5 A (so it protects the wiring but does not melt in normal use) (ECF from (c))
✓ 1
Common error: A fuse rating must be just above the normal current; a rating far too high will not melt in time to protect the wiring.
180IGPC3-4-05r7-12Magnets and magnetic fieldsPaper CMedium6 marks
A beekeeper uses a bar magnet to collect steel pins dropped near the hives.
(a)
Describe how a plotting compass can be used to find the shape of the magnetic field around the bar magnet.
(3)
(b)
State how the field lines show where the magnetic field is strongest.
(1)
(c)
State two differences between magnetically soft iron and magnetically hard steel.
(2)
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Part (a)
Place the magnet on paper and the compass near one pole, then mark the position of both ends of the needle
✓ 1
Move the compass so that one end of the needle is at the previous end mark and mark the new position, repeating this many times
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Join the marks with a smooth line and repeat from other starting points to build up the pattern
✓ 1
Part (b)
Where the field lines are closest together
✓ 1
Part (c)
Soft iron is magnetised easily but steel is harder to magnetise
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Soft iron loses its magnetism easily but steel keeps its magnetism
✓ 1
Exam tip: When describing field plotting, mention that the needle points along the field direction and that you repeat the process to build the line.
181IGPC3-4-05r7-13Charge, current, potential difference, resistance and powerPaper CMedium6 marks
A rechargeable battery in a hospital infusion pump supplies a steady current of 0.20 A for 6.0 hours. The battery has an e.m.f. of 9.0 V and negligible internal resistance.
(a)
Calculate the charge that passes through the pump in 6.0 hours.
(2)
(b)
The resistance of the pump circuit is doubled while the battery stays the same. State what happens to the current and explain your answer.
(2)
(c)
Calculate the energy delivered by the battery during the 6.0 hours when the current is 0.20 A. Give your answer in joules.
(2)
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Part (a)
Q = It with t = 6.0 × 3600 = 21600 s
✓ 1
Q = 0.20 × 21600 = 4320 C
✓ 1
Part (b)
The current halves
✓ 1
I = V / R, so doubling R with the same V halves I
✓ 1
Part (c)
E = QV (or E = IVt)
✓ 1
E = 4320 × 9.0 = 38880 J (ECF from (a))
✓ 1
Common error: Do not leave the time in hours: charge and energy equations need seconds.
182IGPC3-4-06r9-03Electrical safetyPaper CMedium1 mark
A blender in a food-science laboratory has a fuse fitted in the live wire rather than in the neutral wire. Why is the fuse placed in the live wire?
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Correct: C. If the fuse were in the neutral wire, a fault could leave the appliance still connected to the live supply when the fuse melted. A is wrong because the neutral wire carries the same current as the live wire in normal use. B is wrong for the same reason. D is wrong because the earth wire, not the fuse, connects the casing to earth.
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Answer C
Answer: C
Common error: A fuse or switch must be in the live wire so that, once it has operated, the appliance is not left at a high potential.
183IGPC3-4-06r9-08Series and parallel circuitsPaper CMedium1 mark
A mountaineer's head torch has two identical lamps connected in parallel to one cell. One lamp's filament breaks, leaving a gap in its branch. What happens?
Two identical lamps in parallel to one cell, before the fault.
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Correct: D. Each parallel branch has the full cell p.d. and works independently, so the working lamp is unchanged and the cell now supplies only one lamp's current. A treats the lamps as if they were in series. B wrongly thinks the current is shared out. C confuses the change in total resistance with the p.d. across the remaining lamp, which is unchanged.
✓ 1
Answer D
Answer: D
Common error: In a parallel circuit each branch has the same p.d. as the supply, so removing one branch does not change the others.
184IGPC3-4-06r9-11Series and parallel circuitsPaper CMedium5 marks
A telecoms engineer tests a monitoring circuit in which a 15 Ω resistor and a 45 Ω resistor are connected in series to a 12 V battery.
Diagram not to scale
(a)
Calculate the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
Calculate the potential difference across the 45 Ω resistor.
(2)
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Part (a)
15 + 45 = 60 Ω
✓ 1
Part (b)
current = voltage ÷ resistance = 12 ÷ 60
✓ 1
0.20 A
✓ 1
Part (c)
V = IR = 0.20 × 45
✓ 1
9.0 V
✓ 1
Exam tip: In a series circuit the same current passes through every component, so find the current first.
185IGPC3-4-06r9-12Electrical safetyPaper CMedium7 marks
A kettle in a mountain hut kitchen is rated at 2300 W and is connected to the 230 V mains supply. It has a metal casing.
(a)
Explain how the earth wire and the fuse together protect a person if the live wire touches the casing.
(3)
(b)
Calculate the normal current in the kettle.
(2)
(c)
Fuses rated 3 A, 5 A and 13 A are available. State which should be used and give a reason. ECF from (b).
(2)
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Part (a)
a large current flows from the live wire through the casing and the earth wire to earth
✓ 1
the large current melts the fuse
✓ 1
the kettle is then disconnected from the live supply so the casing is no longer live
✓ 1
Part (b)
current = power ÷ voltage = 2300 ÷ 230
✓ 1
10 A
✓ 1
Part (c)
13 A
✓ 1
its rating is just above the normal current, so it will not melt in normal use but will melt in a fault
✓ 1
Common error: The fuse rating must be just above the normal working current; a fuse rated much higher would not protect the wiring.
186IGPC3-4-07r10-05Magnets and magnetic fieldsPaper CMedium1 mark
A simple d.c. motor in a food mixer turns clockwise when connected to a battery. Which single change makes it turn anticlockwise?
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The direction of the force depends on the directions of both the current and the magnetic field. Reversing the current alone reverses the force, so C is correct. Reversing both the current and the field leaves the force direction unchanged, so A is wrong. A larger current or stronger magnets only change the size of the force, not its direction.
✓ 1
Answer C
Answer: C
Common error: Reversing both the current and the field cancels out; only one of them must be reversed to reverse the force.
187IGPC3-4-07r10-14Magnets and magnetic fieldsPaper CMedium6 marks
A climber uses a magnetic compass to navigate on a mountain.
(a)
Explain why the needle of the compass settles pointing towards the north.
(2)
(b)
The needle swings towards a steel ice axe placed close to the compass. Suggest why.
(2)
(c)
The climber wants to find out whether the ice axe head is itself a permanent magnet. Describe a test using a bar magnet with marked poles.
(2)
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Part (a)
the Earth has a magnetic field
✓ 1
the needle is a small magnet that lines up with this field
✓ 1
Part (b)
steel is a magnetic material
✓ 1
the needle's magnet attracts the steel, so the needle turns towards it
✓ 1
Part (c)
bring one marked pole of the bar magnet near each end of the axe head
✓ 1
repulsion at either end shows it is a magnet, because unmagnetised steel is only attracted
✓ 1
Common error: Attraction alone does not show that an object is a magnet; only repulsion does.
188IGPC3-4-07r10-15Electromagnetic induction, motors and transformersPaper CMedium6 marks
A food mixer is driven by a simple d.c. motor. It has a rectangular coil between the poles of two permanent magnets, connected to a battery through a split-ring commutator and brushes.
(a)
State why the coil experiences a force when the current is switched on.
(1)
(b)
State two changes that would make the coil turn faster.
(2)
(c)
Explain the purpose of the split-ring commutator.
(2)
(d)
State what happens when the connections to the battery are reversed.
(1)
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Part (a)
a current-carrying conductor in a magnetic field experiences a force
✓ 1
Part (b)
increase the current
✓ 1
use stronger magnets
✓ 1
(accept more turns on the coil)
—
Part (c)
it reverses the current in the coil every half turn
✓ 1
so the forces keep the coil turning in the same direction
✓ 1
Part (d)
the coil turns in the opposite direction
✓ 1
Common error: Without a commutator the coil would swing back and forth instead of turning continuously.
189IGPC3-4-05r11-03Series and parallel circuitsPaper CMedium1 mark
Two identical lamps in a mountain hut are connected in parallel to a battery of negligible internal resistance. The filament of one lamp breaks. What happens to the brightness of the other lamp and to the current drawn from the battery?
Two identical lamps in parallel, before the filament breaks.
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Each parallel branch has the full battery potential difference across it, so the working lamp keeps the same current and brightness. The battery now supplies only one branch, so its current halves. A is wrong because the lamp does not receive a larger share of the current. C is wrong because the broken branch no longer carries any current. D is wrong because the lamp's potential difference has not fallen.
✓ 1
Answer B
Answer: B
Common error: In a parallel circuit, removing one branch does not change the potential difference across the other branches.
190IGPC3-4-05r11-04Electrical safetyPaper CMedium1 mark
A metal-cased food mixer is connected to the mains supply through a fuse. Why is the fuse fitted in the live wire rather than in the neutral wire?
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If the fuse is in the live wire, melting breaks the connection to the high-potential supply and the appliance is left safe. B is wrong because live and neutral carry the same current in normal use. C is wrong because the earth wire, not the fuse, gives the path to earth. D is wrong because a fuse melts according to the current through it, not which wire it is in.
✓ 1
Answer A
Answer: A
Exam tip: If a fuse is in the neutral wire, the appliance can stay connected to the live supply even after the fuse blows.
191IGPC3-4-05r11-07Electromagnetic induction, motors and transformersPaper CMedium1 mark
A hand-cranked torch used on a mountain trek has a generator with a coil turning between magnets. Which change would increase both the peak e.m.f. and the frequency of the alternating output?
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Faster rotation makes the coil cut field lines more quickly, so the e.m.f. rises, and it completes more turns per second so the frequency rises. A raises the e.m.f. only, because the rotation rate is unchanged. C lowers the e.m.f. D changes only the coil resistance and not the induced e.m.f.
✓ 1
Answer B
Answer: B
Exam tip: Frequency depends only on how many rotations occur each second.
192IGPC3-4-05r11-08Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A food-science laboratory oven is rated at 2.0 kW. It is switched on for 3.5 hours. Electricity costs £0.28 per kWh. What is the cost of the electricity used?
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Energy = 2.0 kW × 3.5 h = 7.0 kWh, so the cost is 7.0 × £0.28 = £1.96. A divides the time by the power. B leaves out the time. C leaves out the power.
✓ 1
Answer D
Answer: D
Exam tip: Cost = power in kW × time in hours × price per kWh.
193IGPC3-4-05r11-12Electrical safetyPaper CMedium6 marks
A yoghurt incubator in a food-science laboratory is rated 460 W when connected to the 230 V mains supply.
(a)
Calculate the current in the incubator when it is working normally.
(2)
(b)
Calculate the resistance of the heating element.
(2)
(c)
The plug can be fitted with a 1 A, 3 A, 5 A or 13 A fuse. State which fuse is most suitable and explain your choice.
(2)
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Part (a)
P = IV so I = P / V = 460 / 230
✓ 1
= 2.0 A
✓ 1
Part (b)
R = V / I = 230 / 2.0
✓ 1
= 115 Ω, ECF from (a)
✓ 1
Part (c)
3 A
✓ 1
it is just above the normal current of 2.0 A, so it does not melt in normal use but melts if a fault causes a larger current
✓ 1
Exam tip: Choose the fuse with the lowest rating that is still above the normal working current.
194IGPC3-4-07r11-03Series and parallel circuitsPaper CMedium1 mark
Two identical lamps on a footbridge are connected in parallel to a 12 V battery that has negligible internal resistance. An ammeter is in the wire next to the battery. The filament of one lamp breaks. What happens?
Two identical lamps in parallel, before the filament breaks.
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B is correct: each parallel branch has the full 12 V, so the working lamp carries the same current and keeps the same brightness, while the total current falls to half. A is wrong because the broken branch no longer carries current. C treats the lamps as if they were in series. D is wrong because the pd across the remaining lamp has not changed.
✓ 1
Answer B
Answer: B
Common error: In a parallel circuit each branch is connected directly across the supply, so removing one branch does not change the others.
195IGPC3-4-07r11-04Electrical safetyPaper CMedium1 mark
A coffee roaster is rated at 2.0 kW when connected to a 230 V mains supply. Fuses rated 3 A, 5 A, 10 A and 13 A are available. Which fuse is the most suitable?
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Normal current I = P / V = 2000 / 230 = 8.7 A. The best fuse is the one rated just above this, so 10 A is correct. 3 A and 5 A are below the normal current and would melt in normal use. 13 A would let a current well above normal flow before melting, so protection is poorer.
✓ 1
Answer C
Answer: C
Exam tip: Choose the fuse rating just above the normal working current, not simply the largest one available.
196IGPC3-4-07r11-11Electrical safetyPaper CMedium7 marks
A metal-cased coffee roaster is rated at 2.76 kW when connected to a 230 V supply through a three-core cable. The plug contains a fuse. Fuses rated 3 A, 5 A and 13 A are available.
(a)
Calculate the normal current in the roaster.
(2)
(b)
State which fuse should be used and give a reason.
(2)
(c)
Explain how the earth wire and the fuse together protect a user if the live wire comes loose and touches the metal casing.
(3)
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Part (a)
P = IV so I = P / V
✓ 1
= 2760 / 230 = 12 A
✓ 1
Part (b)
13 A
✓ 1
it is just above the normal current of 12 A, so it carries the normal current but melts if the current becomes too large
✓ 1
Part (c)
the casing is connected to earth by the earth wire
✓ 1
a very large current flows from the live wire through the casing and earth wire
✓ 1
this melts the fuse, which breaks the circuit so the casing is no longer live
✓ 1
Exam tip: A fuse must be in the live wire so that the appliance is disconnected from the high potential when the fuse melts.
197IGPC3-4-07r11-12Magnets and magnetic fieldsPaper CMedium6 marks
A bar magnet lies on a sheet of paper on a bench. A plotting compass is available.
(a)
Describe how to use the plotting compass to find the shape and direction of the magnetic field of the bar magnet.
(3)
(b)
A soft iron rod is placed with one end near the north pole of the bar magnet. State the polarity induced at the end of the rod nearest the magnet and whether the rod is attracted or repelled.
(2)
(c)
State why steel rather than soft iron is used to make permanent magnets.
(1)
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Notes
Part (a)
place the compass near one pole and mark the positions of both ends of the needle with dots
✓ 1
move the compass so that one end of the needle is on the last dot and mark a new dot; repeat many times and join the dots to give a field line
✓ 1
direction is shown by the north pole of the needle, which points away from the north pole of the bar magnet
✓ 1
Part (b)
south pole
✓ 1
attracted
✓ 1
Part (c)
steel keeps its magnetism / is difficult to demagnetise, whereas soft iron loses its magnetism easily
✓ 1
Exam tip: The north pole of a plotting compass is attracted to a south pole, so it points along the field direction.
198IGPC3-4-07r11-13Charge, current, potential difference, resistance and powerPaper CMedium5 marks
A space probe's battery supplies a constant current of 0.80 A to its instruments for 3.0 hours. The potential difference across the battery is 28 V.
(a)
State what is meant by a potential difference of 28 V.
(1)
(b)
Calculate the charge that passes through the instruments in the 3.0 hours.
(2)
(c)
Calculate the energy transferred by the battery in this time. Give your answer in joules.
(2)
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Part (a)
28 J of energy is transferred for each coulomb of charge passing
✓ 1
Part (b)
t = 3.0 × 3600 = 10 800 s
✓ 1
Q = It = 0.80 × 10 800 = 8640 C
✓ 1
Part (c)
E = QV = 8640 × 28
✓ 1
= 241 920 J (accept 2.4 × 105 J) (ECF from (b))
✓ 1
Common error: Use time in seconds in Q = It; hours give a charge 3600 times too small.
199IGPC3-4-07r11-14Charge, current, potential difference, resistance and powerPaper CMedium6 marks
A temperature sensor on a bridge consists of a 9.0 V battery, a fixed resistor of 150 Ω and a thermistor connected in series. At a certain temperature the thermistor has a resistance of 300 Ω.
Diagram not to scale
(a)
Calculate the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
State what happens to the current as the temperature rises and the resistance of the thermistor falls.
(1)
(d)
Calculate the potential difference across the thermistor.
(2)
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Part (a)
150 + 300 = 450 Ω
✓ 1
Part (b)
I = V / R = 9.0 / 450
✓ 1
= 0.020 A (ECF from (a))
✓ 1
Part (c)
current increases
✓ 1
Part (d)
V = IR = 0.020 × 300
✓ 1
= 6.0 V (ECF from (b))
✓ 1
Exam tip: In a series circuit the pd of the supply is shared in proportion to the resistances, so the larger resistor gets the larger pd.
200IGPC3-4-05r12-03Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A portable ventilator in a clinic draws a steady current of 0.40 A for 2.5 minutes. What charge passes through it in this time?
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Notes
Q = It with t = 2.5 × 60 = 150 s, so Q = 0.40 × 150 = 60 C. 1.0 C comes from using 2.5 without converting to seconds. 6.3 C comes from dividing time by current. 150 C is the time in seconds, not the charge.
✓ 1
Answer B
Answer: B
Common error: Convert minutes to seconds before using Q = It.
201IGPC3-4-05r12-08Charge, current, potential difference, resistance and powerPaper CMedium1 mark
The current through a filament lamp in a laboratory increases less than proportionally when the potential difference across it is raised. What is the correct explanation?
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A larger current heats the filament, so the metal ions vibrate with larger amplitude and the electrons collide more often: resistance increases and I/V falls. A is wrong because the filament does not thin appreciably. C is wrong because the number of free electrons is essentially unchanged. D is wrong because the lamp's own resistance is what changes.
✓ 1
Answer B
Answer: B
Exam tip: Link the change in resistance to temperature and to collisions between electrons and vibrating ions.
202IGPC3-4-05r12-10Magnets and magnetic fieldsPaper CMedium7 marks
A hospital laboratory uses small magnets to hold racks of steel sample trays.
(a)
State what happens when the north pole of one magnet is brought near the north pole of another magnet.
(1)
(b)
A plotting compass is used to draw the magnetic field lines around a bar magnet. Describe how the field line is drawn.
(3)
(c)
State what the direction of a magnetic field at a point shows.
(1)
(d)
The core of an electromagnet is made of soft iron, not steel. Explain why.
(2)
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Part (a)
The magnets repel
✓ 1
Part (b)
Place the compass near one pole and mark with a pencil dot the positions of both ends of the needle
✓ 1
Move the compass so that the end of the needle is on the previous dot and mark the next dot; repeat
✓ 1
Join the dots with a smooth line and add an arrow pointing from the north pole towards the south pole
✓ 1
Part (c)
The direction of the force on a north pole placed at that point
✓ 1
Part (d)
Soft iron is easily magnetised and demagnetised
✓ 1
so the magnetism can be switched off when the current is switched off (steel would stay magnetised)
✓ 1
Common error: Field lines show where a north pole would be pushed, so the arrows always leave a north pole.
203IGPC3-4-05r12-11Charge, current, potential difference, resistance and powerPaper CMedium6 marks
A display lamp in a shop is rated 12 V, 6.0 W. It is on for 10 hours each day.
(a)
Calculate the current in the lamp when it works at its rated power.
(2)
(b)
Calculate the energy transferred by the lamp in one day. Give your answer in kW h.
(2)
(c)
Electrical energy costs €0.30 per kW h. Calculate the cost of running the lamp for 30 days.
(2)
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Notes
Part (a)
I = P / V
✓ 1
I = 6.0 / 12 = 0.50 A
✓ 1
Part (b)
E = P t with P = 0.0060 kW and t = 10 h
✓ 1
E = 0.060 kW h
✓ 1
Part (c)
Energy in 30 days = 0.060 × 30 = 1.8 kW h
✓ 1
Cost = 1.8 × 0.30 = €0.54
✓ 1
Exam tip: Keep power in kW and time in hours to obtain kW h directly.
204IGPC3-4-05r12-13Electrical safetyPaper CMedium7 marks
A hospital steriliser is rated 230 V, 1.38 kW. Its plug may contain a 3 A, 5 A or 13 A fuse.
(a)
Calculate the current in the steriliser during normal use.
(2)
(b)
State which fuse should be used and give a reason.
(2)
(c)
The steriliser has a metal case connected to the earth wire. Explain how the earth wire protects a user if the live wire touches the case.
(3)
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Notes
Part (a)
I = P / V
✓ 1
I = 1380 / 230 = 6.0 A
✓ 1
Part (b)
13 A fuse
✓ 1
3 A and 5 A are below the 6.0 A normal current so they would blow during normal use (ECF from (a))
✓ 1
Part (c)
A large current flows through the low-resistance earth wire
✓ 1
This current is greater than the fuse rating so the fuse melts
✓ 1
which disconnects the live wire so the case is no longer at a high potential
✓ 1
Common error: The earth wire does not carry current in normal use; it provides a low-resistance path during a fault so the fuse blows.
205IGPC3-4-06r13-04Electromagnetic induction, motors and transformersPaper CMedium1 mark
A magnet is pushed into a coil connected to a sensitive voltmeter, and a small reading is obtained. Which change would increase the reading?
The magnet is pushed into the coil. Diagram not to scale
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Notes
Correct: a faster change of field gives a larger induced voltage. Fewer turns and a weaker magnet both reduce it, and a stationary magnet gives zero reading because there is no change.
✓ 1
Answer D
Answer: D
Common error: A magnet at rest inside a coil induces no voltage; the field must be changing.
206IGPC3-4-06r13-06Series and parallel circuitsPaper CMedium1 mark
Two lamps are connected in parallel to a 12 V supply. The ammeter at the supply reads 3.0 A and the current in lamp A is 1.2 A. What is the current in lamp B?
The ammeter at the supply reads 3.0 A; the current in lamp A is 1.2 A.
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Notes
Correct: currents in parallel branches add to the supply current, so 3.0 − 1.2 = 1.8 A. 1.2 A assumes equal sharing; 3.0 A assumes each lamp gets the full current; 4.2 A adds the values.
✓ 1
Answer A
Answer: A
Common error: In a parallel circuit the branch currents add up to the total current; the voltage across each branch is the same.
207IGPC3-4-06r13-10Series and parallel circuitsPaper CMedium6 marks
In an observatory, two resistors of 8.0 Ω and 4.0 Ω are connected in series to a 12 V supply.
Diagram not to scale
(a)
Calculate the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
Calculate the potential difference across the 4.0 Ω resistor.
(2)
(d)
A third resistor is added in series. State what happens to the current.
(1)
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Notes
Part (a)
8.0 + 4.0 = 12 Ω
✓ 1
Part (b)
I = V / R = 12 / 12
✓ 1
= 1.0 A
✓ 1
Part (c)
V = I R = 1.0 × 4.0
✓ 1
= 4.0 V
✓ 1
Part (d)
current decreases
✓ 1
Exam tip: Use the current from the series circuit for each component, since it is the same everywhere.
208IGPC3-4-06r13-12Charge, current, potential difference, resistance and powerPaper CMedium5 marks
A treadmill motor in a gym is rated 1.8 kW and operates from a 230 V supply. It is used for 1.5 hours. Electricity costs £0.30 per kWh.
(a)
Calculate the current in the motor. Give your answer to 2 significant figures.
(2)
(b)
Calculate the energy used in kWh.
(2)
(c)
Calculate the cost of using the treadmill for this time.
(1)
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Notes
Part (a)
I = P / V = 1800 / 230
✓ 1
= 7.8 A
✓ 1
Part (b)
energy = power × time = 1.8 × 1.5
✓ 1
= 2.7 kWh
✓ 1
Part (c)
2.7 × 0.30 = £0.81
✓ 1
Exam tip: Keep power in kW and time in hours when the answer is needed in kWh.
209IGPC3-4-06r15-03Series and parallel circuitsPaper CMedium1 mark
A bicycle light circuit has a 9.0 V battery connected in series to resistors of 2.0 Ω and 4.0 Ω. What is the potential difference across the 4.0 Ω resistor?
Diagram not to scale
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Notes
Total resistance is 6.0 Ω, so I = 9.0 ÷ 6.0 = 1.5 A and V = 1.5 × 4.0 = 6.0 V, so C is correct. A is the p.d. across the 2.0 Ω resistor. B assumes the p.d. is shared equally. D is the whole supply p.d.
✓ 1
Answer C
Answer: C
Common error: In series the p.d. is shared in proportion to resistance, not equally.
210IGPC3-4-06r15-04Electrical safetyPaper CMedium1 mark
A cooling-fan unit in a data centre is rated at 1.1 kW and is used on a 230 V supply. Fuses of 3 A, 5 A, 10 A and 13 A are available. Which fuse is most suitable?
Show mark scheme
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Notes
I = P/V = 1100 ÷ 230 = 4.8 A. The best fuse has a rating just above this, so B (5 A) is correct. A would melt in normal use. C and D would allow a current well above normal before melting.
✓ 1
Answer B
Answer: B
Exam tip: Choose the fuse with the lowest rating that is still above the normal operating current.
211IGPC3-4-06r15-08Charge, current, potential difference, resistance and powerPaper CMedium1 mark
In an archaeological survey a probe passes a current of 0.030 A through soil when a p.d. of 4.5 V is applied. What is the resistance of the soil between the probes?
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Notes
R = V/I = 4.5 ÷ 0.030 = 150 Ω, so B is correct. A is I ÷ V. C is V × I. D is a power-of-ten slip in the current.
✓ 1
Answer B
Answer: B
Exam tip: Write the equation R = V/I and substitute before calculating.
212IGPC3-4-06r15-10Series and parallel circuitsPaper CMedium7 marks
Two cooling fans, P and Q, in a data centre are connected in parallel to a 12 V supply. Fan P has resistance 24 Ω and fan Q has resistance 12 Ω.
Diagram not to scale
(a)
State the potential difference across fan Q.
(1)
(b)
Calculate the current in each fan.
(3)
(c)
Calculate the total current taken from the supply.
(1)
(d)
Give two reasons why the fans are connected in parallel rather than in series.
(2)
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Notes
Part (a)
12 V
✓ 1
Part (b)
I = V/R
✓ 1
fan P: 12 ÷ 24 = 0.50 A
✓ 1
fan Q: 12 ÷ 12 = 1.0 A
✓ 1
Part (c)
0.50 + 1.0 = 1.5 A , ECF from (b)
✓ 1
Part (d)
if one fan fails the other keeps working
✓ 1
each fan receives the full supply p.d.
✓ 1
Exam tip: In a parallel circuit the p.d. across each branch equals the supply p.d.; the branch currents add.
213IGPC3-4-06r15-11Electrical safetyPaper CMedium8 marks
An archaeology team uses a mains floodlight rated at 1.84 kW on a 230 V supply at a dig site.
(a)
State the function of a fuse in a circuit.
(1)
(b)
Explain how an earth wire protects the user if the live wire touches the metal case.
(3)
(c)
Calculate the current in the floodlight.
(2)
(d)
Fuses of 3 A, 5 A and 13 A are available. State which should be used and why.
(2)
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Notes
Part (a)
contains a thin wire that melts when the current exceeds its rating, breaking the circuit
✓ 1
Part (b)
earth wire connects the metal case to earth
✓ 1
a large current flows through the earth wire
✓ 1
the fuse melts and disconnects the supply, so the case cannot remain live
✓ 1
Part (c)
I = P/V = 1840 ÷ 230
✓ 1
8.0 A
✓ 1
Part (d)
13 A , ECF from (c)
✓ 1
its rating is the lowest one above the normal current of 8.0 A
✓ 1
Common error: A fuse goes in the live wire; its rating must be above the normal current.
214IGPC3-4-07r16-03Series and parallel circuitsPaper CMedium1 mark
A fashion-show dress has three identical LEDs connected in parallel to one battery. One LED breaks, leaving a gap in its branch. What happens to the other two LEDs?
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Notes
Each parallel branch is connected directly across the battery, so each remaining LED still has the same p.d. and the same current and keeps its brightness. They would go out only in a series circuit. Dimmer or brighter would need the p.d. across each LED to change, which does not happen when a parallel branch is removed.
✓ 1
Answer C
Answer: C
Exam tip: In a parallel circuit each branch works independently of the others; in a series circuit one break stops everything.
215IGPC3-4-07r16-04Electrical safetyPaper CMedium1 mark
A robot charging unit is connected to the 230 V mains supply and has a power rating of 750 W. Fuses of 3 A, 5 A, 13 A and 30 A are available. Which fuse is most suitable?
Show mark scheme
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Notes
I = P / V = 750 / 230 = 3.3 A (3.26 A). The fuse rating must be just above the normal current, so 5 A is correct. A 3 A fuse is below the normal current and would melt during normal use. 13 A is the standard-plug rating but is far above the normal current, so it would not protect the unit well. 30 A is far too high.
✓ 1
Answer D
Answer: D
Exam tip: Choose the fuse with the smallest rating that is still above the normal working current.
216IGPC3-4-07r16-10Series and parallel circuitsPaper CMedium9 marks
A jacket has two separate light strips, A and B, connected in parallel to one 6.0 V battery pack. Strip A has a resistance of 15 Ω and strip B has a resistance of 30 Ω.
The two light strips are shown as resistors. Diagram not to scale
(a)
Calculate the current in strip A.
(2)
(b)
Calculate the current in strip B and the total current taken from the battery pack.
(2)
(c)
The jacket is worn for 30 minutes with both strips on. Calculate the energy transferred by the battery pack in this time.
(3)
(d)
Strip A is disconnected. State what happens to the current in strip B and give a reason.
(2)
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Notes
Part (a)
M1 I = V / R = 6.0 / 15
✓ 1
A1 0.40 A
✓ 1
Part (b)
A1 current in B = 6.0 / 30 = 0.20 A
✓ 1
A1 total current = 0.40 + 0.20 = 0.60 A (ECF from (a))
✓ 1
Part (c)
M1 E = IVt (or E = Pt with P = IV)
✓ 1
M1 t = 30 × 60 = 1800 s
✓ 1
A1 6480 J (accept 6500 J; ECF from (b))
✓ 1
Part (d)
B1 the current in strip B stays the same (0.20 A)
✓ 1
B1 strip B still has the full 6.0 V of the battery pack across it, because parallel branches work independently
✓ 1
Exam tip: In a parallel circuit the p.d. across each branch equals the supply p.d., and the branch currents add up to the total current.
217IGPC3-4-07r16-13Electrical safetyPaper CMedium8 marks
A fish-smoking oven in a processing plant has a metal case. It is rated at 2300 W and is connected to the 230 V mains supply through a plug containing a fuse.
(a)
Describe how a fuse protects the oven circuit.
(2)
(b)
Calculate the normal current in the oven. Fuses of 3 A, 5 A and 13 A are available; state which one should be used.
(3)
(c)
Explain how connecting the metal case to the earth wire makes the oven safer.
(3)
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Notes
Part (a)
B1 the fuse contains a thin wire that heats up and melts when the current is larger than its rating
✓ 1
B1 this breaks the circuit and stops the current
✓ 1
Part (b)
M1 I = P / V = 2300 / 230
✓ 1
A1 10 A
✓ 1
B1 13 A fuse, the only one with a rating above the normal current
✓ 1
Part (c)
B1 the earth wire gives a low-resistance path from the case to earth
✓ 1
B1 if the live wire touches the case, a large current flows through the earth wire
✓ 1
B1 this melts the fuse and disconnects the oven, so the case does not stay live
✓ 1
Exam tip: The fuse is placed in the live wire so that when it melts the appliance is disconnected from the live supply.
218IGPC3-4-08r21-04Electrical safetyPaper CMedium1 mark
A treadmill in a sports-science laboratory has a metal casing connected to the earth wire. Why is this connection a safety feature?
Show mark scheme
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Notes
If a fault connects the live wire to the casing, the low-resistance earth wire carries a large current, which melts the fuse and disconnects the supply, so the casing does not stay live; D is correct. A, B and C describe effects that an earth wire does not have.
✓ 1
Answer D
Answer: D
Exam tip: The earth wire only carries current when there is a fault.
219IGPC3-4-08r21-07Electromagnetic induction, motors and transformersPaper CMedium1 mark
A speed sensor on a train wheel consists of a coil beside a toothed iron wheel and a fixed magnet. As each tooth passes, a pulse of e.m.f. is induced in the coil. The wheel now turns twice as fast. How do the size of each pulse and the number of pulses per second change?
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Notes
A faster change of magnetic field through the coil induces a larger e.m.f., and twice as many teeth pass each second, so D is correct. A, B and C each leave out one of these two effects.
✓ 1
Answer D
Answer: D
Common error: A faster rate of change of field raises the size of the induced e.m.f. as well as the frequency of the pulses.
220IGPC3-4-08r21-12Charge, current, potential difference, resistance and powerPaper CMedium5 marks
A carriage lighting circuit contains a 9.0 V battery connected in series with a lamp of resistance 12 Ω and a resistor of resistance 6.0 Ω.
Circuit diagram
(a)
Calculate the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(1)
(c)
Calculate the potential difference across the lamp.
(1)
(d)
Calculate the power transferred in the lamp.
(2)
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Notes
Part (a)
12 + 6.0 = 18 Ω
✓ 1
Part (b)
I = V ÷ R = 9.0 ÷ 18 = 0.50 A
✓ 1
Part (c)
V = I R = 0.50 × 12 = 6.0 V
✓ 1
Part (d)
P = I V = 0.50 × 6.0
✓ 1
3.0 W
✓ 1
Exam tip: In a series circuit the same current flows through every component; the battery p.d. is shared.
221IGPC3-4-08r21-13Electrical safetyPaper CMedium6 marks
A gym uses a 1400 W electric fan heater connected to a 230 V mains supply through a three-pin plug.
(a)
Describe how a fuse protects the heater circuit.
(2)
(b)
Explain why the fuse is fitted in the live wire and not in the neutral wire.
(2)
(c)
Calculate the current in the heater in normal use. Give your answer to 3 significant figures.
(2)
Show mark scheme
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Notes
Part (a)
the thin wire in the fuse heats up and melts if the current is larger than its rating
✓ 1
which breaks the circuit and stops the current
✓ 1
Part (b)
when the fuse melts it disconnects the appliance from the live (high-voltage) supply
✓ 1
if it were in the neutral wire the appliance would stay connected to the live wire and could be dangerous to touch
✓ 1
Part (c)
I = P ÷ V = 1400 ÷ 230
✓ 1
6.09 A
✓ 1
Common error: A fuse must be in the live wire; a fuse in the neutral wire leaves the appliance live after it blows.
222IGPC3-4-08r21-15Electromagnetic induction, motors and transformersPaper CMedium6 marks
A bicycle dynamo (generator) has a coil that rotates between the poles of a magnet. It is connected to a lamp.
(a)
Explain why an alternating e.m.f. is induced in the coil as it rotates.
(2)
(b)
The cyclist speeds up so that the coil rotates twice as fast. State how the maximum e.m.f. and the frequency change.
(2)
(c)
At the original speed the e.m.f. has a frequency of 12 Hz. Calculate the period of the e.m.f. after the coil rotates twice as fast. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
the coil cuts the magnetic field lines, so an e.m.f. is induced
✓ 1
the direction of cutting reverses every half turn, so the e.m.f. reverses direction
✓ 1
Part (b)
the maximum e.m.f. increases (it doubles)
✓ 1
the frequency doubles
✓ 1
Part (c)
new frequency = 24 Hz
✓ 1
T = 1 ÷ 24 = 0.0417 s
✓ 1
Exam tip: Period is the reciprocal of frequency: T = 1 ÷ f.
223IGPC3-4-07r23-03Series and parallel circuitsPaper CMedium1 mark
Two identical lamps in a garment are connected in parallel to a 12 V battery with negligible internal resistance. Each lamp carries a current of 0.40 A. One lamp is removed from its holder. Which row gives the new current in the remaining lamp and the new current from the battery?
Circuit diagram
Show mark scheme
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Notes
Each parallel branch has the full battery pd of 12 V, so the remaining lamp still carries 0.40 A. With only one branch left, the battery supplies 0.40 A. The row 0.80 A and 0.80 A wrongly assumes the remaining lamp takes the current of both. The row 0.20 A and 0.20 A wrongly assumes the current is shared between lamps in a way that depends on the number of lamps. The row 0.40 A and 0.80 A forgets that the battery current falls when a branch is removed.
✓ 1
Answer D
Answer: D
Common error: In a parallel circuit, removing one branch does not change the current in the other branches, but it does change the total current.
224IGPC3-4-07r23-04Electrical safetyPaper CMedium1 mark
A heater in a fish-processing room is rated 1.8 kW and works from the 230 V mains supply. Fuses rated 3 A, 5 A, 13 A and 30 A are available. Which fuse is most suitable?
Show mark scheme
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Notes
Current = P ÷ V = 1800 ÷ 230 = 7.8 A. The fuse rating must be just above this, so that it does not melt in normal use but still protects the circuit; that is 13 A. The 3 A and 5 A fuses are below the working current and would melt straight away. A 30 A fuse is far above it, so the wiring could overheat before it melted.
✓ 1
Answer A
Answer: A
Exam tip: Calculate the normal current first, then choose the smallest fuse rating that is larger than it.
225IGPC3-4-07r23-10Charge, current, potential difference, resistance and powerPaper CMedium6 marks
In a fish-tank monitor, a 15 Ω resistor and a 25 Ω resistor are connected in series to a 12 V battery with negligible internal resistance.
Circuit diagram
(a)
Calculate the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
State how the current in the 15 Ω resistor compares with the current in the 25 Ω resistor.
(1)
(d)
Calculate the potential difference across the 25 Ω resistor.
(2)
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Notes
Part (a)
15 + 25 = 40 Ω
✓ 1
Part (b)
I = V / R
✓ 1
= 12 / 40 = 0.30 A (ECF from (a))
✓ 1
Part (c)
They are equal
✓ 1
Part (d)
V = I × R
✓ 1
= 0.30 × 25 = 7.5 V (ECF from (b))
✓ 1
Common error: In a series circuit the current is the same everywhere; the battery pd is shared between the components.
226IGPC3-4-07r23-11Electrical safetyPaper CMedium7 marks
A pressing iron in a clothing workshop is rated 2.1 kW and is used on the 230 V mains supply. The iron has a metal casing and a three-core cable.
(a)
Calculate the current in the iron. Give your answer to 3 significant figures.
(2)
(b)
The plug may be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
(2)
(c)
Explain how the earth wire and the fuse together protect a user if the live wire touches the metal casing.
(3)
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Notes
Part (a)
I = P / V
✓ 1
= 2100 / 230 = 9.13 A
✓ 1
Part (b)
13 A
✓ 1
It is the only rating above the working current of about 9 A, so it does not melt in normal use
✓ 1
Part (c)
The casing is connected to earth through the earth wire, so a large current flows to earth
✓ 1
This large current makes the fuse melt
✓ 1
The appliance is then disconnected from the live supply, so the casing is not at a dangerous pd
✓ 1
Exam tip: Fuses and switches are placed in the live wire so that the appliance is fully disconnected from the dangerous supply.
227IGPC3-4-07r23-16Charge, current, potential difference, resistance and powerPaper CMedium8 marks
A resistor in a LED garment is connected to a 4.5 V battery. The current in the resistor is 0.15 A. The resistance of the resistor does not change.
(a)
Calculate the resistance of the resistor.
(2)
(b)
The battery is replaced by a 9.0 V battery. Calculate the new current in the resistor.
(2)
(c)
Calculate the power of the resistor with the 9.0 V battery.
(2)
(d)
State the name of the meter used to measure the current and how it is connected in the circuit.
(2)
Show mark scheme
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Notes
Part (a)
R = V / I
✓ 1
= 4.5 / 0.15 = 30 Ω
✓ 1
Part (b)
I = V / R
✓ 1
= 9.0 / 30 = 0.30 A (ECF from (a))
✓ 1
Part (c)
P = I × V
✓ 1
= 0.30 × 9.0 = 2.7 W (ECF from (b))
✓ 1
Part (d)
Ammeter
✓ 1
In series with the resistor
✓ 1
Exam tip: Check with P = V² / R as well: 9.0² / 30 = 2.7 W gives the same answer.
228IGPC3-4-06r24-08Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A steriliser in a hospital has a power of 2.4 kW and operates for 30 minutes. Electricity costs €0.20 per kWh. What is the cost of using the steriliser?
Show mark scheme
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Mark
Notes
Energy = 2.4 kW × 0.5 h = 1.2 kWh, and 1.2 × €0.20 = €0.24. €0.48 forgets to multiply by the time. €0.10 forgets to multiply by the power. €14.40 uses the time in minutes instead of hours.
✓ 1
Answer C
Answer: C
Common error: For kilowatt-hours, the power must be in kW and the time in hours.
229IGPC3-4-06r24-10Series and parallel circuitsPaper CMedium5 marks
A hospital monitor circuit has a 12 V battery connected to two resistors in series. Resistor R1 is 8.0 Ω and resistor R2 is 16 Ω.
Circuit diagram
(a)
State the total resistance of the circuit.
(1)
(b)
Calculate the current in the circuit.
(2)
(c)
Calculate the potential difference across R2.
(2)
Show mark scheme
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Notes
Part (a)
24 Ω
✓ 1
Part (b)
I = V ÷ R = 12 ÷ 24 (ECF from (a))
✓ 1
0.50 A
✓ 1
Part (c)
V = IR = 0.50 × 16 (ECF from (b))
✓ 1
8.0 V
✓ 1
Common error: Use the current through the component and the resistance of that component only, not the total resistance, when finding the potential difference across it.
230IGPC3-4-06r24-13Electrical safetyPaper CMedium6 marks
A metal-cased steriliser in a hospital is connected to the mains by a three-core cable containing live, neutral and earth wires. A fuse is fitted in the plug.
(a)
State the colours of the insulation on the earth wire.
(1)
(b)
Explain how the earth wire and the fuse protect a user if the live wire touches the metal casing.
(3)
(c)
State why the fuse is fitted in the live wire rather than the neutral wire.
(1)
(d)
State one advantage of using a circuit breaker instead of a fuse.
(1)
Show mark scheme
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Mark
Notes
Part (a)
green and yellow
✓ 1
Part (b)
a large current flows through the low-resistance earth wire
✓ 1
the fuse melts (or the circuit breaks)
✓ 1
so the casing is no longer connected to the live supply and the user does not receive a shock
✓ 1
Part (c)
so that when the fuse melts the appliance is disconnected from the high voltage
✓ 1
Part (d)
it can be reset (or it switches off faster, or it does not need replacing)
✓ 1
Common error: Explain what happens to the current, then to the fuse, then to the casing; skipping the fuse step loses the mark.
231IGPC3-4-07r24-04Series and parallel circuitsPaper CMedium1 mark
A beekeeper's hive warmer has a 6.0 Ω heating element and a 12 Ω heating element connected in parallel across a 12 V battery with negligible internal resistance. What is the total current drawn from the battery?
Circuit diagram
Show mark scheme
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Notes
Each branch has the full 12 V across it. The current in the 6.0 Ω element is 12/6.0 = 2.0 A and in the 12 Ω element is 12/12 = 1.0 A. The currents in parallel branches add, giving 3.0 A. 0.67 A treats the elements as being in series (12/18). 1.3 A comes from taking the average of the two resistances (9.0 Ω). 2.0 A counts only the current in the 6.0 Ω element.
✓ 1
Answer C
Answer: C
Common error: In parallel the total resistance is smaller than the smallest resistor, so the total current is larger than the current in any one branch.
232IGPC3-4-07r24-07Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A 20 Ω heating strip and a 40 Ω heating strip are connected in series to a 12 V supply of negligible internal resistance, to warm the base of a hive. What is the power dissipated in the 40 Ω strip?
Circuit diagram
Show mark scheme
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Mark
Notes
The total resistance is 60 Ω, so the current is 12/60 = 0.20 A. Power in the 40 Ω strip = I²R = 0.20² × 40 = 1.6 W. 0.80 W is the power in the 20 Ω strip. 3.6 W uses the full 12 V across the 40 Ω strip (12²/40). 7.2 W uses the full 12 V across the 20 Ω strip (12²/20).
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Answer A
Answer: A
Common error: In a series circuit the supply p.d. is shared, so V²/R with the full supply p.d. overestimates the power in one component.
233IGPC3-4-07r24-12Series and parallel circuitsPaper CMedium7 marks
In a Formula 1 team's garage, two identical lamps, each rated 12 V, 36 W, are connected in parallel to a 12 V supply of negligible internal resistance. Both lamps work at their rated power.
Circuit diagram
(a)
Calculate the current in one lamp.
(2)
(b)
State the total current drawn from the supply.
(1)
(c)
The filament of one lamp breaks. State what happens to the current in the other lamp and explain your answer.
(2)
(d)
Calculate the resistance of one lamp when it is working normally.
(2)
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Notes
Part (a)
M1 I = P/V or 36/12
✓ 1
A1 3.0 A
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Part (b)
6.0 A (ECF from (a))
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Part (c)
It stays the same, at 3.0 A
✓ 1
because the other lamp is on its own branch and still has the full 12 V across it
✓ 1
Part (d)
M1 R = V/I or 12/3.0 (ECF from (a))
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A1 4.0 Ω
✓ 1
Exam tip: In parallel, every branch has the same p.d. as the supply, so a fault in one branch does not change the current in the others.
234IGPC3-4-07r24-14Electrical safetyPaper CMedium6 marks
A hospital bed has a motor in a metal casing. The motor is rated 230 V, 750 W, and is connected to the mains supply through a three-pin plug that contains a fuse.
(a)
Explain how the earth wire and the fuse together protect a person if the live wire accidentally touches the metal casing.
(3)
(b)
Calculate the normal operating current of the motor, to 3 significant figures. Fuses of 3 A, 5 A and 13 A are available. State which should be fitted and give a reason.
(3)
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Notes
Part (a)
The casing is connected to earth, giving a low-resistance path to earth
✓ 1
so a large current flows in the live and earth wires
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the fuse melts and breaks the circuit, so the casing is no longer live
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Part (b)
M1 I = P/V = 750/230
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A1 3.26 A
✓ 1
A1 5 A, because it is the lowest rating above the operating current (so it carries the normal current but melts if the current rises)
✓ 1
Common error: The fuse rating must be just above the normal current; a 13 A fuse would not melt for a small fault.
235IGPC3-4-07r26-04Series and parallel circuitsPaper CMedium1 mark
Two identical lamps are connected in parallel across a 12 V supply in a shop window display. The filament of one lamp breaks. Which statement describes what happens?
Circuit diagram
Show mark scheme
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Notes
Each parallel branch has the full 12 V across it, so the working lamp still carries the same current and keeps its brightness. The supply now feeds one branch instead of two, so the supply current halves. A dimmer lamp or a lamp that goes out is the result of thinking as if the lamps were in series. Brighter with an unchanged supply current wrongly assumes the current is redistributed to the remaining lamp.
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Answer C
Answer: C
Common error: Parallel branches work independently: a fault in one branch changes the total current but not the p.d. across the other branches.
236IGPC3-4-07r26-05Electrical safetyPaper CMedium1 mark
A coffee-shop kettle is rated at 1500 W and is connected to the 230 V mains. Fuses rated 3 A, 5 A, 10 A and 13 A are available. Which fuse is most suitable, taking the most suitable fuse to be the one with the lowest rating that does not blow in normal use?
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Notes
Normal current = P / V = 1500 / 230 = 6.5 A. The fuse must be rated above this, and the lowest such rating is 10 A. The 3 A and 5 A fuses are below the normal current so they would blow every time the kettle is used. The 13 A fuse would not blow in normal use but is higher than necessary, so it gives less protection in a fault.
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Answer B
Answer: B
Exam tip: Calculate the normal current with I = P / V, then choose the lowest available fuse rating above it.
237IGPC3-4-07r26-11Series and parallel circuitsPaper CMedium7 marks
A shop's LED sign is supplied with 12 V. It contains two parallel branches. Branch 1 has a 6.0 Ω resistor in series with a 4.0 Ω resistor. Branch 2 is a single 15 Ω resistor. The supply has negligible internal resistance.
Circuit diagram
(a)
Calculate the resistance of branch 1. [1]
(1)
(b)
Calculate the current in the 15 Ω resistor. [2]
(2)
(c)
Calculate the current from the supply. [2]
(2)
(d)
Calculate the potential difference across the 4.0 Ω resistor. [2]
(2)
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Notes
Part (a)
A1 6.0 + 4.0 = 10 Ω
✓ 1
Part (b)
M1 I = V / R = 12 / 15
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A1 0.80 A
✓ 1
Part (c)
M1 current in branch 1 = 12 / 10 = 1.2 A
✓ 1
A1 total current = 1.2 + 0.80 = 2.0 A (ECF from (b))
✓ 1
Part (d)
M1 V = I R = 1.2 × 4.0
✓ 1
A1 4.8 V
✓ 1
Exam tip: The current in a series branch is the same everywhere in that branch, so find it before finding the p.d. across one resistor.
238IGPC3-4-07r26-14Electrical safetyPaper CMedium6 marks
A café's coffee machine has a metal case and is rated at 2.3 kW. It is connected to the 230 V mains supply.
(a)
Calculate the normal current in the machine. [2]
(2)
(b)
The machine's plug can be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason. [2]
(2)
(c)
Explain how the earth wire connected to the metal case makes the machine safe if the live wire touches the case. [2]
(2)
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Notes
Part (a)
M1 I = P / V = 2300 / 230
✓ 1
A1 10 A
✓ 1
Part (b)
A1 13 A
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The rating must be above the normal current of 10 A so that it does not blow in normal use (ECF from (a))
✓ 1
Part (c)
The earth wire gives a low-resistance path to earth, so a large current flows
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The large current melts the fuse, which disconnects the live supply so the case is no longer dangerous
✓ 1
Common error: The fuse goes in the live wire and has a rating just above the normal current, not the highest rating available.
239IGPC3-4-09r30-01Electrical safetyPaper CMedium1 mark
A battery charger for a warehouse robot is rated 460 W and is connected to a 230 V mains supply. The plug can be fitted with a fuse of rating 1 A, 3 A, 5 A or 13 A. Which fuse is the most suitable?
Show mark scheme
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Notes
The normal current is I = P/V = 460/230 = 2.0 A. The fuse must be rated just above this so it does not melt in normal use but still melts soon after a fault, so 3 A is correct. 1 A is below the normal current and would melt straight away. 5 A and 13 A would allow a current well above normal to flow before melting, giving less protection.
✓ 1
Answer B
Answer: B
Exam tip: Work out the normal current first, then choose the smallest fuse rating that is larger than it.
240IGPC3-4-09r30-10Electrical safetyPaper CMedium6 marks
A fashion workshop uses a sewing machine with a metal casing. It is rated 805 W and is connected to the 230 V mains supply through a three-pin plug containing a fuse.
(a)
Explain how a fuse protects the circuit if a fault occurs.
(2)
(b)
Calculate the normal current in the sewing machine.
(2)
(c)
The available fuses are rated 3 A, 5 A and 13 A. State which fuse is most suitable and give a reason.
(2)
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Notes
Part (a)
a fault makes the current larger than normal so the fuse wire heats up and melts
✓ 1
the circuit is broken, so no current flows
✓ 1
Part (b)
M1 I = P/V = 805/230
✓ 1
A1 3.5 A
✓ 1
Part (c)
5 A
✓ 1
just above the normal working current so it does not melt in normal use but melts if the current rises (ECF from (b))
✓ 1
Exam tip: Choose the fuse rating that is just above the normal current; a rating far above it gives little protection.
241IGPC3-4-09r30-14Electromagnetic induction, motors and transformersPaper CMedium7 marks
A student pushes a bar magnet into a coil of wire connected to a sensitive centre-zero meter, to demonstrate electromagnetic induction.
(a)
State what is induced in the coil when the magnet moves.
(1)
(b)
State two changes that would increase the size of the induced e.m.f.
(2)
(c)
Explain why the meter reads zero when the magnet is held still inside the coil.
(2)
(d)
Describe what the meter shows when the magnet is then pulled out of the coil at the same speed it went in.
(2)
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Notes
Part (a)
an e.m.f.
✓ 1
Part (b)
any two from: move the magnet faster; use a stronger magnet; use a coil with more turns
✓ 2
Part (c)
the magnetic field lines are not being cut by the coil (the field linked with the coil is not changing)
✓ 1
so no e.m.f. is induced
✓ 1
Part (d)
the needle deflects in the opposite direction
✓ 1
with the same size of deflection, only while the magnet is moving
✓ 1
Exam tip: An e.m.f. is induced only while the magnetic field linked with the coil is changing.
242IGPC3-4-07r31-02Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A sports-science sensor carries a steady current of 0.30 mA for 2.0 minutes. How much charge passes through the sensor?
Show mark scheme
Marking point
Mark
Notes
Q = It with I = 0.30 × 10−3 A and t = 120 s gives 3.6 × 10−2 C. 36 C comes from forgetting to convert milliamperes to amperes. 6.0 × 10−4 C comes from using the time in minutes instead of seconds. 2.5 × 10−6 C comes from dividing the current by the time instead of multiplying.
✓ 1
Answer C
Answer: C
Common error: Convert mA to A and minutes to seconds before using Q = It.
243IGPC3-4-07r31-03Electrical safetyPaper CMedium1 mark
A fan heater in a rail depot is rated 230 V, 1.8 kW. It is to be fitted with a plug containing one fuse. Which fuse rating is most suitable?
Show mark scheme
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Notes
The normal current is P/V = 1800/230 ≈ 7.8 A, so the fuse must be rated just above this: 13 A allows normal use but melts if the current becomes much larger. A 3 A or 5 A fuse is below the normal current and would melt every time the heater is switched on. A 30 A fuse is so far above the normal current that a fault could overheat the cable before the fuse melted.
✓ 1
Answer D
Answer: D
Exam tip: Work out the normal current first, then choose the next fuse rating above it.
244IGPC3-4-07r31-07Electromagnetic induction, motors and transformersPaper CMedium1 mark
In a simple d.c. motor in a treadmill, the current in the coil and the polarity of the permanent magnets are both reversed at the same time. What happens to the direction of rotation of the coil?
Show mark scheme
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Notes
The direction of the force on the coil depends on both the current direction and the field direction. Reversing one would reverse the force, but reversing both restores the original force direction, so the rotation is unchanged. The rotation does not reverse because that needs only one reversal. The coil does not stop because the force is still present. The speed does not increase because neither the current size nor the field strength changed.
✓ 1
Answer A
Answer: A
Exam tip: Reversing either the current or the field reverses the force; reversing both leaves it unchanged.
245IGPC3-4-07r31-13Electrical safetyPaper CMedium7 marks
An electric kettle in a sports-centre kitchen is rated 230 V, 2.3 kW. Its metal case is connected to the earth wire.
(a)
Calculate the normal current in the kettle.
(2)
(b)
The plug may be fitted with a 3 A, 5 A or 13 A fuse. State which fuse is correct and explain your choice.
(2)
(c)
Explain how the earth wire and the fuse protect a person if the live wire comes loose and touches the metal case.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 I = P/V = 2300/230
✓ 1
A1 10 A
✓ 1
Part (b)
B1 13 A
✓ 1
B1 its rating is just above the normal current of 10 A, so it does not melt in normal use but melts if the current becomes too large (3 A and 5 A would melt immediately)
✓ 1
Part (c)
B1 the earth wire connects the case to earth, so the case cannot stay at a dangerous voltage
✓ 1
B1 a large current flows from the live wire through the case and earth wire
✓ 1
B1 this large current melts the fuse, which disconnects the kettle from the live supply
✓ 1
Exam tip: Fuse and earth work together: the earth wire provides a path for a large current and the fuse then cuts it off.
246IGPC3-4-07r31-16Magnets and magnetic fieldsPaper CMedium5 marks
A horizontal copper rod lies across two rails of a model rail launcher. A uniform magnetic field points vertically downwards through the rod. The current in the rod flows horizontally from west to east.
(a)
State the direction of the force on the rod.
(1)
(b)
State the direction of the force when (i) only the current is reversed, (ii) both the current and the field are reversed.
(2)
(c)
The rod is replaced by a wire that carries current in the same direction as the field lines. State the force on the wire and give a reason.
(2)
Show mark scheme
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Notes
Part (a)
B1 towards the north (horizontally, at right angles to both the field and the current)
✓ 1
Part (b)
B1 (i) towards the south
✓ 1
B1 (ii) towards the north (unchanged)
✓ 1
Part (c)
B1 no force (zero)
✓ 1
B1 the current is parallel to the field, so the wire does not cut field lines
✓ 1
Exam tip: Use Fleming's left-hand rule: the force is at right angles to both the field and the current.
247IGPC3-4-09r31-02Magnets and magnetic fieldsPaper CMedium1 mark
Bar P is a permanent magnet. Bar Q is either a magnet or an unmagnetised piece of iron. Which observation shows that Q must be a magnet?
Show mark scheme
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Notes
Only two magnets can repel, because repulsion needs two like poles. Unmagnetised iron is always attracted to either pole of a magnet through induced magnetism, so attraction at one end, at both ends or to the south pole does not prove Q is a magnet.
✓ 1
Answer C
Answer: C
Exam tip: Repulsion is the only reliable test for a magnet; attraction can happen with unmagnetised iron.
248IGPC3-4-09r31-03Series and parallel circuitsPaper CMedium1 mark
Two heating elements of resistance 6.0 Ω and 3.0 Ω are connected in parallel to a 12 V supply in a water tank. What is the total current from the supply?
Circuit diagram
Show mark scheme
Marking point
Mark
Notes
Each branch has the full 12 V across it, so the currents are 12 ÷ 6.0 = 2.0 A and 12 ÷ 3.0 = 4.0 A. In parallel the currents add, giving 6.0 A. 1.3 A comes from adding the resistances as if they were in series (12 ÷ 9.0). 2.0 A and 4.0 A are the current in only one branch.
✓ 1
Answer D
Answer: D
Common error: In a parallel circuit the branch currents add, and the total resistance is less than the smallest resistor.
249IGPC3-4-09r31-11Magnets and magnetic fieldsPaper CMedium6 marks
A pharmacy removes tiny iron fragments from powdered medicine using an electromagnet.
(a)
Explain why the core of the electromagnet is made of soft iron rather than steel.
(1)
(b)
State two ways of making the electromagnet stronger.
(2)
(c)
Explain why the iron fragments are attracted to either pole of the electromagnet.
(3)
Show mark scheme
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Notes
Part (a)
Soft iron loses its magnetism when the current is switched off, whereas steel stays magnetised
✓ 1
Part (b)
Increase the current in the coil
✓ 1
Increase the number of turns on the coil
✓ 1
(Accept any two correct ways, such as using a larger soft iron core.)
—
Part (c)
The iron fragment becomes magnetised by induction
✓ 1
The end of the fragment nearest the pole becomes the opposite pole
✓ 1
Unlike poles attract, so the fragment is attracted whichever pole is used
✓ 1
Exam tip: 'Explain' answers about induced magnetism should state which pole forms at the nearest end.
250IGPC3-4-09r31-12Series and parallel circuitsPaper CMedium6 marks
Two heating elements in a water heater are connected in parallel to the 230 V mains. One element has resistance 46 Ω and the other has resistance 92 Ω.
Circuit diagram
(a)
Calculate the current in each element.
(2)
(b)
Calculate the total current from the supply.
(1)
(c)
State one advantage of connecting the elements in parallel rather than in series.
(1)
(d)
Calculate the total power of the heater.
(2)
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Notes
Part (a)
M1 I = V ÷ R for each element
✓ 1
A1 5.0 A and 2.5 A
✓ 1
Part (b)
7.5 A (ECF from (a))
✓ 1
Part (c)
Each element has the full 230 V across it, or one can be switched off without affecting the other
✓ 1
Part (d)
M1 P = IV = 7.5 × 230
✓ 1
A1 1725 W (ECF from (b))
✓ 1
Common error: Total power is the sum of the powers of the elements; do not use only one branch current.
251IGPC3-4-10r33-03Series and parallel circuitsPaper CMedium1 mark
Three identical lamps in a music studio are connected in parallel to a mains supply. One lamp is removed from its holder. What happens?
Circuit diagram
Show mark scheme
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Notes
Each parallel branch has the full supply voltage, so the remaining lamps are unchanged in brightness. The supply now provides current to two branches instead of three, so the total current falls. The lamps do not share a fixed current, and the circuit is still complete through the other branches.
✓ 1
Answer D
Answer: D
Common error: Removing one component from a parallel circuit does not affect the other branches; it only reduces the total current.
252IGPC3-4-10r33-08Charge, current, potential difference, resistance and powerPaper CMedium1 mark
A heating wire in an aircraft de-icing circuit is replaced by a wire of the same material. The new wire is longer and has a greater cross-sectional area. Which row shows how each change affects the resistance?
Show mark scheme
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Notes
A longer wire gives charge carriers more collisions along the way, so resistance rises. A wider wire gives more paths for the current, so resistance falls. The other rows reverse one or both of these effects.
✓ 1
Answer B
Answer: B
Exam tip: Resistance rises with length and falls with cross-sectional area (a thicker wire is easier for current).
253IGPC3-4-10r33-10Magnets and magnetic fieldsPaper CMedium6 marks
A music technology student examines a magnet taken from an old loudspeaker.
(a)
Describe how a plotting compass can be used to find the shape of the magnetic field around a bar magnet.
(3)
(b)
State the direction of the magnetic field outside a bar magnet.
(1)
(c)
Soft iron is used in the core of an electromagnet but steel is used for permanent magnets. Explain why.
(2)
Show mark scheme
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Notes
Part (a)
Place the magnet on paper and put the compass near one pole, marking the positions of both ends of the needle
✓ 1
Move the compass so that one end of the needle is at the previous mark and repeat to make a line of points
✓ 1
Join the marks to give a field line, and repeat from different starting positions
✓ 1
Part (b)
From the north pole to the south pole
✓ 1
Part (c)
Soft iron is easily magnetised and demagnetised, so the electromagnet can be switched on and off
✓ 1
Steel keeps its magnetism, which is needed for a permanent magnet
✓ 1
Common error: Field lines never cross, and they are drawn from the north pole to the south pole outside the magnet.
254IGPC3-4-10r33-12Electrical safetyPaper CMedium7 marks
A farm worker connects an electric pump to the 230 V mains. The pump is rated at 1.38 kW. The available fuses are rated 3 A, 5 A and 13 A.
(a)
Describe how a fuse protects a circuit.
(2)
(b)
State why the fuse must be placed in the live wire.
(1)
(c)
Calculate the current in the pump and state which fuse should be used, giving a reason.
(4)
Show mark scheme
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Notes
Part (a)
If the current is too large, the fuse wire heats up and melts
✓ 1
This breaks the circuit
✓ 1
Part (b)
So that when it melts the appliance is disconnected from the live (high voltage) supply
✓ 1
Part (c)
M1 use of I = P / V
✓ 1
A1 I = 6.0 A
✓ 1
A1 13 A fuse
✓ 1
A1 the fuse rating must be above the normal current but as close to it as possible (the 3 A and 5 A fuses would melt)
✓ 1
Exam tip: Choose the fuse with the smallest rating that is still above the normal working current.
255IGPC3-4-10r33-13Series and parallel circuitsPaper CMedium6 marks
Two heating elements in a greenhouse are connected in parallel to a 24 V supply. One element has a resistance of 8.0 Ω and the other has a resistance of 12 Ω.
Circuit diagram
(a)
Calculate the current in each element.
(2)
(b)
Calculate the total current from the supply.
(1)
(c)
The 8.0 Ω element is disconnected. State what happens to the current in the 12 Ω element and explain your answer.
(1)
(d)
Calculate the power of the 8.0 Ω element.
(2)
Show mark scheme
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Notes
Part (a)
A1 8.0 Ω element: I = 24 / 8.0 = 3.0 A
✓ 1
A1 12 Ω element: I = 24 / 12 = 2.0 A
✓ 1
Part (b)
A1 I = 5.0 A (sum of the branch currents); ECF from (a)
✓ 1
Part (c)
The current stays at 2.0 A, because the potential difference across the 12 Ω element is still 24 V
✓ 1
Part (d)
M1 use of P = VI or P = V2 / R
✓ 1
A1 P = 72 W
✓ 1
Common error: In a parallel circuit each branch has the full supply voltage; the branch currents add to give the total.
256IGPC3-4-08r35-06Magnets and magnetic fieldsPaper CMedium1 mark
A straight wire carries a current out of the page between two magnetic poles. The magnetic field points from left to right across the page, and the wire is pushed up the page. The current direction is now reversed and the poles are swapped so that the field points from right to left. What is the direction of the force on the wire?
Plan view
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Notes
The force reverses if either the current or the field is reversed. Reversing both reverses the force twice, so the direction is unchanged and the wire is still pushed up the page. Down the page would follow from reversing only one of them. A force on a current-carrying wire in a field is always at right angles to both, so it cannot point left. The wire still carries current in a field, so the force does not vanish.
✓ 1
Answer D
Answer: D
Exam tip: Reversing one of current or field reverses the force; reversing both leaves it as it was.
257IGPC3-4-08r35-13Electrical safetyPaper CMedium7 marks
A commercial food mixer is rated at 2300 W and is connected to the 230 V mains supply by a three-core cable with a metal case.
(a)
Calculate the current in the mixer when it is working normally.
(2)
(b)
The plug can be fitted with a 3 A, 13 A or 30 A fuse. State which fuse should be used and give a reason.
(2)
(c)
The live wire inside the mixer comes loose and touches the metal case. Explain how the earth wire makes the user safe.
(3)
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Notes
Part (a)
I = P / V
✓ 1
I = 2300 / 230 = 10 A
✓ 1
Part (b)
13 A
✓ 1
It is just above the normal current of 10 A, so it will not melt in normal use but will melt if the current becomes too large
✓ 1
Part (c)
The earth wire connects the metal case to the ground
✓ 1
so a large current flows through the live wire, the case and the earth wire
✓ 1
the fuse melts (the circuit is broken), disconnecting the case from the live supply
✓ 1
Exam tip: In a safety question, explain the whole chain: large current, fuse melts, circuit broken, case no longer live.
258IGPC3-4-08r37-05Electromagnetic induction, motors and transformersPaper CMedium1 mark
In a volcano observatory, a bar magnet is pushed into a coil connected to a sensitive voltmeter, which shows a reading in one direction. Which change would give a larger reading in the same direction?
Show mark scheme
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Mark
Notes
A faster movement means the coil's wires cut the magnetic field more quickly, so a larger e.m.f. is induced in the same direction. A magnet at rest induces no e.m.f. Pulling the magnet out induces an e.m.f. in the opposite direction. Fewer turns would reduce the induced e.m.f.
✓ 1
Answer A
Answer: A
Common error: Believing a strong magnet inside a coil induces an e.m.f.; the e.m.f. only appears while the magnetic field is changing.
259IGPC3-4-08r37-08Series and parallel circuitsPaper CMedium1 mark
A thermistor whose resistance falls as it gets hotter is connected in series with a battery and an ammeter in a probe placed in a volcanic vent. The vent temperature rises. What happens?
Circuit diagram
Show mark scheme
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Mark
Notes
The thermistor's resistance falls as it heats up, so with the same battery p.d. the current rises, since I = V/R. The first alternative has the resistance moving the wrong way. The others pair a resistance change with a current change that does not follow from I = V/R.
✓ 1
Answer D
Answer: D
Exam tip: State the change in resistance first, then use I = V/R at constant p.d. to deduce the change in current.
260IGPC3-4-08r37-10Series and parallel circuitsPaper CMedium7 marks
A ship's cabin lighting circuit has a 12 V battery (ignore internal resistance). A 4.0 Ω resistor is connected in series with a parallel combination of a 12 Ω resistor and a 6.0 Ω resistor.
Circuit diagram
(a)
Calculate the combined resistance of the 12 Ω and 6.0 Ω resistors in parallel.
(2)
(b)
Calculate the total resistance of the circuit.
(1)
(c)
Calculate the current in the battery.
(2)
(d)
Calculate the current in the 12 Ω resistor.
(2)
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Notes
Part (a)
M1 1/R = 1/12 + 1/6
✓ 1
A1 4.0 Ω
✓ 1
Part (b)
A1 8.0 Ω (ECF from (a))
✓ 1
Part (c)
M1 I = V/R = 12 / 8.0
✓ 1
A1 1.5 A
✓ 1
Part (d)
M1 p.d. across the parallel pair = 1.5 × 4.0 = 6.0 V
✓ 1
A1 I = 6.0 / 12 = 0.50 A
✓ 1
Common error: Using the battery p.d. of 12 V across the 12 Ω resistor; the p.d. is shared with the series resistor.
261IGPC3-4-08r37-14Charge, current, potential difference, resistance and powerPaper CMedium6 marks
During an eruption, a lightning flash inside a volcanic ash plume transfers 12 C of charge in 0.040 s. Assume the current is constant during the flash and the potential difference is 2.0 × 107 V.
(a)
Calculate the current in the flash.
(2)
(b)
Calculate the energy transferred by the flash.
(2)
(c)
Name the instrument used to measure current and state how it is connected in a circuit.
(2)
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Part (a)
M1 I = Q/t = 12 / 0.040
✓ 1
A1 300 A
✓ 1
Part (b)
M1 E = IVt = 300 × 2.0 × 107 × 0.040 (ECF from (a))
✓ 1
A1 2.4 × 108 J
✓ 1
Part (c)
A1 ammeter
✓ 1
A1 connected in series with the component
✓ 1
Exam tip: Write the equation, substitute with units, then give the answer with its unit; this keeps the method marks even if the arithmetic slips.
262IGPC3-4-08r37-15Electrical safetyPaper CMedium8 marks
A theatre lamp is rated 1.8 kW and works from a 230 V mains supply.
(a)
Explain how a fuse protects the circuit.
(3)
(b)
State one advantage of using a circuit breaker instead of a fuse.
(1)
(c)
Calculate the current in the lamp. Give your answer to 2 significant figures.
(2)
(d)
The plug may contain a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
(2)
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Part (a)
A1 the fuse is connected in the live wire
✓ 1
A1 if the current is larger than the rating the fuse wire heats up and melts
✓ 1
A1 this breaks the circuit and the current stops
✓ 1
Part (b)
A1 it can be reset after it trips (accept: it acts faster)
✓ 1
Part (c)
M1 I = P/V = 1800 / 230
✓ 1
A1 7.8 A
✓ 1
Part (d)
A1 13 A fuse
✓ 1
A1 3 A and 5 A fuses are below the normal current of 7.8 A so they would melt in normal use; 13 A is the lowest rating above 7.8 A
✓ 1
Common error: Choosing the largest fuse available; the fuse should be just above the normal current, otherwise it will not protect the cable.
263IGPC3-4-07r39-03Series and parallel circuitsPaper CMedium1 mark
Two identical lamps are connected in parallel to a battery with negligible internal resistance. One lamp is removed from its holder. What happens to the brightness of the other lamp and to the current supplied by the battery?
Circuit diagram
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Each parallel branch has the full battery p.d., so the remaining lamp carries the same current and is equally bright. The battery now supplies only one branch, so its current halves. A brighter lamp or a dimmer lamp would need the branch p.d. to change. A doubled battery current would need an extra branch.
✓ 1
Answer A
Answer: A
Common error: Removing one component from a parallel circuit does not change the p.d. across, or current in, the other branches.
264IGPC3-4-07r39-05Magnets and magnetic fieldsPaper CMedium1 mark
A relay in the control circuit of a space probe contains an electromagnet. Why is the core of the electromagnet made of soft iron rather than steel?
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Soft iron magnetises strongly when there is a current in the coil and demagnetises when the current stops, so the relay can switch on and off. Conduction through the core is not the issue. A permanent magnet is the opposite of what a relay needs. Steel would attract the contacts when magnetised, not repel them, and it would also stay magnetised.
✓ 1
Answer D
Answer: D
Common error: Steel is used for permanent magnets; soft iron is used where the magnetism must switch on and off.
265IGPC3-4-07r39-10Magnets and magnetic fieldsPaper CMedium6 marks
A bridge inspector has a steel bar that might be a magnet.
(a)
Describe how a bar magnet of known polarity can be used to show that the steel bar is a magnet and not just unmagnetised iron or steel.
(2)
(b)
Describe the pattern and direction of the magnetic field lines outside a bar magnet.
(2)
(c)
The bar magnet is laid on a bench and a plotting compass is placed beside its north pole. State which pole of the compass needle points towards the north pole of the magnet and explain why.
(2)
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Part (a)
Bring the north pole of the known magnet close to each end of the steel bar in turn
✓ 1
Repulsion at one end shows the bar is a magnet, because unmagnetised iron or steel is only ever attracted
✓ 1
Part (b)
Lines run from the north pole to the south pole, with arrows showing this direction
✓ 1
Lines are closest together at the poles, where the field is strongest, and never cross
✓ 1
Part (c)
The south pole of the needle
✓ 1
Unlike poles attract
✓ 1
Exam tip: Repulsion is the only reliable test for a magnet, because a magnet attracts unmagnetised iron as well as unlike poles.
266IGPC3-4-07r39-13Electrical safetyPaper CMedium7 marks
A coffee roaster has a metal case and is connected to the mains by a three-pin plug. The roaster has a power rating of 2100 W and the mains supply is 230 V.
(a)
Explain how the earth wire and the fuse together protect a user if the live wire inside the roaster touches the metal case.
(3)
(b)
Calculate the current in the roaster in normal use, to 3 significant figures. The plug can be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
(4)
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Part (a)
A large current flows from the live wire through the case and the low-resistance earth wire
✓ 1
This large current makes the fuse melt
✓ 1
The circuit is broken, so the case is no longer connected to the live supply
✓ 1
Part (b)
P = I V so I = 2100 / 230
✓ 1
= 9.13 A
✓ 1
13 A fuse
✓ 1
It is the lowest rating above the normal current, so it does not melt in normal use but still protects the wiring
✓ 1
Exam tip: Choose the fuse with the smallest rating that is still larger than the normal working current.
267IGPC3-4-07r39-14Magnets and magnetic fieldsPaper CMedium5 marks
A straight wire carrying a current is placed between the poles of a magnet and experiences a force. A similar arrangement opens a valve in a probe's fuel line.
(a)
State two ways of increasing the force on the wire.
(2)
(b)
State what happens to the direction of the force when the current is reversed.
(1)
(c)
The wire is now turned so that it lies parallel to the magnetic field lines. State the size of the force and explain your answer.
(2)
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Part (a)
Increase the current in the wire
✓ 1
Use a stronger magnet (accept: use a longer length of wire in the field)
✓ 1
Part (b)
The force reverses direction
✓ 1
Part (c)
The force is zero
✓ 1
The wire is not cutting across the field lines, so the field does not push on it
✓ 1
Exam tip: The force on a wire is largest when the wire is at right angles to the field and zero when it is parallel to the field.
268IGPC3-4-08r42-03Series and parallel circuitsPaper CMedium1 mark
Two identical lamps, each of resistance 24 Ω, are connected in parallel to a 12 V battery of negligible internal resistance. One lamp's filament then breaks. What is the current in the battery after the filament breaks?
Circuit diagram
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Each branch of a parallel circuit has the full 12 V across it, so the working lamp still carries 12 / 24 = 0.50 A. The broken branch carries no current, so the battery now supplies 0.50 A. A current of 0 A would be true only for a series circuit. The value 0.25 A comes from halving the current again. The value 1.0 A is the current before the failure.
✓ 1
Answer D
Answer: D
Common error: In parallel, a break in one branch does not stop the others, because each branch still has the full supply voltage.
269IGPC3-4-08r42-05Charge, current, potential difference, resistance and powerPaper CMedium1 mark
An aircraft landing light is labelled 28 V, 112 W. What is the resistance of its filament when it works normally?
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Current I = P / V = 112 / 28 = 4.0 A, then R = V / I = 28 / 4.0 = 7.0 Ω. The value 4.0 Ω is the current, not the resistance. The value 0.14 Ω comes from using P / V2, the inverse of the correct expression V2 / P. The value 3.1 × 103 Ω comes from multiplying voltage by power.
✓ 1
Answer B
Answer: B
Exam tip: Combine P = IV and V = IR in two steps if you cannot remember R = V²/P.
270IGPC3-4-08r42-08Electromagnetic induction, motors and transformersPaper CMedium1 mark
The pickup of an electric guitar contains a coil and a magnet. A vibrating steel string changes the magnetic field through the coil and induces a voltage. Which change would increase the size of the induced voltage?
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Plucking harder makes the string move faster, so the magnetic field through the coil changes more quickly and a larger voltage is induced. Fewer turns would reduce the voltage. A plastic string is not magnetic, so it would not change the field and no useful voltage would be induced. A weaker magnet produces a smaller change in field.
✓ 1
Answer A
Answer: A
Exam tip: The induced voltage grows with faster change, more turns and a stronger field.
271IGPC3-4-08r42-10Charge, current, potential difference, resistance and powerPaper CMedium6 marks
A light aircraft's cabin has two reading lamps, A and B, connected in parallel to a 12 V supply. The current in lamp A is 0.50 A and the current in lamp B is 0.25 A.
Circuit diagram
(a)
Calculate the total current from the supply.
(1)
(b)
Calculate the resistance of lamp B.
(2)
(c)
Calculate the total energy transferred by the two lamps in 10 minutes.
(3)
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Part (a)
0.50 + 0.25 = 0.75 A
✓ 1
Part (b)
M1 R = V / I = 12 / 0.25
✓ 1
A1 48 Ω
✓ 1
Part (c)
M1 P = IV = 0.75 × 12 (= 9.0 W)
✓ 1
M1 E = Pt = 9.0 × 600
✓ 1
A1 5400 J (ECF from (a))
✓ 1
Exam tip: Change minutes to seconds before using energy = power × time.
272IGPC3-4-08r42-11Electrical safetyPaper CMedium5 marks
A guitar amplifier at an outdoor concert has a metal case. It is connected to the mains supply through a three-core cable that includes a fuse.
(a)
Explain how the fuse protects the circuit if a fault causes a large current.
(2)
(b)
Explain how the earth wire connected to the metal case makes the amplifier safe if the live wire touches the case.
(2)
(c)
State why the fuse and switch are placed in the live wire rather than the neutral wire.
(1)
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Part (a)
The large current heats the fuse wire until it melts
✓ 1
which breaks the circuit and stops the current
✓ 1
Part (b)
A large current flows through the low-resistance earth wire
✓ 1
which melts the fuse and disconnects the supply, so the case does not stay live
✓ 1
Part (c)
So that when the switch is open or the fuse has blown, the appliance is disconnected from the high-voltage live supply
✓ 1
Common error: The fuse must be in the live wire; a fuse in the neutral wire would leave the appliance connected to the live supply.
273IGPC3-4-08r42-16Charge, current, potential difference, resistance and powerPaper CMedium5 marks
A technician tests a length of resistance wire used in a heating element for a violin-maker's glue pot. The readings are: 1.0 V gives 0.10 A; 2.0 V gives 0.20 A; 3.0 V gives 0.30 A.
(a)
Calculate the resistance of the wire when the potential difference is 2.0 V.
(2)
(b)
State whether the resistance is constant, using the data to support your answer.
(1)
(c)
Calculate the power transferred in the wire when the potential difference is 3.0 V.
(2)
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Part (a)
M1 R = V / I = 2.0 / 0.20
✓ 1
A1 10 Ω
✓ 1
Part (b)
Yes, V / I = 10 Ω for each pair of readings
✓ 1
Part (c)
M1 P = IV = 3.0 × 0.30
✓ 1
A1 0.90 W
✓ 1
Exam tip: Show the check for constant resistance by calculating V / I for at least two pairs of readings.
274IGPC-038-03-09Electrical safetyPaper CHard1 mark
An extension lead connected to the 230 V mains contains a 13 A fuse. The appliances below can be plugged into it, and all of them are connected in parallel: heater 2.0 kW, kettle 1.2 kW, toaster 0.80 kW. Which combination, switched on at the same time, will make the fuse melt?
All appliances are connected in parallel. Diagram not to scale
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In parallel, the currents add, so the total current is the total power divided by 230 V. Heater + kettle: 3200/230 = 13.9 A. This is above 13 A, so the fuse melts. Heater + toaster: 2800/230 = 12.2 A. Kettle + toaster: 2000/230 = 8.7 A. Heater only: 2000/230 = 8.7 A. Only the heater and kettle together draw more than 13 A.
✓ 1
Answer A
Answer: A
Exam tip: For parallel appliances, you can add the powers first and then use I = P/V once. This saves time compared with working out each current separately.
275IGPC-038-00-10Magnets and magnetic fieldsPaper CHard1 mark
A solenoid is viewed from its end X. Seen from X, the current flows clockwise around the coil. The north pole of a bar magnet is brought close to end X. Which row is correct?
End view of the solenoid; the arrows show the direction of the current.
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Seen from one end, a clockwise current makes that end a south pole. An anticlockwise current makes it a north pole. So X is a south pole. The magnet's north pole and X are unlike poles, so they attract.
✓ 1
Answer D
Answer: D
Exam tip: Draw the letters S and N with arrows on their ends. The arrows on S go clockwise and those on N go anticlockwise, which helps you remember the end poles of a solenoid.
276IGPC-038-03-15Electrical safetyPaper CHard7 marks
An electric heater with a metal case is connected to the 230 V mains through a plug fitted with a 13 A fuse. A fault occurs and the live wire touches the metal case. With the earth wire correctly connected, the total resistance of the path through the live wire, the case and the earth wire is 0.50 Ω.
(a)
Calculate the current in this path during the fault.
(2)
(b)
State what happens to the fuse.
(1)
(c)
On another heater the earth wire has come loose, and the same fault occurs. A person touches the live case and current passes through their body to the ground. The resistance of this path is 2300 Ω. Calculate the current through the person.
(2)
(d)
Explain why the 13 A fuse does not protect the person touching the case of the heater with the loose earth wire.
(2)
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Part (a)
I = V/R = 230/0.50 ; I = 460 A
✓ 2
Part (b)
The current is far above 13 A, so the fuse melts (almost immediately) and the supply is disconnected
✓ 1
Part (c)
I = V/R = 230/2300 ; I = 0.10 A
✓ 2
Part (d)
The current of 0.10 A is much smaller than the fuse rating of 13 A, so the fuse does not melt ; the case stays live and a current this large through the body can be fatal
✓ 2
Exam tip: A fuse protects the wiring, not people directly. It only melts when the current is large, and the earth wire is what makes the fault current large.
277IGPC-038-00-16Magnets and magnetic fieldsPaper CHard7 marks
Two identical bar magnets are placed in line on a bench, with their north poles facing each other and 8 cm apart.
Diagram not to scale
(a)
Describe the shape and direction of the magnetic field lines in the gap between the two north poles.
(2)
(b)
At the point midway between the two poles the resultant magnetic field is zero. Explain why.
(2)
(c)
A student has two identical-looking steel-coloured bars. One is a magnet and the other is unmagnetised soft iron. She has no other equipment. Describe how she can find out which bar is the magnet.
(3)
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Part (a)
field lines point away from (leave) each north pole ; the lines from the two poles curve sideways away from each other and do not join or cross the gap
✓ 2
Part (b)
the two magnets produce fields of equal strength at the midpoint, because they are identical and the same distance away ; the two fields act in opposite directions, so they cancel
✓ 2
Part (c)
hold the end of one bar against the middle of the other bar ; strong attraction means the bar held by its end is the magnet, OR weak or no attraction means the other bar is the magnet ; because a magnet's field is strongest at its poles and very weak at its centre, so the middle of a magnet hardly attracts iron
✓ 3
Model sketch
Model answer: field lines leave each north pole and bend away from each other; the field is zero at the midpoint. Diagram not to scale
Exam tip: In the two-bar test, say which part of each bar touches the other. Saying only 'see if they attract' earns no marks, because a magnet and a piece of iron attract each other whichever one is the magnet.
278IGPC3-4-04c-16Charge, current, potential difference, resistance and powerPaper CHard8 marks
A lamp rated 6.0 V, 2.4 W is to be run at normal brightness from a 12 V battery. The internal resistance of the battery can be ignored. A resistor is connected in series with the lamp.
Diagram not to scale
(a)
Calculate the current in the lamp when it works normally.
(2)
(b)
State the potential difference across the resistor.
(1)
(c)
Calculate the resistance of the resistor.
(2)
(d)
Calculate the energy transferred in the resistor in 10 minutes.
(3)
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Part (a)
I = P / V = 2.4 / 6.0
✓ 1
I = 0.40 A
✓ 1
Part (b)
12 − 6.0 = 6.0 V
✓ 1
Part (c)
R = V / I = 6.0 / 0.40
✓ 1
R = 15 Ω
✓ 1
Part (d)
power in the resistor = 0.40 × 6.0 = 2.4 W
✓ 1
time = 10 × 60 = 600 s
✓ 1
E = 2.4 × 600 = 1440 J
✓ 1
Exam tip: Series components share the same current, so the current found for the lamp applies to the resistor.
279IGPC3-4-05r7-07Charge, current, potential difference, resistance and powerPaper CHard1 mark
A beekeeper has a heating wire of resistance R. A second wire of the same material has twice the length and a diameter twice as large. What is the resistance of the second wire?
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B is correct: doubling the length doubles the resistance, and doubling the diameter makes the cross-sectional area four times larger, which divides resistance by four; overall 2 / 4 = 1/2, so R/2. A is wrong because it ignores the doubled length. C is wrong because it treats the doubled diameter as only a doubled area (2 / 2). D is wrong because it ignores the larger cross-section.
✓ 1
Answer B
Answer: B
Common error: Area depends on the square of the diameter, so doubling the diameter quadruples the area.
280IGPC3-4-06r9-16Series and parallel circuitsPaper CHard8 marks
A mountaineer designs a heated-glove circuit with a 9.0 V battery. Two identical heating lamps, each rated 6.0 V and 0.50 A, are connected in parallel with each other, and this pair is connected in series with a resistor R so each lamp works at its rating.
Two identical lamps in parallel, in series with the resistor R.
(a)
Calculate the current in the resistor R.
(2)
(b)
Calculate the potential difference across R.
(1)
(c)
Calculate the resistance of R.
(2)
(d)
Calculate the energy transferred by R in 10 minutes.
(3)
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Part (a)
the lamps are in parallel so currents add
✓ 1
0.50 + 0.50 = 1.0 A
✓ 1
Part (b)
9.0 − 6.0 = 3.0 V
✓ 1
Part (c)
R = V ÷ I = 3.0 ÷ 1.0
✓ 1
3.0 Ω
✓ 1
Part (d)
power = VI = 3.0 × 1.0 = 3.0 W
✓ 1
time = 600 s
✓ 1
energy = 1800 J
✓ 1
Exam tip: In a series-parallel circuit, find the p.d. across the series part by subtracting the p.d. across the parallel part from the supply.
281IGPC3-4-07r10-16Electrical safetyPaper CHard8 marks
A mountain refuge has a 230 V supply protected by a 32 A circuit breaker. The refuge has a heater rated at 2.3 kW, an oven rated at 4.6 kW and a kettle rated at 2.3 kW.
Appliances in parallel on the 230 V supply. Diagram not to scale
(a)
Calculate the current in the oven when it is working normally.
(2)
(b)
Calculate the total current when all three appliances are switched on together.
(2)
(c)
Deduce what happens when all three appliances are switched on together.
(2)
(d)
Explain why the circuit breaker must be fitted in the live wire and not the neutral wire.
(2)
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Part (a)
I = P / V
✓ 1
= 4600 / 230 = 20 A
✓ 1
Part (b)
total power = 2.3 + 4.6 + 2.3 = 9.2 kW
✓ 1
I = 9200 / 230 = 40 A
✓ 1
Part (c)
40 A is greater than the 32 A rating (ECF)
✓ 1
the circuit breaker trips and switches the supply off
✓ 1
Part (d)
when it trips, the appliances are disconnected from the live wire
✓ 1
otherwise parts connected to the live wire would stay at 230 V and could give an electric shock
✓ 1
Exam tip: In parallel mains circuits the currents of the appliances add, so total power divided by voltage gives the total current.
282IGPC3-4-06r24-07Electromagnetic induction, motors and transformersPaper CHard1 mark
A technician in a hospital pushes a bar magnet into a coil connected to a sensitive voltmeter. Which change would not increase the induced e.m.f.?
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The induced e.m.f. depends on the rate at which field lines are cut, the strength of the field and the number of turns. The thickness of the wire changes the resistance of the coil but not the e.m.f., so this is the correct choice. Faster movement, a stronger magnet and more turns all increase the e.m.f.
✓ 1
Answer B
Answer: B
Exam tip: Remember the three factors: speed of movement, strength of magnet and number of turns.
283IGPC3-4-07r24-16Charge, current, potential difference, resistance and powerPaper CHard8 marks
A Formula 1 team uses tyre blankets to warm tyres before a race. Each blanket is rated 230 V, 1150 W. Assume the resistance of a blanket stays constant.
(a)
Calculate the resistance of one blanket when it works at its rating.
(2)
(b)
Four blankets are used for 3.0 hours. Calculate the total energy transferred, in kWh.
(2)
(c)
Electrical energy costs £0.28 per kWh. Calculate the cost of the energy in (b), to the nearest penny.
(2)
(d)
The supply voltage falls to 184 V. Calculate the new power of one blanket.
(2)
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Part (a)
M1 I = P/V = 5.0 A and R = V/I, or R = V2/P
✓ 1
A1 46 Ω
✓ 1
Part (b)
M1 1.15 kW × 4 × 3.0
✓ 1
A1 13.8 kWh
✓ 1
Part (c)
M1 13.8 × 0.28 (ECF from (b))
✓ 1
A1 £3.86
✓ 1
Part (d)
M1 P = V2/R = 1842/46
✓ 1
A1 736 W
✓ 1
Common error: Power depends on the square of the p.d. when the resistance is fixed, so a 20% fall in voltage gives about a 36% fall in power, not 20%.
284IGPC3-4-07r26-16Series and parallel circuitsPaper CHard10 marks
A heated display cabinet in a shop has two heating elements with resistances 46 Ω and 92 Ω. They are connected in parallel to the 230 V mains supply.
Circuit diagram
(a)
Calculate the current in each heating element. [2]
(2)
(b)
Calculate the total power of the cabinet in watts. [2]
(2)
(c)
The elements are rewired in series across the same supply. Calculate the current from the supply. Give your answer to 3 significant figures. [2]
(2)
(d)
Calculate the total power in the series arrangement. Use the unrounded current from (c). Give your answer to 3 significant figures. [2]
(2)
(e)
Calculate the power in the series arrangement as a percentage of the power in the parallel arrangement. Use unrounded values and give your answer to 3 significant figures. [2]
(2)
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Part (a)
M1 I = V / R, e.g. 230 / 46
✓ 1
A1 5.0 A in the 46 Ω element and 2.5 A in the 92 Ω element
✓ 1
Part (b)
M1 total current 7.5 A and P = VI = 230 × 7.5
✓ 1
A1 1725 W (ECF from (a))
✓ 1
Part (c)
M1 R = 46 + 92 = 138 Ω so I = 230 / 138
✓ 1
A1 1.67 A
✓ 1
Part (d)
M1 P = VI = 230 × (230 / 138)
✓ 1
A1 383 W
✓ 1
Part (e)
M1 (383.33 / 1725) × 100
✓ 1
A1 22.2 % (ECF from (b) and (d))
✓ 1
Exam tip: For the same supply voltage, P = V² / R shows that a larger total resistance gives a smaller power.
285IGPC3-4-07r31-08Charge, current, potential difference, resistance and powerPaper CHard1 mark
A sports-timing lamp is rated 6.0 V, 3.0 W. It must work at its rated power from a 9.0 V supply, using a resistor connected in series with the lamp. What resistance is needed?
Circuit diagram
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Notes
The lamp current is P/V = 3.0/6.0 = 0.50 A. The resistor must take 9.0 − 6.0 = 3.0 V at 0.50 A, so R = 3.0/0.50 = 6.0 Ω. 1.5 Ω comes from dividing the supply p.d. by the lamp p.d. 12 Ω is the resistance of the lamp itself (6.0/0.50), not of the extra resistor. 18 Ω is the total circuit resistance (9.0/0.50), which includes the lamp.
✓ 1
Answer D
Answer: D
Common error: The series resistor takes only the leftover p.d. (supply minus lamp), not the whole supply p.d.
286IGPC-038-02-03Series and parallel circuitsPaper EEasy1 mark
Two lamps are connected in parallel to a battery. The current in one lamp is 0.30 A and the current in the other lamp is 0.50 A. What is the current in the battery?
Diagram not to scale
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At a junction, the current flowing in equals the current flowing out. The battery current therefore splits into the two branches: 0.30 + 0.50 = 0.80 A. 0.20 A is the difference between the branch currents. 0.40 A is their average. 0.50 A is just the larger branch current.
✓ 1
Answer D
Answer: D
Exam tip: In a parallel circuit, the current from the source is always larger than the current in any single branch.
287IGPC3-4-07r6-06Electromagnetic induction, motors and transformersPaper EEasy1 mark
In a sports-science demonstration a bar magnet is held completely still inside a coil that is connected to a sensitive voltmeter. What does the voltmeter read?
The bar magnet is held still inside the coil. Diagram not to scale
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Notes
D is correct: an emf is induced only while the flux linking the coil changes. A and B assume a strong field alone gives an emf. C confuses a permanent magnet with a changing field.
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Answer D
Answer: D
Common error: A strong magnetic field does not induce an emf by itself; there must be relative motion or a changing field.
288IGPC-038-04-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
A transformer that is 100% efficient steps down a 240 V a.c. supply to 12 V. The current in the secondary coil is 4.0 A. What is the current in the primary coil?
Diagram not to scale
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For a 100% efficient transformer, power in = power out, so IpVp = IsVs. This gives Ip = 4.0 × 12/240 = 0.20 A. Stepping the voltage down steps the current up, so the primary current is smaller. 80 A uses the voltage ratio the wrong way round. 4.0 A assumes the current is the same in both coils. 0.050 A comes from dividing 12 by 240.
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Answer B
Answer: B
Exam tip: Voltage and current change in opposite directions in an ideal transformer. If the voltage goes down by a factor of 20, the current goes up by a factor of 20.
289IGPC-038-04-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
Electrical power is transmitted along a cable of fixed resistance. The power transmitted stays the same, but the transmission voltage is doubled. What happens to the power lost as heat in the cable?
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Since P = IV, doubling V at the same power halves the current. The power loss in the cable is I2R. Halving I therefore reduces the loss to (½)2 = ¼ of its original value. 'Halves' ignores the squared dependence on current. 'Doubles' wrongly applies P = V2/R using the transmission voltage instead of the voltage drop across the cable.
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Answer D
Answer: D
Common error: Do not use V²/R with the transmission voltage to find the cable loss. That voltage is not across the cable. Use I²R with the current in the cable.
290IGPC-038-04-07Electromagnetic induction, motors and transformersPaper EMedium1 mark
A vertical wire carries a current flowing upwards. It is in a uniform horizontal magnetic field directed from west to east. In which direction is the force on the wire?
Side view: the field is horizontal, from west to east.
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Use Fleming's left-hand rule. Point the first finger east (field) and the second finger upwards (current). The thumb then points north, which is the direction of the force. 'South' results from using the right hand. 'East' wrongly assumes the force is along the field. 'Downwards' wrongly assumes the force is along the line of the current. The force is always perpendicular to both the field and the current.
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Answer A
Answer: A
Exam tip: Left hand for motors (force), right hand for generators. Hold your fingers at right angles and say: First finger = Field, seCond finger = Current, thuMb = Motion.
291IGPC-038-01-04Charge, current, potential difference, resistance and powerPaper EMedium1 mark
A current of 0.25 A flows in a torch bulb for 2.0 minutes. How much charge passes through the bulb?
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Q = It, with t in seconds: t = 2.0 × 60 = 120 s, so Q = 0.25 × 120 = 30 C. 0.50 C results from leaving the time in minutes. 480 C comes from dividing time by current, and 0.0021 C from dividing current by time.
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Answer C
Answer: C
Exam tip: Always convert time to seconds before using Q = It.
292IGPC-038-01-06Charge, current, potential difference, resistance and powerPaper EMedium1 mark
A metal wire has a resistance of 8.0 Ω. A second wire is made of the same metal. It has twice the length and twice the diameter of the first wire. What is the resistance of the second wire?
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Resistance is proportional to length and inversely proportional to cross-sectional area. Doubling the length doubles R. Doubling the diameter makes the area 22 = 4 times larger, so R becomes one quarter. New R = 8.0 × 2 ÷ 4 = 4.0 Ω. 8.0 Ω assumes the area doubles, so the two changes appear to cancel. 16 Ω ignores the change in thickness. 2.0 Ω treats length as if it reduced resistance.
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Answer B
Answer: B
Common error: Area depends on the square of the diameter. Doubling the diameter multiplies the area by four, not two.
293IGPC-038-03-10Electrical safetyPaper EMedium1 mark
Cables X and Y are made of copper and have the same length. Cable Y has half the cross-sectional area of cable X. The same current passes through each cable. Which statement is correct?
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Resistance is inversely proportional to cross-sectional area (R ∝ 1/A). Halving the area therefore doubles the resistance. The heating power is P = I2R, and the current is the same in both cables, so doubling R doubles P. The thinner cable gets hotter, which is why cables must be thick enough for the current they carry.
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Answer A
Answer: A
Common error: In P = I²R it is the current that is squared. With the same current, doubling R only doubles the power; it does not multiply it by four.
294IGPC-038-00-06Magnets and magnetic fieldsPaper EMedium1 mark
A student draws the magnetic field lines around a magnet. At point P the lines are 2 mm apart. At point Q they are 6 mm apart. Which statement is correct?
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The spacing of field lines shows the relative strength of the field. Closer lines mean a stronger field. The lines at P are closer than at Q, so the field is stronger at P.
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Answer B
Answer: B
Common error: Widely spaced lines do not mean a 'bigger' field. They mean a weaker field.
295IGPC-038-00-15Magnets and magnetic fieldsPaper EMedium6 marks
This question is about the magnetic fields produced by electric currents.
(a)
Describe the magnetic field pattern around a long straight wire carrying a current.
(2)
(b)
State the effect on this field of (i) increasing the current, (ii) reversing the current.
(2)
(c)
Describe the magnetic field inside a long current-carrying solenoid.
(2)
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Part (a)
concentric circles centred on the wire ; the circles are further apart (the field is weaker) further from the wire
✓ 2
Part (b)
(i) the field becomes stronger (at each point) ; (ii) the direction of the field reverses
✓ 2
Part (c)
the field lines are straight and parallel, along the axis of the solenoid ; the field is strong and (nearly) uniform / the lines are evenly and closely spaced
✓ 2
Model sketch
Model answer: concentric circles, further apart as the distance from the wire increases.
Exam tip: Outside a solenoid, the field is like that of a bar magnet. Inside, it is strong and uniform. Make both points clearly.
296IGPC-038-02-06Series and parallel circuitsPaper EMedium1 mark
A 2.0 kΩ resistor and a 4.0 kΩ resistor are connected in series to a 12 V supply of negligible internal resistance. What is the potential difference across the 4.0 kΩ resistor?
Diagram not to scale
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In a potential divider, the pd divides in the ratio of the resistances. The 4.0 kΩ resistor has 4.0/(2.0 + 4.0) = 2/3 of the total resistance, so its pd is 2/3 × 12 = 8.0 V. 4.0 V is the pd across the 2.0 kΩ resistor. 6.0 V assumes the pd is split equally. 12 V ignores the other resistor.
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Answer C
Answer: C
Common error: The larger resistor in a series pair always has the larger share of the pd. Check that your answer obeys this.
297IGPC-038-02-08Series and parallel circuitsPaper EMedium1 mark
A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel across a 6.0 V battery of negligible internal resistance. What is the current in the battery?
Diagram not to scale
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The combined resistance is R = (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω, so I = V/R = 6.0/2.0 = 3.0 A. You can check this using the branch currents: 1.0 A + 2.0 A = 3.0 A. 0.67 A uses the series total of 9.0 Ω. 1.0 A and 2.0 A are the individual branch currents.
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Answer D
Answer: D
Exam tip: For parallel circuits, check your answer by finding each branch current separately (each branch has the full pd) and then adding them.
298IGPC-038-02-13Series and parallel circuitsPaper EMedium8 marks
A 20 Ω resistor and a 30 Ω resistor are connected in parallel across a 12 V battery of negligible internal resistance.
Diagram not to scale
(a)
Calculate the combined resistance of the two resistors.
(2)
(b)
Calculate the current in the 20 Ω resistor.
(2)
(c)
Calculate the current in the battery.
(2)
(d)
Explain, in terms of charge, why the current in the battery equals the sum of the currents in the two resistors.
(2)
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Part (a)
1/R = 1/20 + 1/30 (or R = (20 × 30)/(20 + 30))
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R = 12 Ω
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Part (b)
The pd across the 20 Ω resistor is 12 V
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I = 12/20 = 0.60 A
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Part (c)
I = 12/12, or 0.60 A + 0.40 A
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= 1.0 A
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Part (d)
Charge is conserved / is not stored or lost at a junction
✓ 1
So the charge flowing into the junction per second equals the charge flowing out per second along the two branches
✓ 1
Exam tip: After using 1/R = 1/R₁ + 1/R₂, remember to take the reciprocal at the end. The answer must be smaller than 20 Ω.
299IGPC-038-02-14Series and parallel circuitsPaper EMedium9 marks
A night-light circuit uses a potential divider. A 10 kΩ fixed resistor and a light-dependent resistor (LDR) are connected in series to a 6.0 V supply of negligible internal resistance. The output voltage is taken across the LDR.
Diagram not to scale
(a)
Describe how the resistance of the LDR changes as the light level decreases.
(1)
(b)
In darkness the resistance of the LDR is 50 kΩ. Calculate the output voltage.
(3)
(c)
In daylight the output voltage is 1.2 V. Calculate the resistance of the LDR in daylight.
(3)
(d)
The output is connected to a switching circuit that switches on a lamp only when its input voltage is greater than 3.0 V. Using your answers to (b) and (c), explain why the lamp is on in darkness and off in daylight.
(2)
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Part (a)
The resistance increases
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Part (b)
Total resistance = 10 + 50 = 60 kΩ
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I = 6.0/60 000 = 1.0 × 10−4 A, or ratio 50/60 used
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V = 5.0 V
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Part (c)
pd across the fixed resistor = 6.0 − 1.2 = 4.8 V
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RLDR/10 kΩ = 1.2/4.8, or I = 4.8/10 000 = 4.8 × 10−4 A
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RLDR = 2.5 kΩ
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Part (d)
In darkness the LDR resistance is high, so it takes a larger share of the supply pd: output 5.0 V is greater than 3.0 V, so the lamp switches on
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In daylight the LDR resistance is low, so the output is only 1.2 V, which is less than 3.0 V, so the lamp stays off
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Common error: Do not use the full supply voltage with just one resistor. In a series potential divider, each component has only its own share of the pd, in proportion to its resistance.
300IGPC-038-01-13Charge, current, potential difference, resistance and powerPaper EMedium5 marks
A mobile phone is charged by a charger that provides a constant current of 1.5 A at a potential difference of 5.0 V. The phone is charged for 40 minutes.
(a)
Define electric current.
(1)
(b)
Calculate the charge that flows into the phone during the 40 minutes.
(2)
(c)
Calculate the energy transferred to the phone during this time.
(2)
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Part (a)
The charge passing a point per unit time, or the rate of flow of charge .
✓ 1
Part (b)
t = 40 × 60 = 2400 s, with Q = It . Q = 1.5 × 2400 = 3600 C .
✓ 2
Part (c)
Uses W = QV, or E = IVt, with correct substitution, e.g. 3600 × 5.0 or 1.5 × 5.0 × 2400 . Energy = 18 000 J (1.8 × 104 J) .
✓ 2
Exam tip: Potential difference is the energy transferred per unit charge, so energy = charge × pd. You can check this result using E = IVt, with the time in seconds.
301IGPC-038-04-16Electromagnetic induction, motors and transformersPaper EMedium6 marks
A simple a.c. generator consists of a rectangular coil rotating at a steady rate between the poles of a magnet. Initially the coil turns at 3000 revolutions per minute.
(a)
State the name of the parts that keep the rotating coil connected to the external circuit without twisting the wires.
(1)
(b)
Calculate the frequency of the alternating e.m.f.
(2)
(c)
Calculate the period of the e.m.f. at this rotation rate.
(1)
(d)
The rotation rate is doubled. State two ways in which the output e.m.f. changes.
(2)
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Part (a)
Slip rings (with brushes)
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Part (b)
3000/60 = 50 revolutions per second
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f = 50 Hz
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Part (c)
T = 1/f = 1/50 = 0.020 s (allow error carried forward from the frequency)
✓ 1
Part (d)
The peak e.m.f. doubles / increases
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The frequency doubles / the period halves
✓ 1
Exam tip: Convert revolutions per minute to revolutions per second by dividing by 60. One revolution of the coil produces one complete cycle of a.c.
302IGPC3-4-04c-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
The coil of a simple d.c. motor turns clockwise. Which change would make it turn anticlockwise?
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The force on the coil depends on the directions of the current and the field. Reversing only one of them reverses the force and the rotation (B). C reverses both, so the direction is unchanged. A and D change the size of the force, not its direction.
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Answer B
Answer: B
Common error: Reversing both the current and the field gives no change in direction.
303IGPC3-4-04c-06Electromagnetic induction, motors and transformersPaper EMedium1 mark
A bar magnet is pushed steadily into a coil connected to a sensitive voltmeter, held still inside the coil, and then pulled steadily out. Which describes the voltmeter reading?
The magnet is moved towards the coil, into it and out again. Diagram not to scale
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An e.m.f. is induced only while the field through the coil is changing. Moving in gives a deflection, a stationary magnet gives zero, and moving out reverses the direction (C). A and D assume a magnet inside the coil keeps inducing an e.m.f. B ignores that withdrawing reverses the change.
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Answer C
Answer: C
Exam tip: No change in the magnetic field means no induced e.m.f., however strong the field is.
304IGPC3-4-04c-13Electromagnetic induction, motors and transformersPaper EMedium6 marks
A small wind turbine turns a coil of wire between the poles of a strong magnet, generating an alternating voltage.
(a)
Explain why a voltage is induced in the coil as it rotates.
(2)
(b)
The wind speed increases and the coil rotates faster. State what happens to the frequency and to the peak voltage of the output.
(2)
(c)
Describe how the generator could be changed to give a direct voltage instead of an alternating voltage.
(2)
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Part (a)
the coil cuts the magnetic field lines
✓ 1
so the magnetic field through the coil changes and induces an e.m.f.
✓ 1
Part (b)
frequency increases
✓ 1
peak voltage increases
✓ 1
Part (c)
replace the slip rings with a split-ring commutator
✓ 1
which reverses the connections every half turn so the output is always in the same direction
✓ 1
Exam tip: An induced e.m.f. needs a changing magnetic field: motion, a changing current, or both.
305IGPC3-4-04c-14Electrical safetyPaper EMedium6 marks
A power station transmits 4.0 MW of electrical power through cables of total resistance 5.0 Ω to a town. 1 MW = 106 W.
Simplified: both cables are shown as one resistor of 5.0 Ω.
(a)
The power is transmitted at 200 kV. Calculate the current in the cables.
(2)
(b)
Calculate the power wasted as thermal energy in the cables.
(2)
(c)
Explain why transmitting at a high voltage reduces the energy wasted.
(2)
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Part (a)
I = P / V = 4.0 × 106 / 2.0 × 105
✓ 1
I = 20 A
✓ 1
Part (b)
P = I2R = 202 × 5.0
✓ 1
P = 2000 W
✓ 1
Part (c)
for the same power, a higher voltage means a smaller current
✓ 1
the power wasted depends on current squared, so it is much smaller
✓ 1
Common error: The wasted power uses the current in the cable, I²R, not the transmission voltage.
306IGPC3-4-06r6-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal transformer in a power supply for a guitar pedal changes the voltage from 230 V to 9.2 V. The current in the primary coil is 0.040 A. What is the current in the secondary coil?
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For an ideal transformer VpIp = VsIs, so Is = 230 × 0.040 / 9.2 = 1.0 A (A). B inverts the ratio, giving a lower current for a lower voltage. C assumes the current is unchanged. D is the voltage ratio 230/9.2 with no use of the primary current.
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Answer A
Answer: A
Common error: A step-down transformer lowers voltage but raises current, because the power stays the same for an ideal transformer.
307IGPC3-4-06r6-06Electromagnetic induction, motors and transformersPaper EMedium1 mark
A bar magnet is pushed into a coil of wire connected to a sensitive voltmeter. Which change will NOT increase the magnitude of the induced e.m.f.?
Bar magnet pushed into the coil. Diagram not to scale
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The induced e.m.f. depends on the rate of cutting of magnetic field lines. It is increased by a stronger magnet (A), more turns (B) and faster movement (C). Thicker wire changes the resistance of the coil but not the e.m.f., so D is correct.
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Answer D
Answer: D
Exam tip: The size of the induced e.m.f. depends on field strength, number of turns and speed of movement, not on the wire thickness.
308IGPC3-4-06r6-15Electromagnetic induction, motors and transformersPaper EMedium7 marks
A dynamic microphone used in a recording studio has a coil attached to a diaphragm. The coil is free to move in the field of a permanent magnet.
(a)
Explain how a sound wave striking the diaphragm produces an e.m.f. in the coil.
(3)
(b)
State what happens to the size of the e.m.f. when the sound becomes louder at the same pitch.
(1)
(c)
A note of frequency 440 Hz is played. State the frequency of the induced e.m.f.
(1)
(d)
Calculate the time for one complete cycle of the e.m.f. Give your answer to 2 significant figures.
(2)
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Part (a)
the sound wave makes the diaphragm and coil vibrate
✓ 1
the coil moves in the magnetic field, so the coil cuts the field lines (the flux changes)
✓ 1
an alternating e.m.f. is induced in the coil, with the same frequency as the sound
✓ 1
Part (b)
the e.m.f. increases (the coil moves faster)
✓ 1
Part (c)
440 Hz
✓ 1
Part (d)
T = 1 / f = 1 / 440
✓ 1
0.00227 s = 2.3×10−3 s
✓ 1
Exam tip: An induced e.m.f. needs relative movement between a coil and a magnetic field; if the movement stops, the e.m.f. falls to zero.
309IGPC3-4-07r6-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal step-down transformer at a rail depot has 2200 turns on the primary coil and 220 turns on the secondary coil. The primary current is 0.50 A. What is the current in the secondary coil?
Diagram not to scale
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For an ideal transformer VpIp = VsIs, so Is = Ip × Np/Ns = 0.50 × 10 = 5.0 A. 0.050 A uses the turns ratio the wrong way round. 0.50 A assumes the current is unchanged. 50 A applies the ratio twice (squared).
✓ 1
Answer D
Answer: D
Common error: When voltage is stepped down, current is stepped up by the same factor in an ideal transformer.
310IGPC3-4-07r6-13Electromagnetic induction, motors and transformersPaper EMedium6 marks
A bicycle lamp is lit by a dynamo. In the dynamo a small magnet is turned by the wheel next to a fixed coil, and the lamp is connected across the coil.
(a)
Explain why an emf is induced in the coil as the magnet turns.
(2)
(b)
State two changes that would increase the peak emf.
(2)
(c)
The cyclist stops. Explain why the lamp goes out.
(2)
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Part (a)
the magnetic flux linking the coil changes (the coil is cut by field lines)
✓ 1
a changing flux induces an emf across the coil
✓ 1
Part (b)
turn the magnet faster
✓ 1
use a stronger magnet or more turns on the coil
✓ 1
Part (c)
there is no relative movement between the magnet and the coil, so the flux does not change
✓ 1
no emf is induced so there is no current in the lamp
✓ 1
Exam tip: For induction answers, name the change in flux first, then say what it causes.
311IGPC3-4-07r6-14Electromagnetic induction, motors and transformersPaper EMedium8 marks
At a rail substation an ideal transformer reduces the voltage from 25 000 V to 1000 V. The secondary coil has 400 turns. The transformer supplies 80 kW to the train line.
Diagram not to scale
(a)
Calculate the number of turns on the primary coil.
(2)
(b)
Calculate the current in the secondary coil.
(2)
(c)
Calculate the current in the primary coil.
(2)
(d)
Explain why the transformer works only with alternating current.
(2)
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Part (a)
Np / Ns = Vp / Vs
✓ 1
Np = 400 × 25 000 / 1000 = 10 000
✓ 1
Part (b)
I = P / V
✓ 1
80 000 / 1000 = 80 A
✓ 1
Part (c)
power in = power out because the transformer is ideal
✓ 1
80 000 / 25 000 = 3.2 A
✓ 1
Part (d)
a changing current in the primary produces a changing magnetic field in the core
✓ 1
the changing flux through the secondary induces an emf; direct current gives no change of flux, so no emf
✓ 1
Exam tip: Check a transformer answer: the coil with fewer turns has the lower voltage and the higher current.
312IGPC3-4-05r7-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
A transformer used for a hospital lamp is assumed to be 100% efficient. The primary coil has 800 turns and the secondary coil has 200 turns. The primary is connected to a 120 V a.c. supply and carries a current of 0.50 A. What is the current in the secondary coil?
Diagram not to scale
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C is correct: Vs = 120 × 200 / 800 = 30 V; power in = power out = 60 W, so Is = 60 / 30 = 2.0 A. A is wrong because it scales the current by the turns ratio the same way as the voltage. B is wrong because it assumes the current is unchanged. D is wrong because 30 is the secondary voltage in volts, not the current.
✓ 1
Answer C
Answer: C
Exam tip: In a step-down transformer the voltage falls and the current rises by the same factor if there are no losses.
313IGPC3-4-05r7-06Electromagnetic induction, motors and transformersPaper EMedium1 mark
A magnet attached to a Formula 1 wheel passes close to a fixed coil connected to a sensor, and an e.m.f. is induced in the coil each time the magnet passes. The car speeds up so that the wheel turns faster. How does the induced e.m.f. change?
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A is correct: a faster-moving magnet cuts field lines at a greater rate, giving a larger e.m.f., and the pulses occur more often. B is wrong because pulses are more frequent when the wheel spins faster. C is wrong because the rate of change of flux depends on speed. D is wrong because the induced e.m.f. depends on the rate of cutting of field lines, not just on the equipment.
✓ 1
Answer A
Answer: A
Common error: Induced e.m.f. depends on how fast the flux changes, not only on the strength of the magnet.
314IGPC3-4-06r9-02Series and parallel circuitsPaper EMedium1 mark
In a telecoms test circuit, a 12 Ω resistor and a 24 Ω resistor are connected in parallel. What is their combined resistance?
Diagram not to scale
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Correct: D. 1/R = 1/12 + 1/24 = 3/24 = 1/8, so R = 8.0 Ω. A adds the resistances as if they were in series. B takes the mean of the two values. C gives 1/R (0.125) and forgets to invert it.
✓ 1
Answer D
Answer: D
Exam tip: The combined resistance of resistors in parallel is always smaller than the smallest individual resistor, which is a quick check on your answer.
315IGPC3-4-06r9-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
A transformer in a telecoms cabinet is connected to a battery that supplies steady direct current to the primary coil. Why is there no output p.d. from the secondary coil?
The primary coil is connected to a battery. Diagram not to scale
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Correct: A. An e.m.f. is induced only when the magnetic field through the secondary coil changes, and a steady current gives a constant field. B is wrong because a coil carries direct current easily. C is wrong because the size of the voltage is not the issue. D is wrong because a magnetised core does not switch off the current; the missing ingredient is a changing field.
✓ 1
Answer A
Answer: A
Exam tip: Induction needs a changing magnetic field, which is why transformers work with a.c. and not with steady d.c.
316IGPC3-4-06r9-13Series and parallel circuitsPaper EMedium6 marks
A food scientist heats samples with two heating elements of resistance 8.0 Ω and 12 Ω connected in parallel to a 24 V d.c. supply.
Diagram not to scale
(a)
Calculate the current in each heating element.
(2)
(b)
Calculate the total current from the supply.
(1)
(c)
A third element is added in parallel. State how the total resistance of the circuit changes.
(1)
(d)
Calculate the combined resistance of the two original elements.
(2)
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Part (a)
8.0 Ω element: 24 ÷ 8.0 = 3.0 A
✓ 1
12 Ω element: 24 ÷ 12 = 2.0 A
✓ 1
Part (b)
3.0 + 2.0 = 5.0 A
✓ 1
Part (c)
it decreases
✓ 1
Part (d)
R = V ÷ I = 24 ÷ 5.0 (or 1/R = 1/8.0 + 1/12)
✓ 1
4.8 Ω
✓ 1
Exam tip: Check a parallel answer against the smallest resistor: the combined resistance must be smaller than it.
317IGPC3-4-06r9-14Electromagnetic induction, motors and transformersPaper EMedium6 marks
A telecoms engineer pushes the north pole of a bar magnet into a coil connected to a sensitive voltmeter.
The north pole of the magnet is pushed into the coil. Diagram not to scale
(a)
Explain why the voltmeter shows a reading while the magnet is moving into the coil.
(2)
(b)
State what the voltmeter shows when the magnet is held still inside the coil.
(1)
(c)
State what happens to the direction of the reading when the magnet is pulled out of the coil.
(1)
(d)
State two ways to increase the size of the induced e.m.f.
(2)
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Part (a)
the magnetic field through the coil is changing (the coil cuts the field lines)
✓ 1
an e.m.f. is induced in the coil
✓ 1
Part (b)
zero reading, since the field is not changing
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Part (c)
the reading reverses direction
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Part (d)
move the magnet faster
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use more turns on the coil / use a stronger magnet
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Exam tip: An e.m.f. is induced only while the magnetic field through the coil is changing.
318IGPC3-4-07r10-03Electromagnetic induction, motors and transformersPaper EMedium1 mark
A hand-cranked generator in a mountaineer's emergency radio produces an alternating e.m.f. Which change would increase the peak e.m.f. produced?
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Notes
Turning faster makes the coil cut the magnetic field more quickly, so the induced e.m.f. is larger: A is correct. Reversing the crank direction only reverses the polarity of the e.m.f. and not its size. A weaker magnet and fewer turns both reduce the e.m.f.
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Answer A
Answer: A
Common error: Direction of rotation changes the sign of the induced e.m.f., not its size; speed, field strength and number of turns change its size.
319IGPC3-4-07r10-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
A step-down transformer in a telecoms cabinet changes a 230 V a.c. mains supply to 9.2 V. Its secondary coil has 150 turns. How many turns are on the primary coil? Assume the transformer is ideal.
Diagram not to scale
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Notes
Np/Ns = Vp/Vs, so Np = 150 × 230 / 9.2 = 3750: D is correct. 6 comes from inverting the ratio (150 × 9.2 / 230). 150 assumes both coils have the same number of turns. 371 comes from adding the voltage difference (230 − 9.2) to 150 instead of using a ratio.
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Answer D
Answer: D
Exam tip: In a step-down transformer the primary always has more turns than the secondary; check your answer against this.
320IGPC3-4-07r10-10Electromagnetic induction, motors and transformersPaper EMedium7 marks
An ideal transformer in a telecoms cabinet has 4600 turns on its primary coil, which is connected to a 230 V a.c. supply. The secondary coil supplies 48 V.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
The secondary current is 6.0 A. Calculate the current in the primary coil. Give your answer to 3 significant figures.
(3)
(c)
Explain why a transformer does not work with a direct current supply.
(2)
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Notes
Part (a)
Ns / Np = Vs / Vp
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Ns = 4600 × 48 / 230 = 960
✓ 1
Part (b)
output power = 48 × 6.0 = 288 W
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for an ideal transformer, input power = output power
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I = 288 / 230 = 1.25 A
✓ 1
Part (c)
a transformer needs a changing magnetic field in the core, which comes from a changing (alternating) current in the primary
✓ 1
a steady d.c. gives a constant field, so no e.m.f. is induced in the secondary
✓ 1
Exam tip: For an ideal transformer, equate input and output power when a current is asked for.
321IGPC3-4-07r10-13Series and parallel circuitsPaper EMedium7 marks
A telecoms test board has a 15 Ω resistor R1 connected in series with two resistors in parallel, R2 = 20 Ω and R3 = 30 Ω. A 27 V supply with negligible internal resistance is connected across the whole combination.
Diagram not to scale
(a)
Calculate the combined resistance of R2 and R3.
(2)
(b)
Calculate the total resistance of the circuit.
(1)
(c)
Calculate the current supplied by the 27 V source.
(1)
(d)
Calculate the potential difference across R1.
(1)
(e)
Calculate the current in R3.
(2)
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Notes
Part (a)
1/R = 1/20 + 1/30
✓ 1
R = 12 Ω
✓ 1
Part (b)
15 + 12 = 27 Ω
✓ 1
Part (c)
I = 27 / 27 = 1.0 A
✓ 1
Part (d)
V = 1.0 × 15 = 15 V (ECF)
✓ 1
Part (e)
p.d. across the parallel pair = 27 − 15 = 12 V
✓ 1
I = 12 / 30 = 0.40 A
✓ 1
Exam tip: The p.d. across a parallel pair is the supply p.d. minus the p.d. across the series resistor.
322IGPC3-4-05r11-05Magnets and magnetic fieldsPaper EMedium1 mark
A winch motor on a mountain rescue vehicle contains a straight wire carrying a current in a uniform magnetic field. The wire experiences an upward force. The direction of the current and the direction of the magnetic field are now both reversed. What is the effect on the force?
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Notes
Reversing the current alone reverses the force, and reversing the field alone also reverses it. Doing both reverses the force twice, leaving it upward and unchanged in size. A counts only one reversal. C is wrong because the wire is still at right angles to the field. D is wrong because the force stays perpendicular to both the current and the field in the same line as before.
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Answer B
Answer: B
Common error: Apply each reversal in turn: two reversals bring the force back to its original direction.
323IGPC3-4-05r11-13Charge, current, potential difference, resistance and powerPaper EMedium7 marks
A telecoms mast on a mountain ridge is supplied through a cable that has a total resistance of 2.5 Ω. The cable carries a current of 8.0 A.
Both cables are shown as one resistor of 2.5 Ω.
(a)
Calculate the power wasted as heat in the cable.
(2)
(b)
Calculate the energy wasted in the cable in 6.0 hours. Give your answer in kWh.
(2)
(c)
Electricity costs £0.25 per kWh. Calculate the cost of the energy wasted in the cable.
(1)
(d)
Explain why transmitting the same power at a higher voltage reduces the energy wasted in the cable.
(2)
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Notes
Part (a)
P = I2R = 8.02 × 2.5
✓ 1
= 160 W
✓ 1
Part (b)
160 W = 0.16 kW
✓ 1
E = 0.16 × 6.0 = 0.96 kWh, ECF from (a)
✓ 1
Part (c)
0.96 × 0.25 = £0.24, ECF from (b)
✓ 1
Part (d)
a higher voltage means a smaller current for the same power (P = IV)
✓ 1
the power wasted is proportional to I2, so less energy is transferred as heat in the cable
✓ 1
Common error: Power wasted in a cable uses the current in the cable and the cable's own resistance, not the supply voltage.
324IGPC3-4-05r11-14Electromagnetic induction, motors and transformersPaper EMedium7 marks
A telecoms hut is supplied through an ideal step-down transformer. The primary coil has 7500 turns and is connected to a 6000 V a.c. supply. The output of the secondary coil is 240 V.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
The current in the secondary coil is 25 A. Calculate the current in the primary coil.
(3)
(c)
Explain why the transformer does not work if the primary coil is connected to a d.c. supply.
(2)
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Notes
Part (a)
Np / Ns = Vp / Vs, so 7500 / Ns = 6000 / 240
✓ 1
Ns = 300
✓ 1
Part (b)
output power = 240 × 25 = 6000 W
✓ 1
input power equals output power because the transformer is ideal
✓ 1
Ip = 6000 / 6000 = 1.0 A
✓ 1
Part (c)
a d.c. current produces a constant magnetic field in the core
✓ 1
there is no changing magnetic field linking the secondary coil, so no e.m.f. is induced
✓ 1
Exam tip: For an ideal transformer, power in equals power out, so the coil with more turns has the smaller current.
325IGPC3-4-07r11-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal transformer has 1200 turns on its primary coil and 150 turns on its secondary coil. The primary coil is connected to a 240 V a.c. supply and the current in the secondary coil is 4.0 A. What is the current in the primary coil?
Diagram not to scale
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Notes
Output pd = 240 × 150 / 1200 = 30 V, output power = 30 × 4.0 = 120 W. For an ideal transformer input power is the same, so I = 120 / 240 = 0.50 A, so B is correct. A divides by the turns ratio twice. C assumes the currents are equal. D multiplies by the turns ratio instead of dividing, as if current rose with the pd.
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Answer B
Answer: B
Common error: In an ideal step-down transformer the current in the secondary is larger than in the primary; power, not current, is conserved.
326IGPC3-4-07r11-15Electromagnetic induction, motors and transformersPaper EMedium6 marks
A transformer at a coffee-roasting plant can be treated as 100% efficient. Its primary coil has 8800 turns and is connected to an 11 000 V a.c. supply. The secondary coil supplies 400 V to the roasting machines.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
The machines together use 36 kW. Calculate the current in the primary coil. Give your answer to 3 significant figures.
(2)
(c)
Explain why the transformer does not work if the primary coil is connected to a d.c. supply.
(2)
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Notes
Part (a)
Np / Ns = Vp / Vs
✓ 1
Ns = 8800 × 400 / 11 000 = 320
✓ 1
Part (b)
input power = output power = 36 000 W
✓ 1
I = P / V = 36 000 / 11 000 = 3.27 A
✓ 1
Part (c)
an alternating current is needed to produce a changing magnetic field (flux) in the core
✓ 1
a d.c. supply gives a constant field so no e.m.f. is induced in the secondary coil
✓ 1
Exam tip: For an ideal transformer, equate input and output power: VpIp = VsIs.
327IGPC3-4-05r12-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
The 400 V output of a wind turbine generator is connected to the primary coil of an ideal step-up transformer with 100 turns. The secondary coil has 2000 turns and carries a current of 3.0 A. What is the current in the primary coil?
Diagram not to scale
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Notes
Secondary voltage = 400 × 2000/100 = 8000 V. For an ideal transformer, input power equals output power: Ip = (8000 × 3.0)/400 = 60 A. 0.15 A divides by the turns ratio instead of multiplying. 3.0 A assumes the current is unchanged. 20 A is the turns ratio, not the current.
✓ 1
Answer D
Answer: D
Exam tip: In an ideal transformer a voltage step-up is a current step-down, and the other way round.
328IGPC3-4-05r12-12Electromagnetic induction, motors and transformersPaper EMedium8 marks
A wind farm generator produces an alternating voltage of 690 V. An ideal transformer with 2200 turns on its secondary coil steps this up to 11 000 V for transmission. The farm delivers a power of 1.5 MW.
Generator output stepped up for transmission. Diagram not to scale
(a)
State two ways of increasing the emf induced in the generator.
(2)
(b)
Explain why the output of the generator is an alternating voltage.
(2)
(c)
Calculate the number of turns on the primary coil.
(2)
(d)
Calculate the current in the primary coil. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
Any one from: stronger magnet, more turns on the coil, faster rotation, larger coil area
✓ 1
A second different way from the same list
✓ 1
Part (b)
As the coil rotates, each side cuts the magnetic field lines first in one direction and then in the opposite direction
✓ 1
so the induced emf reverses its direction every half turn
✓ 1
Part (c)
Np / Ns = Vp / Vs
✓ 1
Np = 2200 × 690 / 11 000 = 138 turns
✓ 1
Part (d)
Power in = power out = 1.5 × 106 W
✓ 1
I = P / V = 1.5 × 106 / 690 = 2174 A, so 2170 A (3 s.f.)
✓ 1
Exam tip: For an ideal transformer, equate the power on the two sides before finding currents.
329IGPC3-4-05r12-14Magnets and magnetic fieldsPaper EMedium7 marks
A laboratory centrifuge for blood samples is driven by a simple d.c. motor with a rectangular coil between the poles of two permanent magnets.
(a)
State what happens to a wire carrying a current at right angles to a magnetic field.
(1)
(b)
State two ways of reversing the direction of rotation of the motor.
(2)
(c)
Explain the function of the split-ring commutator.
(2)
(d)
State two ways of increasing the speed of rotation.
(2)
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Notes
Part (a)
It experiences a force
✓ 1
Part (b)
Reverse the direction of the current
✓ 1
Reverse the direction of the magnetic field (swap the magnet poles)
✓ 1
Part (c)
It reverses the current in the coil every half turn
✓ 1
so the forces on the sides of the coil always turn it in the same direction
✓ 1
Part (d)
Increase the current
✓ 1
Use stronger magnets or more turns on the coil
✓ 1
Exam tip: Without the commutator the coil would swing back and forth and stop in the vertical position.
330IGPC3-4-06r13-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal transformer in an observatory power supply has 1200 turns on the primary coil. The primary voltage is 240 V and the secondary voltage is 12 V. How many turns does the secondary coil have?
Diagram not to scale
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Notes
Correct: Np/Ns = Vp/Vs, so Ns = 1200 × 12 / 240 = 60. 24000 uses the inverted ratio; 1188 subtracts the voltages from the turns; 12 simply copies the voltage.
✓ 1
Answer C
Answer: C
Exam tip: A step-down transformer has fewer secondary turns than primary turns.
331IGPC3-4-06r13-13Electromagnetic induction, motors and transformersPaper EMedium6 marks
A step-down transformer in a railway signalling cabinet has 1150 turns on the primary coil. The primary voltage is 230 V and the secondary voltage is 24 V. Assume the transformer is 100% efficient.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
The secondary current is 9.2 A. Calculate the primary current.
(2)
(c)
Explain why a transformer does not work with a direct current supply.
(2)
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Notes
Part (a)
Np/Ns = Vp/Vs, Ns = 1150 × 24 / 230
✓ 1
= 120
✓ 1
Part (b)
Vp Ip = Vs Is, Ip = 24 × 9.2 / 230
✓ 1
= 0.96 A
✓ 1
Part (c)
a changing current produces a changing magnetic field in the core
✓ 1
only a changing field induces a voltage in the secondary coil; direct current gives constant field so no voltage
✓ 1
Exam tip: For an ideal transformer, power in equals power out, so the higher-voltage side carries the smaller current.
332IGPC3-4-06r15-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
A data-centre power supply contains a step-down transformer with 1200 turns on the primary coil and 60 turns on the secondary coil. The input is 230 V a.c. What is the output p.d.?
Diagram not to scale
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Notes
Vs = Vp × Ns/Np = 230 × 60 ÷ 1200 = 11.5 V, so C is correct. A assumes no change. B inverts the turns ratio. D is the turns ratio Np/Ns, not a p.d.
✓ 1
Answer C
Answer: C
Common error: Put the secondary turns on top when finding the secondary voltage: Vs = Vp × Ns/Np.
333IGPC3-4-06r15-15Electromagnetic induction, motors and transformersPaper EMedium7 marks
The hub of an e-bike contains a d.c. motor. The motor takes a current of 8.0 A from a 36 V battery.
(a)
State what causes the coil of a d.c. motor to turn.
(1)
(b)
Explain the purpose of the commutator.
(2)
(c)
State two ways of increasing the turning effect of the motor.
(2)
(d)
Calculate the electrical power supplied to the motor.
(2)
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Notes
Part (a)
a force acts on a current-carrying wire in a magnetic field
✓ 1
Part (b)
reverses the current in the coil every half turn
✓ 1
so the forces keep turning the coil in the same direction
✓ 1
Part (c)
first valid way, e.g. a larger current
✓ 1
second different valid way, e.g. a stronger magnet or more turns on the coil
✓ 1
Part (d)
P = IV = 8.0 × 36
✓ 1
288 W
✓ 1
Exam tip: Reversing either the current or the field reverses the direction of the force.
334IGPC3-4-07r16-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal transformer in a charging station has 1500 turns on the primary coil. It changes an input of 240 V a.c. to an output of 16 V a.c. How many turns are on the secondary coil?
Diagram not to scale
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Notes
For an ideal transformer Np / Ns = Vp / Vs, so Ns = 1500 × 16 / 240 = 100. 1276 comes from subtracting the voltages from the turns (1500 − 224). 22 500 comes from inverting the ratio (1500 × 240 / 16). 1500 comes from thinking that the turns must stay the same because power is conserved.
✓ 1
Answer A
Answer: A
Common error: Turns and voltages are in direct proportion; do not invert the ratio.
335IGPC3-4-07r16-11Electromagnetic induction, motors and transformersPaper EMedium7 marks
A charging station for warehouse robots contains an ideal transformer. The primary coil has 2000 turns and is connected to a 230 V a.c. supply. The secondary coil gives an output of 11.5 V.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
The secondary coil delivers a current of 4.0 A. Calculate the current in the primary coil. Assume the transformer is 100% efficient.
(3)
(c)
Explain why the transformer does not work if the primary coil is connected to a d.c. supply.
(2)
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Notes
Part (a)
M1 Np / Ns = Vp / Vs, so Ns = 2000 × 11.5 / 230
✓ 1
A1 100 turns
✓ 1
Part (b)
M1 power output = 4.0 × 11.5 = 46 W
✓ 1
M1 power input = power output, so Ip = 46 / 230
✓ 1
A1 0.20 A
✓ 1
Part (c)
B1 a steady d.c. current produces a constant magnetic field in the iron core
✓ 1
B1 no changing magnetic field passes through the secondary coil, so no e.m.f. is induced in it
✓ 1
Exam tip: For an ideal transformer, power in equals power out, so IpVp = IsVs; the coil with more turns has the higher voltage and the lower current.
336IGPC3-4-07r16-16Electromagnetic induction, motors and transformersPaper EMedium8 marks
A small wind generator on a fish-farm island supplies power along a cable to the fish-feeding units. In the generator a magnet rotates close to a fixed coil.
Simplified: the cable's total resistance is shown as one resistor of 0.40 Ω.
(a)
Explain why an e.m.f. is induced across the coil.
(2)
(b)
The generator delivers 12 kW at 240 V. Calculate the current in the cable.
(2)
(c)
The cable has a total resistance of 0.40 Ω. Calculate the power wasted as heat in the cable.
(2)
(d)
Explain why transmitting the same power at a much higher voltage reduces the energy wasted in the cable.
(2)
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Notes
Part (a)
B1 as the magnet turns, the magnetic field through the coil keeps changing (the field lines are cut)
✓ 1
B1 a changing magnetic field linked with the coil induces an e.m.f. across its ends
✓ 1
Part (b)
M1 I = P / V = 12 000 / 240
✓ 1
A1 50 A
✓ 1
Part (c)
M1 P = I²R = 50² × 0.40
✓ 1
A1 1000 W (ECF from (b))
✓ 1
Part (d)
B1 at higher voltage the current for the same power is smaller (I = P / V)
✓ 1
B1 the power wasted is I²R, so it falls much faster than the current does
✓ 1
Exam tip: Power loss in a cable depends on the square of the current, which is why power is transmitted at high voltage.
337IGPC3-4-08r21-06Electromagnetic induction, motors and transformersPaper EMedium1 mark
A step-down transformer in a station building has 6000 turns on its primary coil and 300 turns on its secondary coil. The input is 2400 V a.c. Assuming the transformer is ideal, what is the output p.d.?
Diagram not to scale
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Notes
Vs = Vp × Ns ÷ Np = 2400 × 300 ÷ 6000 = 120 V, so C is correct. B uses the ratio the wrong way round. A subtracts the numbers of turns. D is the turns ratio 6000 ÷ 300 mistaken for a voltage.
✓ 1
Answer C
Answer: C
Common error: A step-down transformer has fewer turns on the secondary coil, so the output voltage must be smaller than the input.
338IGPC3-4-08r21-11Series and parallel circuitsPaper EMedium6 marks
A sports-science laboratory builds a circuit from a 12 V battery of negligible internal resistance, a 20 Ω resistor, and two resistors of 30 Ω and 60 Ω connected in parallel. The 20 Ω resistor is in series with the parallel pair.
Circuit diagram
(a)
Calculate the combined resistance of the 30 Ω and 60 Ω resistors in parallel.
(2)
(b)
Calculate the total resistance of the circuit.
(1)
(c)
Calculate the current supplied by the battery.
(1)
(d)
Calculate the current in the 30 Ω resistor.
(2)
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Part (a)
1/R = 1/30 + 1/60
✓ 1
R = 20 Ω
✓ 1
Part (b)
20 + 20 = 40 Ω (ECF from (a))
✓ 1
Part (c)
I = 12 ÷ 40 = 0.30 A (ECF from (b))
✓ 1
Part (d)
p.d. across the parallel pair = 0.30 × 20 = 6.0 V (ECF)
✓ 1
I = 6.0 ÷ 30 = 0.20 A
✓ 1
Exam tip: Both parallel resistors share the same p.d., so find that p.d. before finding each branch current.
339IGPC3-4-08r21-14Electromagnetic induction, motors and transformersPaper EMedium6 marks
A track-side signal lamp unit uses an ideal transformer. The primary coil has 1840 turns and is connected to a 230 V a.c. supply. The secondary coil supplies 24 V to the lamp unit, which draws a current of 2.5 A.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Explain why the transformer does not work with a d.c. supply.
(2)
(c)
Calculate the current in the primary coil. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
Np ÷ Ns = Vp ÷ Vs, so Ns = 1840 × 24 ÷ 230
✓ 1
192 turns
✓ 1
Part (b)
d.c. produces a constant magnetic field in the core
✓ 1
so the field linking the secondary coil does not change and no e.m.f. is induced
✓ 1
Part (c)
power in = power out = 24 × 2.5 = 60 W
✓ 1
I = 60 ÷ 230 = 0.261 A
✓ 1
Exam tip: For an ideal transformer, the power in the primary equals the power in the secondary.
340IGPC3-4-07r23-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal step-down transformer in a robot charger has 1200 turns on its primary coil. The primary pd is 240 V and the secondary pd is 12 V. How many turns are on the secondary coil?
Diagram not to scale
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Notes
Ns = Np × Vs ÷ Vp = 1200 × 12 ÷ 240 = 60 turns. The value 20 is only the voltage ratio 240 ÷ 12. The value 24 000 comes from inverting the ratio, which would step the voltage up. The value 0.05 is the voltage ratio 12 ÷ 240 without multiplying by the number of primary turns.
✓ 1
Answer C
Answer: C
Common error: A step-down transformer always has fewer turns on the secondary coil than on the primary coil; use this to check your answer.
341IGPC3-4-07r23-06Electromagnetic induction, motors and transformersPaper EMedium1 mark
A bar magnet is pushed into a coil of wire connected to a sensitive voltmeter in a robot's position sensor. Which change would increase the maximum reading on the voltmeter?
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Notes
A faster movement makes the field through the coil change more quickly, so a larger emf is induced. Fewer turns or a weaker magnet each reduce the induced emf. A stationary magnet gives no change in field through the coil, so no emf is induced.
✓ 1
Answer A
Answer: A
Exam tip: Induced emf needs a changing magnetic field through the coil or wire cutting field lines; a stationary magnet gives zero.
342IGPC3-4-07r23-13Electromagnetic induction, motors and transformersPaper EMedium8 marks
A robot charger contains an ideal transformer that changes the 230 V mains supply to 9.0 V. The secondary coil has 45 turns. The output power is 36 W.
Diagram not to scale
(a)
Calculate the number of turns on the primary coil.
(2)
(b)
Calculate the current in the secondary coil.
(2)
(c)
Calculate the current in the primary coil. Give your answer to 3 significant figures.
(2)
(d)
Explain why a transformer does not work with direct current.
(2)
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Notes
Part (a)
Np / Ns = Vp / Vs, or Np = 45 × 230 / 9.0
✓ 1
= 1150 turns
✓ 1
Part (b)
I = P / V
✓ 1
= 36 / 9.0 = 4.0 A
✓ 1
Part (c)
Input power = output power for an ideal transformer, so I = 36 / 230
✓ 1
= 0.157 A
✓ 1
Part (d)
A direct current gives a constant magnetic field in the core, so the field through the secondary coil does not change
✓ 1
No emf is induced in the secondary coil
✓ 1
Exam tip: The current in the primary coil of a step-down transformer is smaller than that in the secondary coil; use this to check your answer.
343IGPC3-4-06r24-03Series and parallel circuitsPaper EMedium1 mark
In a Formula 1 garage, two heating elements of resistance 12 Ω and 4.0 Ω are connected in parallel in a tyre warmer. What is their combined resistance?
Circuit diagram
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Notes
For parallel resistors 1/R = 1/12 + 1/4 = 4/12, so R = 3.0 Ω, which is less than the smaller resistor. 8.0 Ω is the mean of the two values. 16 Ω adds them as if they were in series. 48 Ω multiplies them.
✓ 1
Answer A
Answer: A
Common error: The combined resistance of parallel resistors is always smaller than the smallest one, which gives a quick check.
344IGPC3-4-06r24-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
A step-down transformer in a hospital changes 230 V mains to 12 V for a patient lamp. The secondary coil has 60 turns. How many turns does the primary coil have?
Diagram not to scale
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Notes
For a transformer Vp/Vs = Np/Ns, so Np = 60 × 230 ÷ 12 = 1150. 3 comes from inverting the ratio. 720 multiplies the secondary turns by the secondary voltage. 13 800 multiplies the secondary turns by the primary voltage without dividing.
✓ 1
Answer A
Answer: A
Exam tip: A step-down transformer always has more turns on the primary coil than on the secondary coil.
345IGPC3-4-06r24-11Series and parallel circuitsPaper EMedium7 marks
A Formula 1 team uses a tyre warmer containing two heating elements connected in parallel to a 12 V battery. One element has a resistance of 8.0 Ω and the other has a resistance of 24 Ω.
Circuit diagram
(a)
Calculate the current in each element.
(2)
(b)
Calculate the total current supplied by the battery.
(2)
(c)
Calculate the combined resistance of the two elements.
(2)
(d)
State one advantage of connecting the elements in parallel rather than in series.
(1)
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Notes
Part (a)
I = V ÷ R = 12 ÷ 8.0 = 1.5 A
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I = 12 ÷ 24 = 0.50 A
✓ 1
Part (b)
currents in parallel branches add
✓ 1
1.5 + 0.50 = 2.0 A (ECF from (a))
✓ 1
Part (c)
R = V ÷ I = 12 ÷ 2.0 (ECF from (b))
✓ 1
6.0 Ω
✓ 1
(accept use of 1/R = 1/8.0 + 1/24 with the same answer)
—
Part (d)
if one element fails the other still works (or each element has the full battery voltage)
✓ 1
Exam tip: Check that the combined resistance of parallel resistors is smaller than the smallest one.
346IGPC3-4-07r24-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
A hospital uses an ideal step-down transformer to supply a 23 V device from a 230 V mains supply. The device draws a current of 4.0 A from the secondary coil. What is the current in the primary coil?
Diagram not to scale
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For an ideal transformer the power in the primary equals the power in the secondary. Secondary power = 23 × 4.0 = 92 W, so the primary current is 92/230 = 0.40 A. The primary current is smaller than the secondary current because the voltage is stepped down. 4.0 A assumes the current does not change. 40 A uses the turns ratio the wrong way round. 0.040 A applies the voltage ratio twice.
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Answer B
Answer: B
Exam tip: For an ideal transformer, VpIp = VsIs: if the voltage goes down, the current goes up.
347IGPC3-4-07r24-06Electromagnetic induction, motors and transformersPaper EMedium1 mark
A wheel-speed sensor on a racing car consists of a small permanent magnet fixed to the wheel that passes close to a stationary coil once per revolution, inducing a short pulse of emf each time. The wheel's rotation speed doubles. Which statement describes the pulses?
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The magnet passes the coil twice as fast, so the same change of magnetic flux linkage happens in half the time. The rate of change of flux doubles, so the peak emf doubles. There is one pulse per revolution, so twice as many revolutions per second gives pulses twice as frequent. The other options either leave the peak emf unchanged or reduce it, or keep the pulse rate the same, none of which follow from this.
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Answer D
Answer: D
Common error: Induced emf depends on the rate of change of flux linkage, not on the size of the flux change alone.
348IGPC3-4-07r24-11Electromagnetic induction, motors and transformersPaper EMedium7 marks
A hospital imaging unit uses an ideal transformer. The primary coil has 4600 turns and is connected to the 230 V mains supply. The secondary coil supplies 55 V.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
The current in the secondary coil is 8.0 A. Calculate the current in the primary coil.
(3)
(c)
Explain why a transformer does not work with a direct current supply.
(2)
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Part (a)
M1 Ns/Np = Vs/Vp
✓ 1
A1 1100 turns
✓ 1
Part (b)
M1 power in secondary = 55 × 8.0 = 440 W
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M1 primary power = secondary power, so I = 440/230
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A1 1.91 A
✓ 1
Part (c)
A direct current produces a constant magnetic field in the core
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so there is no changing magnetic flux linking the secondary coil and no emf is induced
✓ 1
Exam tip: A transformer works only on a changing current: explain this using changing magnetic flux in the core, not just 'd.c. does not change'.
349IGPC3-4-07r24-13Series and parallel circuitsPaper EMedium8 marks
A beekeeper's hive heater is made from three resistors. A 3.0 Ω resistor is connected in series with a parallel pair of resistors of 12 Ω and 4.0 Ω. The circuit is connected to a 12 V supply of negligible internal resistance.
Circuit diagram
(a)
Calculate the combined resistance of the 12 Ω and 4.0 Ω resistors in parallel.
(2)
(b)
Calculate the current drawn from the supply.
(2)
(c)
Calculate the potential difference across the parallel pair.
(2)
(d)
Calculate the current in the 12 Ω resistor.
(2)
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Part (a)
M1 1/R = 1/12 + 1/4
✓ 1
A1 3.0 Ω
✓ 1
Part (b)
M1 total resistance = 3.0 + 3.0 = 6.0 Ω and I = V/R (ECF from (a))
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A1 2.0 A
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Part (c)
M1 p.d. across the 3.0 Ω resistor = 2.0 × 3.0 = 6.0 V, so 12 − 6.0 (ECF from (b))
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A1 6.0 V
✓ 1
Part (d)
M1 I = V/R = 6.0/12 (ECF from (c))
✓ 1
A1 0.50 A
✓ 1
Exam tip: Use the unrounded values from earlier parts here: every value in this question comes out exactly.
350IGPC3-4-07r26-06Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal transformer at a wind farm has 1500 turns on its primary coil and 60 turns on its secondary coil. The current in the primary coil is 2.0 A. What is the current in the secondary coil?
Diagram not to scale
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Notes
For an ideal transformer the power in equals the power out, so Ip × Vp = Is × Vs and Is = Ip × (Np / Ns) = 2.0 × (1500 / 60) = 50 A. The 0.080 A option inverts the turns ratio. The 2.0 A option assumes the current is unchanged. The 25 A option uses the turns ratio itself without multiplying by the primary current.
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Answer D
Answer: D
Common error: A step-down transformer lowers the voltage but raises the current by the same factor, because power is conserved.
351IGPC3-4-07r26-08Charge, current, potential difference, resistance and powerPaper EMedium1 mark
A cable at a solar farm has resistance R. A second cable is made of the same material but has twice the length and twice the cross-sectional area of the first. What is the resistance of the second cable?
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Resistance is proportional to length and inversely proportional to cross-sectional area. Doubling both gives a factor of 2 / 2 = 1, so the resistance is unchanged at R. The R / 4 option halves the resistance for both changes. The R / 2 option includes only the effect of the larger area. The 4R option wrongly treats both changes as increasing the resistance.
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Answer C
Answer: C
Common error: A thicker conductor has lower resistance but a longer one has higher resistance, so the two effects can cancel.
352IGPC3-4-07r26-12Electromagnetic induction, motors and transformersPaper EMedium8 marks
A small solar farm generates electrical power at 600 V. An ideal transformer with 250 turns on its primary coil steps the voltage up to 24 kV for transmission. The farm delivers a power of 1.2 MW.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil. [2]
(2)
(b)
Calculate the current in the secondary coil. [2]
(2)
(c)
The transmission cable has a total resistance of 3.0 Ω. Calculate the power wasted as thermal energy in the cable. [2]
(2)
(d)
Explain why transmitting the power at high voltage reduces the power wasted in the cable. [2]
(2)
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Part (a)
M1 Ns / Np = Vs / Vp, so Ns = 250 × 24 000 / 600
✓ 1
A1 10 000 turns
✓ 1
Part (b)
M1 I = P / V = 1 200 000 / 24 000
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A1 50 A
✓ 1
Part (c)
M1 P = I2R = 502 × 3.0
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A1 7500 W (ECF from (b))
✓ 1
Part (d)
For the same power, a higher voltage means a smaller current
✓ 1
The power wasted is proportional to the square of the current, so it falls greatly
✓ 1
Exam tip: Power loss in a cable uses I²R with the current in the cable, not the supply voltage.
353IGPC3-4-07r26-13Series and parallel circuitsPaper EMedium7 marks
A vaccine store has a temperature alarm. An NTC thermistor is connected in series with a fixed 2000 Ω resistor across a 5.0 V supply of negligible internal resistance. A circuit monitors the p.d. across the fixed resistor.
Circuit diagram
(a)
State how the resistance of an NTC thermistor changes as its temperature increases. [1]
(1)
(b)
At 4 °C the resistance of the thermistor is 3000 Ω. Calculate the current in the circuit. [2]
(2)
(c)
At 25 °C the resistance of the thermistor is 500 Ω. Calculate the p.d. across the fixed resistor. [2]
(2)
(d)
Explain why the p.d. across the fixed resistor increases when the temperature rises. [2]
(2)
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Part (a)
A1 It decreases
✓ 1
Part (b)
M1 total resistance = 3000 + 2000 = 5000 Ω so I = 5.0 / 5000
✓ 1
A1 1.0 × 10−3 A
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Part (c)
M1 I = 5.0 / (500 + 2000) = 2.0 × 10−3 A
✓ 1
A1 V = 2.0 × 10−3 × 2000 = 4.0 V
✓ 1
Part (d)
The thermistor resistance falls so the total resistance falls and the current increases
✓ 1
The p.d. across the fixed resistor is V = I R with R constant, so it increases
✓ 1
Common error: The fixed resistor's p.d. rises because the current rises, not because its resistance changes.
354IGPC3-4-09r30-03Series and parallel circuitsPaper EMedium1 mark
A fishing boat's cabin has two resistors of 12 Ω and 4.0 Ω connected in parallel. What is the combined resistance?
Circuit diagram
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For parallel resistors 1/R = 1/12 + 1/4.0 = 4/12, so R = 3.0 Ω, which is less than the smallest resistor. 16 Ω adds the resistances as if in series. 8.0 Ω takes the mean. 48 Ω multiplies them.
✓ 1
Answer A
Answer: A
Exam tip: The combined resistance of resistors in parallel is always smaller than the smallest single resistor, which is a quick check.
355IGPC3-4-09r30-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
A simple a.c. generator on a research trawler turns its coil at a steady rate. The rotation speed of the coil is doubled. What happens to the frequency and the peak induced e.m.f.?
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The coil completes twice as many rotations each second, so the frequency doubles. The coil cuts the field lines twice as fast, so the peak e.m.f. also doubles. Frequency doubling alone ignores the rate of cutting field lines; peak e.m.f. doubling alone ignores that the cycle repeats faster; unchanged values ignore the speed change.
✓ 1
Answer D
Answer: D
Exam tip: Faster rotation means more cycles per second and a larger e.m.f. because field lines are cut more quickly.
356IGPC3-4-09r30-12Series and parallel circuitsPaper EMedium6 marks
A robot has two motors connected in parallel to a 24 V battery. Motor A has a resistance of 8.0 Ω and motor B has a resistance of 12 Ω.
Circuit diagram
(a)
Calculate the current in each motor.
(2)
(b)
Calculate the current supplied by the battery.
(1)
(c)
State one advantage of connecting the motors in parallel rather than in series.
(1)
(d)
Calculate the combined resistance of the two motors.
(2)
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Part (a)
A: I = V/R = 24/8.0 = 3.0 A
✓ 1
B: 24/12 = 2.0 A
✓ 1
Part (b)
3.0 + 2.0 = 5.0 A (ECF from (a))
✓ 1
Part (c)
if one motor fails the other keeps working (accept: each motor has the full battery potential difference)
✓ 1
Part (d)
M1 R = V/I = 24/5.0 OR 1/R = 1/8.0 + 1/12
✓ 1
A1 4.8 Ω
✓ 1
Common error: Do not add the resistances of components in parallel; the combined resistance is smaller than either one.
357IGPC3-4-07r31-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
A step-down transformer in an observatory control room is connected to a 12 V d.c. battery. After the instant of switching on, the output voltage of the secondary coil is zero. What is the reason?
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Notes
An e.m.f. is induced in the secondary coil only when the magnetic field linking it changes. A steady d.c. current gives a constant field, so no e.m.f. is induced. Direct current does flow through a coil, so the idea that it cannot flow is wrong. Direct current magnetises the core steadily rather than demagnetising it. Fewer turns on the secondary would give a smaller, not zero, output voltage if the field were changing.
✓ 1
Answer B
Answer: B
Common error: A transformer needs a changing current; it does not work with a steady d.c. supply.
358IGPC3-4-07r31-05Series and parallel circuitsPaper EMedium1 mark
Two heating elements of resistance 6.0 Ω and 12 Ω are connected in parallel to stop frost forming on a telescope cover. What is the combined resistance of the two elements?
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Notes
For parallel resistors 1/R = 1/6.0 + 1/12 = 3/12, so R = 4.0 Ω. 18 Ω is the series total. 9.0 Ω is the mean of the two resistances, which has no physical meaning here. 0.25 Ω is the value of 1/R before it has been inverted.
✓ 1
Answer C
Answer: C
Exam tip: The combined resistance of parallel resistors is always smaller than the smallest one; use this to check your answer.
359IGPC3-4-07r31-12Series and parallel circuitsPaper EMedium9 marks
A sports-science reaction timer uses a 10 V battery of negligible internal resistance. An 8.0 Ω resistor is connected in series with two resistors, of 20 Ω and 30 Ω, that are connected in parallel with each other.
Circuit diagram
(a)
Calculate the combined resistance of the 20 Ω and 30 Ω resistors.
(2)
(b)
Calculate the current supplied by the battery.
(3)
(c)
Calculate the potential difference across the parallel pair.
(2)
(d)
Calculate the current in the 20 Ω resistor.
(2)
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Part (a)
M1 1/R = 1/20 + 1/30
✓ 1
A1 12 Ω
✓ 1
Part (b)
B1 total resistance = 8.0 + 12 = 20 Ω (ECF from (a))
✓ 1
M1 I = V/R
✓ 1
A1 0.50 A
✓ 1
Part (c)
M1 V = IR = 0.50 × 12 (or 10 − 0.50 × 8.0)
✓ 1
A1 6.0 V
✓ 1
Part (d)
M1 I = V/R = 6.0 / 20
✓ 1
A1 0.30 A
✓ 1
Exam tip: Parallel resistors share the same p.d.; use the p.d. across the pair to find each branch current.
360IGPC3-4-09r31-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal transformer has 2400 turns on the primary coil and 60 turns on the secondary coil. The secondary current is 5.0 A. What is the primary current?
Diagram not to scale
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For an ideal transformer IpNp = IsNs, so Ip = 5.0 × 60 ÷ 2400 = 0.125 A. 200 A comes from inverting the turns ratio. 5.0 A assumes the current is unchanged. 0.025 A comes from 60 ÷ 2400 without multiplying by the secondary current.
✓ 1
Answer B
Answer: B
Exam tip: A step-down transformer lowers the voltage but raises the current, so the primary current is the smaller one.
361IGPC3-4-09r31-08Electromagnetic induction, motors and transformersPaper EMedium1 mark
A horizontal straight wire lies in an east-west direction. Conventional current flows towards the east. A uniform magnetic field acts vertically downwards. In which direction is the force on the wire?
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Using Fleming's left-hand rule with the first finger pointing down (field) and the second finger pointing east (current), the thumb points north, which is the direction of the force. South would result from reversing either the field or the current. The force is always perpendicular to both field and current, so upwards and downwards are impossible.
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Answer A
Answer: A
Common error: The force on a current-carrying wire is perpendicular to both the field and the current, never parallel to either.
362IGPC3-4-09r31-13Electromagnetic induction, motors and transformersPaper EMedium7 marks
The power adapter of a gaming monitor contains a transformer. The primary coil has 1840 turns and is connected to the 230 V mains. The secondary coil supplies a p.d. of 19.0 V. Assume the transformer is 100% efficient.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
The secondary current is 2.0 A. Calculate the primary current, giving your answer to 3 significant figures.
(3)
(c)
Explain why a transformer does not work with a steady direct current.
(2)
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Part (a)
M1 Ns = Np × Vs ÷ Vp = 1840 × 19.0 ÷ 230
✓ 1
A1 152 turns
✓ 1
Part (b)
M1 output power = 19.0 × 2.0 = 38 W
✓ 1
M1 power in = power out, so Ip = 38 ÷ 230
✓ 1
A1 0.165 A
✓ 1
Part (c)
A steady current produces a constant magnetic field in the core
✓ 1
There is no changing flux through the secondary, so no e.m.f. is induced
✓ 1
Exam tip: State that the current must change for the core's magnetic field to change; 'it needs a.c.' alone is not an explanation.
363IGPC3-4-10r33-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
A ground power unit at an airport uses an ideal transformer to step 230 V down to 28 V. The primary coil has 1150 turns. How many turns does the secondary coil have?
Diagram not to scale
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Notes
For an ideal transformer, Ns = Np × Vs / Vp = 1150 × 28 / 230 = 140 turns. 9446 uses the ratio upside down, 1122 subtracts the voltages from the turns, and 1178 adds them.
✓ 1
Answer B
Answer: B
Common error: Turns are in the same ratio as voltages: more turns on the coil with the higher voltage.
364IGPC3-4-10r33-11Electromagnetic induction, motors and transformersPaper EMedium8 marks
A transformer at an airport supplies power to parked aircraft. The primary coil has 2640 turns and is connected to a 6600 V supply. The secondary coil gives an output of 120 V and a power of 13.2 kW. Assume the transformer is 100% efficient.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Calculate the current in the secondary coil.
(2)
(c)
Calculate the current in the primary coil.
(2)
(d)
Explain why this transformer would not work if the primary coil were connected to a steady d.c. supply.
(2)
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Part (a)
M1 use of Ns / Np = Vs / Vp
✓ 1
A1 Ns = 48
✓ 1
Part (b)
M1 use of I = P / V
✓ 1
A1 I = 110 A
✓ 1
Part (c)
M1 power in primary = power in secondary, so I = P / V with 6600 V
✓ 1
A1 I = 2.0 A
✓ 1
Part (d)
A steady d.c. current produces a constant magnetic field in the core
✓ 1
The flux linked with the secondary coil does not change, so no emf is induced
✓ 1
Common error: A transformer works only with a changing current; a constant current does not induce an emf.
365IGPC3-4-08r35-05Electromagnetic induction, motors and transformersPaper EMedium1 mark
A transformer has 4600 turns on its primary coil. It is connected to a 230 V a.c. supply and must provide 5.0 V at its secondary coil. How many turns are needed on the secondary coil?
Diagram not to scale
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Notes
For a transformer, Vp / Vs = Np / Ns, so Ns = 4600 × 5.0 / 230 = 100 turns. 46 is the ratio of the voltages (230 / 5.0) and ignores the number of primary turns. 2300 is half the primary turns, which has no basis in the equation. 211 600 comes from inverting the ratio (4600 × 230 / 5.0).
✓ 1
Answer B
Answer: B
Common error: A step-down transformer always has fewer turns on the secondary coil than on the primary; check your answer against this.
366IGPC3-4-08r35-12Electromagnetic induction, motors and transformersPaper EMedium7 marks
A transformer for a telecoms charger has 2400 turns on its primary coil and 120 turns on its secondary coil. The primary coil is connected to a 230 V a.c. supply. The transformer may be assumed to be 100% efficient.
Diagram not to scale
(a)
Calculate the p.d. across the secondary coil.
(2)
(b)
The secondary current is 2.0 A. Calculate the current in the primary coil.
(2)
(c)
Explain why a transformer does not work with a steady d.c. supply.
(3)
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Part (a)
Vp / Vs = Np / Ns
✓ 1
Vs = 230 × 120 / 2400 = 11.5 V
✓ 1
Part (b)
power in = power out, so Vp Ip = Vs Is
✓ 1
Ip = (11.5 × 2.0) / 230 = 0.10 A (ECF from (a))
✓ 1
Part (c)
A steady d.c. current in the primary coil produces a constant magnetic field in the core
✓ 1
A changing magnetic field (changing flux) in the core is needed to cut the secondary coil
✓ 1
so no e.m.f. is induced in the secondary coil unless the current is changing, as with a.c.
✓ 1
Common error: A step-down transformer lowers the voltage but raises the current; the power stays the same, so the current ratio is the inverse of the voltage ratio.
367IGPC3-4-08r35-14Electromagnetic induction, motors and transformersPaper EMedium8 marks
The generator in a wind turbine at a mountain hut consists of a coil that rotates between the poles of a magnet. The output is an alternating e.m.f.
(a)
Explain why an e.m.f. is induced across the coil as it rotates.
(2)
(b)
State two ways to increase the maximum e.m.f. produced.
(2)
(c)
The coil makes 24 complete rotations in 8.0 seconds. Calculate the frequency of the alternating e.m.f.
(2)
(d)
The turbine speed doubles. State what happens to the frequency and to the maximum e.m.f.
(2)
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Part (a)
The wires of the coil cut the magnetic field lines (the magnetic flux linking the coil changes)
✓ 1
and the changing flux induces an e.m.f. in the coil
✓ 1
Part (b)
Any two from: use a stronger magnet; use more turns on the coil; rotate the coil faster; use a larger coil (area)
✓ 2
Part (c)
frequency = number of cycles / time
✓ 1
24 / 8.0 = 3.0 Hz
✓ 1
Part (d)
frequency doubles
✓ 1
maximum e.m.f. doubles
✓ 1
Exam tip: Faster rotation does two things: it raises the frequency and, because flux changes faster, it also raises the peak e.m.f.
368IGPC3-4-08r37-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal transformer in a stage lighting system changes the 230 V mains supply to 24 V. The primary coil has 4600 turns. How many turns does the secondary coil have?
Diagram not to scale
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Notes
For an ideal transformer Vp/Vs = Np/Ns, so Ns = 4600 × 24 / 230 = 480. A value of 20 is the number of primary turns per volt and has not been multiplied by 24. A value of 4576 comes from subtracting 24 from 4600, which treats the ratio as a difference. A value of 44 083 comes from inverting the ratio.
✓ 1
Answer D
Answer: D
Exam tip: A step-down transformer has fewer turns on the secondary coil, so check your answer is smaller than the primary turns.
369IGPC3-4-08r37-11Electromagnetic induction, motors and transformersPaper EMedium8 marks
A volcano observatory uses an ideal step-down transformer to supply a seismic sensor unit with 12 V from a 230 V mains supply. The primary coil has 1150 turns.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
The sensor unit draws a current of 2.5 A from the secondary coil. Calculate the power delivered to the sensor unit.
(2)
(c)
Calculate the current in the primary coil. Give your answer to 2 significant figures.
(2)
(d)
Explain why the transformer does not work when the primary coil is connected to a battery.
(2)
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Notes
Part (a)
M1 Ns = Np × Vs / Vp = 1150 × 12 / 230
✓ 1
A1 60 turns
✓ 1
Part (b)
M1 P = IV = 2.5 × 12
✓ 1
A1 30 W
✓ 1
Part (c)
M1 power in = power out, so Ip = 30 / 230 (ECF from (b))
✓ 1
A1 0.13 A
✓ 1
Part (d)
A1 a battery gives a constant (direct) current, so the magnetic field in the core does not change
✓ 1
A1 no changing magnetic field is linked with the secondary coil, so no e.m.f. is induced
✓ 1
Exam tip: For an ideal transformer, the power in the primary equals the power in the secondary, so the lower voltage side carries the larger current.
370IGPC3-4-07r39-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
An ideal step-down transformer in a bridge lighting system has a primary coil connected to a 230 V a.c. supply. The secondary coil has 60 turns and gives an output of 12 V. How many turns does the primary coil have?
Diagram not to scale
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Notes
For an ideal transformer Vp/Vs = Np/Ns, so Np = 60 × 230 / 12 = 1150. 3.1 comes from inverting the ratio. 278 comes from adding and subtracting the voltages and turns instead of using a ratio. 13 800 comes from forgetting to divide by 12.
✓ 1
Answer C
Answer: C
Exam tip: A step-down transformer always has more turns on the primary coil than on the secondary coil, so check that your answer fits this.
371IGPC3-4-07r39-06Electromagnetic induction, motors and transformersPaper EMedium1 mark
A magnet is pushed into a coil of 200 turns at a constant speed and induces an e.m.f. of 0.40 V. The experiment is repeated with a coil of 400 turns and the same magnet pushed in at twice the speed. The induced e.m.f. is directly proportional to both the number of turns and the speed. What is the new e.m.f.?
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Notes
Doubling the turns doubles the e.m.f. and doubling the speed doubles it again, so 0.40 × 2 × 2 = 1.6 V. 0.80 V allows for only one of the two changes. 0.40 V assumes constant speed means no change. 3.2 V applies one of the doublings twice.
✓ 1
Answer A
Answer: A
Exam tip: When two factors both change and the quantity is proportional to each, multiply the separate scale factors.
372IGPC3-4-07r39-12Electromagnetic induction, motors and transformersPaper EMedium7 marks
An ideal step-up transformer supplies a ground-station instrument with 112 V a.c. from a 28 V a.c. source. The primary coil has 800 turns.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Explain why a transformer does not work with direct current.
(2)
(c)
The instrument takes 150 W. Assume the transformer is 100% efficient. Calculate the current in the primary coil, giving your answer to 3 significant figures.
(3)
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Part (a)
Vp/Vs = Np/Ns
✓ 1
Ns = 800 × 112 / 28 = 3200
✓ 1
Part (b)
A changing current in the primary coil produces a changing magnetic field in the iron core, which induces an e.m.f. in the secondary coil
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Direct current gives a constant field, so there is no change of field linked with the secondary and no e.m.f. is induced
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Part (c)
Power in = power out, so primary power is 150 W
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I = P / V = 150 / 28
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= 5.36 A
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Exam tip: For an ideal transformer, power in equals power out, so VpIp = VsIs.
373IGPC3-4-08r42-04Electromagnetic induction, motors and transformersPaper EMedium1 mark
A transformer for a farm electric-fence energiser has 1200 turns on its primary coil, connected to a 240 V a.c. supply. The output must be 12 V. How many turns are needed on the secondary coil?
Diagram not to scale
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For a transformer Ns / Np = Vs / Vp, so Ns = 1200 × 12 / 240 = 60. The value 24 000 comes from inverting the ratio, which would step the voltage up. The value 1188 comes from subtracting the voltages from the turns. The value 20 is the turns ratio, not the number of turns.
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Answer B
Answer: B
Exam tip: A step-down transformer always has fewer turns on the secondary than on the primary; use this as a quick check.
374IGPC3-4-08r42-12Electromagnetic induction, motors and transformersPaper EMedium7 marks
A step-down transformer in an airport ground-power unit has 960 turns on its primary coil, connected to a 240 V a.c. supply. The secondary coil supplies 30 V to equipment and carries a current of 8.0 A. Assume the transformer is 100% efficient.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Calculate the power output of the secondary coil.
(1)
(c)
Calculate the current in the primary coil.
(2)
(d)
Explain why the transformer does not work with a steady d.c. supply.
(2)
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Part (a)
M1 Np / Ns = Vp / Vs, so 960 / Ns = 240 / 30
✓ 1
A1 120 turns
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Part (b)
P = IV = 30 × 8.0 = 240 W
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Part (c)
M1 power input = power output, so Ip × 240 = 240
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A1 1.0 A (ECF from (b))
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Part (d)
A steady current gives a constant magnetic field in the core
✓ 1
so there is no changing magnetic field linking the secondary coil and no voltage is induced
✓ 1
Exam tip: For an ideal transformer, VpIp = VsIs; the side with the higher voltage has the lower current.
375IGPC3-4-08r42-15Electromagnetic induction, motors and transformersPaper EMedium5 marks
A small wind turbine on a farm turns a coil of wire between the poles of a magnet to generate electricity.
(a)
Explain why a voltage is induced in the coil.
(2)
(b)
State two changes that would increase the maximum induced voltage.
(2)
(c)
State what happens to the frequency of the alternating voltage when the turbine rotates faster.
(1)
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Notes
Part (a)
The magnetic field through the coil changes (the coil cuts field lines) as it rotates
✓ 1
a changing magnetic field linking the coil induces an e.m.f.
✓ 1
Part (b)
Any one from: turn the coil faster / use a stronger magnet / use more turns on the coil
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a different second change from the same list
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Part (c)
The frequency increases
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Common error: A voltage is induced only while the magnetic field through the coil is changing, not just because a field is present.
376IGPC-038-04-09Electromagnetic induction, motors and transformersPaper EHard1 mark
A rectangular coil rotates at a steady rate in a uniform magnetic field in a simple a.c. generator. At which position is the induced e.m.f. at its maximum value?
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The e.m.f. is largest when the sides of the coil cut the field lines at the greatest rate. This happens when the coil sides move at right angles to the field, that is, when the plane of the coil is parallel to the field. When the plane is perpendicular to the field, the most field lines pass through the coil. At that moment, however, the sides move along the field lines, so the rate of cutting and the e.m.f. are zero. The output alternates, so it is not constant.
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Answer B
Answer: B
Common error: The e.m.f. is not largest when the most field lines pass through the coil. It is largest when the field through the coil is changing fastest.
377IGPC-038-01-08Charge, current, potential difference, resistance and powerPaper EHard1 mark
A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel. This combination is connected in series with a 4.0 Ω resistor. What is the total resistance of the arrangement?
Diagram not to scale
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For the parallel pair: 1/R = 1/6.0 + 1/3.0 = 3/6.0, so R = 2.0 Ω. Adding the series resistor gives 2.0 + 4.0 = 6.0 Ω. 13 Ω treats all three resistors as in series. 1.3 Ω treats all three as in parallel. 8.5 Ω uses the average of 6.0 Ω and 3.0 Ω as the parallel value.
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Answer B
Answer: B
Exam tip: Check that a parallel combination is smaller than its smallest resistor: 2.0 Ω is less than 3.0 Ω.
378IGPC-038-01-15Charge, current, potential difference, resistance and powerPaper EHard7 marks
A 4.0 Ω resistor and a 12 Ω resistor are connected in parallel across a 12 V supply. The supply has negligible internal resistance.
Diagram not to scale
(a)
Calculate the combined resistance of the two resistors.
(2)
(b)
Calculate the current from the supply.
(1)
(c)
Explain why the current in the 4.0 Ω resistor is greater than the current in the 12 Ω resistor.
(2)
(d)
Calculate the power dissipated in the 12 Ω resistor.
(2)
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Part (a)
1/R = 1/4.0 + 1/12 = 4/12 . R = 3.0 Ω .
✓ 2
Part (b)
I = 12 / 3.0 = 4.0 A .
✓ 1
Part (c)
Both resistors have the same pd (12 V) across them because they are in parallel . Since I = V / R, the resistor with the smaller resistance has the larger current .
✓ 2
Part (d)
I = 12 / 12 = 1.0 A . P = IV = 1.0 × 12 = 12 W .
✓ 2
Common error: Do not use the supply current of 4.0 A for a single branch. Each branch has its own current, and the branch currents (3.0 A and 1.0 A) add up to the supply current.
379IGPC-038-03-16Electrical safetyPaper EHard8 marks
A 3.0 kW, 230 V fan heater is connected through a cable on an extension reel. The total resistance of the two current-carrying wires in the cable is 0.40 Ω. The instructions say the cable must be fully unwound from the reel before a high-power appliance is used.
The cable's resistance is shown as a single resistor in series with the heater.
(a)
Calculate the current in the heater.
(2)
(b)
Calculate the power dissipated as heat in the cable.
(2)
(c)
Explain why the cable must be fully unwound before the heater is used.
(2)
(d)
Suggest why a cable with a larger cross-sectional area would be safer for this heater.
(2)
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Part (a)
I = P/V = 3000/230 ; I = 13 A (13.0 A)
✓ 2
Part (b)
P = I2R = 13.02 × 0.40 ; P = 68 W
✓ 2
Part (c)
When coiled, the cable cannot transfer the heat to the surroundings easily, so its temperature rises a lot ; the insulation may melt, causing a fire or exposing live wires that could give a shock
✓ 2
Part (d)
Larger cross-sectional area gives a lower resistance ; so for the same current less power is dissipated (P = I2R) and the cable heats up less
✓ 2
Exam tip: Use the unrounded current (13.04 A) in P = I²R. Rounding to 13 A first gives 67.6 W, which is still about 68 W here, but early rounding can cost marks in other questions.
380IGPC-038-02-09Series and parallel circuitsPaper EHard1 mark
Three identical lamps X, Y and Z are connected to a battery. Lamp X is in series with a parallel pair made of lamps Y and Z. The filament of lamp Z breaks. Assume the lamp resistances stay constant. What happens to the brightness of lamps X and Y?
Circuit before the filament of lamp Z breaks.
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Let each lamp have resistance R and the battery pd be V. Before the break, the total resistance is R + R/2 = 1.5R, so the current in X is V/1.5R ≈ 0.67V/R. Y carries half of this, ≈ 0.33V/R. After Z breaks, X and Y are simply in series with total resistance 2R, so both carry 0.50V/R. The current in X falls, so X dims. The current in Y rises, so Y brightens.
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Answer B
Answer: B
Common error: Removing a parallel branch does not leave the rest of the circuit unchanged. The total resistance rises, so the supply current falls.
381IGPC-038-02-10Series and parallel circuitsPaper EHard1 mark
A thermistor and a 3.0 kΩ fixed resistor are connected in series to a 9.0 V supply of negligible internal resistance. A voltmeter across the fixed resistor reads 6.0 V. What is the resistance of the thermistor?
Diagram not to scale
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The pd across the thermistor is 9.0 − 6.0 = 3.0 V. The current is the same in both components: I = 6.0/3000 = 2.0 × 10−3 A. So the thermistor resistance is R = 3.0/(2.0 × 10−3) = 1500 Ω = 1.5 kΩ. 4.5 kΩ is the total resistance (9.0 V ÷ 2.0 mA), found by forgetting to subtract the fixed resistor. 6.0 kΩ inverts the ratio. 3.0 kΩ assumes the pd is shared equally.
✓ 1
Answer A
Answer: A
Exam tip: In a potential divider, R₁/R₂ = V₁/V₂. Find the pd across the unknown component first by subtracting from the supply pd.
382IGPC-038-01-16Charge, current, potential difference, resistance and powerPaper EHard7 marks
A length of nichrome wire 0.80 m long has a resistance of 4.0 Ω.
(a)
State how the resistance of a wire of a given length and material depends on its cross-sectional area.
(1)
(b)
Calculate the resistance of a 1.4 m length of the same wire.
(2)
(c)
A different nichrome wire is also 1.4 m long, but its diameter is half that of the original wire. Calculate its resistance.
(2)
(d)
The thinner wire is connected to a 7.0 V supply. Calculate the power dissipated in the wire.
(2)
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Part (a)
Resistance is inversely proportional to cross-sectional area (for the same length and material) .
✓ 1
Part (b)
Resistance is proportional to length, or R = 4.0 × 1.4 / 0.80 seen . R = 7.0 Ω .
✓ 2
Part (c)
Halving the diameter makes the cross-sectional area one quarter, so the resistance is 4 times larger . R = 4 × 7.0 = 28 Ω .
✓ 2
Part (d)
I = V / R = 7.0 / 28 = 0.25 A, or P = V2/R = 7.02/28 seen . P = 1.75 W (accept 1.8 W) .
✓ 2
Common error: Halving the diameter does not simply double the resistance. Area depends on the diameter squared, so the area becomes one quarter and the resistance increases four times.
383IGPC-038-00-09Magnets and magnetic fieldsPaper EHard1 mark
A plotting compass is placed 2 cm from a long straight wire carrying a steady current. Which change makes the magnetic field produced by the wire at the position of the compass both stronger and reversed in direction?
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Reversing the direction of the current reverses the direction of the field. The field is stronger closer to the wire. So reversing the current and moving the compass to 1 cm gives a field that is both reversed and stronger. Option A makes the field stronger but does not reverse it. Option B reverses the field but makes it weaker, because the compass is further away. Option D only makes the field weaker.
✓ 1
Answer C
Answer: C
Exam tip: Treat the two factors separately. The direction of the current sets the direction of the field. The size of the current and the distance from the wire set the strength of the field.
384IGPC-038-02-16Series and parallel circuitsPaper EHard10 marks
A 9.0 V supply of negligible internal resistance is connected to a 4.0 Ω resistor. The 4.0 Ω resistor is in series with a parallel combination of a 6.0 Ω resistor and a 12 Ω resistor.
Diagram not to scale
(a)
Calculate the resistance of the parallel combination.
(2)
(b)
Calculate the current in the supply.
(2)
(c)
Calculate the potential difference across the parallel combination.
(2)
(d)
Calculate the current in the 12 Ω resistor.
(1)
(e)
The 12 Ω resistor is removed. Explain what happens to the potential difference across the 4.0 Ω resistor, and calculate its new value.
(3)
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Part (a)
R = (6.0 × 12)/(6.0 + 12), or 1/R = 1/6.0 + 1/12
✓ 1
= 4.0 Ω
✓ 1
Part (b)
Total resistance = 4.0 + 4.0 = 8.0 Ω
✓ 1
I = 9.0/8.0 = 1.1 A (1.125 A)
✓ 1
Part (c)
V = IR = 1.125 × 4.0, or the two equal 4.0 Ω resistances share the 9.0 V equally
✓ 1
= 4.5 V
✓ 1
Part (d)
I = 4.5/12 = 0.38 A (0.375 A)
✓ 1
Part (e)
The resistance of the remaining branch (6.0 Ω) is now greater than 4.0 Ω, so the total resistance rises to 4.0 + 6.0 = 10 Ω
✓ 1
Either: the supply current decreases to 9.0/10 = 0.90 A; or: the 4.0 Ω resistor now has a smaller share of the total resistance, 4.0/10
✓ 1
The pd across the 4.0 Ω resistor decreases from 4.5 V to 0.90 × 4.0 = 9.0 × 4.0/10 = 3.6 V
✓ 1
Common error: In a series–parallel circuit, reduce the parallel section to a single equivalent resistor first. Do not add all three resistances together. Carry the unrounded current (1.125 A) into later parts to avoid rounding errors.
385IGPC3-4-04c-07Charge, current, potential difference, resistance and powerPaper EHard1 mark
A copper wire has resistance R. A second copper wire has twice the length and twice the diameter of the first wire. What is the resistance of the second wire?
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Doubling the length doubles the resistance. Doubling the diameter makes the cross-sectional area four times larger, which divides the resistance by four. Overall 2 / 4 = 0.5R (B). A ignores the longer length. C treats doubling the diameter as doubling the area. D considers only the length.
✓ 1
Answer B
Answer: B
Common error: Area depends on the square of the diameter, so doubling the diameter quadruples the area.
386IGPC3-4-04c-15Electromagnetic induction, motors and transformersPaper EHard8 marks
A model railway uses an ideal transformer to change the 240 V mains voltage to 12 V. The primary coil has 1600 turns. A lamp connected to the secondary coil transfers 36 W.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Calculate the current in the secondary coil.
(2)
(c)
Calculate the current in the primary coil.
(2)
(d)
Explain why the transformer would not work if the primary coil were connected to a battery.
(2)
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Part (a)
Ns = Np × Vs / Vp
✓ 1
Ns = 1600 × 12 / 240 = 80 turns
✓ 1
Part (b)
I = P / V = 36 / 12
✓ 1
I = 3.0 A
✓ 1
Part (c)
for an ideal transformer the input power equals the output power, 36 W
✓ 1
I = 36 / 240 = 0.15 A
✓ 1
Part (d)
a battery gives a steady direct current, so the magnetic field in the core is constant
✓ 1
there is no changing field to induce an e.m.f. in the secondary coil
✓ 1
Common error: In an ideal transformer, power in equals power out, so the coil with more turns carries the smaller current.
387IGPC3-4-06r6-08Series and parallel circuitsPaper EHard1 mark
A 24 V battery of negligible internal resistance is connected to a 6.0 Ω resistor in series with a parallel pair of resistors of 6.0 Ω and 3.0 Ω. What is the current in the 3.0 Ω resistor?
Diagram not to scale
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The parallel pair has resistance 1/(1/6.0 + 1/3.0) = 2.0 Ω, so the total resistance is 8.0 Ω and the battery current is 3.0 A. The p.d. across the parallel pair is 3.0 × 2.0 = 6.0 V, so the current in the 3.0 Ω resistor is 6.0 / 3.0 = 2.0 A (B). A is the current in the 6.0 Ω branch. C is the total current. D is 24 V divided by 6.0 Ω, ignoring the rest of the circuit.
✓ 1
Answer B
Answer: B
Exam tip: Find the total current first, then the p.d. across the parallel section, and finally the current in the branch you want.
388IGPC3-4-06r6-14Electromagnetic induction, motors and transformersPaper EHard8 marks
A transformer in a battery charger for airport baggage trolleys is ideal. The primary coil has 1840 turns and is connected to a 230 V a.c. supply. The secondary coil provides 12 V. The charger supplies a power of 60 W.
The charger (load) takes 60 W from the secondary coil. Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Explain why a transformer does not work with a steady direct current in the primary coil.
(2)
(c)
Calculate the current in the secondary coil.
(2)
(d)
Calculate the current in the primary coil. Give your answer to 2 significant figures.
(2)
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Part (a)
Vp / Vs = Np / Ns
✓ 1
Ns = 1840 × 12 / 230 = 96
✓ 1
Part (b)
a transformer needs a changing magnetic field (flux) in the core to induce an e.m.f. in the secondary coil
✓ 1
a steady d.c. produces a constant magnetic field, so no e.m.f. is induced
✓ 1
Part (c)
I = P / V = 60 / 12
✓ 1
5.0 A
✓ 1
Part (d)
for an ideal transformer, power in = power out, so VpIp = 60 W
✓ 1
Ip = 60 / 230 = 0.26 A (0.2609 unrounded)
✓ 1
Common error: The current in the primary of a step-down transformer is smaller than in the secondary. Use equal power, not equal current.
389IGPC3-4-06r6-16Electrical safetyPaper EHard9 marks
A wind turbine on a farm generates 12 kW. The electrical power is sent to a barn along a cable of total resistance 5.0 Ω. Ignore any other energy losses.
Arrangement for parts (a) and (b). Both cables are shown as one resistor of 5.0 Ω.
(a)
The power is first sent at 600 V. Calculate the current in the cable.
(2)
(b)
Calculate the power wasted as thermal energy in the cable at 600 V.
(2)
(c)
A step-up transformer is used to raise the voltage to 6000 V before transmission. Explain why this reduces the power wasted in the cable.
(2)
(d)
Calculate the power wasted in the cable when the voltage is 6000 V.
(3)
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Part (a)
I = P / V = 12 000 / 600
✓ 1
20 A
✓ 1
Part (b)
P = I2R = 202 × 5.0 (ECF from (a))
✓ 1
2000 W
✓ 1
Part (c)
for the same power, a higher voltage means a smaller current in the cable
✓ 1
wasted power is proportional to I2R, so it falls by a large factor
✓ 1
Part (d)
I = 12 000 / 6000 = 2.0 A
✓ 1
P = I2R = 2.02 × 5.0
✓ 1
20 W
✓ 1
Common error: Power wasted depends on the square of the current, so ten times less current gives one hundred times less power wasted.
390IGPC3-4-07r6-07Electromagnetic induction, motors and transformersPaper EHard1 mark
A straight wire lies between the poles of a magnet. The magnetic field is horizontal, from left to right, and the current is into the page. The force on the wire is downwards. The current is now reversed and the magnet is turned round so that the field points from right to left. What is the force on the wire now?
Side view: the field is horizontal, left to right; the cross shows the current into the page.
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Reversing either the current or the field reverses the force; reversing both restores the original direction. The size is unchanged because neither the current nor the field strength has changed. A treats only one reversal. B and C have no physical basis.
✓ 1
Answer D
Answer: D
Exam tip: Count the reversals: one reversal flips the force, two reversals leave it as it was.
391IGPC3-4-07r6-15Electromagnetic induction, motors and transformersPaper EHard7 marks
A student drops a small bar magnet, north pole first, so that it falls vertically through the centre of a flat coil. The coil is connected to a data logger that records the emf against time. The trace shows one peak as the magnet enters the coil and an opposite peak as it leaves.
The magnet falls vertically, north pole first, through the centre of the coil. Diagram not to scale
(a)
Explain why an emf is induced as the magnet enters the coil.
(2)
(b)
Explain why the second peak is in the opposite direction.
(2)
(c)
The second peak is taller than the first. Explain why.
(2)
(d)
Suggest one change that would reduce the height of both peaks.
(1)
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Part (a)
the magnetic flux linking the coil increases as the magnet enters
✓ 1
a changing flux (field lines cut by the coil) induces an emf
✓ 1
Part (b)
the flux linking the coil is now decreasing instead of increasing
✓ 1
the induced emf opposes the change, so its direction reverses
✓ 1
Part (c)
the magnet is accelerating under gravity so it is moving faster as it leaves
✓ 1
the flux changes at a faster rate, so a larger emf is induced
✓ 1
Part (d)
release the magnet from a lower starting height (accept: weaker magnet, fewer turns)
✓ 1
Common error: A bigger emf comes from a faster rate of flux change, not from a larger total amount of flux.
392IGPC3-4-05r7-15Electromagnetic induction, motors and transformersPaper EHard10 marks
A hospital operating lamp uses 24 V and has a power of 48 W. It is supplied from a 240 V a.c. mains supply by a transformer with 480 turns on its secondary coil. Assume the transformer is 100% efficient.
Diagram not to scale
(a)
Explain why a transformer does not work when it is connected to a d.c. supply.
(3)
(b)
Calculate the number of turns on the primary coil.
(2)
(c)
Calculate the current in the lamp.
(2)
(d)
Calculate the current in the primary coil. Give your answer in amperes.
(3)
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Part (a)
An a.c. supply in the primary coil produces a changing magnetic field in the iron core
✓ 1
The changing field links the secondary coil and induces an e.m.f. in it
✓ 1
A d.c. supply gives a constant field so no e.m.f. is induced
✓ 1
Part (b)
Np / Ns = Vp / Vs
✓ 1
Np = 480 × 240 / 24 = 4800
✓ 1
Part (c)
I = P / V = 48 / 24
✓ 1
I = 2.0 A
✓ 1
Part (d)
Power in = power out = 48 W
✓ 1
I = P / V = 48 / 240
✓ 1
I = 0.20 A
✓ 1
Exam tip: For an ideal transformer, equate power in and power out to find the current on the side not given.
393IGPC3-4-05r7-16Charge, current, potential difference, resistance and powerPaper EHard9 marks
A beekeeper has two heating wires made from the same material. Wire A has a resistance of 4.0 Ω. Wire B is 3.0 times as long as wire A and has one quarter of the cross-sectional area of wire A.
Circuit for parts (c) and (d): wires A and B in series. Diagram not to scale
(a)
State two quantities that affect the resistance of a wire of a given material.
(2)
(b)
Calculate the resistance of wire B.
(3)
(c)
The two wires are connected in series to a 13 V supply. Calculate the current.
(2)
(d)
Calculate the potential difference across wire B. Give your answer in volts.
(2)
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Part (a)
Length of the wire
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Cross-sectional area (or diameter) of the wire
✓ 1
Part (b)
Resistance is proportional to length, so multiply by 3.0
✓ 1
Resistance is inversely proportional to area, so multiply by 4
✓ 1
R = 4.0 × 3.0 × 4 = 48 Ω
✓ 1
Part (c)
Total resistance = 4.0 + 48 = 52 Ω (ECF from (b))
✓ 1
I = 13 / 52 = 0.25 A
✓ 1
Part (d)
V = IR
✓ 1
V = 0.25 × 48 = 12 V (ECF from (b) and (c))
✓ 1
Common error: Resistance falls when the area increases: a smaller area gives a larger resistance.
394IGPC3-4-06r9-07Electromagnetic induction, motors and transformersPaper EHard1 mark
An ideal step-down transformer in a telecoms charger changes 230 V to 11.5 V. The secondary coil delivers a current of 0.60 A. What is the current in the primary coil?
Diagram not to scale
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Correct: C. For an ideal transformer the power in equals the power out, so Ip = VsIs ÷ Vp = 11.5 × 0.60 ÷ 230 = 0.030 A. A multiplies the current by the voltage ratio instead of dividing. B assumes the current is the same in both coils. D divides by the voltage ratio twice.
✓ 1
Answer C
Answer: C
Exam tip: A step-down transformer lowers the voltage and raises the current, so the primary current is the smaller one.
395IGPC3-4-06r9-15Electromagnetic induction, motors and transformersPaper EHard8 marks
A telecoms mast has a step-down transformer with a primary p.d. of 230 V and a secondary p.d. of 12 V. The secondary coil has 60 turns. Power for the mast is brought from a power station, which transmits 200 kW at 20 kV along cables of total resistance 4.0 Ω.
Step-down transformer in the mast. Diagram not to scale
(a)
Calculate the number of turns on the primary coil.
(2)
(b)
Explain why power is transmitted at a high voltage instead of a low voltage.
(3)
(c)
Calculate the power lost as heat in the cables.
(3)
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Part (a)
Np ÷ Ns = Vp ÷ Vs
✓ 1
Np = 60 × 230 ÷ 12 = 1150 turns
✓ 1
Part (b)
for the same power, a higher voltage means a smaller current
✓ 1
power loss in the cables is I2R, so a smaller current means much less energy wasted as heat
✓ 1
the cables can be thinner and cheaper / less heating of cables
✓ 1
Part (c)
current = 200 000 ÷ 20 000 = 10 A
✓ 1
power loss = I2R = 102 × 4.0
✓ 1
400 W
✓ 1
Common error: Use the current in the cables, found from P = IV, in the power-loss equation; the supply voltage alone does not give the loss.
396IGPC3-4-07r10-07Series and parallel circuitsPaper EHard1 mark
A 9.0 V battery with negligible internal resistance is connected to a 3.0 Ω resistor in series with two 6.0 Ω resistors in parallel, as in a food-heating test rig. What is the current in one of the 6.0 Ω resistors?
Diagram not to scale
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Notes
The parallel pair has resistance 3.0 Ω, so the total is 6.0 Ω and the battery current is 9.0/6.0 = 1.5 A. This splits equally between the two identical branches, giving 0.75 A: D is correct. 0.60 A comes from adding all three resistors in series (9.0/15). 1.0 A comes from ignoring one 6.0 Ω resistor (9.0/9.0). 1.5 A is the total current, not the current in one branch.
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Answer D
Answer: D
Common error: After finding the total current, remember to share it between the parallel branches.
397IGPC3-4-07r10-11Electromagnetic induction, motors and transformersPaper EHard7 marks
A student releases a bar magnet from rest above a vertical coil. The coil is connected to a sensitive voltmeter. The magnet falls, with its north pole leading, straight through the coil.
The magnet falls north pole first, vertically through the coil. Diagram not to scale
(a)
State the voltmeter reading while the magnet is held at rest just above the coil.
(1)
(b)
Explain why the voltmeter shows a reading as the magnet enters the coil.
(2)
(c)
The reading as the magnet enters the coil is in one direction. State the direction of the reading as the magnet leaves, and explain your answer.
(2)
(d)
The peak reading as the magnet leaves is larger than the peak reading as it enters. Explain why.
(2)
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Part (a)
zero
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Part (b)
the magnetic field through the coil is changing (the coil turns cut the field lines)
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so an e.m.f. is induced across the coil
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Part (c)
opposite direction
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the field through the coil is now decreasing (the field is cut in the opposite sense)
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Part (d)
the magnet speeds up as it falls under gravity
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so it moves faster through the coil as it leaves, the field changes more rapidly and a larger e.m.f. is induced
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Common error: An e.m.f. is induced only while the field is changing; a stationary magnet in or near the coil gives no reading.
398IGPC3-4-05r11-06Charge, current, potential difference, resistance and powerPaper EHard1 mark
A heating wire in a food dehydrator has a resistance of 12 Ω. The wire is stretched uniformly until it is three times as long. Its volume stays constant. What is its new resistance?
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Tripling the length at constant volume divides the cross-sectional area by 3. Resistance is proportional to length and inversely proportional to area, so it increases by 3 × 3 = 9, giving 12 × 9 = 108 Ω. C accounts for the length only. A divides by 3 and B divides by 9, which reverses the effect of the changes.
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Answer D
Answer: D
Common error: When a wire is stretched, both the length and the area change, so two factors affect the resistance.
399IGPC3-4-05r11-15Series and parallel circuitsPaper EHard8 marks
In a mountain hut, a 12.0 V battery of negligible internal resistance is connected to a 9.0 Ω resistor in series with a parallel combination of a 12 Ω resistor and a 4.0 Ω resistor.
Diagram not to scale
(a)
Calculate the combined resistance of the 12 Ω and 4.0 Ω resistors.
(2)
(b)
Calculate the current in the battery.
(2)
(c)
Calculate the current in the 12 Ω resistor.
(2)
(d)
The 4.0 Ω resistor is removed from the circuit, leaving a gap. Calculate the new current in the battery. Give your answer to 2 significant figures.
(2)
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Part (a)
1/R = 1/12 + 1/4
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R = 3.0 Ω
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Part (b)
total resistance = 9.0 + 3.0 = 12 Ω, ECF from (a)
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I = 12.0 / 12 = 1.0 A
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Part (c)
potential difference across the parallel pair = 1.0 × 3.0 = 3.0 V, ECF from (b)
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I = 3.0 / 12 = 0.25 A
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Part (d)
total resistance = 9.0 + 12 = 21 Ω
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I = 12.0 / 21 = 0.57 A
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Common error: Recalculate the whole circuit after a component is removed; the other currents do not stay the same.
400IGPC3-4-05r11-16Charge, current, potential difference, resistance and powerPaper EHard9 marks
A food dehydrator uses a heating wire. A 2.4 m length of this wire, with cross-sectional area 0.30 mm2, has a resistance of 8.0 Ω. A technician replaces it with a wire of the same material that is 3.0 m long and has cross-sectional area 0.20 mm2.
(a)
Calculate the resistance of the replacement wire.
(3)
(b)
The replacement wire is connected to a 12 V supply. Calculate the current in it.
(2)
(c)
Calculate the power of the replacement wire.
(2)
(d)
Explain, in terms of electrons, why the wire becomes hot when there is a current in it.
(2)
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Part (a)
resistance is proportional to length: factor 3.0 / 2.4 = 1.25
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resistance is inversely proportional to area: factor 0.30 / 0.20 = 1.5
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R = 8.0 × 1.25 × 1.5 = 15 Ω
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Part (b)
I = V / R = 12 / 15
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= 0.80 A, ECF from (a)
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Part (c)
P = IV = 0.80 × 12
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= 9.6 W, ECF from (b)
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Part (d)
electrons collide with the ions (atoms) of the wire as they move through it
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the electrons transfer kinetic energy to the ions, increasing the internal energy and temperature of the wire
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Common error: Both a longer wire and a thinner wire have more resistance, so the two factors multiply.
401IGPC3-4-07r11-06Electromagnetic induction, motors and transformersPaper EHard1 mark
A bar magnet is released from rest above a coil and falls vertically through it with its north pole leading. The ends of the coil are connected to a sensitive voltmeter. Air resistance is negligible. How does the e.m.f. pulse as the magnet leaves the bottom of the coil compare with the pulse as it enters the top?
The magnet falls north pole first, vertically through the coil. Diagram not to scale
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Notes
On entering, the flux linked with the coil increases; on leaving it decreases, so the induced e.m.f. reverses, which rules out A and B. The magnet accelerates, so it moves faster when leaving and the rate of change of flux is greater, giving a larger peak: D is correct. C wrongly assumes constant speed.
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Answer D
Answer: D
Common error: The size of the induced e.m.f. depends on how fast the flux changes, so a falling magnet gives a bigger pulse the faster it moves.
402IGPC3-4-07r11-08Charge, current, potential difference, resistance and powerPaper EHard1 mark
A thin metal wire in a strain gauge on a bridge has a resistance of 120 Ω. It is stretched uniformly to twice its original length. The volume of the metal and its resistivity do not change. What is the new resistance?
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Notes
Doubling the length doubles R, but the volume is fixed so the cross-sectional area halves, which doubles R again. R becomes 120 × 2 × 2 = 480 Ω, so D is correct. A ignores the stretching. B allows for the length only. C wrongly applies a further factor of 2.
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Answer D
Answer: D
Common error: When a wire is stretched, remember the area falls as the length rises, so R changes by the square of the length factor.
403IGPC3-4-05r12-06Electromagnetic induction, motors and transformersPaper EHard1 mark
A bar magnet is pushed at speed v into a coil of N turns and induces an emf E. The same magnet is then pushed into a second, otherwise identical coil of 2N turns at speed 2v. What is the induced emf?
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Notes
The induced emf is proportional to the number of turns and to the rate at which the field lines are cut. Doubling the turns doubles the emf and doubling the speed doubles it again, so 2 × 2 = 4E. E ignores both changes and 2E allows for only one. 8E has no basis: nothing is cubed.
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Answer C
Answer: C
Exam tip: When two factors both change, multiply their effects together.
404IGPC3-4-05r12-15Series and parallel circuitsPaper EHard8 marks
A solar-powered pump for a clinic uses a 12 V supply, with negligible internal resistance. A 15 Ω resistor is in series with two resistors, of 30 Ω and 60 Ω, which are connected in parallel with each other. Use unrounded values from earlier parts in later calculations.
Diagram not to scale
(a)
Calculate the combined resistance of the 30 Ω and 60 Ω resistors.
(2)
(b)
Calculate the current supplied by the 12 V supply. Give your answer to 2 significant figures.
(2)
(c)
Calculate the potential difference across the parallel combination.
(2)
(d)
Calculate the power dissipated in the 60 Ω resistor. Give your answer to 2 significant figures.
(2)
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Part (a)
1/R = 1/30 + 1/60
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R = 20 Ω
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Part (b)
Total resistance = 15 + 20 = 35 Ω (ECF from (a))
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I = 12 / 35 = 0.34 A (0.3429 A unrounded)
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Part (c)
V = I R using I = 0.3429 A and R = 20 Ω
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V = 6.9 V (6.857 V unrounded; ECF from (a) and (b))
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Part (d)
P = V2 / R
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P = 6.8572 / 60 = 0.78 W (accept 0.79 W if 6.9 V is used; ECF from (c))
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Exam tip: Find the resistance of a parallel section first, then treat it as one resistor in series with the rest.
405IGPC3-4-05r12-16Charge, current, potential difference, resistance and powerPaper EHard8 marks
A remote solar farm delivers 50 kW through cables of total resistance 0.80 Ω. At first the power is transmitted at 250 V.
Arrangement for parts (a) and (b). Both cables are shown as one resistor of 0.80 Ω.
(a)
Calculate the current in the cables.
(2)
(b)
Calculate the power wasted as thermal energy in the cables.
(2)
(c)
An ideal step-up transformer is now used to raise the transmission voltage to 5000 V for the same delivered power. Calculate the new power wasted in the cables.
(2)
(d)
Explain why high-voltage transmission reduces the energy wasted in the cables.
(2)
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Part (a)
I = P / V
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I = 50 000 / 250 = 200 A
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Part (b)
P = I2R
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P = 2002 × 0.80 = 32 000 W (ECF from (a))
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Part (c)
New current = 50 000 / 5000 = 10 A
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P = 102 × 0.80 = 80 W
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Part (d)
For the same power, a higher voltage means a smaller current
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The power wasted depends on I2, so it falls much more than the current does
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Common error: Use the current in the cable, not the supply voltage, in P = I²R for the cable loss.
406IGPC3-4-06r13-07Charge, current, potential difference, resistance and powerPaper EHard1 mark
A uniform wire in a sports-science sensor has resistance R. A second wire is made of the same material, with twice the length and twice the diameter. What is the resistance of the second wire?
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Correct: R is proportional to length/area. Doubling the diameter makes the area 4 times larger, so R is multiplied by 2/4, giving R/2. R/4 ignores the length; R assumes the changes cancel; 2R ignores the area.
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Answer B
Answer: B
Common error: Doubling the diameter multiplies the cross-sectional area by four, not two.
407IGPC3-4-06r13-15Charge, current, potential difference, resistance and powerPaper EHard8 marks
In a circuit, a resistor of 6.0 Ω is in series with a parallel pair of resistors of 20 Ω and 30 Ω. The supply voltage is 18 V.
Diagram not to scale
(a)
Calculate the combined resistance of the 20 Ω and 30 Ω resistors in parallel.
(2)
(b)
Calculate the current from the supply.
(2)
(c)
Calculate the current in the 20 Ω resistor.
(2)
(d)
Calculate the power dissipated in the 30 Ω resistor.
(2)
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Part (a)
1/R = 1/20 + 1/30
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R = 12 Ω
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Part (b)
total resistance = 6.0 + 12 = 18 Ω (ECF from (a))
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I = 18 / 18 = 1.0 A
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Part (c)
pd across parallel pair = 18 − 6.0 × 1.0 = 12 V
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I = 12 / 20 = 0.60 A
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Part (d)
P = V2/R = 122 / 30
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= 4.8 W
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Exam tip: The pd across both resistors of a parallel pair is the same; find it by subtracting the pd across the series resistor from the supply.
408IGPC3-4-06r13-16Charge, current, potential difference, resistance and powerPaper EHard8 marks
An electric train takes a power of 2.5 MW from an overhead line at 25 kV. The overhead line has a total resistance of 0.40 Ω.
Simplified: the line's total resistance is shown as one resistor of 0.40 Ω.
(a)
Calculate the current in the line.
(2)
(b)
Calculate the power wasted as thermal energy in the line.
(2)
(c)
Explain why transmitting at a high voltage reduces the power wasted in the line.
(2)
(d)
Calculate the percentage of the 2.5 MW that is wasted in the line.
(2)
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Part (a)
I = P / V = 2.5×106 / 25000
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= 100 A
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Part (b)
P = I2R = 1002 × 0.40
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= 4000 W
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Part (c)
for the same power a higher voltage means a smaller current
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power wasted is proportional to I2, so it falls greatly
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Part (d)
4000 / 2500000 × 100 (ECF from (b))
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= 0.16%
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Common error: Use the current in the line, not the supply voltage, when calculating P = I²R.
409IGPC3-4-06r15-06Series and parallel circuitsPaper EHard1 mark
In a cooling test rig, an 18 V battery of negligible internal resistance is connected to an 8.0 Ω resistor in series with a parallel pair of resistors of 12 Ω and 6.0 Ω. What is the current in the 12 Ω resistor?
Diagram not to scale
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Notes
The parallel pair has resistance 12×6.0/(12+6.0) = 4.0 Ω; the total is 12 Ω so the battery current is 1.5 A. The p.d. across the pair is 1.5 × 4.0 = 6.0 V, so the current in the 12 Ω resistor is 6.0 ÷ 12 = 0.50 A, and C is correct. A is the total current. B assumes the current splits equally. D is the current in the 6.0 Ω resistor.
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Answer C
Answer: C
Exam tip: Find the p.d. across the parallel section first, then divide by each branch's resistance.
410IGPC3-4-06r15-13Electromagnetic induction, motors and transformersPaper EHard8 marks
A data centre uses a step-down transformer. The primary coil has 4800 turns and is connected to a 240 V a.c. supply. The secondary coil has 240 turns and supplies a current of 20 A to a rack of servers. Assume the transformer is 100% efficient.
Diagram not to scale
(a)
Calculate the p.d. across the secondary coil.
(2)
(b)
Calculate the output power.
(2)
(c)
Calculate the current in the primary coil.
(2)
(d)
Explain why a transformer does not work with direct current.
(2)
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Part (a)
Vp/Vs = Np/Ns
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Vs = 240 × 240 ÷ 4800 = 12 V
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Part (b)
P = IV = 12 × 20
✓ 1
240 W
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Part (c)
input power = output power, so IpVp = 240 W , ECF from (b)
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Ip = 240 ÷ 240 = 1.0 A
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Part (d)
d.c. produces a constant magnetic field in the core
✓ 1
no changing flux, so no emf is induced in the secondary coil
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Exam tip: For an ideal transformer, power in equals power out: IpVp = IsVs.
411IGPC3-4-06r15-14Electromagnetic induction, motors and transformersPaper EHard8 marks
A bicycle hub dynamo has a magnet that rotates close to a fixed coil as the wheel turns.
(a)
Explain how an emf is induced in the coil when the wheel turns.
(3)
(b)
State two ways of increasing the emf produced.
(2)
(c)
State the emf produced when the cyclist stops.
(1)
(d)
The dynamo supplies a lamp rated 6.0 V, 2.4 W. Calculate the current in the lamp.
(2)
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Part (a)
the magnet rotates relative to the coil
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magnetic flux linked with the coil changes (field lines are cut)
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a changing flux induces an emf in the coil
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Part (b)
first valid way, e.g. turning the wheel faster, or a stronger magnet
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second different valid way, e.g. more turns on the coil, or a soft iron core
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Part (c)
zero, because the magnet is not moving so the flux does not change
✓ 1
Part (d)
I = P/V = 2.4 ÷ 6.0
✓ 1
0.40 A
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Common error: An emf is induced only while the flux is changing; a stationary magnet gives no emf.
412IGPC3-4-07r16-06Electromagnetic induction, motors and transformersPaper EHard1 mark
The coil of a simple a.c. generator rotates in a uniform magnetic field. The rotation speed is doubled. What happens to the peak e.m.f. and to the period of the output?
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Notes
Faster rotation means the coil cuts the field lines faster, so the peak e.m.f. doubles. Each turn takes half as long, so the period halves. Unchanged e.m.f. ignores the rate of cutting. Period doubling goes with slower rotation, not faster. Halving the e.m.f. has the speed effect reversed.
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Answer A
Answer: A
Common error: A higher rotation speed raises both the size and the frequency of the output, so the period gets shorter.
413IGPC3-4-07r16-08Charge, current, potential difference, resistance and powerPaper EHard1 mark
A fish-tank heater is rated at 120 W when connected to a 24 V supply. It is instead connected to a 12 V supply. Assuming its resistance stays constant, what is its power?
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Notes
Resistance R = V² / P = 24² / 120 = 4.8 Ω. At 12 V, P = V² / R = 144 / 4.8 = 30 W. 2.5 W is the current in amperes (12 / 4.8) given as the answer without using P = IV. 60 W assumes power is proportional to voltage. 120 W assumes the power does not change.
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Answer B
Answer: B
Exam tip: Power is proportional to V² when the resistance is fixed, so halving the p.d. quarters the power.
414IGPC3-4-07r16-12Charge, current, potential difference, resistance and powerPaper EHard10 marks
A heater for a fish hatchery is made from resistance wire. Wire X has length 2.0 m, diameter d and resistance 8.0 Ω. Wire Y is made of the same material and has length 3.0 m and diameter 2d.
Circuit for part (d): wires X and Y in parallel. Diagram not to scale
(a)
State how the resistance of a wire depends on its length and on its cross-sectional area.
(2)
(b)
Calculate the resistance of wire Y.
(3)
(c)
Wire Y alone is connected across a 12 V supply. Calculate the power dissipated in it. Use 3.0 Ω for the resistance.
(2)
(d)
Wire X and wire Y are now connected in parallel across the same 12 V supply. Calculate the total current from the supply. Use 3.0 Ω for the resistance of wire Y.
(3)
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Part (a)
B1 resistance increases as length increases (resistance is proportional to length)
✓ 1
B1 resistance decreases as cross-sectional area increases (inversely proportional to area)
✓ 1
Part (b)
M1 length is 1.5 times greater, so resistance is multiplied by 1.5
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M1 area is proportional to diameter squared, so area is 4 times greater and resistance is divided by 4
✓ 1
A1 8.0 × 1.5 / 4 = 3.0 Ω
✓ 1
Part (c)
M1 P = V² / R, or I = V / R = 4.0 A and then P = IV
✓ 1
A1 48 W (ECF from (b))
✓ 1
Part (d)
M1 current in X = 12 / 8.0 = 1.5 A
✓ 1
M1 current in Y = 12 / 3.0 = 4.0 A (ECF from (b))
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A1 total current = 5.5 A
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Common error: Doubling the diameter multiplies the area by four, not by two.
415IGPC3-4-08r21-03Series and parallel circuitsPaper EHard1 mark
Three identical lamps P, Q and R are connected to a battery of negligible internal resistance. P and Q are in parallel with each other, and this pair is in series with R, as in a train carriage lighting circuit. Lamp P fails and becomes an open circuit. What happens to the brightness of Q and of R?
Three identical lamps; P and Q are in parallel and this pair is in series with R.
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Notes
Let each lamp have resistance r. Before the failure the total resistance is r/2 + r = 1.5r, so R carries 2V/3r and Q carries V/3r. After the failure Q and R are in series with total resistance 2r, so each carries V/2r. The current in Q rises, so Q is brighter; the current in R falls, so R is dimmer. A is correct. B, C and D each ignore that the total resistance rises while the current in Q must now equal the current in R.
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Answer A
Answer: A
Common error: When a branch is removed, find the new total resistance and the new current before comparing lamps.
416IGPC3-4-08r21-08Magnets and magnetic fieldsPaper EHard1 mark
A straight wire carrying a current in a uniform magnetic field experiences a force of size F directed due north. The current is then reversed, the direction of the magnetic field is also reversed, and the current is doubled in size. The strength of the field is unchanged. What is the new force?
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Notes
Reversing the current reverses the force direction, and reversing the field reverses it again, so the direction is unchanged (north). The force is proportional to the current, so doubling the current doubles the force to 2F; C is correct. A and B have the wrong direction. D ignores the doubled current.
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Answer C
Answer: C
Common error: Reversing both the current and the field leaves the force direction unchanged.
417IGPC3-4-08r21-16Charge, current, potential difference, resistance and powerPaper EHard7 marks
A heater wire in an astronomical dome is made of a uniform metal wire of length 2.0 m and cross-sectional area 0.50 mm2. Its resistance is 3.0 Ω. A second wire made from the same metal has length 5.0 m and cross-sectional area 0.25 mm2.
Circuit diagram
(a)
State how the resistance of a uniform wire depends on its length and on its cross-sectional area.
(2)
(b)
Calculate the resistance of the second wire.
(2)
(c)
The two wires are connected in series to a 12 V supply. Calculate the current in the circuit.
(1)
(d)
Calculate the energy transferred in the second wire in 60 s. Use the unrounded current from (c).
(2)
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Part (a)
resistance is directly proportional to length
✓ 1
and inversely proportional to cross-sectional area
✓ 1
Part (b)
resistance increases by a factor 2.5 for the length and a factor 2 for the area
✓ 1
3.0 × 2.5 × 2 = 15 Ω
✓ 1
Part (c)
total resistance = 3.0 + 15 = 18 Ω, so I = 12 ÷ 18 = 0.67 A (ECF from (b))
✓ 1
Part (d)
P = I2R = (12 ÷ 18)2 × 15 = 6.7 W (ECF)
✓ 1
E = P t = 400 J
✓ 1
Exam tip: Compare the new wire with the old one by multiplying by the factor for length and the factor for area.
418IGPC3-4-07r23-07Charge, current, potential difference, resistance and powerPaper EHard1 mark
A thin wire in a robot's sensor has resistance 6.0 Ω. A second wire is made of the same material at the same temperature. It is twice as long and has twice the diameter of the first wire. What is the resistance of the second wire?
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Notes
Resistance is proportional to length ÷ cross-sectional area. Doubling the diameter makes the area four times larger, and doubling the length doubles the resistance, so R = 6.0 × 2 ÷ 4 = 3.0 Ω. The value 1.5 Ω ignores the change in length. The value 6.0 Ω assumes the area only doubles. The value 12 Ω ignores the change in area.
✓ 1
Answer D
Answer: D
Common error: Area depends on the square of the diameter, so doubling the diameter multiplies the area by four, not two.
419IGPC3-4-07r23-14Series and parallel circuitsPaper EHard6 marks
Two heating elements of resistance 12 Ω and 24 Ω are connected in parallel to a 24 V supply with negligible internal resistance in a fish-hatchery tank.
Circuit diagram
(a)
Calculate the current in each element.
(2)
(b)
Calculate the total current from the supply.
(1)
(c)
Calculate the combined resistance of the two elements.
(2)
(d)
Explain why the combined resistance is smaller than the resistance of either element.
(1)
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Part (a)
I = V / R
✓ 1
2.0 A in the 12 Ω element and 1.0 A in the 24 Ω element
✓ 1
Part (b)
2.0 + 1.0 = 3.0 A (ECF from (a))
✓ 1
Part (c)
R = V / I, or 1/R = 1/12 + 1/24
✓ 1
= 24 / 3.0 = 8.0 Ω (ECF from (b))
✓ 1
Part (d)
There are more paths for the current, so for the same pd the total current is larger
✓ 1
Common error: For parallel resistors the combined resistance is always less than the smallest one; if your answer is bigger, check your working.
420IGPC3-4-07r23-15Electromagnetic induction, motors and transformersPaper EHard8 marks
A small generator on a robot wheel has a coil that rotates in the magnetic field between two permanent magnets and produces an alternating emf.
(a)
Explain why an emf is induced in the coil as it rotates.
(2)
(b)
State two changes that would increase the emf produced.
(2)
(c)
The coil rotates 600 times each minute. Calculate the frequency of the alternating emf.
(2)
(d)
The speed of rotation is doubled. State the new frequency and what happens to the peak emf.
(2)
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Part (a)
The wires of the coil cut the magnetic field lines, or the magnetic field through the coil changes
✓ 1
This induces an emf across the coil
✓ 1
Part (b)
Any two from: stronger magnets, more turns on the coil, faster rotation, larger coil area
✓ 2
Part (c)
600 / 60
✓ 1
= 10 Hz
✓ 1
Part (d)
Frequency becomes 20 Hz
✓ 1
The peak emf doubles
✓ 1
Exam tip: One rotation of the coil gives one complete cycle of the alternating emf, so the frequency equals the number of rotations per second.
421IGPC3-4-06r24-12Electromagnetic induction, motors and transformersPaper EHard8 marks
A hospital uses an ideal step-down transformer to change a 230 V alternating supply to 24 V for an examination lamp. The primary coil has 1380 turns. The lamp has a power of 60 W.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Calculate the current in the secondary coil.
(2)
(c)
Calculate the current in the primary coil. Give your answer to 3 significant figures.
(2)
(d)
Explain why the transformer does not work with a direct current supply.
(2)
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Notes
Part (a)
Ns = Np × Vs ÷ Vp = 1380 × 24 ÷ 230
✓ 1
144 turns
✓ 1
Part (b)
I = P ÷ V = 60 ÷ 24
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2.5 A
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Part (c)
power in primary = power in secondary, so I = 60 ÷ 230
✓ 1
0.261 A
✓ 1
Part (d)
a changing current in the primary coil produces a changing magnetic field in the iron core
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a direct current gives a constant field, so no e.m.f. is induced in the secondary coil
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Exam tip: For an ideal transformer, equate the power in the two coils, IpVp = IsVs.
422IGPC3-4-06r24-16Electromagnetic induction, motors and transformersPaper EHard8 marks
A hospital's back-up generator is a simple a.c. generator in which a coil rotates in a magnetic field. The output has a frequency of 50 Hz.
(a)
Explain why an e.m.f. is induced as the coil rotates.
(2)
(b)
Calculate the time for one complete cycle of the output.
(2)
(c)
The generator speed is doubled. State what happens to the peak voltage and to the frequency.
(2)
(d)
Describe how the current is passed from the rotating coil to the external circuit.
(2)
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Notes
Part (a)
the wires of the coil cut the magnetic field lines (or the magnetic field through the coil changes)
✓ 1
this induces an e.m.f. in the coil
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Part (b)
T = 1 ÷ f = 1 ÷ 50
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0.020 s
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Part (c)
peak voltage doubles
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frequency doubles
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Part (d)
slip rings are fixed to the coil and rotate with it
✓ 1
carbon brushes press on the rings and keep a continuous connection to the circuit
✓ 1
Exam tip: The period is the reciprocal of the frequency, so a higher frequency means a shorter period.
423IGPC3-4-07r24-08Series and parallel circuitsPaper EHard1 mark
In a hospital thermometer circuit, a thermistor is connected in series with a 150 Ω fixed resistor across an 8.0 V supply of negligible internal resistance. A voltmeter is connected across the fixed resistor. At 20 °C the resistance of the thermistor is 450 Ω. At 60 °C its resistance has fallen to 50 Ω. What is the voltmeter reading at 60 °C?
Circuit diagram
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Notes
At 60 °C the total resistance is 150 + 50 = 200 Ω, so the current is 8.0/200 = 0.040 A and the p.d. across the fixed resistor is 0.040 × 150 = 6.0 V. 2.0 V is the reading at 20 °C (and also the p.d. across the thermistor at 60 °C), so it ignores the change or reads the wrong component. 4.0 V assumes the supply p.d. is shared equally. 8.0 V assumes the thermistor now has no resistance, so the whole supply p.d. appears across the fixed resistor.
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Answer C
Answer: C
Common error: In a series potential divider the larger share of the p.d. goes across the larger resistance, so when the thermistor resistance falls the fixed resistor's share rises.
424IGPC3-4-07r24-15Electromagnetic induction, motors and transformersPaper EHard9 marks
A hospital's standby generator has a coil that rotates between the poles of permanent magnets. The coil makes 1500 revolutions per minute, and one complete cycle of the output is produced for each revolution.
(a)
Calculate the frequency of the alternating output.
(2)
(b)
Explain why an emf is induced in the coil as it rotates.
(2)
(c)
The coil now rotates twice as fast. State two changes to the output.
(2)
(d)
A student pushes the north pole of a bar magnet towards a coil connected to a sensitive meter. The student has to apply a force to keep the magnet moving. Explain why.
(3)
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Notes
Part (a)
M1 1500/60 = 25 revolutions per second
✓ 1
A1 25 Hz
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Part (b)
The magnetic flux linking the coil changes as it turns
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a changing flux linkage induces an emf in the coil
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Part (c)
The peak (maximum) emf doubles
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The frequency doubles
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Part (d)
A current is induced in the coil
✓ 1
this current creates a magnetic field that opposes the motion of the magnet (so the coil end acts as a north pole facing the magnet)
✓ 1
so work is done against this force, and it is transferred to electrical energy in the coil
✓ 1
Exam tip: In an 'explain' question about induction, link each step: changing flux, induced emf, induced current, opposing force.
425IGPC3-4-07r26-07Electromagnetic induction, motors and transformersPaper EHard1 mark
A student pushes a magnet into a coil of wire at constant speed and records the induced e.m.f. The same arrangement is then changed in the ways listed below. Which change produces the largest induced e.m.f.?
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Notes
The induced e.m.f. is proportional to the rate of change of flux linkage, so it is proportional to the speed × the field strength × the number of turns. Relative to the original, the values are 2 × 1.5 = 3.0, 0.5 × 4 = 2.0, 3 × 0.5 = 1.5 and 2 × 0.5 = 1.0 respectively, so the largest is the first. The other options each include a change that cuts the e.m.f. and less than offsets the gain from the other change.
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Answer A
Answer: A
Exam tip: When several factors change at once, multiply the scaling factors of every quantity the e.m.f. is proportional to.
426IGPC3-4-07r26-15Electromagnetic induction, motors and transformersPaper EHard8 marks
A wind turbine drives a simple generator in which a coil rotates between the poles of a magnet. The average electrical power output of the generator is 2.4 kW over a period of 5.0 hours. Electricity is valued at £0.28 per kWh.
(a)
Explain why an e.m.f. is induced in the coil as it rotates. [2]
(2)
(b)
State two ways of increasing the maximum e.m.f. produced by the generator. [2]
(2)
(c)
Calculate the energy delivered in 5.0 hours. Give your answer in kWh. [2]
(2)
(d)
Calculate the value of this energy. [2]
(2)
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Notes
Part (a)
The wires of the coil cut the magnetic field lines (the magnetic flux linking the coil changes)
✓ 1
A changing flux linkage induces an e.m.f. across the coil
✓ 1
Part (b)
Any valid first way, for example rotating the coil faster
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Any different valid second way, for example using a stronger magnet or more turns on the coil
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Part (c)
M1 E = P × t = 2.4 × 5.0
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A1 12 kWh
✓ 1
Part (d)
M1 12 × 0.28
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A1 £3.36 (ECF from (c))
✓ 1
Exam tip: Keep power in kW and time in hours to get energy directly in kWh.
427IGPC3-4-09r30-04Electromagnetic induction, motors and transformersPaper EHard1 mark
An ideal transformer lowers 230 V mains to 12 V for fashion-shop display lights. The primary coil has 1840 turns. How many turns does the secondary coil have?
Diagram not to scale
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Notes
For an ideal transformer Vs/Vp = Ns/Np, so Ns = 1840 × 12/230 = 96. 35 300 uses the ratio upside down (step-up). 1 622 subtracts the voltage difference (218) from the turns. 8.4 divides turns by the voltage difference.
✓ 1
Answer C
Answer: C
Common error: A step-down transformer always has fewer turns on the secondary, so a secondary larger than the primary must be wrong.
428IGPC3-4-09r30-13Electromagnetic induction, motors and transformersPaper EHard9 marks
A fashion shop uses an ideal transformer to run low-voltage display lights. The primary coil is connected to a 230 V a.c. supply and has 1150 turns. The secondary coil supplies 12 V to lights with a total power of 150 W.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Calculate the current in the secondary coil.
(2)
(c)
Explain why the transformer does not work if the primary coil is connected to a d.c. supply.
(3)
(d)
The transformer is 100% efficient. Calculate the current in the primary coil. Give your answer to 3 significant figures.
(2)
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Notes
Part (a)
M1 Ns = Np × Vs/Vp = 1150 × 12/230
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A1 60 turns
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Part (b)
M1 I = P/V = 150/12
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A1 12.5 A
✓ 1
Part (c)
a d.c. supply gives a constant current in the primary so the magnetic field in the core is constant
✓ 1
the magnetic field through the secondary coil does not change
✓ 1
no e.m.f. is induced in the secondary coil
✓ 1
Part (d)
M1 power in = power out so I = 150/230
✓ 1
A1 0.652 A
✓ 1
Exam tip: For an ideal transformer, power in equals power out, so VpIp = VsIs.
429IGPC3-4-09r30-16Series and parallel circuitsPaper EHard7 marks
A fish-farm heater circuit has a 16 V battery connected to a 5.0 Ω resistor in series with two resistors, of 12 Ω and 4.0 Ω, that are connected in parallel with each other.
Circuit diagram
(a)
Calculate the combined resistance of the 12 Ω and 4.0 Ω resistors.
(2)
(b)
Calculate the total resistance of the circuit and the current from the battery.
(2)
(c)
Calculate the potential difference across the parallel pair.
(1)
(d)
Calculate the current in the 12 Ω resistor.
(2)
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Part (a)
M1 1/R = 1/12 + 1/4.0
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A1 3.0 Ω
✓ 1
Part (b)
total resistance 5.0 + 3.0 = 8.0 Ω
✓ 1
I = 16/8.0 = 2.0 A (ECF from (a))
✓ 1
Part (c)
V = IR = 2.0 × 3.0 = 6.0 V (ECF from (b))
✓ 1
Part (d)
M1 I = V/R = 6.0/12
✓ 1
A1 0.50 A (ECF from (c))
✓ 1
Exam tip: Find the current from the battery first, then the potential difference across the parallel pair, then split into branches.
430IGPC3-4-07r31-11Electromagnetic induction, motors and transformersPaper EHard9 marks
A train takes power from an overhead line at 25 000 V. A transformer in the train gives an output of 750 V at a current of 400 A. The primary coil has 3000 turns. The transformer is 80% efficient.
Diagram not to scale
(a)
Calculate the number of turns on the secondary coil.
(2)
(b)
Calculate the output power of the transformer.
(1)
(c)
Calculate the input power to the transformer.
(2)
(d)
Calculate the current in the primary coil.
(2)
(e)
Explain why the transformer does not work with a steady direct current.
(2)
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Notes
Part (a)
M1 Np/Ns = Vp/Vs, so Ns = 3000 × 750 / 25 000
✓ 1
A1 90 turns
✓ 1
Part (b)
A1 750 × 400 = 300 000 W (300 kW)
✓ 1
Part (c)
M1 input power = output power / efficiency = 300 000 / 0.80
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A1 375 000 W (ECF from (b))
✓ 1
Part (d)
M1 I = P/V = 375 000 / 25 000
✓ 1
A1 15 A (ECF from (c))
✓ 1
Part (e)
B1 a changing current is needed to produce a changing magnetic field in the core
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B1 only a changing magnetic field linked with the secondary coil induces an e.m.f. in it
✓ 1
Common error: With an efficiency below 100%, the input power is larger than the output power, so divide by the efficiency.
431IGPC3-4-09r31-07Series and parallel circuitsPaper EHard1 mark
A lamp is marked 12 V, 36 W. It is connected in series with a resistor to a 24 V supply so that the lamp works at its rated power. What is the resistance of the resistor?
Circuit diagram
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Notes
The lamp current is 36 ÷ 12 = 3.0 A. The resistor must take the remaining 24 − 12 = 12 V, so R = 12 ÷ 3.0 = 4.0 Ω. 8.0 Ω is the total resistance 24 ÷ 3.0. 0.33 Ω comes from 12 ÷ 36. 12 Ω comes from using the 12 V as if it were the resistance.
✓ 1
Answer C
Answer: C
Exam tip: In a series circuit, find the current first and then use the p.d. left over for the extra component.
432IGPC3-4-09r31-15Charge, current, potential difference, resistance and powerPaper EHard8 marks
In a laboratory incubator used to store medicines, a heating wire has a resistance of 8.0 Ω. A second wire is made of the same material and is twice as long with half the cross-sectional area.
Circuit diagram
(a)
Determine the resistance of the second wire.
(2)
(b)
The two wires are connected in series to a 20 V supply. Calculate the current.
(2)
(c)
Calculate the power dissipated in the second wire.
(2)
(d)
Calculate the power dissipated in the first wire.
(2)
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Notes
Part (a)
M1 resistance doubles for double length and doubles again for half the area, so factor of 4
✓ 1
A1 4 × 8.0 = 32 Ω
✓ 1
Part (b)
M1 I = V ÷ R = 20 ÷ (8.0 + 32)
✓ 1
A1 0.50 A
✓ 1
Part (c)
M1 P = I2R = 0.502 × 32
✓ 1
A1 8.0 W
✓ 1
Part (d)
M1 P = I2R = 0.502 × 8.0
✓ 1
A1 2.0 W
✓ 1
Exam tip: For wires of the same material, resistance is proportional to length and inversely proportional to cross-sectional area.
433IGPC3-4-09r31-16Series and parallel circuitsPaper EHard8 marks
The circuit of a medicine-storage unit has a 12 V supply of negligible internal resistance. A 4.0 Ω resistor is in series with a parallel combination of a 6.0 Ω resistor and a 12 Ω resistor.
Circuit diagram
(a)
Calculate the combined resistance of the 6.0 Ω and 12 Ω resistors.
(2)
(b)
Calculate the current from the supply.
(2)
(c)
Calculate the potential difference across the parallel combination.
(2)
(d)
Calculate the current in the 12 Ω resistor.
(2)
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Notes
Part (a)
M1 1/R = 1/6.0 + 1/12
✓ 1
A1 4.0 Ω
✓ 1
Part (b)
M1 total resistance = 4.0 + 4.0 = 8.0 Ω
✓ 1
A1 I = 12 ÷ 8.0 = 1.5 A
✓ 1
Part (c)
M1 p.d. across 4.0 Ω resistor = 1.5 × 4.0 = 6.0 V
✓ 1
A1 12 − 6.0 = 6.0 V
✓ 1
Part (d)
M1 I = V ÷ R = 6.0 ÷ 12
✓ 1
A1 0.50 A (ECF from (c))
✓ 1
Exam tip: The p.d. across each branch of a parallel pair is the same, so find that p.d. before finding branch currents.
434IGPC3-4-10r33-05Electromagnetic induction, motors and transformersPaper EHard1 mark
In an electric guitar pickup, a vibrating steel string above a coil that has a magnet inside it induces a changing emf in the coil. Which change would increase the maximum induced emf?
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Notes
A larger amplitude makes the magnetic flux through the coil change faster, which increases the induced emf. Fewer turns reduces the emf, a nylon string does not disturb the magnetic field, and a greater distance makes the change in flux smaller.
✓ 1
Answer C
Answer: C
Exam tip: Induced emf increases with a faster change of flux, a stronger field, or more turns on the coil.
435IGPC3-4-10r33-06Electromagnetic induction, motors and transformersPaper EHard1 mark
A simple d.c. motor drives a pump on a farm. Both the direction of the current in the coil and the polarity of the permanent magnets are reversed at the same time. What happens to the direction of rotation of the coil?
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Notes
The force on the coil depends on both the current direction and the field direction. Reversing one reverses the force, but reversing both gives the original force direction, so the motor turns the same way. The forces on the two sides do not cancel, and the commutator keeps the turning continuous rather than alternating.
✓ 1
Answer A
Answer: A
Common error: Reversing both the current and the field leaves the force unchanged.
436IGPC3-4-10r33-15Charge, current, potential difference, resistance and powerPaper EHard8 marks
A wind turbine on a farm generates 40 kW at 400 V. The power is sent along a cable of total resistance 0.50 Ω to buildings on the farm. Assume ideal transformers.
(a)
Calculate the current in the cable when the 40 kW is transmitted at 400 V.
(2)
(b)
Calculate the power lost as thermal energy in the cable in (a).
(2)
(c)
A step-up transformer raises the voltage to 4000 V before the cable. Calculate the new current in the cable and the new power lost in the cable.
(3)
(d)
Explain why transmitting at higher voltage wastes less energy.
(1)
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Notes
Part (a)
M1 use of I = P / V
✓ 1
A1 I = 100 A
✓ 1
Part (b)
M1 use of P = I2R
✓ 1
A1 P = 5000 W (5.0 kW); ECF from (a)
✓ 1
Part (c)
M1 use of I = P / V
✓ 1
A1 I = 10 A
✓ 1
A1 power lost = 50 W
✓ 1
Part (d)
The current is smaller, and the power lost depends on the square of the current, so less thermal energy is produced in the cable
✓ 1
Exam tip: For transmission losses, use P = I²R with the current in the cable, not the voltage of the supply.
437IGPC3-4-10r33-16Electromagnetic induction, motors and transformersPaper EHard6 marks
An aircraft has a small generator that uses a coil rotating in a magnetic field to produce an alternating emf.
(a)
State two ways to increase the emf produced by the generator.
(2)
(b)
Explain why the emf produced changes direction every half turn.
(2)
(c)
The speed of rotation of the coil is doubled. State two changes to the emf output.
(2)
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Notes
Part (a)
Any one: rotate the coil faster / use a stronger magnet / use more turns on the coil / use a coil with a larger area
✓ 1
Any second different way from the list above
✓ 1
Part (b)
The sides of the coil cut the magnetic field lines in the opposite direction after half a turn
✓ 1
So the direction of the induced current (and emf) reverses
✓ 1
Part (c)
The peak emf doubles
✓ 1
The frequency doubles (the period halves)
✓ 1
Common error: Doubling the rotation speed changes both the size and the frequency of the alternating emf, not just the size.
438IGPC3-4-08r35-08Electromagnetic induction, motors and transformersPaper EHard1 mark
A telecoms mast is supplied with the same power through a long cable. The supply voltage is reduced by a factor of 10 while the power supplied stays the same. By what factor does the power wasted as heat in the cable change?
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Notes
For constant power, the current is I = P / V, so a voltage 10 times smaller means a current 10 times larger. Power wasted in the cable is I2R, so it increases by 102 = 100 times. A factor of 10 forgets that the current is squared. A decrease is the opposite of what happens: it is the effect of raising the voltage, which is why high-voltage transmission is used.
✓ 1
Answer C
Answer: C
Common error: Power loss depends on the square of the current, so a tenfold change in current changes the loss a hundredfold.
439IGPC3-4-08r35-16Charge, current, potential difference, resistance and powerPaper EHard9 marks
A remote mountain radio mast needs 6.0 kW and is supplied through a cable of total resistance 4.0 Ω. The supply voltage is 2400 V.
(a)
Calculate the current in the cable.
(2)
(b)
Calculate the power wasted as heat in the cable.
(2)
(c)
The power loss lasts for 30 days (720 hours). Electricity costs €0.30 per kWh. Calculate the cost of the energy wasted in the cable.
(3)
(d)
Explain why transmitting at a higher voltage reduces the wasted power, and state what is needed at the mast.
(2)
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Notes
Part (a)
I = P / V
✓ 1
I = 6000 / 2400 = 2.5 A
✓ 1
Part (b)
P = I2R
✓ 1
2.52 × 4.0 = 25 W (ECF from (a))
✓ 1
Part (c)
energy wasted = 0.025 kW × 720 h
✓ 1
= 18 kWh
✓ 1
cost = 18 × 0.30 = €5.40 (ECF from (b))
✓ 1
Part (d)
At higher voltage the same power needs a smaller current, and the power wasted is I2R, so much less
✓ 1
A step-down transformer is needed at the mast
✓ 1
Common error: Power loss in a cable is I²R using the current in the cable, not the supply voltage squared divided by the cable resistance.
440IGPC3-4-08r37-06Charge, current, potential difference, resistance and powerPaper EHard1 mark
A ship in port receives 600 kW of power through a cable of total resistance 4.0 Ω from the shore, supplied at 20 kV. The supply voltage is raised to 60 kV while the same power is delivered to the ship. How does the power wasted as heat in the cable change?
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Notes
The current is P/V, so it falls from 30 A to 10 A. The power wasted is I2R, so it falls from 3600 W to 400 W, one ninth of the original. A drop to one third forgets that the current is squared in the power loss. Staying the same wrongly assumes that the delivered power controls the heating. An increase to nine times has the factor inverted.
✓ 1
Answer C
Answer: C
Exam tip: Power loss in a cable depends on the square of the current, so cutting the current to one third cuts the loss to one ninth.
441IGPC3-4-08r37-16Charge, current, potential difference, resistance and powerPaper EHard8 marks
A sensor at the top of a ship's mast is connected by a long copper cable. A 5.0 m length of the cable has a resistance of 0.80 Ω.
(a)
Calculate the resistance of a 40 m length of the same cable.
(2)
(b)
A second cable has the same length and is made of the same material but has twice the cross-sectional area. Calculate its resistance for a 40 m length and explain your answer.
(2)
(c)
The 40 m cable (6.4 Ω) is in series with a sensor of resistance 17.6 Ω. The supply p.d. is 24 V. Calculate the current in the circuit.
(2)
(d)
Calculate the power wasted as heat in the cable.
(2)
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Notes
Part (a)
M1 resistance is proportional to length, 40 / 5.0 = 8 times larger
✓ 1
A1 6.4 Ω
✓ 1
Part (b)
A1 3.2 Ω
✓ 1
A1 resistance is inversely proportional to cross-sectional area, so doubling the area halves the resistance
✓ 1
Part (c)
M1 total resistance 6.4 + 17.6 = 24.0 Ω
✓ 1
A1 I = 24 / 24.0 = 1.0 A
✓ 1
Part (d)
M1 P = I2R = 1.02 × 6.4, or V = IR = 6.4 V then P = IV (ECF from (c))
✓ 1
A1 6.4 W
✓ 1
Exam tip: Cable resistance rises with length and falls as the cross-sectional area increases, so thick short cables waste the least energy.
442IGPC3-4-07r39-07Series and parallel circuitsPaper EHard1 mark
A 6.0 V battery of negligible internal resistance is connected to an 8.0 Ω resistor in series with a parallel pair made of a 12 Ω resistor and a 6.0 Ω resistor. What is the potential difference across the 12 Ω resistor?
Circuit diagram
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Notes
The parallel pair has resistance 1/(1/12 + 1/6) = 4.0 Ω, so the total is 12 Ω and the current is 6.0/12 = 0.50 A. The p.d. across the pair, and so across the 12 Ω resistor, is 0.50 × 4.0 = 2.0 V. 4.0 V is the p.d. across the 8.0 Ω resistor. 3.0 V comes from sharing 6.0 V equally between the two sections. 6.0 V ignores the series resistor.
✓ 1
Answer C
Answer: C
Common error: The p.d. across a parallel group is the current in the whole group multiplied by the combined resistance of the group, not by one branch resistance.
443IGPC3-4-07r39-15Series and parallel circuitsPaper EHard7 marks
A coffee roaster's temperature monitor uses a 9.0 V supply connected in series with a 3.0 kΩ fixed resistor and an NTC thermistor. At 180 °C the thermistor has a resistance of 1.5 kΩ. A high-resistance voltmeter is connected across the thermistor.
Circuit diagram
(a)
Calculate the total resistance of the circuit at 180 °C, in kΩ.
(1)
(b)
Calculate the current in the circuit at 180 °C. Give your answer in amperes.
(2)
(c)
Calculate the voltmeter reading at 180 °C. Use the unrounded current from (b).
(2)
(d)
The temperature rises further. Explain what happens to the voltmeter reading.
(2)
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Part (a)
1.5 + 3.0 = 4.5 kΩ
✓ 1
Part (b)
I = V / R with R = 4500 Ω
✓ 1
= 2.0×10−3 A
✓ 1
Part (c)
V = I R = 2.0×10−3 × 1500
✓ 1
= 3.0 V
✓ 1
Part (d)
The resistance of the NTC thermistor decreases as temperature rises
✓ 1
The thermistor takes a smaller share of the 9.0 V, because the p.d. divides in proportion to resistance, so the reading falls
✓ 1
Common error: In a series circuit the p.d. divides in proportion to the resistances; a fall in one resistance shrinks that component's share.
444IGPC3-4-07r39-16Charge, current, potential difference, resistance and powerPaper EHard8 marks
Power of 40 kW is sent along a cable of total resistance 2.0 Ω to a remote bridge sensor station. Assume the power entering the cable is 40 kW.
(a)
Calculate the current in the cable when the power is sent at 8000 V.
(2)
(b)
Calculate the power lost in the cable in (a).
(2)
(c)
The same power is instead sent at 400 V. Calculate the power lost in the cable.
(2)
(d)
Use your answers to explain why electrical power is transmitted at high voltage.
(2)
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Part (a)
I = P / V
✓ 1
= 40 000 / 8000 = 5.0 A
✓ 1
Part (b)
P = I²R
✓ 1
= 5.0² × 2.0 = 50 W
✓ 1
Part (c)
I = 40 000 / 400 = 100 A
✓ 1
P = 100² × 2.0 = 2.0×104 W (accept 20 000 W)
✓ 1
Part (d)
For the same power, a higher voltage means a smaller current
✓ 1
Power lost in the cable depends on the square of the current, so the loss is much smaller
✓ 1
Exam tip: Power loss in a cable is I²R, so halving the current cuts the loss to a quarter.
445IGPC3-4-08r42-07Series and parallel circuitsPaper EHard1 mark
In a test circuit for a guitar amplifier, a 12 V supply is connected to a 20 Ω resistor in series with a parallel pair made from a 30 Ω resistor and a 60 Ω resistor. What is the current in the 60 Ω resistor?
Circuit diagram
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Notes
The parallel pair has resistance 1 / (1/30 + 1/60) = 20 Ω, so the total resistance is 40 Ω and the supply current is 0.30 A. The potential difference across the parallel pair is 0.30 × 20 = 6.0 V, so the current in the 60 Ω resistor is 6.0 / 60 = 0.10 A. The value 0.30 A is the total current. The value 0.20 A is the current in the 30 Ω resistor. The value 0.15 A comes from assuming the total current splits equally.
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Answer D
Answer: D
Common error: Current divides in inverse proportion to resistance in parallel branches; it does not split equally unless the resistors are equal.
446IGPC3-4-08r42-13Charge, current, potential difference, resistance and powerPaper EHard7 marks
An electric fence on a farm uses a long uniform wire. A 150 m length of this wire has a resistance of 3.0 Ω.
(a)
State how the resistance of a wire depends on its length when its cross-sectional area and temperature are constant.
(1)
(b)
Calculate the resistance of 450 m of this wire.
(2)
(c)
A second wire is made of the same material and is also 450 m long, but its cross-sectional area is twice as large. Determine its resistance.
(2)
(d)
The second wire is connected across a 12 V supply. Calculate the power dissipated in the wire.
(2)
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Part (a)
Resistance is directly proportional to length
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Part (b)
M1 450 m is 3 times 150 m, so resistance is 3 times larger
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A1 9.0 Ω
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Part (c)
M1 resistance is inversely proportional to cross-sectional area, so it is halved
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A1 4.5 Ω (ECF from (b))
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Part (d)
M1 P = V2 / R = 144 / 4.5 (or I = V / R then P = IV)
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A1 32 W (ECF from (c))
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Common error: A wire with a larger cross-sectional area has a smaller resistance; do not multiply by the area ratio.
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