Original exam-style questions with full mark schemes. Use the filters to choose level and difficulty, and tick questions to build a printable test.
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Original questions in the style of each exam boardEvery answer re-solved independently before publishingCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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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.
✓ 1
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?
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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.
✓ 1
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?
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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.
✓ 1
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.
✓ 1
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.
✓ 1
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.
✓ 1
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.
✓ 1
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.
✓ 1
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.
✓ 1
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.
✓ 1
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
✓ 2
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.
✓ 1
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?
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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?
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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?
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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.
(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.
30IGPC-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.
31IGPC-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.
32IGPC-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.
33IGPC-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.
34IGPC-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.
(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.
35IGPC-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.
36IGPC-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.
37IGPC-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.
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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.
38IGPC-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.
39IGPC-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'.
40IGPC-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.
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Answer C
Answer: C
Exam tip: 'Repulsion is the only true test for a magnet' is a statement that earns marks.
41IGPC-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?
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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.
42IGPC-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?
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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.
43IGPC-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.
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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.
44IGPC-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?
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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.
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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.
45IGPC-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.
(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.
46IGPC-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.
(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.
47IGPC-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.
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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.
48IGPC-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?
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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.
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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.
49IGPC-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?
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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.
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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.
50IGPC-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.
51IGPC-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.
(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
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.
52IGPC-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?
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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.
53IGPC-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?
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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.
✓ 1
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.
54IGPC-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.
✓ 1
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.
55IGPC-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?
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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.
✓ 1
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.
56IGPC-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.
✓ 1
Answer C
Answer: C
Exam tip: Always convert time to seconds before using Q = It.
57IGPC-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.
✓ 1
Answer B
Answer: B
Common error: Area depends on the square of the diameter. Doubling the diameter multiplies the area by four, not two.
58IGPC-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.
✓ 1
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.
59IGPC-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.
✓ 1
Answer B
Answer: B
Common error: Widely spaced lines do not mean a 'bigger' field. They mean a weaker field.
60IGPC-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
Exam tip: Outside a solenoid, the field is like that of a bar magnet. Inside, it is strong and uniform. Make both points clearly.
61IGPC-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?
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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.
✓ 1
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.
62IGPC-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?
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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.
✓ 1
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.
63IGPC-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.
(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))
✓ 1
R = 12 Ω
✓ 1
Part (b)
The pd across the 20 Ω resistor is 12 V
✓ 1
I = 12/20 = 0.60 A
✓ 1
Part (c)
I = 12/12, or 0.60 A + 0.40 A
✓ 1
= 1.0 A
✓ 1
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.
64IGPC-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.
(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
✓ 1
Part (b)
Total resistance = 10 + 50 = 60 kΩ
✓ 1
I = 6.0/60 000 = 1.0 × 10−4 A, or ratio 50/60 used
✓ 1
V = 5.0 V
✓ 1
Part (c)
pd across the fixed resistor = 6.0 − 1.2 = 4.8 V
✓ 1
RLDR/10 kΩ = 1.2/4.8, or I = 4.8/10 000 = 4.8 × 10−4 A
✓ 1
RLDR = 2.5 kΩ
✓ 1
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
✓ 1
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
✓ 1
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.
65IGPC-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.
66IGPC-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)
✓ 1
Part (b)
3000/60 = 50 revolutions per second
✓ 1
f = 50 Hz
✓ 1
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
✓ 1
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.
67IGPC-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.
✓ 1
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.
68IGPC-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?
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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.
✓ 1
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.
69IGPC-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.
(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 .
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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.
70IGPC-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.
(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.
71IGPC-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?
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Notes
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.
✓ 1
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.
72IGPC-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?
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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.
73IGPC-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 .
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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.
74IGPC-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.
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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.
75IGPC-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.
(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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Notes
Part (a)
R = (6.0 × 12)/(6.0 + 12), or 1/R = 1/6.0 + 1/12
✓ 1
= 4.0 Ω
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Part (b)
Total resistance = 4.0 + 4.0 = 8.0 Ω
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I = 9.0/8.0 = 1.1 A (1.125 A)
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Part (c)
V = IR = 1.125 × 4.0, or the two equal 4.0 Ω resistances share the 9.0 V equally
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= 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.
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