127 free, high-quality past-paper-style questions on Current and circuits for IB Physics SL, each with a full mark scheme. Covers: Current & charge; Conductors, insulators & resistance; Ohmic & non-ohmic conductors; and more. For when you have finished the past papers, or want to practise one topic at a time.B.5 at Higher Level →Read the B.5 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.691 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1B-1A-20Current & chargePaper 1AEasy1 mark
In a time of 8.0 s, a charge of 24 C flows through a heating element and 288 J of electrical energy is transferred in the element.
Which row gives the current in the element and the potential difference across it?
CurrentPotential difference
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Step 1I = Δq/Δt = 24 / 8.0 = 3.0 A.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2V = W/q = 288 / 24 = 12 V.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe current has been calculated as Δt/Δq. Current is charge per unit time.
BTwo errors: Δt/Δq for the current, and energy per unit time (the power, 36 W) for the potential difference.
C36 is the energy transferred per second — the power in watts. Potential difference is energy per unit charge.
DCorrect: I = 24/8.0 = 3.0 A and V = 288/24 = 12 V.
2B-1A-41Conductors, insulators & resistancePaper 1AEasy1 mark
Why is copper a much better electrical conductor than glass?
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Step 1Conduction depends on the availability and mobility of charge carriers. Copper has about 10²⁹ free electrons per cubic metre that can drift through the ionic lattice; in glass essentially every electron is bound.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AAtomic size is not what determines conduction. What matters is whether charge carriers are free to move.
BCopper is in fact much denser than glass, and density is not what governs conductivity.
CCorrect: in a metal the outer electrons are delocalised and drift readily under a potential difference. In glass the electrons are all bound to particular atoms.
DA metal is a lattice of positive ions surrounded by free electrons. It is the mobility of those electrons, not the absence of ions, that makes it conduct.
3B-1A-59Ohmic & non-ohmic conductorsPaper 1AEasy1 mark
The potential difference V across a filament lamp is varied from a large negative value to a large positive value, and the current I in the lamp is measured.
Which graph shows the variation of I with V?
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Step 1As the current increases the filament gets hotter and its resistance V/I increases, so the gradient of the I–V graph decreases. The filament conducts in the same way for either direction of current, so the graph has the same shape for negative V.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AA straight line through the origin is the graph of an ohmic conductor at constant temperature. The filament heats up and its resistance rises.
BHere the gradient increases, so the resistance falls as the current rises. A filament's resistance increases as it gets hotter.
CA filament conducts equally well in both directions: reversing V reverses I with the same size, so the graph cannot be zero for negative V.
DCorrect: the gradient decreases as |V| increases (resistance rising with temperature), with the same shape for both polarities.
6B-1A-102Conductors, insulators & resistancePaper 1AEasy1 mark
The copper winding of an electric motor warms from 20 °C to 80 °C while the motor is running. The change in the dimensions of the wire is negligible.
Which row describes the number of free electrons per unit volume in the copper and the resistance of the winding?
Free electrons per unit volumeResistance of the winding
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Step 1In a metal every atom already provides free electrons, so heating does not change their number per unit volume significantly. The lattice ions vibrate with a larger amplitude, so the drifting electrons collide with them more often and the resistance increases.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThis describes a semiconductor such as an NTC thermistor, in which heating releases extra charge carriers. A metal already has a very large number of free electrons.
BCorrect: the number of carriers is essentially fixed; more vigorous lattice vibrations cause more frequent collisions, so the resistance rises.
CThe rise in resistance has been put down to the loss of free electrons. Heating does not bind the electrons; the resistance rises because they collide with the ions more often.
DThe effect of the larger lattice vibrations has been ignored. The resistance of a metal increases significantly as its temperature rises.
7B-1A-107Series & parallel circuitsPaper 1AEasy1 mark
A signal lamp on a model railway is rated 6.0 V, 0.30 A. It is to work at its rated values from a 9.0 V battery of negligible internal resistance, using a resistor connected in series with the lamp.
What is the resistance of the resistor?
Circuit diagram
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Step 1The pd across the resistor is 9.0 − 6.0 = 3.0 V, and the current in it is the lamp current, 0.30 A.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2R = 3.0/0.30 = 10 Ω.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the series resistor takes the remaining 3.0 V at the same current of 0.30 A.
BThis is the resistance of the lamp itself at its rated values, 6.0/0.30.
CThis is the total resistance of the circuit, 9.0/0.30. The lamp provides 20 Ω of it.
DThe two pds have been added instead of subtracted. In series the pds across the components add up to the battery pd.
8B-1A-110Ohmic & non-ohmic conductorsPaper 1AEasy1 mark
Which statement is Ohm's law?
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Step 1Ohm's law states that the current in a metallic conductor is proportional to the potential difference across it, provided that the temperature and other physical conditions stay constant.
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Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThis is the definition of resistance, R = V/I, which applies to every conductor, ohmic or not. Ohm's law is the statement that this ratio is constant.
BThe condition has been reversed. A metal obeys Ohm's law only while its temperature is constant; its resistance rises as it gets hotter.
CV = IR is proportional to R only when the current is fixed. Ohm's law is about the proportionality of I and V for one conductor.
DCorrect: proportionality of I and V together with the condition of constant temperature.
9IBP2S-016-03-01Current & chargePaper 1AEasy1 mark
A defibrillator passes an average current of 20 A through a patient's chest for 5.0 ms. What is the charge that flows through the chest?
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Charge is current multiplied by time: Δq = IΔt = 20 × 5.0×10−3 = 0.10 C. The answer of 100 C comes from leaving the time in milliseconds. The answer of 4.0 C comes from dividing current by time, and 2.5×10−4 C comes from dividing time by current.
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Answer A
Answer: A
Common error: Convert milliseconds to seconds before multiplying. A defibrillator pulse carries only a fraction of a coulomb.
A copper wire has a thick section joined to a thin section. The two sections are in series and carry a steady current. Which statement is correct?
Diagram not to scale
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Charge is conserved, so the same current flows through every part of a series path. From I = nAvq, with I, n and q fixed, v is inversely proportional to A. The electrons therefore drift faster in the thin section. The other options either treat current as being 'used up' or 'squeezed', or reverse the inverse relationship between drift speed and area.
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Answer B
Answer: B
Exam tip: In series, fix the current first. Then use I = nAvq to compare drift speeds.
11IBP2S-016-03-06Conductors, insulators & resistancePaper 1AEasy1 mark
Polyethylene is used to insulate high-voltage cables. Which statement best explains why polyethylene is an electrical insulator?
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Every material contains electrons. In an insulator, almost all of them are bound to atoms or molecules, so the number density of mobile charge carriers is tiny. A potential difference therefore produces a negligible current. Density is not the deciding factor; many metals are denser than polyethylene. The speed of the electrons is irrelevant.
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Answer B
Answer: B
Exam tip: Explain conducting and insulating behaviour in terms of the number density of free (mobile) charge carriers.
Rearranging R = ρL/A gives ρ = RA/L. The units are Ω × m2/m = Ω m. The unit Ω m−1 is resistance per unit length, which is a different quantity. The unit Ω m−2 comes from inverting the area term. The unit Ω is resistance itself.
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Answer A
Answer: A
Common error: Resistivity is a property of the material and is measured in Ω m. Resistance per unit length, in Ω m⁻¹, is a different quantity.
13IBP2S-016-01-01Conductors, insulators & resistancePaper 1AEasy1 mark
A negative temperature coefficient (NTC) thermistor is used as a coolant temperature sensor in a car engine. A constant potential difference is maintained across the thermistor. As the engine warms up, what happens to the current in the thermistor?
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In an NTC thermistor, a rise in temperature releases many more charge carriers. This outweighs the extra collisions, so the resistance falls. With the pd held constant, I = V/R, so the current increases. Option B applies only to metals. Option C assumes the resistance is fixed. Option D ignores the large increase in the number of charge carriers.
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Answer A
Answer: A
Common error: A metal's resistance rises with temperature, but an NTC thermistor's resistance falls. Check which component you are dealing with before applying the rule.
14IBP2S-016-01-02Cells & sources of emfPaper 1AEasy1 mark
The battery pack of a camera drone is made from four identical cells connected in series. Each cell has emf 3.7 V and internal resistance 0.050 Ω. What are the emf and the internal resistance of the pack?
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For cells in series, both the emfs and the internal resistances add. The emf is 4 × 3.7 = 14.8 V and the internal resistance is 4 × 0.050 = 0.20 Ω. Option B is the rule for cells in parallel. Options C and D mix the series rule with the parallel rule.
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Answer A
Answer: A
Exam tip: Internal resistances are simply resistors inside the cells. Combine them with the same series and parallel rules you use for external resistors.
15IBP2S-016-01-10Current & chargePaper 1AEasy1 mark
A heart pacemaker draws a mean current of 10 μA from its battery. How much charge flows from the battery in one day?
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Use Q = It with the time in seconds. Q = 10×10−6 × 24 × 3600 = 0.864 C ≈ 0.86 C. Option A uses the time in hours. Option B uses the time in minutes. Option D forgets the micro prefix.
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Answer C
Answer: C
Common error: Q = It gives coulombs only when I is in amperes and t is in seconds. Convert both before substituting.
16IBP2S-016-00-01emf & internal resistancePaper 1AEasy1 mark
A car has its headlights switched on while the engine is off. When the driver operates the starter motor, the headlights become noticeably dimmer. The headlights and the starter motor are connected in parallel across the battery. What is the best explanation for the dimming?
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Terminal pd V = ε − Ir. The starter motor draws a very large current I, so the 'lost volts' Ir across the battery's internal resistance become large and V falls. The headlights are in parallel across the terminals, so the pd across them falls and their power V²/R decreases. The emf is a property of the cell and does not change (A). The components are in parallel, not series (C). In a parallel branch, the current depends only on the pd across that branch, so the headlights dim only because that pd falls (D).
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Answer B
Answer: B
Common error: Emf is not 'used up' by a large current. What changes is the terminal pd, because of the Ir term across the internal resistance.
17IBP2S-016-00-06Current & chargePaper 1AEasy1 mark
The undersoil heating system of a football stadium operates at a power of 450 kW for 8.0 hours before a match. What is the energy transferred?
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E = Pt = 450 × 103 W × 8.0 × 3600 s = 1.3 × 1010 J. Option A is the energy in kWh (3600 kWh) wrongly labelled as joules. Option B omits the factor of 103 in kW. Option C converts the time to minutes instead of seconds.
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Answer D
Answer: D
Common error: To get joules, power must be in watts and time in seconds. 1 kWh = 3.6 × 10⁶ J.
18IBP2S-016-02-02Thermistors, LDRs & dividersPaper 1AEasy1 mark
An egg incubator uses a negative temperature coefficient (NTC) thermistor connected in series with a fixed resistor across a battery. The output pd is taken across the fixed resistor. The temperature inside the incubator increases. What happens to the resistance of the thermistor and to the output pd?
Circuit diagram
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For an NTC thermistor, resistance falls as temperature rises. The total resistance falls, so the current rises. The fixed resistor therefore takes a larger share of the supply pd, and the output across it increases. The other options either reverse the thermistor behaviour or assume the output is taken across the thermistor.
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Answer C
Answer: C
Exam tip: Always check which component the output is taken across. The pd across the thermistor and the pd across the fixed resistor always change in opposite directions, because they add up to the supply pd.
19IBP2S-016-02-05Cells & sources of emfPaper 1AEasy1 mark
An electric bicycle is powered by a rechargeable lithium-ion battery. Which statement about cells is correct?
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A secondary cell is recharged by forcing a current through it in the reverse direction. This reverses the chemical reactions and restores the stored chemical energy. Charge is not used up: the same charge carriers circulate, and what is used up is chemical energy. Primary cells also have internal resistance. Emf is energy transferred per unit charge, not a force.
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Answer A
Answer: A
Common error: Cells store chemical energy, not charge. Emf is measured in volts (J C⁻¹), so it cannot be a force.
20IBP2S-016-02-09Conductors, insulators & resistancePaper 1AEasy1 mark
A thermistor in a car engine coolant sensor has a resistance that decreases as its temperature increases. Which statement explains this behaviour?
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In a semiconductor thermistor, a rise in temperature releases many more charge carriers. This increase in number density outweighs the extra collisions with the more strongly vibrating lattice, so the resistance falls. Lattice vibrations increase, not decrease, with temperature. Thermal expansion is far too small to matter. A lower drift speed would increase the resistance, not decrease it.
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Answer D
Answer: D
Common error: Metals and NTC thermistors respond oppositely to heating. In both, the lattice vibrates more; in the thermistor, the large rise in carrier number density dominates.
21B-1A-22emf & internal resistancePaper 1AMedium1 mark
A cell is connected to a variable resistor. The graph shows the variation of the potential difference V across the terminals of the cell with the current I in the cell.
Which row gives the emf and the internal resistance of the cell?
Graph drawn to scale
emfInternal resistance
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Step 1V = ε − Ir, so the intercept on the V axis (where I = 0) is the emf: ε = 6.0 V.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The gradient is −r: r = 6.0 / 3.0 = 2.0 Ω.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
A3.0 is the intercept on the current axis — the current in amperes when the terminals are short-circuited — not the emf.
BThe reciprocal of the gradient (ΔI/ΔV) has been taken. The internal resistance is −ΔV/ΔI.
CCorrect: the intercept on the V axis gives ε = 6.0 V and the magnitude of the gradient gives r = 2.0 Ω.
DV/I at a single point (2.0 V / 2.0 A) gives the resistance of the external circuit at that current, not the internal resistance.
22B-1A-42Cells & sources of emfPaper 1AMedium1 mark
Chemical cells and solar cells are both used as sources of emf in circuits.
Which statement is correct?
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Step 1A solar cell transfers the energy of incident electromagnetic radiation directly to electrical energy, so its output depends on the intensity of the radiation. A chemical cell transfers chemical energy to electrical energy while it drives a current.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AEmf is the energy transferred per unit charge (J C⁻¹ = V), not per unit current.
BCorrect: a solar cell's source of energy is the incident radiation, so less intense light gives a smaller output.
CThis is the reverse transfer, which happens while a rechargeable cell is being charged. A cell driving a current transfers chemical energy to electrical energy.
DA solar cell stores no energy; with no light falling on it there is no output. Energy for the night needs a separate storage cell — one disadvantage of solar cells.
24B-1A-103emf & internal resistancePaper 1AMedium1 mark
A cell of emf ε and internal resistance r is connected to a variable resistor of resistance R. A voltmeter measures the potential difference V across the terminals of the cell.
Which graph shows how V varies with R as R is increased from zero? The dashed line in each graph shows V = ε.
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Step 1I = ε/(R + r), so V = IR = εR/(R + r).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2V = 0 when R = 0, V rises steeply at first and then levels off, approaching ε when R ≫ r.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: V = εR/(R + r) is zero for a short circuit and tends to ε for a very large external resistance.
BThis is the straight-line graph of V = ε − Ir against current, redrawn with R on the horizontal axis. V increases, not decreases, as R increases.
CThis treats the current as constant, so that V = IR is proportional to R. The current falls as R increases, and V can never exceed ε.
DThis is the potential difference across the internal resistance, Ir = εr/(R + r), which falls as R increases. The terminal pd is ε minus this.
25B-1A-104Electrical power & energyPaper 1AMedium1 mark
Two filament lamps X and Y of the same type are used in a 12 V garden-lighting kit. X is rated 12 V, 20 W and Y is rated 12 V, 5.0 W. The lamps are connected first in parallel and then in series to a 12 V supply of negligible internal resistance. Assume that the resistance of each lamp is constant and that the brighter lamp is the one that dissipates the greater power.
Which row identifies the brighter lamp in each arrangement?
Brighter lamp in parallelBrighter lamp in series
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Step 1R = V2/P: RX = 122/20 = 7.2 Ω and RY = 122/5.0 = 28.8 Ω.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Parallel: same pd, so P = V2/R is larger for X (20 W against 5.0 W). Series: same current, 0.33 A, so P = I2R is larger for Y (3.2 W against 0.80 W).
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe power rating has been taken to fix the brightness in any circuit. A rating applies only with 12 V across the lamp; in series the lamps carry the same current and the larger resistance dissipates more power.
BBoth comparisons have been reversed. In parallel each lamp has the full 12 V, so the lamp of smaller resistance, X, takes the larger current and power.
CP = I2R has been applied to the parallel arrangement as if the lamps carried the same current. In parallel they have the same pd, so P = V2/R favours the smaller resistance, X.
DCorrect: in parallel X dissipates 20 W and Y 5.0 W; in series the current is 12/36 = 0.33 A, so X dissipates 0.80 W and Y 3.2 W.
26B-1A-105emf & internal resistancePaper 1AMedium1 mark
A bicycle lamp is connected across a battery that has a significant internal resistance. A second, identical lamp is then switched on in parallel with the first. Assume that the resistance of each lamp is constant.
Which row gives the change in the brightness of the first lamp and the change in the current in the battery?
Lamp 2 is initially disconnected; switch shown open
Brightness of first lampCurrent in battery
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Step 1Adding a lamp in parallel reduces the external resistance, so I = ε/(R + r) increases.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The terminal pd ε − Ir falls, so the pd across the first lamp falls and it becomes dimmer.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe larger battery current has been assumed to pass through the first lamp. It is shared between the two lamps, and the pd across each is smaller.
BCorrect: the current in the battery rises, so more of the emf is used across the internal resistance and the terminal pd falls.
CThe internal resistance has been ignored. That answer would be correct only for a battery of negligible internal resistance.
DAdding a component in parallel has been taken to increase the total resistance. A parallel branch provides an extra path, so the total resistance falls.
27B-1A-106Series & parallel circuitsPaper 1AMedium1 mark
Three resistors are connected in a triangle. The resistor between corners X and Y has resistance 2R; each of the other two has resistance R.
What is the resistance between X and Y?
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Step 1Between X and Y there are two paths: the 2R resistor, and the two R resistors in series via Z (2R).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Two 2R paths in parallel: (1/2R + 1/2R)−1 = R.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AAll three resistors have been treated as connected in parallel between X and Y. The two R resistors are in series with each other on the path through Z.
BCorrect: the path through Z (R + R = 2R) is in parallel with the direct 2R resistor, giving R.
COnly the resistor joining X and Y directly has been counted. Charge can also flow from X to Y through Z.
DAll three resistors have been added as if in series. Corner Z is not on the direct path, and the direct resistor provides a separate parallel path.
28B-1A-109Thermistors, LDRs & dividersPaper 1AMedium1 mark
A lamp of resistance 24 Ω is controlled by a potentiometer whose track has a total resistance of 24 Ω. The supply has an emf of 12.0 V and negligible internal resistance. In arrangement P the track is used as a potential divider; in arrangement Q it is used as a variable resistor. Assume the lamp's resistance is constant.
The slider is moved over the whole track. Which row gives the smallest pd across the lamp in each arrangement?
Diagram NOT accurately drawn
Arrangement PArrangement Q
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Step 1P: with the slider at the lower end of the track, the lamp is connected between two points at the same potential, so its pd is 0 V.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Q: the most resistance that can be put in series is the whole 24 Ω track, so the lamp pd is at least 12.0 × 24/(24 + 24) = 6.0 V.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the potential divider can reduce the lamp pd to zero, but the variable resistor can only share the 12.0 V equally with the 24 Ω lamp.
BThe variable resistor has been assumed able to stop the current. Its largest resistance is 24 Ω, so the current cannot fall below 0.25 A.
CThe two arrangements have been interchanged: it is the potential divider that gives the full range from 0 V to 12.0 V.
DThe potential divider has been treated like the series arrangement. With the slider at the lower end, the lamp has no pd across it.
29IBP2S-016-03-03Conductors, insulators & resistancePaper 1AMedium1 mark
A metal wire carries a constant current I. Which change, with I kept constant, doubles the drift speed of the conduction electrons?
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From v = I/(nAq), at constant I the drift speed is inversely proportional to nA. Halving A therefore doubles v. Halving the diameter quarters the area, so v becomes four times larger. Length does not appear in v = I/(nAq) when I is fixed. Doubling n halves v.
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Answer A
Answer: A
Common error: Area depends on diameter squared. Halving the diameter changes the area by a factor of four, not two.
A metal wire in a strain gauge has resistance R. The wire is stretched uniformly to twice its original length. Its volume and resistivity do not change. What is its new resistance?
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The volume is constant, so doubling the length halves the cross-sectional area. Then R' = ρ(2L)/(A/2) = 4ρL/A = 4R. The answer 2R ignores the change in area. The answer R wrongly assumes that constant volume means constant resistance.
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Answer C
Answer: C
Exam tip: When a wire is stretched or drawn out, use constant volume (LA = constant) to find the new area.
31IBP2S-016-03-05Current & chargePaper 1AMedium1 mark
An ion thruster on a space probe emits a beam of doubly charged xenon ions, each with charge +2e. The beam current is 2.4 A. How many ions leave the thruster each second? (e = 1.6×10−19 C)
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The charge leaving per second is 2.4 C. Each ion carries 2 × 1.6×10−19 = 3.2×10−19 C. The number per second is 2.4/3.2×10−19 = 7.5×1018. The value 1.5×1019 ignores the double charge. The two very small values come from multiplying the current by the charge instead of dividing.
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Answer A
Answer: A
Common error: Check the charge on each carrier. Ions may carry more than one elementary charge.
32IBP2S-016-03-07Conductors, insulators & resistancePaper 1AMedium1 mark
A semiconductor thermistor is used to monitor the temperature of a lithium-ion battery pack. As the temperature of the thermistor rises, its resistance falls. What is the main reason for this?
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In a semiconductor, a rise in temperature releases many more charge carriers. The increase in n outweighs the increase in collisions with the lattice, so the resistance falls. Lattice vibrations actually increase with temperature. Thermal expansion is far too small to explain the large change in resistance.
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Answer A
Answer: A
Common error: Do not use the explanation for metals. In a metal the carrier density is roughly constant and resistance rises with temperature. In a semiconductor the carrier density rises sharply.
33IBP2S-016-03-09Series & parallel circuitsPaper 1AMedium1 mark
A smartphone battery is rated at 3000 mA h. What is the total charge it can deliver?
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3000 mA h = 3.0 A × 3600 s = 1.08×104 C ≈ 1.1×104 C. The value 1.1×107 C leaves the current in mA. The value 1.8×102 C converts hours into minutes rather than seconds. The value 3.0 C is simply the figure in A h.
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Answer A
Answer: A
Exam tip: The ampere-hour is a unit of charge: 1 A h = 3600 C.
34IBP2S-016-01-03Electrical power & energyPaper 1AMedium1 mark
A filament lamp in a bicycle headlight draws a current of 0.50 A when the potential difference across it is 6.0 V. What is the most likely current when the potential difference is 3.0 V?
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At 6.0 V the resistance is 6.0/0.50 = 12 Ω. At 3.0 V the filament is cooler, so its resistance is less than 12 Ω. The current must therefore be more than 3.0/12 = 0.25 A. A smaller pd cannot give a larger current, so the current is also less than 0.50 A. Only 0.32 A fits, corresponding to about 9.4 Ω. Option B assumes ohmic behaviour. Option A assumes the resistance rises as the lamp cools. Option D assumes the current stays constant.
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Answer C
Answer: C
Common error: Halving the pd halves the current only for an ohmic conductor. For a filament, find which side of the ohmic value the answer lies on by thinking about the filament temperature.
35IBP2S-016-01-04Series & parallel circuitsPaper 1AMedium1 mark
A car rear-window demister has eight identical heating strips connected in parallel across a 12 V supply of negligible internal resistance. One strip breaks. Which statement is correct?
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Each parallel branch still has the full 12 V across it. Its resistance is unchanged, so its current and its power P = V²/R are unchanged. The broken strip now contributes nothing, so the total power falls to 7/8 of its original value. Option A treats the strips as being in series. Options B and C wrongly assume that current is redistributed among the branches or shared out differently.
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Answer D
Answer: D
Exam tip: With an ideal supply, parallel branches are independent of each other. Removing one branch does not change the pd across the others.
36IBP2S-016-01-05Cells & sources of emfPaper 1AMedium1 mark
A cell of emf 1.5 V powers a TV remote control. While the remote is operating, the potential difference across the cell terminals is 1.2 V. How much energy is transferred inside the cell, to its internal resistance, for each coulomb of charge that passes through it?
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The emf is the energy transferred per coulomb by the cell, which is 1.5 J C−1. Of this, 1.2 J C−1 is delivered to the external circuit, as shown by the terminal pd. The remainder, 1.5 − 1.2 = 0.30 J per coulomb, is dissipated in the internal resistance. Option C is the emf, option B is the terminal pd, and option D adds the two instead of subtracting.
✓ 1
Answer A
Answer: A
Exam tip: Read ε = V + Ir as an energy-per-coulomb statement: energy supplied = energy delivered outside + energy lost inside.
37IBP2S-016-01-06Cells & sources of emfPaper 1AMedium1 mark
A solar cell is modelled as a source of emf in series with an internal resistance. Its terminal pd on open circuit is 0.60 V. When its terminals are connected by a wire of negligible resistance, the current is 3.0 A. What is the internal resistance of the cell?
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On open circuit there is no current, so the terminal pd equals the emf: ε = 0.60 V. On short circuit the only resistance in the circuit is r, so r = ε/I = 0.60/3.0 = 0.20 Ω. Option A confuses zero external resistance with zero total resistance. Option C multiplies V by I. Option D divides I by V.
✓ 1
Answer B
Answer: B
Common error: A short circuit removes the external resistance only. The internal resistance is what limits the current.
38IBP2S-016-01-07Series & parallel circuitsPaper 1AMedium1 mark
Three identical resistors X, Y and Z are connected to a battery. Y and Z are connected in parallel with each other, and this combination is connected in series with X. What is the ratio \(\frac{\text{power dissipated in X}}{\text{power dissipated in Y}}\)?
Circuit diagram
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All of the current I passes through X. It then splits equally between Y and Z, so Y carries I/2. With equal resistances, P = I²R gives PX/PY = I²/(I/2)² = 4. Option B assumes power is proportional to current. Option A assumes identical resistors always dissipate equal power. Option D inverts the ratio.
✓ 1
Answer C
Answer: C
Common error: Power depends on the square of the current. Halving the current in a resistor quarters its power.
39IBP2S-016-01-08emf & internal resistancePaper 1AMedium1 mark
A car battery has a significant internal resistance and is supplying the headlamps. The starter motor is then switched on in parallel with the headlamps. What happens to the terminal pd of the battery and to the power dissipated in its internal resistance?
Starter motor switched on after the headlamps
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Adding the motor in parallel lowers the external resistance, so the current drawn from the battery increases. The lost volts Ir increase, so the terminal pd V = ε − Ir falls, which is why the headlamps dim. The internal power I²r also rises with the larger current. Option B treats the terminal pd as always equal to the emf. Option C forgets that the current has increased. Option D gets the direction of the change in pd wrong.
✓ 1
Answer A
Answer: A
Exam tip: Start every internal-resistance problem with the current. Once you know whether I goes up or down, the changes in Ir and V follow directly.
A metal strain gauge wire bonded to a steel bridge girder is stretched uniformly. Its length becomes 1.5 times its original length. The volume and resistivity of the wire stay constant. By what factor does the resistance of the wire change?
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R = ρL/A. The volume LA is constant, so if L increases by a factor of 1.5, A decreases by a factor of 1.5. R therefore changes by a factor of 1.5 × 1.5 = 2.25. Option B ignores the reduction in cross-sectional area. Option A wrongly assumes that a constant volume means a constant resistance. Option D adds the two effects instead of multiplying them.
✓ 1
Answer C
Answer: C
Exam tip: When a wire is stretched at constant volume, both the length and the area change. Resistance then scales with the square of the length factor.
41IBP2S-016-00-03Cells & sources of emfPaper 1AMedium1 mark
A battery of emf 12 V and internal resistance 0.50 Ω powers a resistive heater at a polar research station. The heater resistance is chosen so that the power delivered to the heater is as large as possible. Which row gives the heater resistance and the power delivered to it?
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Maximum power is transferred to the load when R = r = 0.50 Ω. The current is then I = ε/(R + r) = 12/1.0 = 12 A, so the power in the heater is I²R = 144 × 0.50 = 72 W. Option B is the total power εI supplied by the battery, half of which is dissipated inside the battery. Option C gives the largest current, but with R = 0 no power reaches the load: all 288 W is dissipated internally. Option D uses R = 2r, which gives less than the maximum.
✓ 1
Answer A
Answer: A
Exam tip: At maximum power transfer the terminal pd is ε/2 and the efficiency is only 50%. Half of the energy is dissipated inside the source.
Two copper cables, A and B, are used in an electric bicycle. Cable B has twice the length and twice the diameter of cable A. What is the ratio \(\frac{R_B}{R_A}\)?
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R = ρL/A, and A is proportional to d². Doubling the diameter increases the area by a factor of 4. Doubling the length increases R by a factor of 2. So RB/RA = 2/4 = 0.5. Option C assumes that doubling the diameter doubles the area. Option D ignores the change in area. Option A ignores the change in length.
✓ 1
Answer B
Answer: B
Common error: Area depends on diameter squared. Doubling d gives four times the area, not twice.
43IBP2S-016-00-07emf & internal resistancePaper 1AMedium1 mark
A cell in a bicycle lamp is connected to a variable resistor. When the current is 0.20 A the terminal potential difference is 1.40 V. When the current is 0.80 A the terminal potential difference is 1.10 V. What are the emf ε and the internal resistance r of the cell?
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V = ε − Ir, so r = ΔV/ΔI = (1.40 − 1.10)/(0.80 − 0.20) = 0.30/0.60 = 0.50 Ω. Then ε = V + Ir = 1.40 + 0.20 × 0.50 = 1.50 V. Option B treats a reading taken while current flows as the emf. Option C inverts the gradient (ΔI/ΔV). Option D calculates V/I, which is the external resistance.
✓ 1
Answer A
Answer: A
Exam tip: On a graph of V against I, the gradient is −r and the V-intercept is ε. Two data points are enough to find both.
44IBP2S-016-00-08Series & parallel circuitsPaper 1AMedium1 mark
A car rear-window demister has two identical heating elements. They can be switched from a series arrangement to a parallel arrangement across the same supply. The supply has negligible internal resistance. What is \(\frac{\text{total power in parallel}}{\text{total power in series}}\)?
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With element resistance R, the series resistance is 2R and the parallel resistance is R/2. For a fixed V, P = V²/Rtotal. The ratio is therefore (V²/(R/2)) / (V²/(2R)) = 2R/(R/2) = 4. Option C considers only one change of a factor of 2. Options A and B invert the relationship by using P = I²R with a fixed current.
✓ 1
Answer D
Answer: D
Common error: When the supply pd is fixed, use P = V²/R. A lower total resistance means more power, not less.
45IBP2S-016-00-10emf & internal resistancePaper 1AMedium1 mark
A cell with internal resistance is connected to an external resistor. The resistance of the external resistor is increased steadily from zero. Which statement is correct?
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I = ε/(R + r), so the current decreases as R increases (A is wrong). The terminal pd V = ε − Ir increases as I falls. The efficiency is V/ε = R/(R + r), which also increases with R, so C is correct. The power in R rises to a maximum at R = r and then falls (B is wrong). The emf is constant (D is wrong).
✓ 1
Answer C
Answer: C
Exam tip: Maximum power and maximum efficiency do not occur at the same load. Maximum power occurs at R = r, while efficiency keeps rising as R increases.
46IBP2S-016-02-01Circuit diagrams & metersPaper 1AMedium1 mark
A student builds a circuit to test a filament lamp. The circuit contains a battery of emf 6.0 V with negligible internal resistance. By mistake, the student connects a high-resistance digital voltmeter in series with the lamp instead of in parallel with it. What happens?
The circuit as built by the student
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The voltmeter has a very large resistance, so the current in the series circuit is almost zero and the lamp does not light. The pd across the lamp is then almost zero (V = IR with I ≈ 0), so almost the whole emf of 6.0 V appears across the voltmeter. The options where the lamp lights wrongly assume a meter has no effect on the circuit. The option of 0 V wrongly assumes that a tiny current means a tiny reading.
✓ 1
Answer B
Answer: B
Common error: A tiny current does not mean a tiny voltmeter reading. In a series circuit the emf is shared in proportion to resistance, so the component with by far the largest resistance takes almost all of the pd.
47IBP2S-016-02-03Thermistors, LDRs & dividersPaper 1AMedium1 mark
A street-lamp controller contains a light-dependent resistor (LDR) in series with an 8.0 kΩ fixed resistor. The pair is connected across a 9.0 V supply of negligible internal resistance. The LDR has a resistance of 40 kΩ in darkness and 2.0 kΩ in bright daylight. What is the pd across the LDR in darkness and in daylight?
Circuit diagram
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VLDR = 9.0 × RLDR/(RLDR + 8.0 kΩ). Darkness: 9.0 × 40/48 = 7.5 V. Daylight: 9.0 × 2.0/10 = 1.8 V. The option 1.5 V / 7.2 V gives the pd across the fixed resistor. The option 1.8 V / 7.5 V swaps the light and dark values. The option 4.5 V / 4.5 V assumes the pd is shared equally in a series circuit.
✓ 1
Answer D
Answer: D
Exam tip: Use the ratio form Vout = Vin × R/(Rtotal). Units of kΩ cancel, so there is no need to convert to ohms.
48IBP2S-016-02-04emf & internal resistancePaper 1AMedium1 mark
A zinc–air hearing-aid cell has an emf of 1.40 V. When it supplies a current of 20 mA, the pd across its terminals is 1.30 V. What is the internal resistance of the cell?
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The lost pd is ε − V = 1.40 − 1.30 = 0.10 V. r = 0.10/0.020 = 5.0 Ω. 65 Ω is the load resistance (1.30/0.020). 70 Ω is the total resistance (1.40/0.020). 0.50 Ω comes from a mistake converting mA to A.
✓ 1
Answer B
Answer: B
Common error: Internal resistance is found from the 'lost volts' (ε − V) divided by the current, not from either the emf or the terminal pd on its own.
49IBP2S-016-02-06Thermistors, LDRs & dividersPaper 1AMedium1 mark
Two 10 kΩ resistors are connected in series across a 12 V supply of negligible internal resistance. A data logger with an input resistance of 10 kΩ is connected across one of the resistors. What pd does the data logger record?
Circuit diagram
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The logger in parallel with one resistor gives a combined resistance of 5.0 kΩ. The pd across this combination is 12 × 5.0/(10 + 5.0) = 4.0 V. 6.0 V assumes an ideal (infinite-resistance) meter. 8.0 V is the pd across the other resistor. 3.0 V wrongly halves the ideal value of 6.0 V.
✓ 1
Answer B
Answer: B
Exam tip: A voltmeter whose resistance is similar to the component it measures 'loads' the circuit and lowers the reading. Replace the meter and component with their parallel combination first.
50IBP2S-016-02-08Thermistors, LDRs & dividersPaper 1AMedium1 mark
A student investigates how the current in an LDR varies with the pd across it. The student wants data starting from 0 V. Why is a potential divider (potentiometer) arrangement preferred to a variable resistor connected in series with the LDR?
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With a series variable resistor, the pd across the LDR can never fall to zero, because the rheostat has a finite maximum resistance. A potentiometer taps off any fraction of the supply pd, from 0 V up to the full value. It actually draws more power, because current flows continuously through the whole potentiometer. It does not remove internal resistance effects. It has no effect on meter precision.
✓ 1
Answer B
Answer: B
Exam tip: Remember the key advantage of the potential divider circuit: the full range from 0 V upwards. Its disadvantage is the extra energy wasted in the divider.
Wires X and Y are made of the same metal and are at the same temperature. Y is twice as long as X and has twice the diameter of X. The two wires are connected in series to a battery.
R is the resistance of a wire and P is the power dissipated in it. Which row gives RY/RX and PY/PX?
RY / RXPY / PX
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Step 1R = ρL/A with A ∝ d²: RY/RX = 2 / 2² = 1/2.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2In series the wires carry the same current, so P = I²R ∝ R: PY/PX = 1/2.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe change in cross-section has been ignored, so only the doubled length has been counted.
BThe area has been taken as proportional to the diameter instead of the diameter squared, so the two factors of 2 cancel.
CP = V²/R has been used as if the wires had the same potential difference. In series they carry the same current, so the smaller resistance dissipates less.
DCorrect: R ∝ L/d² gives 2/4 = 1/2, and with equal currents P = I²R gives the same ratio.
52B-1A-43Thermistors, LDRs & dividersPaper 1AHard1 mark
A thermistor is connected in series with a fixed resistor R, an ideal ammeter and a cell of negligible internal resistance. An ideal voltmeter is connected across the thermistor. The resistance of the thermistor decreases as its temperature increases.
The temperature of the thermistor is increased. Which row gives the change in the ammeter reading and the change in the voltmeter reading?
Diagram NOT accurately drawn
Ammeter readingVoltmeter reading
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Step 1The thermistor's resistance falls, so the total resistance falls and the current I = ε / (R + RT) increases.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The pd across R, IR, increases. The two pds add up to the constant emf, so the pd across the thermistor decreases.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
ATwo errors: the thermistor's resistance has been taken to rise with temperature, like a metal wire, and V = IR has then been applied to the thermistor using the fall in current alone.
BThe thermistor's resistance has been taken to rise with temperature, like a metal wire. Here it falls, so the current rises and the thermistor takes a smaller share of the emf.
CCorrect: the total resistance falls, so the current rises; R then takes a larger share of the emf and the thermistor a smaller one.
DV = IR has been applied to the thermistor using the rise in current alone, ignoring the fall in its resistance. The pd across R rises, so the pd across the thermistor falls.
53B-1A-44Series & parallel circuitsPaper 1AHard1 mark
Resistor X, of resistance R, is connected in series with a parallel combination of resistor Y, of resistance R, and resistor Z, of resistance 2R. The supply has emf V and negligible internal resistance.
What is the current in Z?
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Step 1Y and Z in parallel: (1/R + 1/(2R))−1 = 2R/3. Total resistance = R + 2R/3 = 5R/3, so the supply current is 3V / (5R).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The pd across the parallel pair is 3V/(5R) × 2R/3 = 2V/5, so the current in Z is (2V/5) / 2R = V / (5R).
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: Z, with twice Y's resistance, takes one third of the supply current: ⅓ × 3V/(5R) = V/(5R).
BThe supply current has been split equally between Y and Z. Parallel branches share the current in inverse proportion to their resistances.
CThe larger share of the current has been given to the larger resistance (⅔ × 3V/(5R)). More current flows through the branch with less resistance.
DThe whole supply emf has been taken across Z. Part of the emf is across X, so the pd across Z is only 2V/5.
54B-1A-61Series & parallel circuitsPaper 1AHard1 mark
Three identical lamps P, Q and R are connected to a cell of negligible internal resistance as shown. Lamp R is then removed from its holder. Assume that the resistance of each lamp does not change.
Which row gives the change in the brightness of lamp P and of lamp Q?
Diagram NOT accurately drawn
Lamp PLamp Q
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Step 1With R removed, the total resistance rises from 1.5Rlamp to 2Rlamp, so the current from the cell, which is the current in P, falls: P is dimmer.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The p.d. across Q rises from ε/3 to ε/2 (Q now takes the whole current, which is larger than its earlier half-share), so Q is brighter.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
ARemoving a lamp has been taken to reduce the total resistance. Removing a parallel branch increases the total resistance, so the cell current falls.
BThe fall in the cell current has been applied to Q as well. Q now carries the whole of the current instead of half of it, and its share of the emf rises from one third to one half.
CThe current from the cell has been treated as fixed. The emf is fixed; the current falls because the total resistance increases.
DCorrect: the current in P falls (larger total resistance), while the p.d. across Q rises from ε/3 to ε/2.
55B-1A-62Thermistors, LDRs & dividersPaper 1AHard1 mark
A potentiometer with a track of total resistance 12.0 kΩ is connected across a 6.0 V supply of negligible internal resistance. The slider S is at the midpoint of the track. A voltmeter of resistance 12.0 kΩ is connected between S and the lower end of the track.
What is the reading on the voltmeter?
Diagram NOT accurately drawnShow mark scheme
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Step 1The lower half of the track (6.0 kΩ) is in parallel with the voltmeter (12.0 kΩ): (1/6.0 + 1/12.0)−1 = 4.0 kΩ.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2This is in series with the upper half (6.0 kΩ): reading = 6.0 × 4.0/(4.0 + 6.0) = 2.4 V.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe parallel combination, 4.0 kΩ, has been found, but the total resistance has been kept at 12.0 kΩ: 6.0 × 4.0/12.0 = 2.0 V. With the voltmeter connected the total resistance is 4.0 + 6.0 = 10.0 kΩ.
BCorrect: the voltmeter in parallel with the lower half gives 4.0 kΩ, which takes 4.0/10.0 of the 6.0 V.
CThis treats the voltmeter as ideal. Here its resistance is stated and is comparable with the track, so it draws a significant current.
DThis is the p.d. across the upper half of the track (6.0 × 6.0/10.0), not the part the voltmeter is connected across.
A salinity probe passes a direct current through sea water between two metal plates. Each second, 1.0 × 1016 sodium ions (Na+) cross a plane between the plates towards the negative plate, and 1.5 × 1016 chloride ions (Cl−) cross the same plane towards the positive plate.
What is the current in the sea water?
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Step 1Positive charge moving one way and negative charge moving the other way both transfer charge in the same direction (towards the negative plate in conventional terms).
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All 2 steps must be completed — there is no mark for a part-answer.
AThe charges of the two kinds of ion have been subtracted, as if they cancelled. A negative ion moving towards the positive plate is equivalent to positive charge moving towards the negative plate, so the two flows add.
BOnly the positive ions have been counted, as if current were carried by positive charge alone. The chloride ions are charge carriers too.
COnly the chloride ions have been counted, as if negative carriers were the only ones that move in a conductor. In a solution both kinds of ion are mobile.
DCorrect: both carriers contribute to the same conventional current: 2.5 × 1016 × 1.60 × 10−19 = 4.0 mA.
57B-1A-108Cells & sources of emfPaper 1AHard1 mark
Four identical cells, each of emf ε and internal resistance r, are available to drive a current through a load whose resistance is negligible compared with r.
Which arrangement of the cells gives the largest current in the load?
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Step 1With the load resistance negligible, I ≈ (total emf)/(total internal resistance).
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All 2 steps must be completed — there is no mark for a part-answer.
58B-1A-111Ohmic & non-ohmic conductorsPaper 1AHard1 mark
The graph shows how the current I in a filament lamp varies with the potential difference V across it. The dashed line is the tangent to the curve at point P.
Which row gives the resistance of the lamp at P and the change in its resistance as V increases?
Graph drawn to scale
Resistance at PAs V increases, resistance
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Step 1R = V/I at the point: 6.0/0.30 = 20 Ω.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2V/I grows along the curve (at 10 V: 10/0.395 ≈ 25 Ω), so the resistance increases as the filament gets hotter.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThe gradient of the tangent, ΔI/ΔV = 0.030 A V−1, has been taken as the resistance, and its fall along the curve read as a fall in resistance.
BThe value at P is right, but a decreasing gradient of an I–V graph has been read as decreasing resistance. A smaller gradient means less current per volt, so more resistance.
CCorrect: R = V/I = 20 Ω at P, and V/I increases along the curve.
DThe resistance has been found from the tangent, 1/0.030 ≈ 33 Ω. Resistance is the ratio V/I at the point, not the reciprocal of the gradient.
Two wires, A and B, are made of the same metal and connected in parallel across a battery. Wire B is twice as long as wire A and has twice the diameter. What is the ratio \(\frac{\text{drift speed in A}}{\text{drift speed in B}}\)?
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Both wires have the same potential difference V. Write v = I/(nAq) and I = V/R = VA/(ρL). Combining these gives v = V/(nqρL), so the area cancels and v ∝ 1/L. Wire A is half as long as wire B, so vA/vB = 2. The answer 1/2 is the ratio of the currents (IB = 2IA). The answer 1 assumes the same metal means the same drift speed. The answer 4 considers the areas only.
✓ 1
Answer A
Answer: A
Exam tip: Combine I = nAvq with R = ρL/A before substituting. Quantities that cancel make the comparison quick.
60IBP2S-016-01-09Circuit diagrams & metersPaper 1AHard1 mark
In a medical monitoring circuit, two 10 kΩ resistors are connected in series to a 9.0 V supply of negligible internal resistance. A voltmeter with resistance 10 kΩ is connected across one of the resistors. What does the voltmeter read?
Circuit diagram
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The voltmeter in parallel with one 10 kΩ resistor forms a combination of 5.0 kΩ. This is in series with the other 10 kΩ resistor, so the supply pd divides in the ratio 5 : 10. The reading is 9.0 × 5/15 = 3.0 V. Option B assumes an ideal voltmeter of infinite resistance. Option C is the pd across the other resistor. Option D is the full supply pd.
✓ 1
Answer A
Answer: A
Exam tip: A voltmeter whose resistance is similar to the component it measures will load the circuit. Include its resistance in the calculation.
61IBP2S-016-00-09Cells & sources of emfPaper 1AHard1 mark
A backup battery for a planetary rover uses six identical cells, each of emf 1.5 V and internal resistance 0.60 Ω. The cells form two parallel rows, and each row has three cells in series. What are the emf and internal resistance of the battery?
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Each row has an emf of 3 × 1.5 = 4.5 V and an internal resistance of 3 × 0.60 = 1.8 Ω. Two identical rows in parallel have the same emf, 4.5 V, and the combined internal resistance is 1.8/2 = 0.90 Ω. Option A treats all six cells as being in series. Option B adds the two row resistances as if the rows were in series. Option D wrongly halves the emf for a parallel connection.
✓ 1
Answer C
Answer: C
Exam tip: Identical sources in parallel keep the same emf, but their internal resistance is divided by the number of branches. This lets the battery supply larger currents.
62IBP2S-016-02-07Thermistors, LDRs & dividersPaper 1AHard1 mark
An oven temperature monitor uses two potential dividers connected in parallel across the same supply. • Divider 1: a 2.0 kΩ resistor (upper) in series with a 3.0 kΩ resistor (lower). • Divider 2: a thermistor (upper) in series with a 6.0 kΩ resistor (lower). A voltmeter is connected between the two junction points. What is the resistance of the thermistor when the voltmeter reads zero?
Circuit diagram
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The voltmeter reads zero when both junctions are at the same potential. This happens when the ratio of upper to lower resistance is the same in each divider: RT/6.0 = 2.0/3.0, so RT = 4.0 kΩ. 9.0 kΩ uses the inverted ratio. 2.0 kΩ simply matches the upper resistor of divider 1. 5.0 kΩ is the total resistance of divider 1.
✓ 1
Answer B
Answer: B
Exam tip: A balanced pair of dividers needs equal resistance ratios, not equal resistances. The supply pd does not affect the balance condition.
63IBP2S-016-02-10Cells & sources of emfPaper 1AHard1 mark
A small satellite panel contains 12 identical solar cells. Under constant illumination, each cell is modelled as having an emf of 0.60 V and an internal resistance of 0.40 Ω. The cells are arranged as two parallel strings, with each string made of 6 cells in series. What are the emf and the internal resistance of the panel?
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Each string has an emf of 6 × 0.60 = 3.6 V and a resistance of 6 × 0.40 = 2.4 Ω. Two identical strings in parallel keep the emf at 3.6 V but halve the resistance to 1.2 Ω. 7.2 V / 4.8 Ω treats all 12 cells as in series. 3.6 V / 2.4 Ω forgets to halve the resistance for the parallel strings. 3.6 V / 4.8 Ω adds all twelve internal resistances.
✓ 1
Answer C
Answer: C
Exam tip: Identical sources in parallel share the current: the emf stays the same, and the internal resistance divides by the number of branches.
64B-1B-19Thermistors, LDRs & dividersPaper 1BEasy9 marks
A light-dependent resistor (LDR) is placed next to the sensor of a light meter in a dark box lit by an adjustable lamp. For each illuminance E the resistance R of the LDR is measured three times with a digital ohmmeter.
The graph shows the mean resistance for six of the seven illuminances.
E / lx
R1 / kΩ
R2 / kΩ
R3 / kΩ
mean R / kΩ
50
13.4
13.0
13.3
13.2
100
8.2
7.8
8.1
200
5.0
4.8
4.9
4.9
300
3.6
3.7
3.6
3.6
400
2.8
3.0
2.9
2.9
600
2.2
2.2
2.1
2.2
800
1.8
1.7
1.8
1.8
Graph drawn to scale
(a)
State the resolution of the ohmmeter.
(1)
(b)
Calculate the mean resistance for E = 100 lx and its absolute uncertainty.
(2)
(c)
Plot your mean from (b) on the graph and draw the curve of best fit.
(2)
(d)
Using your curve, determine R when E = 250 lx.
(1)
(e)
The LDR is connected in series with a 10.0 kΩ resistor across a 6.0 V supply of negligible internal resistance. Calculate the p.d. across the LDR when E = 250 lx.
(2)
(f)
Suggest one improvement to the range or distribution of the measurements.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.1 kΩ
✓ 1
Accept ±0.1 kΩ.
Part (b)
mean = 8.0 kΩ
✓ 1
Accept 8.03.
uncertainty = ½ × range = ±0.2 kΩ
✓ 1
Do not accept ±0.1 kΩ «the resolution»: the spread of the repeats is larger.
Part (c)
point plotted correctly
✓ 1
Allow ECF from (b).
smooth curve through the points, steep at low E and levelling off at high E; not dot-to-dot
✓ 1
Part (d)
≈ 4.1 kΩ
✓ 1
Accept 3.8–4.5 kΩ.
Part (e)
V = 6.0 × R / (R + 10.0)
✓ 1
= 1.74 V
✓ 1
Allow ECF from (d). Accept 1.6–1.9 V.
Part (f)
take more readings at low illuminance «below 200 lx» where R changes most rapidly / extend the range below 50 lx
✓ 1
Do not accept “repeat the readings” alone.
Answers: (b) (8.0 ± 0.2) kΩ · (d) ≈ 4.1 kΩ · (e) 1.74 V (the remaining parts are explanations — see the table above)
A fully charged rechargeable AA cell is labelled “1.2 V, 2000 mA h”. A student connects it across a resistor of resistance 3.30 Ω ± 1 %, and a data logger records the terminal p.d. V, to the nearest 0.01 V, every 0.5 h. The current is calculated from I = V/R.
The graph shows I against time t for every reading except the first.
t / h
V / V
I / A
0.0
1.32
0.5
1.25
0.379
1.0
1.24
0.376
1.5
1.23
0.373
2.0
1.22
0.370
2.5
1.21
0.367
3.0
1.20
0.364
3.5
1.19
0.361
4.0
1.19
0.361
4.5
1.18
0.358
5.0
1.17
0.355
5.5
1.13
0.342
6.0
0.87
0.264
Graph drawn to scale
(a)
Show that the current at t = 0 is about 0.40 A, and plot this point on the graph.
(1)
(b)
Determine the percentage uncertainty in the current at t = 0.
(1)
(c)
Draw the curve of best fit for the data.
(1)
(d)
Outline why the area under the graph is equal to the charge that flows through the resistor.
(1)
(e)
Estimate, in mA h, the total charge that flows through the resistor between t = 0 and t = 6.0 h.
(2)
(f)
Comment on the label on the cell.
(1)
(g)
Estimate the energy transferred to the resistor between t = 0 and t = 6.0 h.
(1)
(h)
Suggest one improvement to the way the data were recorded.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
I = 1.32/3.30 = 0.400 A AND point plotted
✓ 1
Must see the substitution OR an answer to at least 3 s.f.
Part (b)
0.01/1.32 × 100 + 1 % = 1.8 %
✓ 1
Accept 1.8 % or 2 %. Accept 1.4 % using half the resolution.
Part (c)
smooth curve through the points: a short initial fall, then almost level and falling slowly, then falling steeply after about 5.5 h
✓ 1
Not dot-to-dot.
Part (d)
I = Δq/Δt, so Δq = IΔt is the area of a thin strip under the graph; adding the strips gives the total charge
✓ 1
OWTTE.
Part (e)
area found by counting squares / trapezia / a rectangle of mean height × 6.0 h
✓ 1
≈ 2170 mA h
✓ 1
Accept 2060–2270 mA h. Award [1] max for an answer in A h without conversion.
Part (f)
the charge delivered, ≈ 2170 mA h, is at least «about 8 % more than» the 2000 mA h on the label, so the label is consistent with the data «the label gives the minimum charge the cell should deliver»
✓ 1
Allow ECF from (e).
Part (g)
E ≈ mean V × Q ≈ 1.19 × 2170 × 10−3 × 3600 ≈ 9.3 × 103 J
✓ 1
Accept 8 × 103 to 1.0 × 104 J. Accept ∑I²RΔt. Allow ECF from (e).
Part (h)
record the p.d. more often «e.g. every 5 minutes» after about 5 h, where the current falls rapidly «so the area there is estimated more accurately»
✓ 1
Accept ‘continue recording until the current is close to zero’.
Answers: (a) 0.400 A · (b) 1.8 % · (e) ≈ 2170 mA h · (g) ≈ 9.3 × 103 J (the remaining parts are explanations — see the table above)
66B-1B-47Series & parallel circuitsPaper 1BEasy10 marks
A student tests the rule for resistors in parallel. Six resistors are each labelled “10.0 Ω ± 1 %”. The student connects n of them in parallel on a breadboard and measures the total resistance Rtotal with a digital ohmmeter, using the same two leads each time. The ohmmeter displays resistance to 0.01 Ω.
The graph shows Rtotal against 1/n for five of the six readings.
n
1/n
Rtotal / Ω
1
1.000
10.51
2
0.500
5.48
3
0.333
3.83
4
0.250
3.00
5
0.200
2.51
6
0.167
2.18
Graph drawn to scale
(a)
State the absolute uncertainty in each reading of Rtotal due to the resolution of the ohmmeter.
(1)
(b)
Show that n identical resistors, each of resistance R, connected in parallel have a total resistance of R/n.
(1)
(c)
Plot the point for n = 4 and draw the line of best fit for the data.
(2)
(d)
Determine the gradient of your line.
(2)
(e)
Comment on whether your answer to (d) is consistent with the label on the resistors.
(1)
(f)
The line does not pass through the origin. Determine the intercept on the Rtotal axis and suggest what it represents.
(2)
(g)
Suggest how the student could correct each reading for the effect identified in (f).
67B-1B-07Ohmic & non-ohmic conductorsPaper 1BMedium10 marks
A student investigates how the current I in a 12 V filament lamp depends on the potential difference V across it. The voltmeter has a resolution of 0.01 V and the ammeter a resolution of 0.001 A.
The student suggests that I = kVn, where k and n are constants, and plots lg I against lg V. The graph shows six of the seven points.
V / V
I / A
lg(V / V)
lg(I / A)
1.00
0.160
0.000
−0.796
2.00
0.238
0.301
−0.623
3.50
0.332
0.544
−0.479
5.00
0.406
7.00
0.497
0.845
−0.304
9.50
0.589
0.978
−0.230
12.00
0.677
1.079
−0.169
Graph drawn to scale
(a)
Calculate the percentage uncertainty in the current when V = 1.00 V.
(1)
(b)
Calculate the missing values in the table for V = 5.00 V and plot the point on the graph.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine n.
(2)
(e)
Determine k.
(2)
(f)
Explain how your value of n shows that the filament is a non-ohmic conductor, and why it behaves in this way.
(2)
(g)
Another student claims that the current is proportional to √V. Comment on this claim.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.001 / 0.160 × 100 = 0.63 %
✓ 1
Accept 0.6 % or 0.625 %.
Part (b)
0.699 AND −0.391 AND plotted correctly
✓ 1
Accept 2 to 4 decimal places.
Part (c)
single straight line of best fit through the points
✓ 1
Part (d)
lg I = n lg V + lg k, so n = gradient
✓ 1
Must see the gradient identified with n.
n = 0.58
✓ 1
Accept 0.56–0.60.
Part (e)
lg k = intercept on the lg I axis «at lg V = 0» ≈ −0.80
✓ 1
Or substitute a point on the line.
k = 10−0.80 = 0.16 «A V−0.58»
✓ 1
Accept 0.15–0.17. Allow ECF.
Part (f)
for an ohmic conductor I ∝ V, so n would be 1; n < 1 means V/I increases with V
✓ 1
a larger current raises the temperature of the filament; the lattice ions vibrate more so the electrons collide more often and the resistance increases
✓ 1
OWTTE.
Part (g)
the claim predicts n = 0.5 but n ≈ 0.58; the points lie so close to the line that a gradient of 0.5 cannot be drawn through them, so the claim is not supported
✓ 1
Allow ECF from (d). Accept a comparison with the candidate’s range of acceptable gradients.
Answers: (a) 0.63 % · (b) 0.699, −0.391 · (d) 0.58 · (e) 0.16 (the remaining parts are explanations — see the table above)
68B-1B-08emf & internal resistancePaper 1BMedium10 marks
A student investigates a used 1.5 V zinc–carbon cell. The cell is connected in series with an ammeter, a rheostat and a switch, and a voltmeter is connected across the terminals of the cell. For each setting of the rheostat the switch is closed only long enough to read the meters.
The uncertainties are ±0.02 V in V and ±0.01 A in I, as shown by the error bars.
I / A
0.10
0.20
0.30
0.40
0.50
0.60
0.70
V / V
1.41
1.30
1.23
1.15
1.04
0.98
0.88
Graph drawn to scale
(a)
Outline why the terminal p.d. decreases as the current increases.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the emf ε of the cell.
(1)
(d)
Determine the internal resistance r of the cell.
(2)
(e)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in r, and write r with its uncertainty.
(3)
(f)
The manufacturer states that a new cell of this type has an internal resistance of 0.20 Ω. Comment on the student’s result in the light of your answer to (e).
(1)
(g)
Explain why the switch was closed only briefly for each reading.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
V = ε − Ir: more energy per unit charge is transferred in the internal resistance as I increases
✓ 1
OWTTE.
Part (b)
single straight line through all the error bars
✓ 1
Part (c)
ε = intercept on the V axis = 1.49 V
✓ 1
Accept 1.47–1.51 V.
Part (d)
r = −gradient, from a large triangle
✓ 1
r = 0.86 Ω
✓ 1
Accept 0.82–0.90 Ω.
Part (e)
steepest and shallowest lines through all error bars «gradients ≈ −0.95 and −0.79 Ω»
✓ 1
Δr = ½(max − min) ≈ 0.08 Ω
✓ 1
Accept 0.05–0.12 Ω. Allow ECF.
r = (0.86 ± 0.08) Ω
✓ 1
MP3 is for matching the precision of value and uncertainty. Accept (0.9 ± 0.1) Ω.
Part (f)
0.20 Ω is well outside the range of r, so the internal resistance of the cell has increased as it has been used
✓ 1
Allow ECF from (e).
Part (g)
to avoid heating/discharging the cell «and the rheostat» during the experiment, which would change ε and r
✓ 1
OWTTE.
Answers: (c) 1.49 V · (d) 0.86 Ω · (e) (0.86 ± 0.08) Ω (the remaining parts are explanations — see the table above)
A student determines the resistivity ρ of constantan. The diameter d of the wire is measured with a digital micrometer (resolution 0.001 mm) at five places. The wire is then taped to a metre rule and its resistance R is measured with an ohmmeter connected by crocodile clips at different lengths L from one end. Each value of R has an uncertainty of ±0.05 Ω.
Reading
1
2
3
4
5
d / mm
0.315
0.318
0.312
0.316
0.314
L / m
R / Ω
0.200
1.54
0.350
2.44
0.500
3.44
0.650
4.41
0.800
5.31
0.950
6.29
Graph drawn to scale
(a)
Calculate the mean diameter of the wire and its absolute uncertainty.
(2)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine ρ using the gradient of your line.
(3)
(d)
The uncertainty in the gradient is ±2 %. Determine the absolute uncertainty in ρ.
(2)
(e)
The literature value for constantan is 4.9 × 10−7 Ω m. Comment on the student’s result.
(1)
(f)
The line does not pass through the origin. Suggest a reason for this and explain why it does not affect the value of ρ found in (c).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
mean d = 0.315 mm
✓ 1
uncertainty = ½ × range = ±0.003 mm
✓ 1
Do not accept ±0.001 mm: the spread of the readings is larger than the resolution.
Part (b)
single straight line through all the error bars, not forced through the origin
✓ 1
Part (c)
gradient = 6.35 Ω m−1
✓ 1
Accept 6.16–6.54 Ω m−1.
A = πd2/4 = 7.79 × 10−8 m2
✓ 1
Allow ECF from (a).
ρ = gradient × A = 4.95 × 10−7 Ω m
✓ 1
Allow ECF. Unit needed.
Part (d)
Δρ/ρ = 2 % + 2 × 0.95 % = 3.9 %
✓ 1
The diameter term is doubled because A ∝ d2.
Δρ = ±2 × 10−8 Ω m
✓ 1
Allow ECF. Accept ±0.19 × 10−7 Ω m.
Part (e)
4.9 × 10−7 lies within the range 4.75–5.14 × 10−7 Ω m, so the result is consistent with the literature value
✓ 1
Allow ECF from (d).
Part (f)
the resistance of the leads/crocodile clips/contacts is included in every reading
✓ 1
OWTTE.
this adds the same resistance to every value, shifting the line up without changing its gradient, and only the gradient is used for ρ
✓ 1
Answers: (a) (0.315 ± 0.003) mm · (c) 4.95 × 10−7 Ω m · (d) ±2 × 10−8 Ω m (the remaining parts are explanations — see the table above)
A manufacturer’s data table gives the resistance per unit length R/L at 20 °C of nichrome heating wire for several standard diameters d. A student uses these secondary data to test whether R/L is inversely proportional to the cross-sectional area of the wire, and plots lg(R/L) against lg d. The graph shows six of the seven points.
d / mm
(R/L) / Ω m−1
lg(d / mm)
lg(R/L / Ω m−1)
0.200
34.8
−0.699
1.542
0.250
21.8
−0.602
1.338
0.315
14.0
−0.502
1.146
0.400
8.54
0.500
5.56
−0.301
0.745
0.630
3.42
−0.201
0.534
0.800
2.16
−0.097
0.334
Graph drawn to scale
(a)
Calculate the missing values in the table for d = 0.400 mm and plot the point on the graph.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine the gradient of your line.
(2)
(d)
Explain how your answer to (c) supports the hypothesis that R/L is inversely proportional to the cross-sectional area.
(2)
(e)
Using the graph, determine the resistivity of nichrome.
(2)
(f)
The wire in an electric heater operates at about 600 °C. Suggest why its resistance is not correctly predicted by the table.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
−0.398 AND 0.931 AND plotted correctly
✓ 1
Accept 2 to 4 decimal places.
Part (b)
single straight line of best fit through the points
✓ 1
Part (c)
gradient from a large triangle «negative sign shown»
✓ 1
gradient = −2.00
✓ 1
Accept −2.06 to −1.94.
Part (d)
A = πd2/4 so R/L ∝ 1/A means R/L ∝ d−2
✓ 1
so the graph of lg(R/L) against lg d should be a straight line of gradient −2, as found
✓ 1
Allow ECF from (c).
Part (e)
at lg d = 0 «d = 1.00 mm», lg(R/L) ≈ 0.14 so R / L ≈ 1.38 Ω m−1
✓ 1
Any point read from the line may be used.
ρ = (R/L) × πd2/4 = 1.1 × 10−6 Ω m
✓ 1
Accept 1.0–1.2 × 10−6 Ω m. Allow ECF.
Part (f)
the resistivity «resistance» increases with temperature «greater lattice vibration», so the resistance is larger than the 20 °C value
✓ 1
OWTTE.
Answers: (a) −0.398, 0.931 · (c) −2.00 · (e) 1.1 × 10−6 Ω m (the remaining parts are explanations — see the table above)
71B-1B-28Chemical & solar cellsPaper 1BMedium10 marks
A student investigates how the output power of a small solar panel depends on the resistance R of the load connected to it. The panel has an area of 6.0 × 10−3 m² and is lit by a lamp that gives a constant intensity of 900 W m−2 at the panel. The p.d. V across the load is measured with a voltmeter of resolution 0.01 V and the current I with an ammeter of resolution 0.001 A.
The output power is P = VI. The graph shows P against R for nine of the ten readings.
R / Ω
V / V
I / A
P / W
5.0
0.85
0.170
0.145
10.0
1.70
0.170
0.289
20.0
3.40
0.170
0.578
25.0
4.23
0.169
0.715
30.0
4.92
0.164
0.807
35.0
5.29
0.151
40.0
5.44
0.136
0.740
50.0
5.65
0.113
0.638
70.0
5.81
0.083
0.482
100.0
5.90
0.059
0.348
Graph drawn to scale
(a)
Calculate the missing value of P for R = 35.0 Ω and plot the point on the graph.
(1)
(b)
Determine the absolute uncertainty in your answer to (a), using the resolutions of the meters.
(2)
(c)
Draw the curve of best fit for the data.
(1)
(d)
Determine the maximum output power of the panel and the load resistance for which it occurs.
(2)
(e)
The manufacturer states that the maximum efficiency of the panel is 15 %. Determine whether the student's results support this statement.
(2)
(f)
Suggest one improvement to the choice of values of R that would allow the maximum power to be found more precisely.
(1)
(g)
Outline why the output power is small both when R is very small and when R is very large.
Use of half the resolution throughout is acceptable.
ΔP ≈ ±0.007 W
✓ 1
Allow ECF from (a). Accept ±0.006 to ±0.007 W, or ±0.003 W using half the resolutions.
Part (c)
smooth curve through the points rising to a single maximum and then falling more slowly; not dot-to-dot
✓ 1
Part (d)
Pmax ≈ 0.81 W
✓ 1
Accept 0.80–0.83 W from the candidate's curve.
at R ≈ 32 Ω
✓ 1
Accept 27–37 Ω.
Part (e)
incident power = 900 × 6.0 × 10−3 = 5.4 W so efficiency = 0.81/5.4 = 15 %
✓ 1
Allow ECF from (d).
close to / consistent with the stated 15 % «given the uncertainty in reading the maximum from the curve» so the results support the statement
✓ 1
Conclusion must be consistent with the candidate's value; accept ‘not supported’ only if their value differs from 15 % by more than about 1 percentage point.
Part (f)
take more readings for values of R close to the maximum «between about 25 Ω and 45 Ω» / use smaller steps of R there
✓ 1
Do not accept ‘repeat the readings’ alone.
Part (g)
at small R the current cannot exceed about 0.17 A, so V = IR is small; at large R the p.d. cannot exceed about 6 V, so I = V/R is small; in both cases the product P = VI is small
✓ 1
Both extremes needed. OWTTE.
Answers: (a) 0.799 W · (b) ±0.007 W · (d) ≈ 0.81 W at ≈ 32 Ω · (e) ≈ 15 % (the remaining parts are explanations — see the table above)
72B-1B-45Electrical power & energyPaper 1BMedium11 marks
A student tests the relation P = I2R for a coil of manganin wire, whose resistance hardly changes with temperature. The coil is immersed in 0.150 kg of oil in a thin metal can and the oil is stirred continuously. For each current I the oil starts at room temperature and its temperature rise Δθ in 120 s is measured. The ammeter reading has an uncertainty of ±0.01 A and each value of Δθ has an uncertainty of ±0.2 K.
Specific heat capacity of the oil = 1.97 × 103 J kg−1 K−1. The graph shows Δθ against I2 for four of the six readings.
I / A
I2 / A2
Δθ / K
1.02
1.04
1.9
1.49
3.8
2.01
4.04
7.3
2.48
6.15
10.8
3.00
16.1
3.53
12.46
22.0
Graph drawn to scale
(a)
Calculate the percentage uncertainty in I2 when I = 1.02 A.
(2)
(b)
Complete the table and plot the two missing points on the graph.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine the gradient of your line, giving its unit.
(2)
(e)
Assuming that all the energy dissipated in the coil heats the oil, Δθ = (Rt/mc) I2, where t = 120 s. Determine R using your answer to (d).
(2)
(f)
The resistance of the coil measured with an ohmmeter is (4.70 ± 0.05) Ω. Comment on your value in (e).
(1)
(g)
Suggest the cause of the difference and one change to the experiment that would reduce it.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
fractional uncertainty in I = 0.01/1.02 = 0.98 %
✓ 1
uncertainty in I2 = 2 × 0.98 ≈ 2.0 %
✓ 1
Award [2] for CNA. Accept 2 %. Award [1 max] for 1.0 % «power not doubled».
Part (b)
2.22 AND 9.00 «A2» AND both points plotted correctly
✓ 1
Accept 2 or 3 s.f. Plotting to within half a small square.
Part (c)
single straight line passing through every error bar and close to the origin
✓ 1
Do not accept a line joining the first and last points only.
Part (d)
gradient from a large triangle «at least half the line»
✓ 1
gradient = 1.77 K A−2
✓ 1
Accept 1.75–1.80 K A−2. Unit needed.
Part (e)
R = gradient × mc/t
✓ 1
R = 1.77 × 0.150 × 1.97 × 103/120 = 4.36 Ω
✓ 1
Accept 4.32–4.42 Ω. Allow ECF from (d). Award [2] for CNA.
Part (f)
the value from (e) is smaller than 4.65 Ω, the lower limit of the ohmmeter value, so the two do not agree «a systematic error is present»
✓ 1
Allow ECF from (e). A conclusion must be supported by a comparison with the range 4.65–4.75 Ω.
Part (g)
some of the energy dissipated in the coil heats the metal can «and coil» or is lost to the surroundings, so Δθ is smaller than if it all heated the oil, and R is underestimated
✓ 1
OWTTE.
include the heat capacity of the can (and coil) in the calculation / insulate the can and fit a lid / start below room temperature so that gains and losses cancel
✓ 1
The improvement must address the energy not received by the oil. Do not accept 'repeat the readings'.
Answers: (a) 2.0 % · (d) 1.77 K A−2 · (e) 4.36 Ω (the remaining parts are explanations — see the table above)
73B-1B-48Chemical & solar cellsPaper 1BMedium11 marks
A student investigates a silicon solar cell of area 2.5 × 10−3 m2 lit by an adjustable LED array. The irradiance G (power of the light per unit area) at the cell is measured with a meter that is accurate to ±3 %. For each value of G the student measures the pd V across the cell with a voltmeter of very high resistance connected on its own, and then the current Isc when the cell is connected directly to an ammeter of negligible resistance. The readings were taken in order of increasing G.
The uncertainty in each value of V is ±0.003 V and in each value of Isc is ±0.005 A. The student suggests that V = a + b lg G, where a and b are constants. The graph shows V against lg G for six of the seven readings.
G / W m−2
lg(G / W m−2)
V / V
Isc / A
100
2.000
0.547
0.092
150
2.176
0.557
0.133
250
2.398
0.577
0.228
400
0.588
0.357
600
2.778
0.604
0.544
800
2.903
0.611
0.716
1000
3.000
0.621
0.897
Graph drawn to scale
(a)
Explain why V is equal to the emf of the cell.
(1)
(b)
Calculate the missing value of lg G for G = 400 W m−2 and plot the point on the graph.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine b, giving its unit.
(2)
(e)
Deduce, using the readings for G = 100 W m−2 and G = 1000 W m−2, whether Isc is proportional to G. Ignore the uncertainty in G for this part.
(2)
(f)
The maximum power that the cell can deliver to a load is 0.78 VIsc. The manufacturer states that the efficiency of the cell is 20 %. Determine whether the student's data at G = 1000 W m−2 support this statement.
(2)
(g)
The LED array also warms the cell, and the emf of a silicon cell decreases as its temperature rises. Suggest the effect of this on the value of b found in (d) and one way of reducing it.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the voltmeter draws «almost» no current, so there is no pd across the internal resistance «V = ε − Ir = ε»
✓ 1
OWTTE.
Part (b)
2.602 AND point plotted correctly
✓ 1
Accept 2.60. Within half a small square.
Part (c)
single straight line through all the error bars
✓ 1
Part (d)
b = gradient, from a large triangle
✓ 1
b = 0.074 V
✓ 1
Accept 0.071–0.077 V. Unit needed.
Part (e)
Isc/G = (9.2 ± 0.5) × 10−4 and (8.97 ± 0.05) × 10−4 A W−1 m2
✓ 1
Uncertainties from 0.005/0.092 «5 %» and 0.005/0.897 «0.6 %».
the ranges overlap, so the data are consistent with Isc ∝ G
✓ 1
MP2 only scores if a comparison including the uncertainties is made.
uncertainty ≈ 4 % of the value, «±0.7 percentage points», so even the upper limit, 18 %, is below 20 %: the data do not support the statement
✓ 1
Allow ECF from MP1. Accept a qualitative argument that the ±3 % uncertainty in G cannot account for the difference.
Part (g)
the cell is hotter at the higher irradiances «readings taken later», so V is reduced most at large G: the line is less steep and b is too small
✓ 1
OWTTE.
e.g. allow the cell to cool between readings / mount it on a heat sink or cool it with a fan / measure its temperature and keep it constant / take the readings in a random order
✓ 1
Do not accept 'repeat the readings' alone.
Answers: (b) 2.602 · (d) b = 0.074 V · (f) 17.4 % (the remaining parts are explanations — see the table above)
74B-1B-49Ohmic & non-ohmic conductorsPaper 1BMedium10 marks
A student tests whether the copper coil of an electromagnet in a door lock obeys Ohm's law. The coil, at 20 °C, is connected to a variable dc supply with an ammeter in series and a voltmeter in parallel with the coil. For each current I the student waits two minutes before reading the pd V. The ammeter has an uncertainty of ±0.005 A and the voltmeter ±0.01 V.
The graph shows R = V/I against I for five of the seven readings.
I / A
V / V
R / Ω
0.200
0.48
0.400
0.98
2.45
0.600
1.50
2.50
0.800
2.07
2.59
1.000
2.70
1.200
3.44
2.87
1.400
4.31
3.08
Graph drawn to scale
(a)
Calculate R for I = 0.200 A and its absolute uncertainty.
(3)
(b)
Calculate R for I = 1.000 A and plot the points for (a) and (b) on the graph, without error bars.
(1)
(c)
Draw the curve of best fit for the data.
(1)
(d)
Determine whether the coil obeys Ohm's law over the whole range of currents.
(2)
(e)
The resistance of copper increases by 0.39 % of its value at 20 °C for every kelvin rise in temperature. Use the graph to estimate the temperature of the coil when I = 1.400 A.
(2)
(f)
Suggest one change to the procedure that would allow the student to test whether the copper itself obeys Ohm's law.
MP3 is for matching the precision of value and uncertainty. Accept (2.40 ± 0.11) Ω.
Part (b)
2.70 «Ω» AND both points plotted correctly
✓ 1
Within half a small square. Allow ECF from (a).
Part (c)
smooth curve through the error bars, almost level at small currents and rising increasingly steeply; not dot-to-dot
✓ 1
Part (d)
at 0.200 A, R = (2.4 ± 0.1) Ω; at 1.400 A, R = (3.08 ± 0.02) Ω
✓ 1
Any two readings whose ranges do not overlap.
the ranges do not overlap, so R is not constant and the coil does not obey Ohm's law over the whole range
✓ 1
MP2 only scores if MP1 scores. Accept 'obeys Ohm's law only for currents up to about 0.6 A'.
Part (e)
R0 ≈ 2.40 Ω from the curve at I = 0; increase = 3.08/2.40 − 1 = 0.283 «28 %»
✓ 1
Accept R0 from 2.38 to 2.42 Ω.
ΔT = 28.3/0.39 ≈ 72 K, so the temperature ≈ 92 °C
✓ 1
Accept 90–95 °C. Award [1 max] for a temperature rise given as the answer.
Part (f)
keep the coil at constant temperature, e.g. immerse it in a stirred water/oil bath, or take each reading immediately after switching on «and switch off between readings»
✓ 1
OWTTE.
Answers: (a) (2.4 ± 0.1) Ω · (b) 2.70 Ω · (e) ≈ 92 °C (the remaining parts are explanations — see the table above)
75B-1B-10Thermistors, LDRs & dividersPaper 1BHard11 marks
The resistance R of a negative temperature coefficient (NTC) thermistor is measured with an ohmmeter while the thermistor is in a stirred water bath at temperature θ. The manufacturer states that R = R0 eB/T, where T is the absolute temperature and R0 and B are constants.
The student plots ln(R/kΩ) against 1/T. The graph shows six of the seven points.
θ / °C
R / kΩ
(1/T) / 10−3 K−1
ln(R / kΩ)
10.0
20.3
3.534
3.011
20.0
12.5
3.413
2.526
30.0
8.06
3.300
2.087
40.0
5.30
50.0
3.57
3.096
1.273
60.0
2.50
3.003
0.916
70.0
1.75
2.915
0.560
Graph drawn to scale
(a)
Calculate the two missing values in the table for θ = 40.0 °C and plot the point on the graph.
(2)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine B, giving its unit.
(2)
(d)
The uncertainty in the gradient is ±3 %. The manufacturer quotes B = 3950 K ± 1 %. Comment on the agreement.
(1)
(e)
The thermistor is connected in series with a fixed 4.70 kΩ resistor across a 5.00 V supply of negligible internal resistance. Calculate the p.d. across the fixed resistor when the thermistor is at 40.0 °C.
(2)
(f)
State and explain how the p.d. across the fixed resistor changes as the temperature rises.
(2)
(g)
The ohmmeter passes a small current through the thermistor. Suggest how this could cause a systematic error in R.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
1/T = 1/313 K = 3.195 × 10−3 K−1
✓ 1
Temperature must be converted to kelvin. Accept 3.19 × 10−3.
ln(R/kΩ) = ln 5.30 = 1.668 AND point plotted
✓ 1
Accept 1.67.
Part (b)
single straight line of best fit through the points
✓ 1
Part (c)
ln R = ln R0 + B(1/T) so B = gradient, allowing for the 10−3 in the axis
✓ 1
B = 3.96 × 103 K
✓ 1
Accept 3.8 × 103–4.1 × 103 K. Unit needed. Award [1] max for 3.9 «power of ten missing».
Part (d)
the student’s range (3840–4070 K) overlaps the manufacturer’s range (3910–3990 K) so the values agree
✓ 1
Allow ECF from (c).
Part (e)
V = 5.00 × 4.70 / (4.70 + 5.30)
✓ 1
Uses the table value of R at 40 °C.
= 2.35 V
✓ 1
Accept 2.3–2.4 V.
Part (f)
the resistance of the thermistor decreases «NTC»
✓ 1
so the fixed resistor takes a larger share of the supply p.d.: the p.d. across it increases
✓ 1
Part (g)
the current heats the thermistor above the bath temperature, so R is lower than the value at the recorded temperature
✓ 1
OWTTE.
Answers: (a) 3.195 × 10−3 K−1, 1.668 · (c) 3.96 × 103 K · (e) 2.35 V (the remaining parts are explanations — see the table above)
76B-1B-18emf & internal resistancePaper 1BHard10 marks
A zinc strip and a copper strip are pushed into a lemon to make a cell. The cell is connected in series with a resistance box of resistance R and a digital microammeter. No voltmeter is used. The current I is recorded for several values of R.
The graph shows 1/I against R for five of the six readings.
R / kΩ
I / μA
(1/I) / mA−1
1.0
278.1
3.60
2.0
211.9
4.72
3.0
174.9
5.72
5.0
126.5
7.0
100.7
9.93
10.0
75.4
13.26
Graph drawn to scale
(a)
Show that, for a cell of emf ε and internal resistance r, 1/I = R/ε + r/ε.
(1)
(b)
Calculate the missing value of 1/I and plot the point on the graph.
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine ε.
(2)
(e)
Determine r, assuming the microammeter is ideal.
(2)
(f)
The microammeter has a resistance of 0.10 kΩ. Explain how this affects your answer to (e) and state a corrected value of r.
(2)
(g)
The current from the lemon cell slowly decreases while the circuit is connected. Suggest one change to the procedure that would reduce the effect of this on the results.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
ε = I(R + r) ⇒ 1/I = (R + r)/ε = R/ε + r/ε
✓ 1
Must see the starting equation and the division by εI.
Part (b)
1000 / 126.5 = 7.91 «mA−1» AND plotted
✓ 1
Accept 7.9.
Part (c)
single straight line of best fit through the points
✓ 1
Part (d)
gradient = 1/ε ≈ 1.07 «mA−1 kΩ−1 =» V−1
✓ 1
Accept 1.04–1.10.
ε = 0.94 V
✓ 1
Accept 0.91–0.97 V. Allow ECF.
Part (e)
intercept = r/ε ≈ 2.54 mA−1
✓ 1
Accept 2.39–2.69.
r = intercept × ε = 2.4 kΩ
✓ 1
Allow ECF from (d) and their intercept.
Part (f)
the total resistance in the circuit is R + r + RA, so the intercept gives r + RA and (e) overestimates r «a systematic error»
✓ 1
r = 2.4 − 0.1 = 2.3 kΩ
✓ 1
Allow ECF from (e).
Part (g)
disconnect the circuit between readings / take each reading quickly after connecting / take readings in a random or reversed order and average
✓ 1
OWTTE.
Answers: (b) 7.91 mA−1 · (d) 0.94 V · (e) 2.4 kΩ · (f) 2.3 kΩ (the remaining parts are explanations — see the table above)
A student uses a manufacturer's datasheet for a 12 V sealed lead–acid battery labelled “7.0 A h at the 20 h rate”. The datasheet gives the time t for which a fully charged battery can deliver a constant current I before its terminal pd falls to 10.5 V. The manufacturer states that each time is correct to within ±5 %.
The student suggests that t = aI−k, where a and k are constants, and plots lg t against lg I. The graph shows six of the seven points.
I / A
t / h
lg(I / A)
lg(t / h)
0.350
20.0
−0.456
1.301
0.650
9.63
−0.187
0.984
1.20
4.51
0.079
0.654
2.10
2.36
4.20
0.989
0.623
−0.005
7.00
0.555
0.845
−0.256
14.0
0.237
1.146
−0.625
Graph drawn to scale
(a)
Show that the charge delivered at the 20 h rate is about 2.5 × 104 C.
(1)
(b)
Calculate the two missing values in the table for I = 2.10 A and plot the point on the graph.
(2)
(c)
Show that the error bar on each value of lg t is about ±0.02.
(1)
(d)
Draw the line of best fit for the data.
(1)
(e)
Determine k.
(2)
(f)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in k.
(2)
(g)
A customer claims that the battery always delivers the same total charge, whatever the current. Comment on this claim, using your answers to (e) and (f).
(1)
(h)
Determine, as a percentage of the 7.0 A h on the label, the charge delivered at a current of 7.00 A.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
q = It = 0.350 × 20 × 3600 = 2.52 × 104 C
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)
lg I = 0.322 AND lg t = 0.373
✓ 1
Accept 2 to 4 decimal places.
point plotted correctly
✓ 1
Allow ECF from the calculated values. Within half a small square.
Part (c)
lg 1.05 = 0.021 OR ½(lg 1.05 − lg 0.95) = 0.022
✓ 1
Must see a calculation, not just the statement.
Part (d)
single straight line through all the error bars
✓ 1
Part (e)
gradient = −k «lg t = lg a − k lg I», found from a large triangle
✓ 1
k = 1.20
✓ 1
Accept 1.17–1.23. Do not accept a negative value for k.
Part (f)
steepest and shallowest lines that pass through all the error bars
✓ 1
«gradients ≈ −1.23 and −1.18»
Δk = ½(max − min) ≈ 0.03
✓ 1
Accept 0.02–0.05. Allow ECF.
Part (g)
a constant charge means It = constant, so k = 1; the range of k from (f) does not include 1 «the charge falls as the current increases», so the claim is not supported
✓ 1
Allow ECF from (e) and (f). The comparison with k = 1 is needed.
Part (h)
charge = 7.00 × 0.555 = 3.9 A h
✓ 1
Accept a value of t read from the line.
percentage = 3.9/7.0 × 100 ≈ 56 %
✓ 1
Accept 53–59 %. Award [2] for CNA.
Answers: (a) 2.52 × 104 C · (b) 0.322, 0.373 · (e) k = 1.20 · (f) ±0.03 · (h) 56 % (the remaining parts are explanations — see the table above)
78B-2-09Conductors, insulators & resistancePaper 2Easy11 marks
Before a new cable is installed, an electrician tests its insulation. The cable has a copper core surrounded by a layer of PVC plastic. A metal foil is wrapped tightly around the outside of the PVC. An insulation tester applies a direct potential difference V between the copper core and the foil and measures the very small current I that passes through the PVC.
The table gives the results for two cables. Cable 2 has a damaged section into which water has entered. The graph shows the results for cable 1.
V / V
I / μA (cable 1)
I / μA (cable 2)
100
0.40
0.50
250
1.00
1.40
500
2.00
3.40
1000
4.00
11.0
Graph drawn to scale
(a)
State Ohm's law.
(1)
(b)
Deduce, using the data, whether the PVC of cable 1 obeys Ohm's law.
(2)
(c)
A cable is acceptable only if the resistance of its insulation is at least 1.0 × 108 Ω at every test potential difference. Determine whether cable 2 is acceptable.
(2)
(d)
Draw, on the graph, the line of best fit for cable 1. Sketch, on the same axes, the graph of I against V for cable 2.
(2)
(e)
Outline, in terms of charge carriers, why the PVC is an electrical insulator whereas the copper core is a good electrical conductor.
(2)
(f)
Calculate the number of electrons that flow each second along the wire connecting the tester to the copper core of cable 1 when V = 1000 V.
(1)
(g)
Suggest why the tester applies a large potential difference, such as 1000 V, rather than the 1.5 V of a single cell.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the current in a conductor is (directly) proportional to the potential difference across it, provided the temperature «and other physical conditions» stay constant
✓ 1
Both proportionality and the condition needed.
Part (b)
V/I = 2.5 × 108 Ω for every reading OR doubling V doubles I
✓ 1
At least two readings must be tested.
V/I is constant «I ∝ V», so the PVC of cable 1 obeys Ohm's law «is ohmic»
✓ 1
MP2 only scores if MP1 scores.
Part (c)
at 1000 V: R = 1000/(11.0 × 10−6) = 9.1 × 107 Ω
✓ 1
Values at the other p.d.s: 2.0 × 108, 1.8 × 108 and 1.5 × 108 Ω.
9.1 × 107 Ω is less than 1.0 × 108 Ω, so cable 2 is not acceptable
✓ 1
MP2 needs a comparison with a calculated value at 1000 V and a consistent conclusion.
Part (d)
cable 1: a single ruled straight line through the origin and through all four points
✓ 1
cable 2: a curve through the origin, lying above the line for cable 1, whose gradient increases with V and that passes close to (1000 V, 11.0 μA)
✓ 1
Accept the four points of cable 2 plotted and joined by a smooth curve.
Part (e)
copper contains a very large number of free «delocalised» electrons that can drift through the metal when a potential difference is applied
✓ 1
in PVC the electrons are bound to their atoms/molecules, so there are almost no mobile charge carriers «and the current is tiny»
at 1.5 V the current would be only about 6 × 10−9 A, too small to be measured accurately «so a large V gives a measurable current / tests the insulation under the conditions of use»
✓ 1
Accept a reference to the very large resistance requiring a large p.d. for a measurable current.
Answers: (b) 2.5 × 108 Ω, ohmic · (c) 9.1 × 107 Ω — not acceptable · (f) 2.5 × 1013 s−1(the remaining parts are explanations — see the table above)
79B-2-12Thermistors, LDRs & dividersPaper 2Easy10 marks
A wind vane turns the slider S of a rotary potentiometer. The circular track of the potentiometer has a total resistance of 10.0 kΩ and is connected across a supply of emf 5.00 V and negligible internal resistance. The output is the potential difference between S and the end E of the track. The resistance between E and S is proportional to the angle θ through which the vane has turned from north; θ = 360° corresponds to the whole track.
Diagram NOT accurately drawnAxes for (b)
(a)
(i)
Calculate the current in the track.
(1)
(ii)
Show that the output potential difference is about 1.7 V when θ = 120°.
(1)
(iii)
Calculate the power dissipated in the track.
(1)
(b)
Draw, on the axes, a graph of the output potential difference against θ for 0 ≤ θ ≤ 360°.
(2)
(c)
The output is connected to a data logger of very large resistance that accepts a maximum input of 4.00 V. A fixed resistor is connected in series between the supply and the upper end of the track.
(i)
Calculate the resistance of the fixed resistor needed for the output to be 4.00 V when θ = 360°.
(2)
(ii)
Calculate the output potential difference when θ = 270°.
(1)
(iii)
Explain why the output potential difference is still proportional to θ.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
I = V/R = 5.00 / 10000 = 5.0 × 10−4 A
✓ 1
Part (a)(ii)
V = 5.00 × 120/360 = 1.67 V
✓ 1
Must see the fraction of the track OR the answer to 3 s.f.
Part (a)(iii)
P = V2/R = 5.002/10000 = 2.5 × 10−3 W
✓ 1
Accept IV with ECF from (a)(i).
Part (b)
straight line through the origin
✓ 1
reaching 5.00 V at 360° (passing through about 1.7 V at 120°)
✓ 1
Ruled line needed.
Part (c)(i)
current in the track = 4.00/(10.0 × 103) = 4.0 × 10−4 A AND pd across the fixed resistor = 5.00 − 4.00 = 1.00 V
✓ 1
ALTERNATIVE: potential-divider ratio R/10.0 kΩ = 1.00/4.00.
the logger draws negligible current, so the current in the track is the same for every position of S
✓ 1
the resistance between E and S is proportional to θ, so V = IRES is proportional to θ
✓ 1
OWTTE.
Answers: (a)(i) 5.0 × 10−4 A · (a)(ii) 1.67 V · (a)(iii) 2.5 × 10−3 W · (c)(i) 2.5 kΩ · (c)(ii) 3.00 V (the remaining parts are explanations — see the table above)
A portable power bank contains lithium-ion cells with a combined emf of 3.7 V. It is labelled “10 000 mA h”, which means that when fully charged it can deliver a current of 10 000 mA for 1 hour.
(a)
State what is meant by the emf of a cell.
(1)
(b)
(i)
Show that the charge that the fully charged power bank can deliver is about 3.6 × 104 C.
(1)
(ii)
Calculate the energy stored in the fully charged power bank.
(2)
(c)
The power bank is recharged by a portable solar panel of area 0.060 m² and efficiency 18 %, in sunlight of intensity 650 W m−2. Of the electrical energy produced by the panel, 85 % ends up stored in the power bank.
(i)
Calculate the electrical power output of the solar panel.
(1)
(ii)
Determine whether 5.0 hours of this sunlight is enough to charge the power bank fully from empty.
(3)
(d)
Later in the day the Sun's rays make an angle of 30° with the plane of the panel. The intensity of the sunlight, measured perpendicular to the rays, is still 650 W m−2. Determine the electrical power output of the panel now.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the energy transferred «from chemical to electrical energy» per unit charge passing through the cell
✓ 1
Accept “work done per unit charge by the cell”. Do not accept “the voltage of the cell”.
Part (b)(i)
q = It = 10.0 × 3600 = 3.6 × 104 C
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
W = qε = 3.6 × 104 × 3.7
✓ 1
W = 1.3 × 105 J
✓ 1
Award [2] for CNA.
Part (c)(i)
P = 0.060 × 650 × 0.18 = 7.0 W
✓ 1
Part (c)(ii)
energy produced in 5.0 h = 7.0 × 18 000 = 1.26 × 105 J
✓ 1
Allow ECF from (c)(i).
energy stored = 0.85 × 1.26 × 105 = 1.07 × 105 J
✓ 1
1.07 × 105 J is less than 1.33 × 105 J, so it is not enough
✓ 1
ALTERNATIVE: time needed = 1.33 × 105/(0.85 × 7.0) = 2.2 × 104 s ✓✓ ≈ 6.2 h, longer than 5.0 h ✓. MP3 requires a comparison. Allow ECF from (b)(ii).
Part (d)
area of the panel projected perpendicular to the rays = 0.060 × sin 30° = 0.030 m²
✓ 1
OWTTE: the panel intercepts half as much radiation.
P = 0.030 × 650 × 0.18 = 3.5 W
✓ 1
Award [1 max] for 6.1 W «cos 30° used».
Answers: (b)(i) 3.6 × 104 C · (b)(ii) 1.3 × 105 J · (c)(i) 7.0 W · (c)(ii) 1.07 × 105 J stored: no · (d) 3.5 W (the remaining parts are explanations — see the table above)
81B-2-52Ohmic & non-ohmic conductorsPaper 2Easy13 marks
A platinum resistance thermometer contains a coil of fine platinum wire. Its resistance is 100.0 Ω at 0 °C and increases by 0.385 Ω for each kelvin rise in temperature.
A supply keeps the current in the coil at 1.00 mA, and an ideal voltmeter measures the potential difference across the coil.
Axes for (e)
(a)
Calculate the resistance of the coil at 37.0 °C.
(1)
(b)
The voltmeter reads 0.1270 V. Determine the temperature of the coil.
(2)
(c)
State whether the platinum coil, kept at constant temperature, is an ohmic conductor, and explain why the current in the thermometer is kept small.
(3)
(d)
The temperature of the coil rises by 0.40 K above that of its surroundings for every 1.0 mW of power dissipated in it. Determine whether, at 37.0 °C, the error caused by this heating would be less than 0.10 K if the current were increased to 5.0 mA.
(3)
(e)
Sketch, on the axes, graphs to show how the resistance varies with temperature from 0 °C to 100 °C for the platinum coil (label it P) and for a negative temperature coefficient (NTC) thermistor that has a resistance of 100 Ω at 25 °C (label it N).
(2)
(f)
Suggest one advantage of the platinum thermometer and one advantage of the thermistor as temperature sensors.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
R = 100.0 + 0.385 × 37.0 = 114.2 Ω
✓ 1
Accept 114 Ω.
Part (b)
R = V/I = 0.1270 / 1.00 × 10−3 = 127.0 Ω
✓ 1
θ = (127.0 − 100.0)/0.385 = 70.1 °C
✓ 1
Award [2] for CNA.
Part (c)
ohmic: at constant temperature the current is proportional to the potential difference
✓ 1
the current dissipates energy in the coil «P = I2R», which raises its temperature
✓ 1
the coil would then be hotter than the object being measured, giving a reading that is too high; a small current makes this heating negligible
✓ 1
OWTTE
Part (d)
P = I2R = (5.0 × 10−3)2 × 114.2 = 2.86 × 10−3 W
✓ 1
Allow ECF from (a).
temperature rise = 2.86 × 0.40 = 1.14 K
✓ 1
this is more than 0.10 K, so the error would not be less than 0.10 K
✓ 1
MP3 needs a comparison and a consistent conclusion.
Part (e)
P: straight line with positive gradient from 100 Ω at 0 °C to about 138.5 Ω at 100 °C
✓ 1
N: curve decreasing through 100 Ω at 25 °C, with a gradient whose magnitude decreases as θ increases
✓ 1
Do not accept a straight line for N.
Part (f)
platinum: its resistance varies linearly with temperature, so calibration is simple / it can be used over a wide range of temperature
✓ 1
thermistor: its resistance changes much more per kelvin, so it is more sensitive / it is cheaper
✓ 1
Answers: (a) 114.2 Ω · (b) 70.1 °C · (d) 1.14 K — not less than 0.10 K (the remaining parts are explanations — see the table above)
82B-2-83Series & parallel circuitsPaper 2Easy11 marks
A pair of heated cycling gloves is powered by a USB power bank that acts as a source of emf 5.0 V with negligible internal resistance. Each glove contains a heating element made of conductive thread. The two elements are connected in parallel to the power bank and each element is designed to dissipate 3.0 W. A single switch turns both gloves on and off.
Diagram NOT accurately drawn
(a)
Complete the circuit diagram to show the two elements connected in parallel to the power bank, the switch, and an ammeter that measures the total current from the power bank.
(2)
(b)
(i)
Show that the resistance of each element is about 8 Ω.
(1)
(ii)
Calculate the current in the power bank when both gloves are switched on.
(1)
(c)
The thread has a resistance of 12 Ω per metre of its length. Calculate the length of thread needed for one element.
(1)
(d)
The fully charged power bank can deliver a charge of 1.8 × 104 C. Calculate, in hours, the time for which both gloves can be used.
(2)
(e)
The thread in one glove breaks. State and explain the effect on the power dissipated in the other glove.
(2)
(f)
A cyclist suggests connecting the two elements in series instead, so that the power bank lasts twice as long. Evaluate this suggestion.
(2)
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Notes
Part (a)
the two elements connected in parallel with each other across the power bank terminals
✓ 1
Ignore the positions of the switch and ammeter for MP1.
switch AND ammeter in series with the power bank, in the part of the circuit that carries the total current
✓ 1
Correct circuit symbols needed for the switch and ammeter. Do not award MP2 if either is in one branch only.
Part (b)(i)
R = V2/P = 5.02/3.0 = 8.33 Ω
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
I = 2 × 3.0/5.0 = 1.2 A
✓ 1
Accept 2 × 5.0/8.33. Allow ECF from (b)(i).
Part (c)
L = 8.33/12 = 0.69 m
✓ 1
Allow ECF from (b)(i). Accept 0.67 m from 8 Ω.
Part (d)
t = Q/I = 1.8 × 104/1.2 = 1.5 × 104 s
✓ 1
Allow ECF from (b)(ii).
= 4.2 h
✓ 1
Award [2] for CNA. Award [1 max] for an answer in seconds only.
Part (e)
the power in the other glove is unchanged «3.0 W»
✓ 1
MP1 only scores if MP2 scores.
each element is connected directly across the power bank, which has negligible internal resistance, so the pd across the other element is still 5.0 V «and its current is unchanged»
✓ 1
OWTTE.
Part (f)
in series the current is halved to 0.30 A, so the power in each element falls to a quarter, 0.75 W
✓ 1
Accept P = V2/R with 2.5 V across each element.
the power bank would last four times as long «about 17 h», not twice, but each glove would give only a quarter of the heating, so the gloves would probably be too cold «the suggestion is not a good one»
✓ 1
A judgement supported by the change in power is needed for MP2.
Answers: (b)(i) 8.33 Ω · (b)(ii) 1.2 A · (c) 0.69 m · (d) 4.2 h (the remaining parts are explanations — see the table above)
83B-2-86Thermistors, LDRs & dividersPaper 2Easy11 marks
In a car fuel gauge, a float on the fuel moves the slider of a potentiometer track that is used as a variable resistor. Its resistance varies linearly with the fraction of the tank filled, from 110 Ω when the tank is empty to 10 Ω when it is full. The track is in series with the gauge, a meter of resistance 40 Ω whose needle deflection is proportional to the current in it, and a battery of emf 12.0 V and negligible internal resistance.
Diagram NOT accurately drawnAxes for (c)
(a)
Calculate the current in the gauge when the tank is full and when it is empty.
(2)
(b)
Show that the current in the gauge is 0.12 A when the tank is half full.
(1)
(c)
Sketch, on the axes, the variation of the current in the gauge with the fraction of the tank filled.
(2)
(d)
Explain why the marks on the scale of the gauge for equal fractions of a tankful are not evenly spaced.
(2)
(e)
Calculate the power dissipated in the track when the tank is empty.
(1)
(f)
When the engine is running the emf of the battery rises to 14.4 V. The gauge was calibrated at 12.0 V. Determine the fraction of the tank that the gauge indicates when the tank is actually half full.
(3)
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Notes
Part (a)
full: I = 12.0/(40 + 10) = 0.240 A
✓ 1
empty: I = 12.0/(40 + 110) = 0.080 A
✓ 1
Part (b)
track = 60 Ω, so I = 12.0/(40 + 60) = 0.120 A
✓ 1
Must see the track resistance of 60 Ω.
Part (c)
curve rising from 0.08 A at 0 to 0.24 A at 1.0, through about 0.12 A at 0.5
✓ 1
Allow ECF from (a).
gradient increasing with the fraction filled «curve below the straight line joining the end points»
✓ 1
Do not accept a straight line.
Part (d)
equal changes in fuel give equal changes in the resistance of the track, but I = ε/(Rgauge + Rtrack) is not proportional to the resistance «it is inversely proportional to the total resistance»
✓ 1
so the current changes by more for each equal step in fuel when the tank is nearly full than when it is nearly empty «the marks are crowded at the empty end»
✓ 1
OWTTE.
Part (e)
P = I2R = 0.0802 × 110 = 0.70 W
✓ 1
Allow ECF from (a).
Part (f)
I = 14.4/100 = 0.144 A
✓ 1
at 12.0 V this current needs a total resistance of 12.0/0.144 = 83.3 Ω, i.e. a track resistance of 43.3 Ω
✓ 1
fraction indicated = (110 − 43.3)/100 = 0.67 «about two-thirds of a tank»
✓ 1
Award [3] for CNA. Accept reading the fraction for 0.144 A from the candidate's graph in (c) for MP2 and MP3.
Answers: (a) 0.240 A and 0.080 A · (e) 0.70 W · (f) 0.67 (the remaining parts are explanations — see the table above)
84B-2-88Conductors, insulators & resistancePaper 2Easy11 marks
A technician maintains a battery storage unit at a solar farm. Its terminals are at a potential difference of 600 V (dc), and its negative terminal is connected to earth. The resistance of the technician's body between a hand and the ground is 2.0 × 103 Ω. A current greater than 30 mA through the body is dangerous. The technician wears rubber gloves.
(a)
Outline, in terms of charge carriers, why rubber is an electrical insulator whereas sweat, which is salty water, is a conductor.
(2)
(b)
Calculate the current through the technician's body if a bare hand touches the positive terminal while the technician stands on the ground.
(1)
(c)
The rubber of the glove has a resistivity of 1.0 × 1013 Ω m and a thickness of 2.0 mm. The area of the glove in contact with the terminal is 4.0 × 10−4 m2.
(i)
Show that the resistance of the rubber between the terminal and the hand is about 5 × 1013 Ω.
(1)
(ii)
Determine whether the technician is safe when touching the terminal with the gloved hand.
(2)
(d)
The glove has a pinhole, which fills with sweat of resistivity 0.80 Ω m. The sweat forms a column of length 2.0 mm and cross-sectional area 2.0 × 10−7 m2 through the rubber.
(i)
Explain why the resistance of the rubber around the pinhole can be ignored.
(1)
(ii)
Determine whether the technician is now safe.
(3)
(e)
A 12 V car battery can deliver a current of several hundred amperes. Suggest why touching its terminals is not dangerous for the technician.
(1)
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Notes
Part (a)
in rubber the electrons are bound to their atoms/molecules, so there are «almost» no mobile charge carriers
✓ 1
Do not accept 'rubber has no electrons'.
sweat contains dissolved ions «Na+, Cl−» that are free to move through the liquid and carry charge
Must see full substitution OR answer to at least 3 s.f.
Part (c)(ii)
I = 600/(5.0 × 1013 + 2.0 × 103) ≈ 1.2 × 10−11 A
✓ 1
The body resistance may be ignored.
this is far less than 30 mA, so the technician is safe
✓ 1
MP2 requires a comparison and a consistent conclusion.
Part (d)(i)
the rubber is in parallel with the sweat column and its resistance is about 6 × 109 times larger, so the combined resistance is «almost exactly» that of the sweat «almost all the current passes through the pinhole»
✓ 1
OWTTE.
Part (d)(ii)
resistance of the sweat column = 0.80 × 2.0 × 10−3/2.0 × 10−7 = 8.0 × 103 Ω
✓ 1
I = 600/(8.0 × 103 + 2.0 × 103) = 0.060 A «60 mA»
✓ 1
Allow ECF. Award [2 max] for 0.075 A «body resistance omitted» with a consistent conclusion.
this is greater than 30 mA, so the technician is not safe
✓ 1
MP2 requires a comparison and a consistent conclusion.
Part (e)
the current through the body is limited by the body's resistance: 12/2.0 × 103 = 6 mA, well below 30 mA «the battery's ability to supply large currents is irrelevant»
✓ 1
OWTTE.
Answers: (b) 0.30 A · (c)(ii) 1.2 × 10−11 A, safe · (d)(ii) 8.0 × 103 Ω; 60 mA, not safe (the remaining parts are explanations — see the table above)
86IBP2S-016-01-11emf & internal resistancePaper 2Easy6 marks
An electric bicycle is powered by a battery of emf 36 V and internal resistance 0.15 Ω. When the bicycle climbs a hill, the motor draws a steady current of 12 A from the battery.
(a)
Calculate the potential difference across the battery terminals.
(2)
(b)
Calculate the power delivered to the motor.
(2)
(c)
Determine the percentage of the total power generated by the battery's emf that is delivered to the motor.
(2)
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Notes
Part (a)
Lost volts: Ir = 12 × 0.15 = 1.8 V
✓ 1
V = ε − Ir = 36 − 1.8 = 34.2 V
✓ 1
Part (b)
P = VI = 34.2 × 12, ECF from (a)
✓ 1
P ≈ 410 W
✓ 1
Part (c)
Total power = εI = 36 × 12 = 432 W, so fraction = 410/432, or V/ε = 34.2/36, ECF from (a) or (b)
✓ 1
= 0.95, i.e. 95% (accept 94% from rounded values)
✓ 1
Exam tip: The fraction of power delivered to the external circuit reduces to V/ε, because the same current passes through both the internal and the external resistance.
87IBP2S-016-01-12Ohmic & non-ohmic conductorsPaper 2Easy7 marks
A student investigates a small component taken from a toy torch. The table shows the current I in the component for different potential differences V across it.
V / V: 0, 1.0, 2.0, 3.0, 4.0 I / A: 0, 0.20, 0.32, 0.40, 0.46
Data from the table
(a)
State what is meant by an ohmic conductor.
(1)
(b)
Calculate the resistance of the component when the potential difference is 1.0 V and when it is 4.0 V.
(2)
(c)
State, with a reason, whether the component is ohmic.
(2)
(d)
Suggest what the component is, and explain the trend in its resistance.
(2)
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Notes
Part (a)
A conductor in which the current is directly proportional to the pd across it, i.e. the resistance is constant, at constant temperature
✓ 1
Part (b)
R = 1.0/0.20 = 5.0 Ω
✓ 1
R = 4.0/0.46 = 8.7 Ω
✓ 1
Part (c)
Not ohmic
✓ 1
Its resistance (V/I) increases as V increases, or equivalently the current is not proportional to the pd, ECF from (b)
✓ 1
Part (d)
A filament lamp
✓ 1
A larger current raises the filament temperature, so the lattice ions vibrate more and collisions with the electrons are more frequent, which increases the resistance
✓ 1
Common error: Resistance is V/I at a point, not the gradient of the I–V graph. Using the gradient gives the wrong values for a non-ohmic component.
88IBP2S-016-00-11Series & parallel circuitsPaper 2Easy7 marks
A pair of heated ski-boot insoles is powered by a rechargeable battery. The battery has emf 7.4 V and internal resistance 0.20 Ω. The heating element has a resistance of 3.5 Ω.
Circuit diagram
(a)
Calculate the current in the heating element.
(2)
(b)
Calculate the terminal potential difference of the battery.
(1)
(c)
Calculate the power dissipated in the heating element.
(2)
(d)
The battery can deliver a charge of 9.0 × 103 C before it needs recharging. Determine the time for which the insoles can be heated.
(2)
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Notes
Part (a)
I = ε/(R + r) = 7.4/(3.5 + 0.20)
✓ 1
I = 2.0 A
✓ 1
Part (b)
V = ε − Ir = 7.4 − 2.0 × 0.20 = 7.0 V, or V = IR = 2.0 × 3.5 = 7.0 V; ECF from (a)
✓ 1
Part (c)
P = I²R or P = VI = 2.02 × 3.5
✓ 1
P = 14 W; ECF from (a) and (b)
✓ 1
Part (d)
t = Q/I = 9.0 × 103/2.0
✓ 1
t = 4.5 × 103 s (75 minutes); ECF from (a)
✓ 1
Exam tip: Include the internal resistance in the total resistance when finding the current, and only then use the current for the other parts.
A hot-wire foam cutter used in a model-making workshop has a nichrome wire of length 0.45 m and diameter 0.30 mm. The resistivity of nichrome is 1.1 × 10−6 Ω m.
(a)
Show that the resistance of the wire is about 7 Ω.
(3)
(b)
The wire is connected to a 12 V supply of negligible internal resistance. Calculate the power dissipated in the wire.
(2)
(c)
As the wire heats up, the resistivity of nichrome increases slightly. Explain the effect on the power dissipated in the wire.
(2)
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Notes
Part (a)
A = π(0.15 × 10−3)2 = 7.07 × 10−8 m2
✓ 1
R = ρL/A = 1.1 × 10−6 × 0.45 / 7.07 × 10−8
✓ 1
R = 7.00 Ω, giving at least one more significant figure than the value shown
✓ 1
Part (b)
P = V2/R = 122/7.0
✓ 1
P = 20.6 W (≈ 21 W); allow 20–21 W; ECF from (a)
✓ 1
Part (c)
The resistance R increases because R is proportional to ρ for fixed L and A
✓ 1
With V constant, P = V2/R, so the power decreases (slightly)
✓ 1
Common error: Using the diameter instead of the radius in A = πr² makes the area four times too large and the resistance four times too small.
91IBP2S-016-02-12Circuit diagrams & metersPaper 2Easy6 marks
A hospital blood-warmer contains a nichrome heating element. A technician checks the element using an ammeter and a voltmeter connected to a dc supply.
(a)
Outline how the ammeter and the voltmeter should be connected. Explain why the voltmeter should have a very high resistance.
(2)
(b)
The voltmeter reads 24.0 V and the ammeter reads 1.60 A. Calculate the resistance of the element.
(1)
(c)
The element is a nichrome wire of diameter 0.40 mm. The resistivity of nichrome is 1.1×10−6 Ω m. Determine the length of the wire.
(3)
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Notes
Part (a)
Ammeter in series with the element, and voltmeter in parallel across the element
✓ 1
A very high resistance means the voltmeter draws negligible current, so the circuit/readings are not altered
✓ 1
Part (b)
R = 24.0/1.60 = 15.0 Ω
✓ 1
Part (c)
A = π(0.20×10−3)2 = 1.26×10−7 m2
✓ 1
L = RA/ρ = 15.0 × 1.26×10−7/1.1×10−6
✓ 1
L = 1.7 m
✓ 1
ECF from (b)
—
Common error: Use the radius, not the diameter, in A = πr², and convert mm to m. Using the diameter makes the length four times too large.
An engineer designs the heating element of a 230 V, 1.00 kW electric heater. The element is to be made from nichrome wire of diameter 0.40 mm. At the working temperature of the element the resistivity of nichrome is 1.10 × 10−6 Ω m.
The heater contains two identical elements of this kind. A switch connects to the supply either one element alone, or the two elements in series, or the two elements in parallel. Assume that the resistance of each element is the same on every setting.
(a)
Outline, in terms of charge carriers, the origin of the electrical resistance of a metal.
(2)
(b)
(i)
Show that the resistance of the element at its working temperature must be about 53 Ω.
(1)
(ii)
Determine the length of nichrome wire needed.
(3)
(c)
(i)
Determine the three power outputs that the heater can provide.
(3)
(ii)
The heater is to be used with an extension cable rated for a maximum current of 6.0 A. Determine the highest setting that can be used safely.
(2)
(iii)
The engineer considers using nichrome wire of twice the diameter. State and explain how the length of wire needed would change.
(2)
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Notes
Part (a)
free electrons (charge carriers) drift through the metal under the potential difference
✓ 1
they collide with the (vibrating) lattice ions, transferring energy to them, which opposes their flow and heats the metal
✓ 1
OWTTE.
Part (b)(i)
R = V2/P = 2302 / 1000 = 52.9 Ω
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
A = π(0.20 × 10−3)2 = 1.26 × 10−7 m2
✓ 1
Award [0] for this mark if the diameter is used as the radius.
L = RA/ρ = 52.9 × 1.26 × 10−7 / 1.10 × 10−6
✓ 1
L = 6.0 m
✓ 1
Accept 6.0–6.1 m. Award [3] for CNA. A diameter–radius error gives 24 m: award [2].
ALTERNATIVE: series — the same p.d. across twice the resistance halves the power; parallel — each element has the full 230 V, so the power doubles.
series: P = 2302/106 = 500 W
✓ 1
Allow ECF from (b)(i).
parallel: P = 2302/26.5 = 2.00 × 103 W; with one element alone 1.00 × 103 W
✓ 1
Award [3] for 500 W, 1.00 kW and 2.00 kW with some working.
Part (c)(ii)
I = P/V: 2.00 kW setting 8.70 A; 1.00 kW setting 4.35 A
✓ 1
Allow ECF from (c)(i). Accept I = V/R with the resistances from (c)(i).
8.70 A > 6.0 A but 4.35 A < 6.0 A, so the highest safe setting is one element alone «1.00 kW»
✓ 1
MP2 requires a comparison of calculated currents with 6.0 A.
Part (c)(iii)
the length must be four times as great
✓ 1
the cross-sectional area ∝ d2 is four times as great, and R = ρL/A must stay the same, so L must also increase four times
✓ 1
Answers: (b)(i) 52.9 Ω · (b)(ii) 6.0 m · (c)(i) 500 W, 1.00 × 103 W, 2.00 × 103 W · (c)(ii) one element «1.00 kW» · (c)(iii) 4 times longer (the remaining parts are explanations — see the table above)
A student coats a steel key with copper. The key and a copper plate are dipped into copper sulfate solution and connected in the circuit shown. The supply has an emf of 6.0 V and negligible internal resistance, and the meters are ideal.
In the solution, charge is carried by positive copper ions Cu2+, each of charge +2e, and by negative sulfate ions. Each copper ion that reaches the key gains two electrons and is deposited on the key as a copper atom. The current is kept constant at 0.250 A for 40.0 minutes.
Molar mass of copper = 0.0635 kg mol−1. Density of copper = 8.96 × 103 kg m−3. Surface area of the key = 18.0 cm2.
Diagram NOT accurately drawn
(a)
Outline how charge flows in the connecting wires and how it flows in the solution.
(2)
(b)
(i)
Calculate the charge that flows through the circuit.
(1)
(ii)
Show that about 1.9 × 1021 copper ions are deposited on the key.
(1)
(iii)
Calculate the mass of copper deposited on the key.
(2)
(iv)
Determine the thickness of the copper coating, assuming that it is uniform.
(2)
(c)
(i)
The voltmeter reads 1.5 V. Calculate the resistance of the fixed resistor R.
(2)
(ii)
Calculate the energy transferred in the solution and electrodes during the coating.
(1)
(iii)
The student now coats a larger key, of twice the surface area, with a coating of the same thickness in the same time. The voltmeter reading stays the same. Determine the resistance that the fixed resistor must now have.
(2)
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Notes
Part (a)
in the wires: free electrons drift «from the negative terminal of the supply towards the key and from the copper plate to the positive terminal»
✓ 1
in the solution: positive «copper» ions move towards the key and negative «sulfate» ions move in the opposite direction
✓ 1
OWTTE. Do not accept 'electrons flow through the solution'.
Part (b)(i)
q = It = 0.250 × 2400 = 600 C
✓ 1
Part (b)(ii)
N = q/2e = 600/(2 × 1.60 × 10−19) = 1.88 × 1021
✓ 1
Must see full substitution OR answer to at least 3 s.f. The factor 2 must be seen.
Award [1 max] for 24 Ω «6.0 V used». Award [2] for CNA.
Part (c)(ii)
W = qV = 600 × 1.5 = 900 J
✓ 1
Allow ECF from (b)(i).
Part (c)(iii)
twice the mass must be deposited in the same time, so the current must double to 0.50 A
✓ 1
R = 4.5/0.50 = 9.0 Ω
✓ 1
Allow ECF from (c)(i). Award [1 max] for 36 Ω «resistance doubled».
Answers: (b)(i) 600 C · (b)(iii) 2.0 × 10−4 kg · (b)(iv) 1.2 × 10−5 m · (c)(i) 18 Ω · (c)(ii) 900 J · (c)(iii) 9.0 Ω (the remaining parts are explanations — see the table above)
94B-2-18Ohmic & non-ohmic conductorsPaper 2Medium13 marks
A 230 V, 60 W filament lamp is replaced by a 230 V LED lamp of power 8.5 W that gives the same light output. When the filament lamp is cold, its resistance is 70 Ω.
Axes for (b)
(a)
(i)
Show that the resistance of the filament at its normal working temperature is about 880 Ω.
(1)
(ii)
Explain why the resistance of the filament at its working temperature is much greater than when it is cold.
(3)
(b)
Sketch, on the axes, the variation of the current I in the filament lamp with the potential difference V across it, for both positive and negative values of V.
(2)
(c)
The lamp is switched on when the filament is cold. Determine the ratio of the current at the instant of switching on to the current in normal operation.
(2)
(d)
(i)
The lamp is used for 5.0 hours every day for a year of 365 days, and electrical energy costs €0.30 per kW h. Calculate the money saved in a year by using the LED lamp instead of the filament lamp.
(3)
(ii)
Deduce how many times more efficient the LED lamp is than the filament lamp, and outline what happens to most of the energy supplied to the filament lamp.
(2)
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Notes
Part (a)(i)
R = V2/P = 2302 / 60 = 882 Ω
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
the current heats the filament to a very high temperature
✓ 1
the lattice ions vibrate with a greater amplitude
✓ 1
so the conduction electrons collide with them more often, which increases the resistance
✓ 1
OWTTE.
Part (b)
curve through the origin whose gradient decreases as V increases (current increases less than in proportion)
✓ 1
the curve for negative V is the same shape, rotated by 180° about the origin (third quadrant)
✓ 1
Part (c)
switch-on current = 230/70 = 3.29 A; normal current = 60/230 = 0.261 A
✓ 1
ALTERNATIVE: ratio = Rhot/Rcold = 882/70.
ratio = 12.6
✓ 1
Accept 12–13. Award [2] for CNA.
Part (d)(i)
power saved = 60 − 8.5 = 51.5 W
✓ 1
energy saved = 0.0515 kW × 1825 h = 94 kW h
✓ 1
saving = 94 × 0.30 = €28
✓ 1
Accept €28.
Part (d)(ii)
same useful output, so efficiency ratio = 60/8.5 ≈ 7
✓ 1
Accept 7.1.
most of the energy supplied to the filament lamp increases the internal energy of the filament and is transferred to the surroundings as thermal energy (infrared radiation)
✓ 1
Answers: (a)(i) 882 Ω · (c) 12.6 · (d)(i) €28 · (d)(ii) about 7 times (the remaining parts are explanations — see the table above)
95B-2-31Series & parallel circuitsPaper 2Medium12 marks
Four identical filament lamps are each rated “6.0 V, 3.0 W”. They are to be connected to a battery of emf 12 V and negligible internal resistance so that every lamp operates at its rated values. The figure shows the battery.
Diagram NOT accurately drawn
(a)
(i)
Calculate the current in a lamp when it operates at its rated values.
(1)
(ii)
Calculate the resistance of a lamp when it operates at its rated values.
(1)
(b)
Draw, on the figure, a circuit diagram to show how the four lamps can be connected to the battery so that each lamp operates at its rated values.
(2)
(c)
(i)
Calculate the current in the battery.
(1)
(ii)
Show that the total resistance of the four lamps is equal to the resistance of one lamp.
(1)
(d)
In the arrangement used, lamps P and Q are in series in one branch, and lamps R and S are in series in a second branch connected in parallel with the first. A fifth identical lamp T is connected between the junction of P and Q and the junction of R and S. State and explain whether lamp T lights.
(2)
(e)
Lamp T is removed. Lamp P is then replaced by a lamp of resistance 6.0 Ω. Assume that the resistance of every lamp stays constant.
(i)
Determine whether lamp Q now operates above its rated power.
(3)
(ii)
Outline why the brightness of lamp R does not change.
(1)
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Notes
Part (a)(i)
I = P/V = 3.0/6.0 = 0.50 A
✓ 1
Part (a)(ii)
R = V/I = 6.0/0.50 = 12 Ω
✓ 1
Accept R = V²/P. Allow ECF from (a)(i).
Part (b)
two lamps in series in each of two branches
✓ 1
Correct lamp symbols needed.
the two branches connected in parallel across the battery
✓ 1
ALTERNATIVE: two pairs of lamps in parallel, the two pairs connected in series ✓✓. Award [0] for all four lamps in series or all four in parallel.
Part (c)(i)
I = 2 × 0.50 = 1.0 A OR I = 4 × 3.0/12 = 1.0 A
✓ 1
Part (c)(ii)
each branch is 24 Ω and two equal branches in parallel give 12 Ω OR 12 V/1.0 A = 12 Ω
✓ 1
Part (d)
T does not light
✓ 1
No mark for a bare statement without reasoning. MP1 only scores if MP2 scores.
each junction is midway between two equal resistances, so both junctions are at the same potential «6.0 V above the negative terminal»; there is no potential difference across T and so no current in it
✓ 1
OWTTE.
Part (e)(i)
current in the branch = 12/(6.0 + 12) = 0.67 A
✓ 1
Allow ECF from (a)(ii).
power in Q = 0.67² × 12 = 5.3 W
✓ 1
5.3 W is greater than 3.0 W, so Q operates above its rated power «and may fail»
✓ 1
ALTERNATIVE: pd across Q = 0.67 × 12 = 8.0 V ✓✓, greater than 6.0 V ✓. MP3 requires a comparison of calculated values.
Part (e)(ii)
R and S form a separate branch connected directly across the battery, which has negligible internal resistance, so the pd across R is still 6.0 V «and its current is unchanged»
✓ 1
OWTTE.
Answers: (a)(i) 0.50 A · (a)(ii) 12 Ω · (c)(i) 1.0 A · (e)(i) 5.3 W > 3.0 W: yes (the remaining parts are explanations — see the table above)
96B-2-34Electrical power & energyPaper 2Medium13 marks
An electric car is charged from a direct-current (dc) fast charger that delivers a power of 50 kW. The charging cable is 4.0 m long and contains two copper conductors, one carrying the current to the car and one returning it. Each conductor has a cross-sectional area of 35 mm². The resistivity of copper is 1.7 × 10−8 Ω m.
The battery of car X charges at a potential difference of 400 V; the battery of car Y charges at 800 V.
(a)
(i)
Show that the total resistance of the two conductors is about 4 mΩ.
(1)
(ii)
Calculate the current in the cable while car X is charging.
(1)
(iii)
Calculate the power dissipated in the cable while car X is charging.
(2)
(iv)
Calculate the charge that flows through the cable in 30 minutes while car X is charging.
(1)
(b)
Deduce, without calculating the current, the power dissipated in the cable while car Y is charging at 50 kW.
(2)
(c)
The cable may safely dissipate no more than 50 W per metre of its length. A new charger delivers 150 kW. Determine whether this cable can be used safely with the new charger for car X and for car Y.
(4)
(d)
Explain why, for the same power loss, car Y can be charged at a given power through a thinner cable than car X.
Must see L = 8.0 m and A in m². Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
I = P/V = 50 000/400 = 125 A
✓ 1
Part (a)(iii)
P = I²R = 125² × 3.9 × 10−3
✓ 1
Allow ECF from (a)(i) and (a)(ii).
P = 61 W
✓ 1
Accept 59–63 W.
Part (a)(iv)
q = It = 125 × 1800 = 2.25 × 105 C
✓ 1
Accept 2.3 × 105 C.
Part (b)
the same power at twice the potential difference needs half the current
✓ 1
P = I²R, so the power dissipated is a quarter: about 15 W
✓ 1
Allow ECF from (a)(iii).
Part (c)
currents at 150 kW: X 375 A; Y 188 A
✓ 1
power dissipated with car X = 375² × 3.9 × 10−3 = 5.5 × 102 W
✓ 1
Allow ECF from (a)(i).
power dissipated with car Y = 188² × 3.9 × 10−3 = 1.4 × 102 W
✓ 1
greatest safe power = 4.0 × 50 = 200 W, so the cable is not safe for car X but is safe for car Y
✓ 1
ALTERNATIVE for MP4: per metre, X 137 W m−1 and Y 34 W m−1, compared with 50 W m−1. MP4 requires both conclusions supported by values.
Part (d)
at twice the potential difference the current is halved, so for the same resistance the power loss I²R is a quarter
✓ 1
R = ρL/A, so the cable resistance can be four times larger, i.e. the cross-sectional area can be a quarter, for the same power loss
✓ 1
Answers: (a)(i) 3.9 × 10−3 Ω · (a)(ii) 125 A · (a)(iii) 61 W · (a)(iv) 2.25 × 105 C · (b) 15 W · (c) X: 550 W > 200 W, no; Y: 140 W < 200 W, yes (the remaining parts are explanations — see the table above)
97B-2-48Thermistors, LDRs & dividersPaper 2Medium12 marks
An automatic garden light is controlled by a potential divider made from a light-dependent resistor (LDR) and a fixed resistor connected to a 6.0 V supply of negligible internal resistance. The light switches on when the output potential difference Vout is 4.0 V or more.
The table gives the resistance R of the LDR at different illuminances E (intensities of the light falling on it), measured in lux (lx).
E / lx
1
3
10
30
100
300
1000
R / kΩ
120
52.6
21.3
9.36
3.79
1.66
0.675
Diagram NOT accurately drawn
(a)
The light must switch on when it becomes dark. Complete the circuit diagram by drawing the circuit symbols for the LDR and the fixed resistor in the gaps X and Y.
(2)
(b)
Outline why the resistance of the LDR decreases as the illuminance increases.
(1)
(c)
Calculate the resistance of the fixed resistor needed for the light to switch on at an illuminance of 10 lx.
(2)
(d)
The designer uses a 10.0 kΩ resistor instead. Determine whether the light is already on when the illuminance has fallen to 10 lx.
(2)
(e)
The output is connected to a switching circuit with an input resistance of 50 kΩ. Determine Vout at an illuminance of 10 lx when the switching circuit is connected and the fixed resistor is 10.0 kΩ.
(3)
(f)
Hence state the effect of connecting the switching circuit on the illuminance at which the light switches on.
(1)
(g)
Suggest one advantage of using a potentiometer instead of the fixed resistor.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
LDR symbol drawn in gap Y «across the output»
✓ 1
The LDR symbol must include the arrows.
fixed resistor symbol drawn in gap X
✓ 1
Part (b)
more light frees more charge carriers «electrons» in the LDR material, so more mobile charge carriers are available to carry current
✓ 1
OWTTE
Part (c)
at 10 lx RLDR = 21.3 kΩ and the pd across the fixed resistor is 2.0 V
✓ 1
R = 21.3 × 2.0/4.0 = 10.7 kΩ
✓ 1
Award [2] for CNA. Award [1] max for 42.6 kΩ (LDR in position X).
Part (d)
Vout = 6.0 × 21.3/(21.3 + 10.0) = 4.08 V
✓ 1
this is more than 4.0 V, so the light is already on
✓ 1
MP2 needs a comparison and a consistent conclusion.
Part (e)
resistance of the LDR in parallel with 50 kΩ = (1/21.3 + 1/50)−1 = 14.9 kΩ
✓ 1
Vout = 6.0 × 14.9/(14.9 + 10.0)
✓ 1
= 3.6 V
✓ 1
Award [3] for CNA. Allow ECF.
Part (f)
the light switches on at a lower illuminance «later in the evening, when it is darker»
✓ 1
Allow ECF from (e).
Part (g)
the illuminance at which the light switches on can be adjusted «without changing components»
✓ 1
OWTTE
Answers: (c) 10.7 kΩ · (d) 4.08 V — yes · (e) 3.6 V (the remaining parts are explanations — see the table above)
98B-2-49emf & internal resistancePaper 2Medium13 marks
The battery pack of a delivery drone contains 12 identical lithium-ion cells, each of emf 3.70 V and internal resistance 0.025 Ω. The cells are connected as two identical strings in parallel; each string consists of 6 cells in series.
Each cell can deliver a charge of 5.0 A h (a current of 5.0 A for 1 hour) before it is discharged.
(a)
(i)
Calculate the emf of the battery pack.
(1)
(ii)
Show that the internal resistance of the battery pack is 0.075 Ω.
(1)
(b)
When the drone hovers, the current in the battery pack is 40 A.
(i)
Calculate the potential difference between the terminals of the pack.
(1)
(ii)
Calculate the efficiency with which the pack transfers energy to the motors.
(2)
(iii)
Determine the maximum time for which the drone can hover.
(2)
(c)
The motor controller shuts down if the terminal potential difference falls below 18.0 V. To climb quickly, the motors need an electrical power of 1.20 kW. Determine whether the pack can supply this power.
(3)
(d)
State and explain one advantage of using two parallel strings of 6 cells rather than a single string of 6 cells.
(2)
(e)
On a very cold day the internal resistance of each cell is larger. Suggest one effect of this on the performance of the drone.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
6 × 3.70 = 22.2 V
✓ 1
Part (a)(ii)
one string: 6 × 0.025 = 0.15 Ω; two equal strings in parallel: 0.15/2 = 0.075 Ω
✓ 1
Both steps must be seen.
Part (b)(i)
V = ε − Ir = 22.2 − 40 × 0.075 = 19.2 V
✓ 1
Part (b)(ii)
useful power = 19.2 × 40 = 768 W and total power = 22.2 × 40 = 888 W
✓ 1
ALTERNATIVE: efficiency = V/ε.
efficiency = 768/888 = 0.865 «86.5 %»
✓ 1
Award [2] for CNA.
Part (b)(iii)
charge available = 2 × 5.0 A h = 10 A h = 3.6 × 104 C
✓ 1
The two strings in parallel each deliver 5.0 A h; the cells in series do not add charge.
t = 3.6 × 104 / 40 = 900 s «15 min»
✓ 1
Award [1] max for 450 s (one string only) or 5400 s (12 × 5.0 A h).
Part (c)
largest current with V ≥ 18.0 V: I = (22.2 − 18.0)/0.075 = 56 A
✓ 1
largest power = 18.0 × 56 = 1.01 × 103 W
✓ 1
this is less than 1.20 kW, so the pack cannot supply the power
✓ 1
ALTERNATIVE: (22.2 − 0.075I)I = 1200 gives I = 71 A and V = 16.9 V, which is below 18.0 V. MP3 needs a consistent conclusion.
Part (d)
the internal resistance is halved / the pack can deliver twice the charge
✓ 1
so less energy is wasted inside the pack and the terminal pd is higher for the same current / so the drone can fly for twice as long
✓ 1
MP2 must explain the advantage stated in MP1.
Part (e)
the terminal pd falls more for the same current, so the controller shuts down sooner / less power is available for climbing / more energy is wasted in the cells
✓ 1
OWTTE
Answers: (a)(i) 22.2 V · (b)(i) 19.2 V · (b)(ii) 86.5 % · (b)(iii) 900 s · (c) maximum 1.01 × 103 W — no (the remaining parts are explanations — see the table above)
The load cell of a digital kitchen scale contains a strain gauge bonded to a metal beam. The gauge is a thin foil of constantan formed into a long zig-zag track. The track has a rectangular cross-section 20 μm wide and 5.0 μm thick, and its resistance is 120 Ω. Resistivity of constantan = 4.9 × 10−7 Ω m.
Circuit for part (c)
(a)
(i)
Calculate the cross-sectional area of the track.
(1)
(ii)
Show that the length of the track is about 24 mm.
(1)
(b)
When a load is placed on the scale, the beam bends and the gauge is stretched, so that the length of the track increases by 0.10 %. The volume of the track does not change. Show that the resistance of the gauge increases by about 0.20 %.
(2)
(c)
The gauge is connected in series with a fixed 120 Ω resistor across a 5.00 V supply of negligible internal resistance. The output is the potential difference across the gauge. Calculate the change in the output when the gauge is stretched as in (b).
(3)
(d)
The increase in length of 0.10 % is produced by a load of mass 5.0 kg, and the increase in length is proportional to the load. The output is measured by a voltmeter with a resolution of 1 mV. Determine whether this arrangement could display the mass of a load to the nearest gram.
(2)
(e)
The resistivity of constantan hardly changes with temperature, but that of copper increases significantly. Suggest why a gauge made of copper would give false readings as the kitchen warms up.
(3)
(f)
In practice the fixed resistor is replaced by an identical gauge bonded to a part of the beam that does not stretch. Suggest why.
(1)
Show mark scheme
Marking point
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Notes
Part (a)(i)
A = 20 × 10−6 × 5.0 × 10−6 = 1.0 × 10−10 m²
✓ 1
Part (a)(ii)
L = RA/ρ = 120 × 1.0 × 10−10 / 4.9 × 10−7 = 2.45 × 10−2 m «24.5 mm»
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)
the volume is constant, so the cross-sectional area decreases by the same factor «A → A/1.0010»
✓ 1
R = ρL/A ∝ L2, so R increases by a factor 1.00102 = 1.0020 «0.20 %»
✓ 1
Award [0] for an answer that changes only the length (0.10 %).
Part (c)
resistance of the stretched gauge = 120 × 1.0020 = 120.24 Ω and the initial output is 2.500 V
✓ 1
new output = 5.00 × 120.24/(120.24 + 120) = 2.5025 V
✓ 1
change = 2.5 × 10−3 V «2.5 mV»
✓ 1
Award [3] for CNA. Award [2] max for 1.25 mV (an increase of 0.10 % used).
Part (d)
change in output per gram = 2.5 × 10−3 / 5000 = 5 × 10−7 V
✓ 1
Allow ECF from (c).
this is far smaller than the 1 mV resolution «even the full 5.0 kg gives only 2.5 mV», so it could not
✓ 1
MP2 needs a comparison and a consistent conclusion.
Part (e)
the resistance of a copper gauge increases when its temperature rises, even with no change in load
✓ 1
this change can be comparable to or larger than the 0.20 % change produced by the load
✓ 1
the scale interprets the change in resistance as a change in load «the zero drifts»
✓ 1
OWTTE
Part (f)
a temperature change affects both gauges equally, so the ratio of their resistances «and the output» changes only because of the strain
✓ 1
OWTTE
Answers: (a)(i) 1.0 × 10−10 m² · (a)(ii) 24.5 mm · (c) 2.5 mV · (d) 5 × 10−7 V per gram — no (the remaining parts are explanations — see the table above)
In the X-ray tube of a medical CT scanner, electrons leave a hot cathode with negligible kinetic energy and are accelerated across a vacuum by a potential difference of 120 kV. They strike a metal anode. The current in the tube is 200 mA. Of the energy carried by the electrons, 99 % becomes internal energy of the anode.
(a)
Calculate the number of electrons that strike the anode each second.
(1)
(b)
Show that the kinetic energy of an electron arriving at the anode is about 1.9 × 10−14 J.
(1)
(c)
In the cable that connects the tube to its supply, the conduction electrons do not gain kinetic energy continuously as they do in the vacuum of the tube. Outline why.
(2)
(d)
A scan lasts 4.0 s. The anode has a mass of 1.8 kg and a specific heat capacity of 150 J kg−1 K−1. Determine the rise in temperature of the anode during the scan, assuming no energy leaves the anode in this time.
(3)
(e)
Between scans, oil is pumped past the tube and removes energy at an average rate of 3.5 kW. The specific heat capacity of the oil is 1.9 × 103 J kg−1 K−1 and its temperature must rise by no more than 12 K. Calculate the minimum mass of oil that must flow past the tube each second.
Must see full substitution OR answer to at least 3 s.f.
Part (c)
in the metal the electrons collide repeatedly with the «vibrating» lattice ions
✓ 1
transferring the energy gained from the field to the lattice «as internal energy», so their drift speed stays small/constant; in the vacuum there are no collisions, so all the work done on them becomes kinetic energy
✓ 1
OWTTE.
Part (d)
power delivered by the electrons = IV = 0.200 × 1.2 × 105 = 2.4 × 104 W
✓ 1
Accept N × Ek from (a) and (b).
Q = 0.99 × 2.4 × 104 × 4.0 = 9.50 × 104 J
✓ 1
Accept use of 100 % «9.6 × 104 J, 356 K».
ΔT = Q/mc = 9.50 × 104/(1.8 × 150) = 352 K
✓ 1
Award [3] for CNA. Accept 350–360 K. Allow ECF.
Part (e)
mass per second = P/(cΔT) = 3500/(1.9 × 103 × 12)
✓ 1
= 0.15 kg s−1
✓ 1
Award [2] for CNA. Accept 0.15–0.16 kg s−1.
Part (f)
so that the electrons do not collide with gas molecules «losing energy and ionising the gas» and reach the anode with energy eV
✓ 1
OWTTE.
Answers: (a) 1.25 × 1018 s−1 · (b) 1.92 × 10−14 J · (d) 352 K · (e) 0.15 kg s−1(the remaining parts are explanations — see the table above)
101B-2-87Chemical & solar cellsPaper 2Medium11 marks
A solar racing car has a solar array of area 4.0 m2 and efficiency 24 %, and a lithium-ion battery of emf 125 V. On a level road the car travels at a constant speed of 22.0 m s−1 against a total resistive force of 75 N. The motor and its controller transfer electrical energy to useful mechanical work with an efficiency of 95 %. The sunlight has an intensity of 1000 W m−2 at the array. The array and the battery together supply the motor.
(a)
Calculate the electrical power output of the array.
(1)
(b)
Show that the electrical power needed by the motor is about 1.7 kW.
(1)
(c)
The battery supplies the rest of the power needed. The pd across its terminals is then 123.4 V.
(i)
Show that the current in the battery is about 6.3 A.
(1)
(ii)
Calculate the internal resistance of the battery.
(2)
(d)
The battery can transfer 5.0 kW h of chemical energy before it is flat. Determine whether the car can complete a 400 km stage at this speed, assuming the intensity of the sunlight stays the same.
(3)
(e)
The array consists of strings of solar cells connected in series. Explain why a shadow falling on just one cell of a string reduces the power output of the whole string.
(2)
(f)
Suggest why the assumption made in (d) about the sunlight makes your conclusion unreliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
P = 4.0 × 1000 × 0.24 = 960 W
✓ 1
Part (b)
P = Fv/0.95 = 75 × 22.0/0.95 = 1737 W
✓ 1
Must see full substitution OR answer to at least 3 s.f. Award [0] for 1650 W alone.
Part (c)(i)
I = (1737 − 960)/123.4 = 6.30 A
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (c)(ii)
pd across the internal resistance = 125 − 123.4 = 1.6 V
✓ 1
r = 1.6/6.30 = 0.25 Ω
✓ 1
Award [2] for CNA. Allow ECF from (c)(i).
Part (d)
time = 4 × 105/22.0 = 1.82 × 104 s «5.05 h»
✓ 1
chemical energy used = εIt = 125 × 6.30 × 1.82 × 104 = 1.43 × 107 J = 4.0 kW h
✓ 1
Accept VIt = 3.9 kW h. Allow ECF from (c)(i).
this is less than 5.0 kW h, so the car can complete the stage
✓ 1
MP3 requires a comparison of calculated values and a consistent conclusion.
Part (e)
cells in series carry the same current
✓ 1
the shaded cell can produce only a small current «it acts as a large resistance», which limits the current in every cell of the string, so the power of the whole string falls
✓ 1
OWTTE.
Part (f)
the intensity changes with the time of day, the angle of the Sun and cloud cover, so the array output «and hence the energy the battery must provide» changes
✓ 1
OWTTE. Accept that the race lasts more than 5 hours, so the Sun's position changes significantly.
Answers: (a) 960 W · (b) 1740 W · (c)(i) 6.30 A · (c)(ii) 0.25 Ω · (d) 4.0 kW h, yes (the remaining parts are explanations — see the table above)
102B-2-89Ohmic & non-ohmic conductorsPaper 2Medium11 marks
The rear light of a bicycle uses a red light-emitting diode (LED) in series with a resistor. The battery has an emf of 4.5 V and negligible internal resistance. The LED is designed to operate with a current of 20 mA. The graph shows how the current I in the LED varies with the potential difference V across it.
Graph drawn to scale
(a)
Show, using the graph, that the resistance of the LED is about 100 Ω when the current is 20 mA.
(1)
(b)
Explain, with reference to the graph, whether the LED is an ohmic conductor.
(2)
(c)
Calculate the resistance of the series resistor needed for the LED to operate at 20 mA.
(2)
(d)
Calculate the fraction of the power supplied by the battery that is dissipated in the LED.
(2)
(e)
As the battery runs down its emf falls to 3.9 V; the resistor from (c) is unchanged. By drawing a suitable line on the graph, determine the current in the LED.
(3)
(f)
Suggest, with reference to the graph, why the LED must not be connected directly across the 4.5 V battery.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
at 20 mA, V = 2.00 V, so R = 2.00/0.020 = 100 Ω
✓ 1
Accept V read as 1.95–2.05 V «98–103 Ω». Must see the value read from the graph.
Part (b)
the graph is not a straight line through the origin: the current is almost zero below about 1.7 V and then rises steeply
✓ 1
so V/I is not constant «e.g. about 380 Ω at 5 mA and 100 Ω at 20 mA»: the LED is non-ohmic
✓ 1
MP2 only scores if MP1 scores. Accept any two correctly calculated values of V/I.
Part (c)
pd across resistor = 4.5 − 2.00 = 2.50 V
✓ 1
Allow ECF from (a).
R = 2.50/0.020 = 125 Ω
✓ 1
Award [2] for CNA. Award [1 max] for 225 Ω «emf/current» or 100 Ω.
Part (d)
PLED = 2.00 × 0.020 = 0.040 W AND Pbattery = 4.5 × 0.020 = 0.090 W
✓ 1
ALTERNATIVE: fraction = VLED/ε.
fraction = 0.44
✓ 1
Award [2] for CNA. Accept 44 %.
Part (e)
straight line meeting the V axis at 3.9 V «I = (3.9 − V)/125»
✓ 1
with gradient −1/125 Ω−1 «−8 mA V−1», e.g. passing through (0, 31 mA) or (2.0 V, 15.2 mA)
✓ 1
Allow ECF from (c).
intersection with the LED curve: I ≈ 15 mA
✓ 1
Accept 14–16 mA. ALTERNATIVE: trial values of V that make (3.9 − V)/125 equal to the current on the graph ✓✓✓.
Part (f)
the curve is very steep: a pd of 4.5 V across the LED would produce an extremely large current «nothing limits the current», which would overheat and destroy the LED
✓ 1
OWTTE.
Answers: (a) 100 Ω · (c) 125 Ω · (d) 0.44 · (e) ≈ 15 mA (the remaining parts are explanations — see the table above)
A high-voltage direct-current (HVDC) overhead line transmits a power of 600 MW at a potential difference of 400 kV over a route of length 150 km. The current flows to the distant end along one conductor and returns along a second, identical conductor. Data for two possible conductor materials at 20 °C are given in the table.
Material
Resistivity / Ω m
Density / kg m−3
aluminium
2.8 × 10−8
2.70 × 103
copper
1.7 × 10−8
8.96 × 103
(a)
Calculate the current in the line.
(1)
(b)
The power dissipated in the two conductors at 20 °C must be no more than 2.0 % of the power transmitted.
(i)
Show that the total resistance of the two conductors must be no more than about 5 Ω.
(1)
(ii)
Determine the minimum cross-sectional area of each conductor if they are made of aluminium.
(2)
(c)
Determine the ratio (mass of copper conductors) / (mass of aluminium conductors) for two sets of conductors that have the same resistance and length.
(3)
(d)
Suggest one reason, other than cost, why aluminium is chosen for overhead lines.
(1)
(e)
On a hot day, while the line still transmits 600 MW at 400 kV, the conductors reach 70 °C. The resistivity of aluminium increases by 0.40 % of its value at 20 °C for each kelvin rise in temperature.
(i)
Outline why the resistivity of a metal increases with temperature.
(1)
(ii)
Determine the power dissipated in the conductors as a percentage of the power transmitted, assuming the conductors have the area found in (b)(ii).
(2)
Show mark scheme
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Mark
Notes
Part (a)
I = P/V = 6.0 × 108/4.0 × 105 = 1500 A
✓ 1
Part (b)(i)
R = 0.020 × 6.0 × 108/15002 = 5.33 Ω
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
total length of conductor = 2 × 150 km = 3.0 × 105 m
✓ 1
Award [1 max] for 7.9 × 10−4 m2 «route length used».
A student tests constantan wire for the heating element of a small 3D-printer bed. The wire has diameter 0.40 mm. The student measures the resistance R for different lengths L.
L / m
0.20
0.40
0.60
0.80
1.00
R / Ω
0.79
1.55
2.35
3.11
3.92
Data from the table
(a)
Outline why a graph of R against L for this wire is expected to be a straight line through the origin.
(2)
(b)
Determine the gradient of the graph of R against L.
(2)
(c)
Calculate the resistivity of constantan.
(2)
(d)
State why the diameter should be measured at several points along the wire.
(1)
Show mark scheme
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Notes
Part (a)
R = ρL/A, and ρ and A are constant for the wire
✓ 1
so R is directly proportional to L, giving a straight line with gradient ρ/A through the origin
✓ 1
Part (b)
Uses a wide range of data, e.g. (3.92 − 0.79)/(1.00 − 0.20)
✓ 1
gradient ≈ 3.9 Ω m−1 (accept 3.85–3.95)
✓ 1
Part (c)
A = π(0.20×10−3)2 = 1.26×10−7 m2
✓ 1
ρ = gradient × A = 3.9 × 1.26×10−7 ≈ 4.9×10−7 Ω m (ECF from (b))
✓ 1
Part (d)
The wire may not be uniform, so averaging gives a better value of A and reduces random error
✓ 1
Common error: Use the radius, not the diameter, in A = πr². Using the diameter makes the resistivity four times too large.
A high-voltage direct-current (HVDC) link uses an aluminium conductor of cross-sectional area 5.0×10−4 m2. It carries a current of 400 A. Aluminium has 1.8×1029 free electrons per cubic metre. (e = 1.6×10−19 C)
(a)
Calculate the drift speed of the electrons in the conductor.
(2)
(b)
The drift speed is very small. Explain why equipment at the far end of the link responds almost instantly when the circuit is completed.
(2)
(c)
The conductor is 20 km long. Estimate the time taken for one electron to drift the full length.
(1)
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Notes
Part (a)
v = I/(nAe) = 400/(1.8×1029 × 5.0×10−4 × 1.6×10−19)
✓ 1
v = 2.8×10−5 m s−1
✓ 1
Part (b)
The conductor is already full of free electrons, and the electric field is set up along the conductor at close to the speed of light
✓ 1
so electrons along the whole length of the conductor begin drifting almost simultaneously
✓ 1
Part (c)
t = L/v = 2.0×104/2.8×10−5 ≈ 7.2×108 s (about 23 years); accept 7.1–7.2×108 s; allow ECF from (a)
✓ 1
Exam tip: Keep two ideas separate: the slow drift of the charge carriers and the fast spread of the electric field (the signal).
106IBP2S-016-03-14Conductors, insulators & resistancePaper 2Medium7 marks
A sensor monitors a saline drip by measuring the electrical resistance of the saline in a plastic tube. The saline column is 0.80 m long, the tube has an internal diameter of 3.0 mm, and the resistivity of saline is 0.70 Ω m.
(a)
Outline how charge is carried in the saline and how this differs from conduction in a copper wire.
(2)
(b)
Show that the resistance of the saline column is about 8×104 Ω.
(2)
(c)
The tube is replaced by one of half the internal diameter, with the same length of saline. State and explain the effect on the resistance.
(2)
(d)
A potential difference of 9.0 V is applied across the original saline column. Calculate the current.
(1)
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Notes
Part (a)
In saline, the charge carriers are positive and negative ions moving in opposite directions
✓ 1
in copper, the charge carriers are free (delocalised) electrons only, while the positive ions stay fixed in the lattice
✓ 1
Part (b)
A = π(1.5×10−3)2 = 7.07×10−6 m2
✓ 1
R = ρL/A = 0.70 × 0.80 / 7.07×10−6 = 7.9×104 Ω
✓ 1
Part (c)
The cross-sectional area becomes one quarter of the original value, since A ∝ d2
✓ 1
R ∝ 1/A, so the resistance becomes four times larger, about 3.2×105 Ω
✓ 1
Part (d)
I = V/R = 9.0 / 7.92×104 = 1.14×10−4 A ≈ 1.1×10−4 A (allow ECF from (b))
✓ 1
Common error: Convert the diameter from mm to m and halve it to get the radius before finding the area.
107IBP2S-016-01-13Electrical power & energyPaper 2Medium7 marks
A spacecraft keeps its instruments warm using three identical heaters, each of resistance 24 Ω. The heaters are connected to a 28 V supply of negligible internal resistance. A switching unit can connect the heaters in different arrangements.
(a)
Calculate the total power delivered when all three heaters are connected in parallel.
(2)
(b)
Calculate the total power delivered when all three heaters are connected in series.
(2)
(c)
In a third arrangement, two heaters are connected in parallel and this combination is connected in series with the third heater. Determine the power dissipated in the third heater.
(3)
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Part (a)
Total resistance = 24/3 = 8.0 Ω
✓ 1
P = V²/R = 28²/8.0 = 98 W
✓ 1
Part (b)
Total resistance = 3 × 24 = 72 Ω
✓ 1
P = 28²/72 = 11 W (10.9 W)
✓ 1
Part (c)
Total resistance = 12 + 24 = 36 Ω
✓ 1
Current = 28/36 = 0.78 A
✓ 1
P = I²R = 0.778² × 24 ≈ 15 W (14.5 W), ECF for the current
✓ 1
Exam tip: To find the power in one component of a mixed network, first find the current in that component or the pd across it.
108IBP2S-016-01-14emf & internal resistancePaper 2Medium7 marks
A cell from a portable medical thermometer is connected to a variable resistor. The table shows the terminal pd V of the cell for different currents I.
I / A: 0.10, 0.20, 0.30, 0.40 V / V: 1.36, 1.22, 1.08, 0.94
Data from the table
(a)
Explain why a graph of V against I for this cell is a straight line with a negative gradient.
(2)
(b)
Determine the internal resistance of the cell.
(2)
(c)
Determine the emf of the cell.
(1)
(d)
Calculate the current if the cell terminals were short-circuited.
(2)
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Notes
Part (a)
ε = V + Ir, so V = −rI + ε
✓ 1
ε and r are constant, so the graph is linear (y = mx + c) with gradient −r, which is negative
✓ 1
Part (b)
Gradient = (0.94 − 1.36)/(0.40 − 0.10) = −1.4 V A−1, or any correct pair of points
✓ 1
r = 1.4 Ω
✓ 1
Part (c)
ε = V + Ir = 1.36 + 0.10 × 1.4 = 1.50 V; also accept the V-intercept found by extrapolating to I = 0; ECF from (b)
✓ 1
Part (d)
I = ε/r, since the external resistance is zero
✓ 1
I = 1.50/1.4 = 1.07 A (accept 1.1 A); ECF from (b) and (c)
✓ 1
Exam tip: Calculate the gradient from points that are far apart in the data. The emf is the intercept on the V-axis, found by extrapolating to I = 0.
109IBP2S-016-00-13emf & internal resistancePaper 2Medium10 marks
A lithium-ion cell from a racing drone is tested by measuring its terminal potential difference V for different currents I. I / A: 2.0 | 4.0 | 6.0 | 8.0 | 10.0 V / V: 3.76 | 3.72 | 3.68 | 3.64 | 3.60
Data from the table
(a)
Outline why the terminal potential difference decreases as the current increases.
(2)
(b)
Determine the internal resistance of the cell.
(2)
(c)
Determine the emf of the cell.
(2)
(d)
The drone uses a pack of four of these cells in series. During a climb the motors draw a current of 25 A. Calculate the rate at which energy is dissipated inside the pack.
(2)
(e)
Determine the efficiency with which the pack transfers energy to the motors during the climb.
(2)
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Notes
Part (a)
The cell has internal resistance, so energy is dissipated inside the cell / there is a pd Ir across r
✓ 1
V = ε − Ir, so a larger I means more 'lost volts' and a smaller V
✓ 1
Part (b)
r = −gradient = ΔV/ΔI, e.g. (3.76 − 3.60)/(10.0 − 2.0)
✓ 1
r = 0.020 Ω
✓ 1
Part (c)
Uses ε = V + Ir with one data pair, e.g. 3.76 + 2.0 × 0.020, or extrapolates to I = 0
✓ 1
ε = 3.80 V; ECF from (b)
✓ 1
Part (d)
Total internal resistance = 4 × 0.020 = 0.080 Ω
✓ 1
P = I²r = 252 × 0.080 = 50 W; ECF from (b)
✓ 1
Part (e)
Terminal pd = 4 × 3.80 − 25 × 0.080 = 13.2 V, or useful power = 15.2 × 25 − 50 = 330 W
✓ 1
Efficiency = 13.2/15.2, or 330/380, = 0.87 (87%); ECF from (c) and (d)
✓ 1
Exam tip: For cells in series, add both the emfs and the internal resistances. Efficiency can then be found directly as V/ε.
110IBP2S-016-00-14Electrical power & energyPaper 2Medium9 marks
A home electric-vehicle charger delivers 7.2 kW at 240 V. It is connected to the household supply by a cable 25 m long that contains two copper conductors, so the current travels through 50 m of copper in total. Each conductor has a cross-sectional area of 6.0 mm2. The resistivity of copper is 1.7 × 10−8 Ω m.
(a)
Calculate the current in the cable.
(1)
(b)
Calculate the total resistance of the copper in the cable.
(2)
(c)
Calculate the power dissipated in the cable.
(2)
(d)
A full charge takes 6.0 hours. Calculate the energy dissipated in the cable during a full charge, in kW h.
(2)
(e)
Explain why chargers with higher power ratings must be connected using cables with a larger cross-sectional area.
(2)
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Notes
Part (a)
I = P/V = 7200/240 = 30 A
✓ 1
Part (b)
R = ρL/A = (1.7 × 10−8 × 50) / (6.0 × 10−6), with the area converted to m2
✓ 1
R = 0.14 Ω (0.142 Ω)
✓ 1
Part (c)
P = I2R = 302 × 0.142
✓ 1
P = 130 W (127.5 W); accept 126–130 W; ECF from (a) and (b)
✓ 1
Part (d)
E = P × t = 0.1275 kW × 6.0 h
✓ 1
E = 0.77 kW h (0.765 kW h); accept 0.75–0.78 kW h; ECF from (c)
✓ 1
Part (e)
A higher power at the same supply pd needs a larger current (I = P/V), and the power loss I2R increases with the square of the current
✓ 1
A larger area gives a smaller resistance (R = ρL/A), which reduces the heating and keeps the cable temperature safe
✓ 1
Common error: 1 mm² = 1 × 10⁻⁶ m², not 10⁻³ m². Also remember that the current flows out and back, so the total conductor length is twice the cable length.
111IBP2S-016-02-13Thermistors, LDRs & dividersPaper 2Medium8 marks
A greenhouse ventilation fan is switched by a potential divider. An NTC thermistor is connected in series with a fixed resistor R across a 6.0 V supply of negligible internal resistance. The output pd is taken across R. The fan switches on when the output pd reaches 3.6 V. The thermistor's resistance RT varies with temperature θ as follows:
Explain why the output pd increases as the temperature in the greenhouse rises.
(2)
(b)
Determine the value of R needed for the fan to switch on at 30 °C.
(2)
(c)
Using your value of R, calculate the output pd when the temperature is 10 °C.
(2)
(d)
The fan's switching circuit is replaced by one with an input resistance similar in size to R. Explain the effect on the temperature at which the fan switches on.
(2)
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Notes
Part (a)
The thermistor resistance decreases as the temperature rises
✓ 1
So R takes a larger share of the 6.0 V / the current increases, so the pd across R (= IR) increases
✓ 1
Part (b)
pd across thermistor = 6.0 − 3.6 = 2.4 V, so R/4.8 = 3.6/2.4
✓ 1
R = 7.2 kΩ
✓ 1
Part (c)
Vout = 6.0 × 7.2/(7.2 + 12.0)
✓ 1
= 2.3 V (2.25 V)
✓ 1
ECF from (b)
—
Part (d)
The input resistance in parallel with R lowers the effective resistance across which the output is taken, so the output pd is lower at each temperature
✓ 1
So the thermistor resistance must fall further, and the fan switches on at a higher temperature
✓ 1
Exam tip: In a potential divider, the pds are in the same ratio as the resistances. Finding the pd across the other component first gives a quick ratio.
112IBP2S-016-02-14Cells & sources of emfPaper 2Medium8 marks
An engineer tests a single solar cell for a CubeSat under constant illumination. The terminal pd V is measured for several currents I, and the following data are obtained:
I / mA : 20 | 40 | 60 | 80 V / V : 0.56 | 0.52 | 0.48 | 0.44
Over this range, the cell may be modelled as a source of constant emf with a constant internal resistance.
Data from the table
(a)
Outline a circuit that could be used to obtain these data.
(2)
(b)
Determine the internal resistance and the emf of the cell.
(3)
(c)
State the energy transfer that takes place in the solar cell.
(1)
(d)
The cell is connected to an 8.0 Ω load. Calculate the power delivered to the load.
(2)
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Notes
Part (a)
The cell is connected in series with a variable resistor and an ammeter
✓ 1
A voltmeter is connected across the terminals of the cell, and the variable resistor is adjusted to change I
✓ 1
Part (b)
Use V = ε − Ir with the data. Either take the gradient of V against I as −r, or write two simultaneous equations such as 0.56 = ε − 0.020r and 0.44 = ε − 0.080r
✓ 1
r = (0.56 − 0.44)/(0.080 − 0.020) = 2.0 Ω
✓ 1
ε = 0.56 + 0.020 × 2.0 = 0.60 V
✓ 1
Part (c)
Light (electromagnetic radiation) energy is transferred to electrical energy
✓ 1
Part (d)
I = 0.60/(8.0 + 2.0) = 0.060 A
✓ 1
P = I2R = 0.0602 × 8.0 = 0.029 W (29 mW)
✓ 1
Allow ECF from (b).
—
Exam tip: When V–I data fit V = ε − Ir, the emf is the intercept at I = 0 and the internal resistance is minus the gradient. Convert mA to A before finding the gradient.
113IBP2S-016-02-15Thermistors, LDRs & dividersPaper 2Medium6 marks
A sprint-timing gate uses a laser beam directed onto an LDR. The LDR is connected in series with a 4.5 kΩ fixed resistor across a 5.0 V supply of negligible internal resistance. The pd across the LDR is fed to a timer, which triggers when its input exceeds 2.5 V. The timer's input resistance is very large. The LDR has a resistance of 500 Ω when illuminated by the beam and 20 kΩ when the beam is blocked.
Circuit diagram
(a)
Calculate the pd across the LDR when the beam is unbroken.
(2)
(b)
Calculate the pd across the LDR when an athlete blocks the beam.
(2)
(c)
Explain whether the timer triggers correctly as the athlete crosses the line.
(2)
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Notes
Part (a)
V = 5.0 × 0.50/(0.50 + 4.5)
✓ 1
= 0.50 V
✓ 1
Part (b)
V = 5.0 × 20/(20 + 4.5)
✓ 1
= 4.1 V
✓ 1
Part (c)
With the beam unbroken, 0.50 V < 2.5 V, so the timer does not trigger
✓ 1
With the beam blocked, 4.1 V > 2.5 V, so the timer triggers when the athlete crosses
✓ 1
ECF from (a) and (b)
—
Common error: Keep track of whether the output is across the LDR or the fixed resistor. Here, more light means a lower pd across the LDR.
114B-2-10Thermistors, LDRs & dividersPaper 2Hard12 marks
The heated mirror on a car door contains a thin heating element made of a positive temperature coefficient (PTC) thermistor material, whose resistance increases as its temperature rises. The element is connected directly across the 12.0 V battery of the car, which has negligible internal resistance.
The graph shows how the power P dissipated in the element varies with the temperature θ of the mirror. The mirror and its element have a total mass of 0.15 kg and a mean specific heat capacity of 750 J kg−1 K−1.
Graph drawn to scale
(a)
(i)
Calculate, using the graph, the resistance of the element at 0 °C.
(1)
(ii)
Describe how the resistance of the element varies as θ increases from 0 °C to 70 °C.
(2)
(b)
On a frosty morning the air and the mirror are at −10 °C when the heater is switched on. Estimate the initial rate of increase of the temperature of the mirror.
(2)
(c)
The mirror transfers energy to the surrounding air at a rate of 0.50 W for every kelvin by which the mirror is hotter than the air.
(i)
Draw, on the graph, a line to show the rate of energy transfer from the mirror to the air when the air is at −10 °C.
(2)
(ii)
Hence determine the temperature at which the mirror reaches thermal equilibrium.
(1)
(iii)
On a mild day the air is at 20 °C. Determine the new equilibrium temperature of the mirror.
(2)
(d)
The manufacturer could instead have used a fixed resistor of resistance equal to that of the PTC element at 0 °C. Explain, with reference to your answers to (c), one advantage of the PTC element.
(2)
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Notes
Part (a)(i)
R = V2/P = 12.02/34.6 = 4.16 Ω
✓ 1
Accept 4.1–4.2 Ω «P read as 34.3–35.1 W».
Part (a)(ii)
from 0 °C to about 40 °C the resistance increases only slowly «from about 4.2 Ω to about 5.3 Ω, as P falls slowly»
✓ 1
above about 40–45 °C it increases very rapidly, to about 79 Ω at 70 °C «P falls steeply»
✓ 1
Accept any value from 60 to 100 Ω at 70 °C. Accept 'about 20 times larger'.
Part (b)
P at −10 °C = 36 W; the mirror is at the temperature of the air, so «initially» no energy is transferred to the air
✓ 1
Accept 35–36 W.
Δθ/Δt = P/mc = 36/(0.15 × 750) = 0.32 K s−1
✓ 1
Accept 0.31–0.32 K s−1. Award [2] for CNA.
Part (c)(i)
straight line starting at (−10 °C, 0)
✓ 1
with gradient 0.50 W K−1, e.g. passing through (30 °C, 20 W) and (50 °C, 30 W)
✓ 1
± half a small square.
Part (c)(ii)
intersection of the line with the curve: θ ≈ 42 °C
✓ 1
Accept 40–45 °C. Allow ECF from the candidate's line.
Part (c)(iii)
a parallel line «same gradient» starting at (20 °C, 0) drawn or used
✓ 1
intersection at θ ≈ 52 °C
✓ 1
Accept 50–55 °C. ALTERNATIVE: trial values from the graph that balance P and 0.50(θ − 20).
Part (d)
with the PTC element a 30 K rise in air temperature raises the equilibrium temperature by only about 10 K, because as the mirror warms above about 40 °C its resistance rises sharply and P = V2/R falls
✓ 1
a fixed resistor would dissipate about 35 W at all temperatures, giving equilibrium temperatures of about 59 °C and 89 °C: the PTC mirror cannot overheat «and wastes less energy on mild days»
✓ 1
OWTTE. Accept a qualitative comparison for MP2 if the fixed resistor's temperature rise ∝ air temperature is clearly explained.
Answers: (a)(i) 4.16 Ω · (b) 0.32 K s−1 · (c)(ii) ≈ 42 °C · (c)(iii) ≈ 52 °C (the remaining parts are explanations — see the table above)
115B-2-32emf & internal resistancePaper 2Hard14 marks
A car battery has an emf of 12.6 V. When the driver operates the starter motor with the headlamps switched off, the battery supplies a current of 150 A to the motor and the potential difference between the battery terminals falls to 9.9 V.
The headlamps are connected directly across the battery terminals. Each headlamp has a power of 60 W when the terminal potential difference is 12.6 V.
(a)
(i)
Show that the internal resistance of the battery is about 0.02 Ω.
(1)
(ii)
Calculate the rate at which thermal energy is generated inside the battery while the starter motor operates.
(2)
(b)
Deduce the fraction of the energy transferred from chemical energy by the battery that is delivered to the starter motor.
(2)
(c)
The headlamps are now switched on while the starter motor operates. Assume that the headlamps have a constant resistance and that the terminal potential difference stays at 9.9 V.
(i)
Determine the percentage reduction in the power of a headlamp while the starter motor operates.
(3)
(ii)
Suggest why the actual reduction in the power of a headlamp is smaller than your answer to (c)(i).
(2)
(d)
On a very cold morning the internal resistance of the battery rises to 0.030 Ω but its emf is unchanged. The headlamps are switched off. Treat the starter motor as a fixed resistance. The motor needs a potential difference of at least 9.5 V across it to turn the engine.
(i)
Determine whether the motor can turn the engine on this morning.
(3)
(ii)
Suggest why car batteries are designed to have a very small internal resistance.
(1)
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Notes
Part (a)(i)
r = (12.6 − 9.9)/150 = 0.018 Ω
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
P = I²r = 150² × 0.018 OR P = I(ε − V) = 150 × 2.7
✓ 1
P = 405 W «4.1 × 102 W»
✓ 1
Accept 4.0–4.5 × 102 W «450 W from r = 0.02 Ω». Award [2] for CNA.
Part (b)
power from chemical energy = εI = 12.6 × 150 = 1890 W AND power delivered to the motor = VI = 9.9 × 150 = 1485 W
✓ 1
ALTERNATIVE: fraction = VI/εI = V/ε ✓.
fraction = 1485/1890 = 0.79
✓ 1
Accept 79 %. Award [2] for CNA.
Part (c)(i)
R = V²/P = 12.6²/60 = 2.65 Ω
✓ 1
P = 9.9²/2.65 = 37 W
✓ 1
reduction = (60 − 37)/60 ≈ 38 %
✓ 1
ALTERNATIVE: P ∝ V², so P2/P1 = (9.9/12.6)² = 0.617 ✓✓, reduction 38 % ✓. Award [1 max] for 21 % «P ∝ V used».
Part (c)(ii)
at the lower potential difference the filament is cooler
✓ 1
so its resistance is lower and P = V²/R is larger than predicted «the filament is non-ohmic»
✓ 1
Part (d)(i)
resistance of motor = 9.9/150 = 0.066 Ω
✓ 1
I = 12.6/(0.066 + 0.030) = 131 A
✓ 1
Allow ECF.
V = 131 × 0.066 = 8.7 V, less than 9.5 V, so the motor cannot turn the engine
✓ 1
MP3 requires a comparison with 9.5 V.
Part (d)(ii)
so that the very large current needed by the starter motor causes only a small drop in terminal potential difference «and little energy is wasted in the battery»
✓ 1
OWTTE.
Answers: (a)(i) 0.018 Ω · (a)(ii) 405 W · (b) 0.79 · (c)(i) 38 % · (d)(i) 8.7 V: no (the remaining parts are explanations — see the table above)
116B-2-33Circuit diagrams & metersPaper 2Hard12 marks
A student determines the resistance of a resistor X using the circuit shown. The battery has an emf of 3.0 V and negligible internal resistance. The voltmeter is ideal. The ammeter has a resistance of 0.50 Ω. The student calculates the resistance of X as (voltmeter reading)/(ammeter reading). The resistance of X is 4.0 Ω.
Diagram NOT accurately drawn
(a)
State what is meant by an ideal ammeter.
(1)
(b)
(i)
Calculate the reading on the ammeter.
(2)
(ii)
State the reading on the voltmeter.
(1)
(iii)
Show that the student's value for the resistance of X is about 4.5 Ω.
(1)
(c)
Explain why the student's value is larger than the resistance of X.
(2)
(d)
The student uses the same circuit to measure other resistors. Determine the smallest resistance that can be measured in this way with an error of no more than 1 %.
(3)
(e)
Suggest how the student should change the circuit so that the value obtained equals the resistance of X, and explain why.
(2)
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Notes
Part (a)
an ammeter with zero resistance «so that there is no potential difference across it and it does not change the current»
✓ 1
OWTTE.
Part (b)(i)
total resistance = 4.0 + 0.50 = 4.5 Ω
✓ 1
I = 3.0/4.5 = 0.67 A
✓ 1
Award [1 max] for 0.75 A «ammeter resistance ignored».
Part (b)(ii)
3.0 V «it is connected across the battery terminals and the internal resistance is negligible»
✓ 1
Part (b)(iii)
R = 3.0/0.667 = 4.50 Ω
✓ 1
Must see full substitution OR answer to at least 3 s.f. Allow ECF from (b)(i).
Part (c)
the voltmeter measures the potential difference across X AND the ammeter
✓ 1
there is a potential difference IRA across the ammeter, so V/I = RX + RA = 4.5 Ω
✓ 1
OWTTE.
Part (d)
the measured value is always larger than the true value by the ammeter resistance, 0.50 Ω
✓ 1
fractional error = 0.50/R ≤ 0.01
✓ 1
R ≥ 50 Ω
✓ 1
Award [3] for CNA. Award [2 max] for a correct check of a single value «e.g. 100 Ω gives 0.5 %» without finding the smallest resistance.
Part (e)
connect the voltmeter directly across X only «between the junction of the ammeter and X and the other end of X»
✓ 1
the voltmeter then reads only the pd across X, and as it is ideal it takes no current, so the ammeter still reads the current in X and V/I = RX
✓ 1
OWTTE.
Answers: (b)(i) 0.67 A · (b)(ii) 3.0 V · (b)(iii) 4.50 Ω · (d) 50 Ω (the remaining parts are explanations — see the table above)
117B-2-50Chemical & solar cellsPaper 2Hard14 marks
A remote weather station is powered by a small solar panel of area 0.090 m². The graph shows how the current I in the panel varies with the potential difference V across its terminals when sunlight of intensity 1000 W m−2 is incident normally on the panel.
Graph drawn to scale
(a)
Explain, with reference to the graph, why the panel cannot be modelled as a cell of constant emf and constant internal resistance.
(2)
(b)
State the emf of the panel.
(1)
(c)
The panel is connected to a resistor of resistance 20 Ω.
(i)
Draw, on the graph, a line to show how the current in the resistor would vary with the potential difference across it.
(2)
(ii)
Hence determine the power delivered to the resistor.
(2)
(d)
Determine whether a resistor of resistance 40 Ω connected to the panel would receive more power than the 20 Ω resistor.
(4)
(e)
The maximum power that the panel can deliver in these conditions is 10.3 W. Calculate the efficiency of the panel.
(1)
(f)
Outline one advantage and one disadvantage of a solar panel, compared with a chemical cell, as the source of energy for the weather station.
(2)
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Part (a)
for a cell of constant ε and r, V = ε − Ir, so the graph of I against V would be a straight line
✓ 1
the graph is curved: the current is almost constant up to about 15 V and then falls steeply, so the «effective» internal resistance is not constant
✓ 1
OWTTE
Part (b)
21 V «the pd when the current is zero»
✓ 1
Accept 20.8–21.2 V.
Part (c)(i)
straight line through the origin
✓ 1
passing through (10 V, 0.50 A) / gradient 0.050 A V−1
✓ 1
Any correct point on the line.
Part (c)(ii)
intersection at about 12.4 V and 0.62 A
✓ 1
Allow ECF from (c)(i).
P = 12.4 × 0.62 = 7.7 W
✓ 1
Accept 7.4–7.9 W.
Part (d)
line for 40 Ω drawn, through the origin and (20 V, 0.50 A)
✓ 1
Accept calculation of V/40 at points on the curve instead of a line.
intersection at about 19.2 V and 0.48 A
✓ 1
P = 19.2 × 0.48 = 9.2 W
✓ 1
Accept 8.9–9.5 W.
this is more than the power to the 20 Ω resistor, so yes
✓ 1
Allow ECF from (c)(ii). MP4 needs a comparison and a consistent conclusion.
A tram takes its electrical supply from an overhead contact wire and returns the current through the rails. Two substations, A and B, are 2.00 km apart. Each acts as a source of emf 750 V with negligible internal resistance. The contact wire has a cross-sectional area of 1.07 × 10−4 m2 and is made of a copper alloy of resistivity 1.8 × 10−8 Ω m. The rails have a resistance of 0.020 Ω per km. The tram takes a constant current of 600 A.
Diagram NOT accurately drawnAxes for (d)
(a)
Show that the resistance of 1.00 km of the contact wire is about 0.17 Ω.
(1)
(b)
At first only substation A supplies the tram, which is 1.50 km from A.
(i)
Calculate the potential difference between the contact wire and the rails at the tram.
(2)
(ii)
Calculate the power dissipated in the contact wire and rails.
(1)
(c)
Substation B is now also connected, so that the tram is supplied from both ends.
(i)
Explain why the current supplied by B is three times the current supplied by A.
(2)
(ii)
Show that the current supplied by B is 450 A.
(1)
(iii)
Calculate the potential difference between the contact wire and the rails at the tram.
(1)
(iv)
Determine the power dissipated in the contact wire and rails, and compare it with your answer to (b)(ii).
(2)
(d)
Sketch, on the axes, how the potential difference at the tram varies as it travels from A to B with both substations connected.
Must see full substitution OR answer to at least 3 s.f.
Part (b)(i)
resistance of wire and rails = 1.50 × (0.168 + 0.020) = 0.282 Ω
✓ 1
Award [0] for this mark if the rails are omitted «0.252 Ω».
V = 750 − 600 × 0.282 = 581 V
✓ 1
Award [2] for CNA. Allow ECF.
Part (b)(ii)
P = I2R = 6002 × 0.282 = 1.0 × 105 W
✓ 1
Accept I(ε − V). Allow ECF.
Part (c)(i)
both paths run from a point at 750 V «the supply» to the same point at the tram, so they have the same pd across them «they are in parallel»
✓ 1
the path from B is 0.50 km long, so its resistance is one third of that from A, and I = V/R is three times as large
✓ 1
OWTTE.
Part (c)(ii)
IA + IB = 600 with IB = 3IA, so IA = 150 A and IB = 450 A
✓ 1
Must see both conditions used.
Part (c)(iii)
V = 750 − 450 × 0.094 = 708 V
✓ 1
Accept 750 − 150 × 0.282. Accept 707–709 V.
Part (c)(iv)
P = 1502 × 0.282 + 4502 × 0.094 = 2.5 × 104 W
✓ 1
Allow ECF from (c)(ii).
this is a quarter of the power dissipated with one substation «1.0 × 105 W»
✓ 1
Accept 'about 75 % less'. Allow ECF from (b)(ii).
Part (d)
curve starting and ending at 750 V «at 0 and 2.0 km», symmetrical, with a single minimum at 1.0 km
✓ 1
Do not accept straight-line segments meeting at a point.
minimum value about 694 V «between 685 V and 700 V», passing close to (1.5 km, 708 V)
✓ 1
Allow ECF from (c)(iii) for the point at 1.5 km.
Answers: (a) 0.168 Ω · (b)(i) 581 V · (b)(ii) 1.0 × 105 W · (c)(iii) 708 V · (c)(iv) 2.5 × 104 W (the remaining parts are explanations — see the table above)
119B-2-91emf & internal resistancePaper 2Hard12 marks
A camper van has a starter battery of emf 12.7 V and internal resistance 0.020 Ω, and a partly discharged leisure battery of emf 12.1 V and internal resistance 0.040 Ω. The two batteries are connected in parallel by link cables of total resistance 0.010 Ω, positive terminal to positive terminal and negative to negative, as shown. Nothing else is connected.
Diagram NOT accurately drawn
(a)
Draw, on the diagram, an arrow to show the direction of the conventional current in the upper link cable.
(1)
(b)
Show that the current in the link cables is about 8.6 A.
(2)
(c)
(i)
Calculate the potential difference between the terminals of the starter battery.
(1)
(ii)
The potential difference between the terminals of the leisure battery is greater than its emf. Explain why.
(2)
(d)
(i)
Calculate the rate at which chemical energy is transferred to electrical energy in the starter battery.
(1)
(ii)
Calculate the rate at which electrical energy is transferred to chemical energy in the leisure battery.
(1)
(e)
Hence determine the total rate at which thermal energy is generated in the circuit, and show that your answer agrees with the resistances of the circuit.
(2)
(f)
Later, both batteries are fully charged to an emf of 12.7 V and the link is left in place. Explain why the starter motor, connected across the terminals of the starter battery, then receives a larger current than it would from the starter battery alone.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
arrow on the upper cable pointing from the starter battery towards the leisure battery
✓ 1
Part (b)
the emfs act in opposite senses round the loop, so the resultant emf = 12.7 − 12.1 = 0.6 V
✓ 1
Award [0] for adding the emfs.
I = 0.6/(0.020 + 0.040 + 0.010) = 8.57 A
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (c)(i)
V = ε − Ir = 12.7 − 8.57 × 0.020 = 12.53 V
✓ 1
Allow ECF from (b).
Part (c)(ii)
the current enters the leisure battery at its positive terminal «it is being charged»: each coulomb must deliver energy ε to the chemical store and also energy Ir to its internal resistance
✓ 1
so V = ε + Ir = 12.1 + 8.57 × 0.040 = 12.44 V, greater than 12.1 V
✓ 1
OWTTE. MP2 can be awarded for the calculation with a correct reason.
Part (d)(i)
εI = 12.7 × 8.57 = 108.9 W
✓ 1
Part (d)(ii)
εI = 12.1 × 8.57 = 103.7 W
✓ 1
Do not accept VI for the leisure battery.
Part (e)
difference = 108.9 − 103.7 = 5.14 W
✓ 1
Allow ECF from (d).
I2Rtotal = 8.572 × 0.070 = 5.14 W, the same
✓ 1
Both values and the comparison are needed for MP2.
Part (f)
the batteries have equal emfs, so they act as one source of emf 12.7 V whose internal resistance is that of the two «battery paths» in parallel, which is less than 0.020 Ω
✓ 1
«0.0143 Ω including the link cable»
so for the same motor resistance the total resistance is smaller, the current is larger and less of the emf is «wasted» across the internal resistance
✓ 1
OWTTE.
Answers: (b) 8.57 A · (c)(i) 12.53 V · (d)(i) 108.9 W · (d)(ii) 103.7 W · (e) 5.14 W (the remaining parts are explanations — see the table above)
120B-2-92Ohmic & non-ohmic conductorsPaper 2Hard12 marks
A 10 A blade fuse in a car contains a zinc strip of length 6.0 mm and rectangular cross-section 0.25 mm × 0.20 mm. The fuse is connected in the wire from the battery to a headlamp.
Data for zinc: resistivity at 20 °C = 5.9 × 10−8 Ω m; density = 7.14 × 103 kg m−3; specific heat capacity = 390 J kg−1 K−1; melting point = 420 °C; specific latent heat of fusion = 1.12 × 105 J kg−1.
(a)
Outline why the fuse is connected in series with the headlamp.
(1)
(b)
(i)
Show that the resistance of the strip at 20 °C is about 7 mΩ.
(1)
(ii)
Calculate the power dissipated in the fuse when the current is 10 A.
(1)
(c)
A fault in the headlamp wiring makes the current rise suddenly to 100 A. The strip is initially at 20 °C.
(i)
Determine the energy needed to raise the strip to its melting point and melt it.
(3)
(ii)
Estimate the time taken for the fuse to melt, assuming the resistance of the strip stays at its value at 20 °C and no energy leaves the strip.
(2)
(iii)
The resistance of zinc increases as its temperature rises. State and explain whether the actual time is longer or shorter than your estimate in (c)(ii).
(2)
(d)
Deduce the effect on the time to melt if the fault current were 200 A instead of 100 A.
(1)
(e)
At the rated current of 10 A the fuse dissipates energy continuously but does not melt. Explain why.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
all the current to the headlamp passes through the fuse, so when it melts the circuit is broken and the current stops
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
P = I2R = 102 × 7.08 × 10−3 = 0.71 W
✓ 1
Allow ECF from (b)(i). Accept 0.70 W from 7 mΩ.
Part (c)(i)
m = density × volume = 7.14 × 103 × 3.0 × 10−10 = 2.14 × 10−6 kg
✓ 1
mcΔT = 2.14 × 10−6 × 390 × 400 = 0.334 J
✓ 1
mL = 0.240 J, so total = 0.57 J
✓ 1
Award [3] for CNA. Award [2 max] if the latent heat is omitted «0.33 J».
Part (c)(ii)
P = 1002 × 7.08 × 10−3 = 70.8 W
✓ 1
Allow ECF from (b)(i).
t = 0.57/70.8 ≈ 8.1 × 10−3 s
✓ 1
Allow ECF from (c)(i). Accept 8 ms. Award [2] for CNA.
Part (c)(iii)
shorter
✓ 1
MP1 only scores if MP2 scores.
the current is fixed «by the rest of the circuit», so as R increases the power I2R dissipated in the strip increases and the energy is supplied more quickly
✓ 1
OWTTE. Accept a mention that losses to the surroundings in so short a time are negligible.
Part (d)
P = I2R is four times as large, so the time is about a quarter «≈ 2 ms»
✓ 1
Allow ECF from (c)(ii).
Part (e)
the strip warms until it transfers energy to its surroundings «holder, air, wires» at the same rate as it is dissipated «0.71 W», which happens at a temperature below its melting point
✓ 1
OWTTE.
Answers: (b)(i) 7.08 × 10−3 Ω · (b)(ii) 0.71 W · (c)(i) 0.57 J · (c)(ii) 8.1 × 10−3 s (the remaining parts are explanations — see the table above)
An aircraft engineer compares copper and aluminium for a 2.0 m power feeder cable. The cable must have a resistance of 1.0×10−4 Ω.
Metal
Resistivity / Ω m
Density / kg m−3
Copper
1.7×10−8
8900
Aluminium
2.8×10−8
2700
(a)
Show that the cross-sectional area of a copper cable meeting the requirement is about 3×10−4 m2.
(2)
(b)
Calculate the cross-sectional area of an aluminium cable with the same length and resistance.
(1)
(c)
Determine the ratio \(\frac{\text{mass of aluminium cable}}{\text{mass of copper cable}}\).
(3)
(d)
Outline why aluminium is preferred for this application despite its higher resistivity.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A = ρL/R
✓ 1
A = 1.7×10−8 × 2.0/1.0×10−4 = 3.4×10−4 m2
✓ 1
Part (b)
A = 2.8×10−8 × 2.0/1.0×10−4 = 5.6×10−4 m2
✓ 1
Part (c)
Mass = density × A × L, and L is the same for both cables
✓ 1
ratio = (2700 × 5.6×10−4)/(8900 × 3.4×10−4) (ECF from (a) and (b))
✓ 1
ratio = 0.50
✓ 1
Part (d)
For the same resistance, the aluminium cable has about half the mass, and saving mass is critical in an aircraft
✓ 1
Exam tip: For equal length and resistance, A ∝ ρ, so the mass is proportional to ρ × density. Comparing this product for each metal gives the ratio directly.
122IBP2S-016-03-16Current & chargePaper 2Hard6 marks
A cardiac pacemaker delivers 70 pulses per minute. Each pulse is a current of 5.0 mA lasting 0.50 ms. Between pulses, and throughout operation, the pacemaker's electronics draw a steady current of 10 μA. The battery has a capacity of 1.0 A h.
(a)
Calculate the charge delivered in one pulse.
(1)
(b)
Show that the average current drawn from the battery is about 1.3×10−5 A.
(2)
(c)
Calculate the battery capacity in coulombs.
(1)
(d)
Estimate the lifetime of the battery in years.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
q = 5.0×10−3 × 0.50×10−3 = 2.5×10−6 C
✓ 1
Part (b)
Charge per minute = 70 × 2.5×10−6 + 10×10−6 × 60 = 7.75×10−4 C (ECF from (a))
✓ 1
average I = 7.75×10−4/60 = 1.29×10−5 A
✓ 1
Part (c)
1.0 A × 3600 s = 3.6×103 C
✓ 1
Part (d)
t = 3600/1.29×10−5 = 2.8×108 s (ECF from (b) and (c))
✓ 1
2.8×108/3.15×107 ≈ 9 years
✓ 1
Common error: Include the steady background current as well as the pulses. Here the background current accounts for most of the charge used.
123IBP2S-016-01-15Ohmic & non-ohmic conductorsPaper 2Hard8 marks
A light-emitting diode (LED) in a phone-charging indicator operates correctly with a pd of 2.0 V across it and a current of 20 mA. It is connected in series with a resistor R to a 5.0 V USB supply of negligible internal resistance.
Circuit diagram
(a)
Calculate the resistance of R needed for correct operation.
(2)
(b)
The LED is a non-ohmic component. Explain why the series resistor is needed.
(2)
(c)
The same LED and the resistor from (a) are now connected to a battery of emf 4.5 V and internal resistance 6.0 Ω. Assume the pd across the LED remains 2.0 V. Calculate the new current.
(2)
(d)
Determine the fraction of the energy transferred by the battery in (c) that is delivered to the LED.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
pd across R = 5.0 − 2.0 = 3.0 V
✓ 1
R = 3.0/0.020 = 150 Ω
✓ 1
Part (b)
The LED conducts very little until its pd reaches about 2 V, then its current rises very steeply with small increases in pd, so it is not proportional to the pd
✓ 1
Without R, the 5.0 V supply would drive an excessive current that overheats or destroys the LED, so R limits the current
✓ 1
Part (c)
I = (ε − VLED)/(R + r) = (4.5 − 2.0)/(150 + 6.0), ECF from (a)
✓ 1
I = 0.016 A, i.e. 16 mA
✓ 1
Part (d)
Fraction = (VLED × I)/(ε × I) = 2.0/4.5, because the same current passes through every component
✓ 1
= 0.44 (44%)
✓ 1
Common error: Do not apply Ohm's law to the LED with a single fixed resistance. Treat it as having a fixed operating pd and apply Ohm's law to the resistors only.
124IBP2S-016-01-16emf & internal resistancePaper 2Hard10 marks
An electric car battery pack is built from identical cells, each of emf 3.6 V and internal resistance 0.020 Ω. Four cells are connected in parallel to form a module. Ninety-six modules are then connected in series to form the pack.
(a)
Show that a module has an emf of 3.6 V and an internal resistance of 0.0050 Ω.
(2)
(b)
Calculate the emf and the internal resistance of the complete pack.
(2)
(c)
During rapid acceleration the motor draws 150 A from the pack. Calculate the terminal pd of the pack.
(2)
(d)
Determine the rate at which thermal energy is generated in a single cell during this acceleration.
(2)
(e)
An alternative design uses a single cell in place of each module, giving 96 single cells in series with the same emf. Explain one advantage of the four-cell parallel modules over this alternative design when the motor draws 150 A.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Identical cells in parallel share the same pd, so the module emf equals one cell's emf, 3.6 V
✓ 1
The internal resistances combine in parallel: 1/r = 4/0.020, giving r = 0.0050 Ω
✓ 1
Part (b)
emf = 96 × 3.6 ≈ 346 V (345.6 V)
✓ 1
r = 96 × 0.0050 = 0.48 Ω
✓ 1
Part (c)
V = ε − Ir = 345.6 − 150 × 0.48, ECF from (b)
✓ 1
V ≈ 274 V
✓ 1
Part (d)
Each module carries 150 A, which divides equally among its four cells, so each cell carries 37.5 A
✓ 1
P = I²r = 37.5² × 0.020 ≈ 28 W
✓ 1
Part (e)
Any one advantage, identified with a valid comparison: the pack internal resistance is 0.48 Ω rather than 96 × 0.020 = 1.92 Ω, OR each cell carries 37.5 A rather than 150 A, OR the pack stores four times the charge
✓ 1
A matching consequence: a smaller Ir drop, so a higher terminal pd and less energy wasted as heat; OR less heating per cell (P ∝ I²), so a lower risk of overheating; OR a greater range per charge
✓ 1
Exam tip: Check your answer to (d) by multiplying by the number of cells. Here 384 × 28.1 W should equal the total internal power I²r of the pack, 150² × 0.48 = 10.8 kW.
125IBP2S-016-00-15Chemical & solar cellsPaper 2Hard9 marks
A thermoelectric generator on a deep-space probe is modelled as a source of emf ε = 30 V with internal resistance r = 2.0 Ω. It supplies a load of resistance R.
(a)
Show that the power delivered to the load is \(P = \frac{\varepsilon^2 R}{(R+r)^2}\).
(2)
(b)
State the value of R for which P is a maximum.
(1)
(c)
Calculate the maximum power that can be delivered to the load.
(2)
(d)
Calculate the power delivered to the load when R = 6.0 Ω.
(2)
(e)
Determine the efficiency of energy transfer to the load when R = 6.0 Ω, and compare it with the efficiency when R = r.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Current I = ε/(R + r)
✓ 1
P = I²R, which leads to P = ε²R/(R + r)²
✓ 1
Part (b)
R = r = 2.0 Ω
✓ 1
Part (c)
P = 302 × 2.0/(4.0)2, or I = 7.5 A and P = 7.52 × 2.0
This is higher than the 50% efficiency at R = r. Only 25% of the energy supplied is dissipated in the internal resistance (28 W of 112.5 W), compared with 50% at R = r (112.5 W of 225 W). ECF from (d)
✓ 1
Exam tip: Maximum power transfer and maximum efficiency are not the same condition. Efficiency depends on the fraction of the total power dissipated in r, not on the absolute power lost there.
An engineer is designing a heated car seat. The heating element is a constantan wire. It is powered by a battery of emf 12.6 V and internal resistance 0.030 Ω, through connecting leads of total resistance 0.070 Ω. The element must dissipate 48 W. The resistivity of constantan is 4.9 × 10−7 Ω m. The resistivity of copper is 1.7 × 10−8 Ω m.
Circuit diagram
(a)
Explain why the terminal potential difference of the battery is less than its emf when the heater is operating.
(2)
(b)
Show that the resistance R of the heating element must be about 3.1 Ω.
(3)
(c)
The wire has a diameter of 0.40 mm. Calculate the length of wire needed.
(2)
(d)
Suggest two reasons why constantan is used for the element rather than copper.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
There is a current through the internal resistance of the battery
✓ 1
Some of the energy per unit charge is dissipated inside the battery (a pd Ir across r), so V = ε − Ir < ε
✓ 1
Part (b)
Uses the total circuit resistance: I = 12.6/(R + 0.10)
✓ 1
Uses P = I2R for the element only, with P = 48 W: 48 = (12.6/(R + 0.10))2 × R
✓ 1
Either solves 48R2 − 149.16R + 0.48 = 0 to give R = 3.10 Ω and rejects the ≈ 0.003 Ω root as impractical, or substitutes R = 3.1 Ω to give I = 3.94 A and P = 48.1 W ≈ 48 W
✓ 1
Part (c)
A = π(0.20 × 10−3)2 = 1.26 × 10−7 m2
✓ 1
L = RA/ρ = 3.10 × 1.26 × 10−7/4.9 × 10−7 = 0.80 m; allow ECF from (b)
✓ 1
Part (d)
Copper's resistivity is about 30 times lower, so an impractically long (about 20 m) or very thin wire would be needed for the same resistance
✓ 1
The resistivity of constantan stays almost constant with temperature, so the resistance and heating power stay stable as the element warms up
✓ 1
Common error: Forgetting the lead resistance and the internal resistance when setting up the equation for R. The 48 W refers only to the element, but the current depends on the total circuit resistance.
127IBP2S-016-02-16emf & internal resistancePaper 2Hard8 marks
A drone's control board is powered by a battery of emf 9.0 V and internal resistance 2.0 Ω. The battery is connected to a potential divider made from an 18 Ω resistor in series with a 30 Ω resistor.
Circuit diagram
(a)
Show that the current in the battery is 0.18 A.
(2)
(b)
Calculate the pd across the 30 Ω resistor.
(1)
(c)
A sensor of resistance 30 Ω is now connected in parallel with the 30 Ω resistor. Determine the new pd across the 30 Ω resistor.
(3)
(d)
Explain why the terminal pd of the battery decreases when the sensor is connected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Total resistance = 2.0 + 18 + 30 = 50 Ω
✓ 1
I = 9.0/50 = 0.18 A
✓ 1
Part (b)
V = 0.18 × 30 = 5.4 V
✓ 1
Part (c)
The parallel combination is 15 Ω
✓ 1
I = 9.0/(2.0 + 18 + 15) = 0.257 A
✓ 1
V = 0.257 × 15 = 3.9 V
✓ 1
ECF for the current
—
Part (d)
The total resistance of the circuit decreases, so the current in the battery increases
✓ 1
The lost pd Ir increases, so the terminal pd ε − Ir decreases (from 8.64 V to 8.49 V)
✓ 1
Exam tip: Connecting any load to a divider changes the current everywhere, including in the battery. Recalculate from the total resistance; do not assume that the earlier current still applies.
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