84 free, high-quality past-paper-style questions on Current and circuits for IB Physics HL, each with a full mark scheme. Covers: Conductors and insulators; Circuit diagrams and meters; Origin of resistance and the heating effect; and more. For when you have finished the past papers, or want to practise one topic at a time.B.5 at Standard 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.396 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1B-1A-21Conductors and insulatorsPaper 1AEasy1 mark
Copper is a good electrical conductor and polythene is an electrical insulator.
Which statement explains this difference?
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Step 1A current needs mobile charge carriers.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In a metal, one or more electrons per atom are delocalized (free electrons) and drift through the lattice when a potential difference is applied.
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Step 3In polythene the electrons are held in covalent bonds, so there are almost no mobile charge carriers and the resistivity is enormous.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: conduction depends on the number of mobile charge carriers, and polythene has almost none.
BCopper does contain more electrons per unit volume (about 8 times as many), but this factor is far too small to explain a difference in resistivity of more than 1020; what matters is how many electrons are free to move, not how many there are.
CThe ions in a metal vibrate about fixed lattice positions; the charge carriers in a metal are the free electrons.
DThe drift speed of electrons in a metal is very small (of order 10−4 m s−1); the difference lies in the number of free carriers.
3B-1A-53Origin of resistance and the heating effectPaper 1AEasy1 mark
When there is a current in a metal resistor, the resistor becomes warm.
Which statement explains this heating effect?
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Step 1The pd sets up an electric field in the metal; the free electrons are accelerated by it and gain kinetic energy.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The electrons repeatedly collide with the vibrating lattice ions, which impede their drift — this is the origin of resistance.
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Step 3In each collision energy is transferred to the ions, which vibrate with greater amplitude: the internal energy and temperature of the resistor rise.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
ACharge is conserved: the same number of electrons leave the resistor as enter it, so the current is not "used up".
BCollisions keep the drift speed of the electrons constant along a uniform wire; the energy goes into the lattice, not into ever-faster electrons.
CResistance is a property of the bulk material (collisions with lattice ions throughout the volume), not surface friction.
DCorrect: work done by the field on the electrons is transferred to the lattice in collisions.
4IBP2H-017-01-01Chemical cells as energy sourcesPaper 1AEasy1 mark
A cardiac pacemaker delivers 70 current pulses per minute. Each pulse has a current of 4.0 mA and lasts 0.50 ms. What is the total charge delivered in one minute?
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Charge per pulse q = It = (4.0×10−3)(0.50×10−3) = 2.0×10−6 C. Over 70 pulses the charge is 70 × 2.0×10−6 = 1.4×10−4 C. Option A is the charge of a single pulse only. Option C wrongly multiplies by a further 60. Option D comes from leaving the pulse duration in milliseconds.
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Answer B
Answer: B
Common error: Convert both mA and ms to SI units before multiplying. Then check whether the question asks for one pulse or for many.
Two copper wires X and Y carry the same current. The diameter of Y is twice the diameter of X. What is \(\frac{\text{drift speed in Y}}{\text{drift speed in X}}\)?
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From I = nAvq, the current I, the charge carrier density n and the charge q are the same for both copper wires. So v ∝ 1/A. Area is proportional to d2, so doubling the diameter makes the cross-sectional area four times larger. The drift speed therefore falls to one quarter, 0.25 (D). Distractors: C (0.5) treats v as proportional to 1/d instead of 1/d2. B (2) wrongly takes v as proportional to d. A (4) wrongly takes v as proportional to A.
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Answer D
Answer: D
Exam tip: Area depends on diameter squared. A factor of 2 in diameter is a factor of 4 in area.
6IBP2H-017-02-02Conductors and insulatorsPaper 1AEasy1 mark
At room temperature, copper is a good electrical conductor and PVC is an electrical insulator. What best explains the difference?
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Conductivity depends on how many charge carriers per unit volume are free to move. In a metal, roughly one electron per atom is delocalised. In an insulator, almost all electrons are bound to their atoms. Drift speeds in metals are actually very small, so the second option is wrong. PVC does contain electrons, but they are bound, so the third option is wrong. Resistivity is a property of the material and does not depend on sample thickness, so the fourth option is wrong.
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Answer A
Answer: A
Common error: Do not confuse resistance (which depends on the dimensions of a sample) with resistivity (which depends only on the material and its temperature).
7IBP2H-017-02-03Solar cells as energy sourcesPaper 1AEasy1 mark
A solar panel powers a remote weather station. The panel has an area of 1.6 m2 and an efficiency of 20%. Sunlight of intensity 800 W m−2 is incident normally on the panel. What is the electrical power output of the panel?
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Incident power = intensity × area = 800 × 1.6 = 1280 W. Electrical output = 0.20 × 1280 = 256 W. The value 160 W ignores the area. The value 1280 W ignores the efficiency. The value 6400 W divides by the efficiency instead of multiplying by it.
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Answer B
Answer: B
Exam tip: Check that your answer makes sense: the useful output must always be less than the incident power.
8IBP2H-017-02-08Chemical cells as energy sourcesPaper 1AEasy1 mark
A cardiac pacemaker is powered by a lithium cell rated at 1.2 A h, with a terminal potential difference of 2.8 V. What is the approximate total energy the cell can deliver?
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Charge Q = 1.2 A × 3600 s = 4320 C. Energy = QV = 4320 × 2.8 ≈ 1.2 × 104 J = 12 kJ. The value 3.4 J omits the time entirely. The value 200 J converts hours to minutes instead of seconds. The value 3.4 kJ treats 1.2 A h as 1200 A s.
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Answer A
Answer: A
Common error: The A h is a unit of charge, not energy. Convert it to coulombs (1 A h = 3600 C) and then multiply by the potential difference.
9IBP2H-017-00-01Circuit diagrams and metersPaper 1AEasy1 mark
A technician testing a bicycle lamp powered by a cell of small internal resistance connects an ammeter in parallel with the lamp by mistake, instead of in series with it. What is the most likely result?
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An ammeter has a very small resistance. In parallel with the lamp it provides an almost zero-resistance path, so it short-circuits the lamp. Almost no current passes through the lamp, so it goes out. The total circuit resistance is now little more than the internal resistance of the cell, so a large current passes through the ammeter. Option B treats the ammeter as having a high resistance, like a voltmeter. Option C wrongly assumes that an extra parallel path increases the lamp's share of the current. Option D ignores the meter's effect on the circuit.
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Answer A
Answer: A
Common error: Ammeters (near-zero resistance) go in series and voltmeters (near-infinite resistance) go in parallel. Swapping them either short-circuits a component or blocks the current.
10IBP2H-017-00-08Chemical cells as energy sourcesPaper 1AEasy1 mark
Which of the following is a correct definition of the emf of a battery?
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Emf is the energy converted into electrical energy (for example from chemical energy) per unit charge that passes through the source. Options A and D both describe the terminal potential difference, which is less than the emf whenever current flows through the internal resistance. Option C is wrong because emf is energy per unit charge, not a force.
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Answer B
Answer: B
Common error: Despite its name, electromotive force is not a force. Its unit is J C⁻¹ = V.
12B-1A-12emf and internal resistancePaper 1AMedium1 mark
A cell is connected to a variable resistor. The graph shows how the terminal potential difference V of the cell varies with the current I.
The terminals of the cell are then connected by a wire of negligible resistance. What is the current in the wire?
Terminal pd against current for the cell (drawn to scale). The V axis does not start at zero.Show mark scheme
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Step 1The intercept on the V axis (at I = 0) is the emf: ε = 1.50 V.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The internal resistance is minus the gradient: r = (1.50 − 1.00)/(0.40 − 0) = 1.25 Ω. Note that the horizontal axis is drawn at V = 1.00 V, not at V = 0.
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Step 3With zero external resistance ε = I(0 + r), so I = 1.50/1.25 = 1.2 A.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis is where the line meets the drawn current axis, which is at V = 1.00 V, not V = 0 (false origin).
BThis uses V/I = 1.00/0.40 = 2.5 Ω for the point where the line meets the axis as the internal resistance: 1.50/2.5 = 0.60 A. That ratio is the external resistance at that point.
CCorrect: I = ε/r = 1.50/1.25 = 1.2 A.
DThis takes the internal resistance as ΔI/ΔV = 0.80 Ω, the inverted gradient: 1.50/0.80 = 1.9 A.
15B-1A-29Solar cells as energy sourcesPaper 1AMedium1 mark
A solar cell of area 2.0 × 10−2 m² is illuminated normally by sunlight of intensity 900 W m−2. When nothing is connected to the cell, the potential difference between its terminals is 0.60 V. The cell is then connected to a resistor of resistance 0.10 Ω, and the potential difference across the resistor is 0.50 V.
What is the efficiency of the solar cell when it is connected to the resistor?
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Step 1Input power = intensity × area = 900 × 2.0 × 10−2 = 18 W.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Output power to the resistor = V²/R = 0.50²/0.10 = 2.5 W (the current is 5.0 A).
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Step 3η = 2.5/18 = 0.14 = 14 %. The 0.60 V is the emf of the cell; when there is a current, 0.10 V is across the internal resistance of the cell, so it is not the pd across the resistor.
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Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis divides the output power by the intensity (in W m−2) instead of by the input power: 2.5/900 = 0.28 %. The intensity must be multiplied by the area.
BCorrect: 2.5 W of the 18 W incident on the cell is delivered to the resistor.
CThis uses the emf instead of the terminal pd: 0.60²/0.10 = 3.6 W and 3.6/18 = 20 %. Part of the emf is across the internal resistance of the cell.
DThis takes V/R = 5.0, which is the current in amperes, as the output power: 5.0/18 = 28 %.
16B-1A-54Chemical cells as energy sourcesPaper 1AMedium1 mark
A portable power bank contains a rechargeable chemical cell of emf 3.7 V. The cell is rated at 10 000 mA h, meaning that it can supply a current of 10 000 mA for one hour.
What is the maximum energy that the cell can transfer?
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Step 1Charge: q = IΔt = 10.0 A × 3600 s = 3.6 × 104 C.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2The emf is the energy transferred per unit charge, so W = qε.
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Step 3W = 3.6 × 104 × 3.7 = 1.3 × 105 J.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis treats the rating 10 A h as 10 C, forgetting to convert hours to seconds: 10 × 3.7 = 37 J.
BThis uses the number 10 000 (the rating in mA h) as if it were a charge in coulombs: 10 000 × 3.7 = 3.7 × 104 J.
CCorrect: 3.6 × 104 C × 3.7 V = 1.3 × 105 J.
DThis takes the current as 10 000 A instead of 10 000 mA: 10 000 × 3600 × 3.7 = 1.3 × 108 J.
A copper track on a circuit board has a rectangular cross-section of width 0.25 mm and thickness 35 μm, and is 8.0 cm long. The resistivity of copper is 1.7 × 10−8 Ω m.
The power dissipated in the track must not exceed 0.50 W. What is the maximum current in the track?
A lamp of resistance 6.0 Ω is to be dimmed using an 18 Ω variable resistor. Circuit P uses the resistor as a variable series resistor; circuit Q uses it as a potentiometer.
Which row gives the range of potential difference across the lamp that can be obtained in each circuit?
Both cells have negligible internal resistance. The lamp may be assumed to have a constant resistance of 6.0 Ω.
Range in PRange in Q
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Step 1In P, with the variable resistor at 0 Ω the lamp has the full 12 V; at 18 Ω the lamp takes 12 × 6.0/(6.0 + 18) = 3.0 V. The pd can never reach zero.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In Q, with the slider at the bottom of the track the lamp is connected between two points at the same potential: pd = 0.
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Step 3With the slider at the top the lamp is connected directly across the 12 V supply: pd = 12 V. So Q gives the full range 0 to 12 V.
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Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the series resistor limits the minimum to 3.0 V; the potentiometer gives 0 to 12 V.
BThis assumes the potentiometer behaves like the series resistor; in Q the lamp can be connected across a zero-length section of the track.
C9.0 V is the pd across the 18 Ω resistor in P, not across the lamp.
20B-1A-58Light-dependent resistor sensorPaper 1AMedium1 mark
The circuit shown is used to switch on a garden light at dusk. At dusk the resistance of the light-dependent resistor (LDR) is 2.0 kΩ, and the output Vout must then be 4.0 V.
Which row gives the required value of R and the change in Vout as it becomes darker?
The cell has negligible internal resistance and the voltmeter is ideal.
Value of RChange in Vout as it becomes darker
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Step 1The pd across R must be 4.0 V, so the pd across the LDR is 6.0 − 4.0 = 2.0 V.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2In series the pds are in the ratio of the resistances: R/2.0 kΩ = 4.0/2.0, so R = 4.0 kΩ.
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Step 3As it becomes darker the resistance of the LDR increases, so the LDR takes a larger share of the 6.0 V and Vout across R decreases.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
A1.0 kΩ would give 4.0 V across the LDR, not across R (the ratio has been inverted).
BThis analyses the circuit as if the output were taken across the LDR: both entries would then follow, but the voltmeter is across R.
CCorrect: R = 4.0 kΩ, and the LDR resistance rises in the dark, so Vout falls.
DThe resistance of an LDR increases, not decreases, as the light intensity falls.
A student determines the resistance of a resistor R using the circuit shown. The voltmeter is not ideal: its resistance is 1.0 kΩ. The voltmeter reads 6.0 V and the ammeter reads 0.040 A.
What is the resistance of R?
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Step 1The ammeter measures the total current in R and the voltmeter together.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2Current in the voltmeter = 6.0/1000 = 0.0060 A, so the current in R = 0.040 − 0.0060 = 0.034 A.
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Step 3R = 6.0/0.034 = 176 Ω ≈ 180 Ω.
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Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis adds the voltmeter current instead of subtracting it: 6.0/(0.040 + 0.0060) = 130 Ω.
BThis is V/I from the meter readings; it is the combined resistance of R in parallel with the voltmeter, not R itself.
CCorrect: only 0.034 A passes through R, so R = 6.0/0.034 ≈ 180 Ω.
DThis subtracts the measured value from the voltmeter resistance, 1000 − 150 = 850 Ω, as if the combined 150 Ω were a difference of resistances rather than a parallel combination.
22B-1A-71Current, charge and potential differencePaper 1AMedium1 mark
In a gas-discharge tube, free electrons move towards the positive electrode and positive ions, each of charge +e, move towards the negative electrode.
In a time of 2.0 s, 6.0 × 1018 electrons and 1.0 × 1018 positive ions cross a plane perpendicular to the tube. What is the current in the tube?
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Step 1Electrons (negative) moving one way and positive ions moving the opposite way both transfer charge in the same direction, so their contributions to the current add.
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All 3 steps must be completed — there is no mark for a part-answer.
AThis counts only the positive ions: 1.0 × 1018 × 1.60 × 10−19/2.0 = 0.080 A, ignoring the electrons.
BThis subtracts the two flows, as if charges moving in opposite directions cancelled: (6.0 − 1.0) × 1018 × 1.60 × 10−19/2.0 = 0.40 A. Opposite charges moving in opposite directions carry current in the same direction.
CThis counts only the electrons: 6.0 × 1018 × 1.60 × 10−19/2.0 = 0.48 A, as if only electrons could be charge carriers.
DCorrect: both flows carry charge in the same direction, so 7.0 × 1018 elementary charges pass in 2.0 s, giving 0.56 A.
23IBP2H-017-01-03Origin of resistance and the heating effectPaper 1AMedium1 mark
The resistance of the tungsten filament in a projector lamp increases as the potential difference across it increases. Which statement best explains this?
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A larger pd drives a larger current, so more energy is dissipated and the filament temperature rises. The lattice ions then vibrate with greater amplitude. This increases the rate of collisions between the free electrons and the lattice, which reduces the drift speed achieved for a given field, so resistance rises. In a metal, n stays essentially constant (A). The I–V gradient actually decreases for a filament lamp (C). The charge of an electron is fixed (D).
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Answer B
Answer: B
Exam tip: Link the resistance of a metal to the rate of electron–ion collisions, not to a change in the number of charge carriers.
24IBP2H-017-01-04Resistance of a non-ohmic componentPaper 1AMedium1 mark
A semiconductor sensor used in a smartwatch has a current of 50 mA when the pd across it is 2.0 V. It has a current of 90 mA when the pd is 2.2 V. What is the resistance of the sensor when the pd is 2.2 V?
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Resistance is defined as R = V/I at the operating point: R = 2.2/0.090 = 24 Ω (2 s.f.). Option A (ΔV/ΔI = 0.20/0.040 = 5.0 Ω) is the reciprocal of the local gradient of the I–V graph. That is not the resistance of a non-ohmic device. Option C (2.0/0.050 = 40 Ω) is the resistance at 2.0 V, not at 2.2 V. Option D (0.090/2.2 = 0.041) inverts the ratio and calculates I/V.
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Answer B
Answer: B
Common error: For a non-ohmic conductor, calculate R = V/I using the values at the point in question. Do not use ΔV/ΔI between two points; that is the reciprocal of the I–V gradient, not the resistance.
25IBP2H-017-01-05Internal resistance and lamp brightnessPaper 1AMedium1 mark
Three identical resistors X, Y and Z are connected to a cell of negligible internal resistance. X is in series with a parallel combination of Y and Z. Resistor Z then breaks and becomes an open circuit. What happens to the pd across X and the pd across Y?
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With resistance R for each resistor, the initial total resistance is R + R/2 = 1.5R. The pd across X is then 2ε/3 and the pd across Y is ε/3. After Z breaks, X and Y are simply in series with total 2R, so each has ε/2. The pd across X falls from 2ε/3 to ε/2, and the pd across Y rises from ε/3 to ε/2. Option A assumes the extra current is diverted through X. Option C ignores the fact that the pds must still sum to ε. Option D assumes parallel branches are independent of each other.
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Answer B
Answer: B
Exam tip: Removing a parallel branch increases the total resistance, so the circuit current falls. Re-share the emf between the series sections.
In microchip manufacture, a gold bonding wire of resistance R is drawn out so that its diameter halves. Its volume and resistivity remain constant. What is its new resistance?
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When the diameter halves, the area falls to A/4. Constant volume then requires the length to become 4L. From R = ρL/A, the new resistance is ρ(4L)/(A/4) = 16R. Option B accounts for only one of the two changes. Option A assumes the length doubles and ignores the area change. Option C combines a doubled length with a quartered area.
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Answer D
Answer: D
Common error: When a wire is stretched at constant volume, the length and the area both change. Both changes increase the resistance.
A copper wire of resistance R is pulled through a die so that its diameter is halved. The volume and resistivity of the copper are unchanged. What is the new resistance of the wire?
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Halving the diameter reduces the cross-sectional area by a factor of 4. Because the volume AL is constant, the length increases by a factor of 4. Since R = ρL/A, the resistance increases by 4 × 4 = 16. The option 4R uses only the area change and ignores the increase in length. The option 2R treats resistance as inversely proportional to diameter. The option 8R assumes the length only doubles.
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Answer D
Answer: D
Common error: When a wire is stretched or drawn at constant volume, both L and A change. Use AL = constant to link them before you apply R = ρL/A.
A touch-screen uses a thin transparent conducting film. A square of film with side L and thickness t has resistance R between two opposite edges. A second square is made of the same material, with side 3L and thickness t/2. What is its resistance between opposite edges?
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For a square of side L, R = ρL/(Lt) = ρ/t, so the side length cancels. Halving the thickness doubles the resistance, giving 2R. The option 6R triples the length but ignores the increase in width. The option 2R/3 includes the extra width but ignores the extra length. The option 2R/9 treats resistance as inversely proportional to the face area L2.
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Answer C
Answer: C
Exam tip: Always identify the length along which the current flows and the cross-section it flows through. For a film, that cross-section is (width × thickness).
29IBP2H-017-02-06Solar cells as energy sourcesPaper 1AMedium1 mark
A small solar panel is modelled as 40 identical cells. Each cell has emf 0.60 V and internal resistance 0.20 Ω. The cells are arranged as 4 parallel strings, and each string has 10 cells in series. What are the emf and the internal resistance of the panel?
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Each string has emf 10 × 0.60 = 6.0 V and internal resistance 10 × 0.20 = 2.0 Ω. Four identical strings in parallel keep the emf at 6.0 V. Their combined resistance is 2.0/4 = 0.50 Ω. The option 6.0 V and 2.0 Ω ignores the parallel combination. The option 24 V adds the emfs of all 40 cells. The option 8.0 Ω adds all the internal resistances as if every cell were in series.
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Answer C
Answer: C
Common error: Identical sources in parallel do not add their emfs. Connecting them in parallel only reduces the internal resistance, which allows a larger current.
30IBP2H-017-02-07Conductors and insulatorsPaper 1AMedium1 mark
The temperature of a metal wire and of a sample of pure silicon (a semiconductor) is increased. Which statement is correct?
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In a metal, the number of free electrons is essentially fixed. Larger lattice vibrations scatter the electrons more often, so the resistance rises. In a semiconductor, the extra thermal energy frees many more charge carriers. This effect outweighs the increased scattering, so the resistance falls. The first option reverses both trends. The second option ignores the change in carrier density in silicon. The third option gives the wrong reason for the metal.
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Answer D
Answer: D
Exam tip: Explain resistance changes in terms of two factors: the number density of charge carriers and how often they collide. State which factor dominates.
31IBP2H-017-00-02PotentiometersPaper 1AMedium1 mark
A battery of emf ε and internal resistance r is connected to a variable resistor of resistance R. The resistor is adjusted until R = r. Which statement is correct for this setting?
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The power in the external resistor is P = ε2R/(R + r)2, and this is greatest when R = r. The current is the same through R and r, so the two dissipate equal power and the efficiency is R/(R + r) = 50%. Option A would need zero current. Option B describes R = 0, which is a short circuit. Option D is wrong because efficiency keeps increasing as R increases, approaching 100% for R ≫ r.
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Answer C
Answer: C
Exam tip: Maximum power transfer and maximum efficiency are different conditions. Maximum power occurs at R = r, where the efficiency is only 50%.
32IBP2H-017-00-03Light-dependent resistor sensorPaper 1AMedium1 mark
A greenhouse grow-lamp controller uses a potential divider made of a light-dependent resistor (LDR) in series with a fixed resistor, connected across a constant-voltage supply. The output voltage is taken across the fixed resistor. Evening falls and the light level decreases. What happens to the output voltage, and why?
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In lower light the LDR resistance increases, so the total resistance increases and the current decreases. The pd across the fixed resistor is IRfixed, so it decreases. Equivalently, the LDR now takes a larger share of the supply voltage. Option A gives the correct reason for the change in the pd across the LDR, not across the fixed resistor. Option C ignores how the supply voltage is shared. Option D gets the direction of the current change wrong.
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Answer B
Answer: B
Common error: Always check which component the output is taken across. When one resistance rises, the pd across that component rises and the pd across the other component falls.
33IBP2H-017-00-04Power in series and parallelPaper 1AMedium1 mark
Three identical resistors X, Y and Z are connected to a battery. Y and Z are in parallel with each other, and this pair is in series with X. What is the ratio of the power dissipated in X to the power dissipated in Y?
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All of the current I passes through X. It then splits equally, so Y carries I/2. Using P = I2R with equal resistances, PX : PY = I2 : (I/2)2 = 4 : 1. Option C (2 : 1) assumes power is proportional to current rather than current squared. Options A and B invert the ratio.
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Answer D
Answer: D
Exam tip: For components with equal resistance, use P = I²R. Halving the current quarters the power.
34IBP2H-017-00-06A thermistor thermometerPaper 1AMedium1 mark
Two 10 kΩ resistors are connected in series across a 12 V supply of negligible internal resistance. A voltmeter of resistance 10 kΩ is connected across one of the resistors. What does the voltmeter read?
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The voltmeter in parallel with one 10 kΩ resistor gives a combined resistance of 5 kΩ. The divider is now 5 kΩ in series with 10 kΩ, so the reading is 12 × 5/15 = 4.0 V. Option B (6.0 V) treats the voltmeter as ideal. Option C (8.0 V) is the pd across the other resistor. Option D (3.0 V) comes from wrongly dividing 12 V into four equal shares.
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Answer A
Answer: A
Exam tip: A real voltmeter with a resistance comparable to the component it measures lowers the reading. It should have a resistance much greater than that component.
Two 1.0 kΩ resistors in series are connected across a 12 V supply of negligible internal resistance. A 1.0 kΩ load is then connected across the lower resistor.
What is the potential difference across the load?
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Step 1Without the load the divider gives 6.0 V; connecting the load changes the ratio.
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All 4 steps must be completed — there is no mark for a part-answer.
Step 2The load and the lower resistor in parallel: (1.0 × 1.0)/(1.0 + 1.0) = 0.50 kΩ.
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Step 3The divider is now 1.0 kΩ above 0.50 kΩ: Vout = 12 × 0.50/(1.0 + 0.50).
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Step 4Vout = 4.0 V.
✓ 1
Answer B
Answer: B · 4 stages of work, one mark
Every option, and why
AThis uses the original total resistance 2.0 kΩ instead of 1.5 kΩ: 12 × 0.50/2.0 = 3.0 V.
BCorrect: the loaded lower section is 0.50 kΩ, taking 1/3 of the 12 V.
CThis ignores the effect of the load on the divider ratio.
37B-1A-52Ohmic and non-ohmic conductorsPaper 1AHard1 mark
The graph shows the I–V characteristics of a filament lamp and of a resistor.
The lamp and the resistor are connected in series to a cell of emf 3.0 V and negligible internal resistance. What is the current in the circuit?
I–V characteristics of the lamp and the resistor (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1In series the current is the same in both components and their potential differences add to 3.0 V.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Find the current at which Vlamp + Vresistor = 3.0 V: at 0.20 A the lamp has 1.0 V and the resistor 2.0 V.
—
Step 3So the current is 0.20 A (check: at 0.30 A the resistor alone already needs 3.0 V).
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis treats the lamp as a fixed 10 Ω, its resistance at the crossing point (4.0 V, 0.40 A): 3.0/20 = 0.15 A. The lamp's resistance at low current is smaller.
BCorrect: the only current for which the two pds add to 3.0 V.
CThis ignores the lamp: 3.0 V across the 10 Ω resistor alone.
DThis reads the lamp curve at 3.0 V, as if the lamp had the whole emf across it.
38B-1A-60Internal resistance and lamp brightnessPaper 1AHard1 mark
A cell of emf 6.0 V and internal resistance 2.0 Ω is connected to two identical lamps, L1 and L2, as shown. Each lamp has a constant resistance of 4.0 Ω. Switch S is initially open and is then closed.
What is the ratio (power in L1 with S closed)/(power in L1 with S open)?
The dashed box represents the cell, with its internal resistance shown separately. Each lamp has a constant resistance of 4.0 Ω.Show mark scheme
Marking point
Mark
Notes
Step 1S open: I = 6.0/(4.0 + 2.0) = 1.0 A, so the power in L1 = 1.0² × 4.0 = 4.0 W.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2S closed: the lamps in parallel have 2.0 Ω, so I = 6.0/(2.0 + 2.0) = 1.5 A and each lamp carries 0.75 A.
—
Step 3Power in L1 = 0.75² × 4.0 = 2.25 W; ratio = 2.25/4.0 = 9/16.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis assumes the total current stays at 1.0 A and simply divides between the lamps, giving 0.50 A in L1: (0.50/1.0)² = 1/4.
BCorrect: the terminal pd falls from 4.0 V to 3.0 V, so the power in L1 falls by (3.0/4.0)² = 9/16.
CThis is the ratio of the currents in L1 (0.75/1.0); power depends on the square of the current.
DThis ignores the internal resistance: with an ideal cell the pd across L1 would stay at 6.0 V and its power would not change.
39B-1A-70emf and internal resistancePaper 1AHard1 mark
A cell is connected to a variable resistor. The graph shows how the power P dissipated in the variable resistor varies with its resistance R.
What is the emf of the cell?
Power dissipated in the variable resistor against its resistance (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1Two readings: R = 1.0 Ω, P = 1.00 W gives I = √(P/R) = 1.00 A and terminal pd 1.00 V; R = 4.0 Ω, P = 1.00 W gives I = 0.50 A and terminal pd 2.00 V.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2ε = V + Ir for both: 1.00 + 1.00r = 2.00 + 0.50r, so r = 2.0 Ω.
—
Step 3ε = 1.00 + 1.00 × 2.0 = 3.0 V. (Check at the maximum: I = √(1.125/2.0) = 0.75 A and 0.75 × (2.0 + 2.0) = 3.0 V.)
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis is the terminal pd at the maximum of the graph (√(1.125 × 2.0)); at that point half of the emf is across the internal resistance.
BThis takes R = r = 2.0 Ω at the maximum but then uses P = ε2/(R + r), treating the power in the resistor as the total power: √(1.125 × 4.0) = 2.1 V.
CThis takes the terminal pd at the largest resistance shown, 16 Ω, as the emf: √(0.44 × 16) = 2.7 V. The current there is still 0.17 A, so 0.33 V is across the internal resistance.
DCorrect: two points on the graph give two equations ε = I(R + r), which give r = 2.0 Ω and ε = 3.0 V.
40B-1A-72Ohmic and non-ohmic conductorsPaper 1AHard1 mark
The graph shows the I–V characteristic of a filament lamp.
Two identical lamps with this characteristic are connected in series to a supply of emf 8.0 V and negligible internal resistance. One of the lamps is then connected on its own to the same supply.
What is (total power dissipated by the two lamps in series)/(power dissipated by the single lamp)?
Current against potential difference for a filament lamp (drawn to scale)Show mark scheme
Marking point
Mark
Notes
Step 1In series the identical lamps share the emf equally: 4.0 V each. From the graph the current is 0.20 A, so the total power = 8.0 × 0.20 = 1.6 W.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2On its own the lamp has 8.0 V across it. From the graph the current is 0.28 A, so the power = 8.0 × 0.283 = 2.26 W.
—
Step 3Ratio = 1.6/2.26 = 0.71. (If the resistance were constant the ratio would be 0.50; the lamps in series are cooler and have a lower resistance, so they take relatively more current.)
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis treats the resistance as constant and gives the power in only one of the two lamps: (4.0 V)2/R ÷ (8.0 V)2/R = 0.25.
BThis uses the graph correctly but includes the power of only one lamp of the series pair: 4.0 × 0.20/2.26 = 0.35.
CThis treats the lamp resistance as constant: two equal resistances in series take half the power of one, V2/2R ÷ V2/R = 0.50. The cooler lamps in series have a smaller resistance.
DCorrect: the graph gives 0.20 A at 4.0 V and 0.28 A at 8.0 V, so the ratio is 8.0 × 0.20/(8.0 × 0.283) = 0.71.
41IBP2H-017-01-07Power in series and parallelPaper 1AHard1 mark
A 1000 W heater and a 500 W heater are both rated for operation at 230 V. They are connected in series across a 230 V supply. Assume their resistances are constant. What is the total power dissipated?
Show mark scheme
Marking point
Mark
Notes
The resistances are R1 = V2/1000 and R2 = V2/500 = 2R1. In series the total resistance is 3R1 = 3V2/1000. The power is therefore V2/(3V2/1000) = 1000/3 ≈ 333 W, which is less than either rating. Option A adds the powers as if the heaters were in parallel. Option B averages the ratings. Option C assumes the smaller heater limits the power.
✓ 1
Answer D
Answer: D
Exam tip: Power ratings apply only at the rated voltage. Work out each resistance first, then analyse the new circuit.
42IBP2H-017-01-08Origin of resistance and the heating effectPaper 1AHard1 mark
Wires P and Q have identical dimensions and are connected in series, so they carry the same current. P has twice the free-electron number density of Q and half the resistivity of Q. Which row gives the drift speed ratio \(\frac{v_P}{v_Q}\) and the electric field ratio \(\frac{E_P}{E_Q}\)?
Show mark scheme
Marking point
Mark
Notes
From I = nAvq with the same I and A, v ∝ 1/n, so vP/vQ = 1/2. The field in each wire is E = V/L = IR/L = Iρ/A, so E ∝ ρ and EP/EQ = 1/2. Option B assumes more electrons move faster. Option C assumes a slower drift needs a stronger field. Option D assumes the same current means identical conditions in both wires.
✓ 1
Answer A
Answer: A
Exam tip: Series wires share the current, not the drift speed or the field. Use I = nAvq and R = ρL/A separately.
43IBP2H-017-02-04emf and internal resistancePaper 1AHard1 mark
A chemical cell of emf ε and internal resistance r is connected to a variable resistor of resistance R. The resistance R is reduced steadily from a very large value to zero. What happens to the terminal potential difference of the cell, and to the power dissipated in the variable resistor?
Show mark scheme
Marking point
Mark
Notes
The terminal pd is V = εR/(R + r). This falls from ε to zero as R falls. The power in the resistor is P = ε2R/(R + r)2. This is close to zero when R is very large, because the current is tiny. It is also zero when R = 0, because then there is no pd across the resistor. In between, it reaches a maximum when R = r. The first option misses this maximum. The third option ignores the lost volts across r. The fourth option has the pd trend reversed.
✓ 1
Answer B
Answer: B
Exam tip: Maximum power is transferred to the external resistor when R = r. At that point only half of the cell's energy reaches the load, so the efficiency is 50%.
44IBP2H-017-00-05Ohmic and non-ohmic conductorsPaper 1AHard1 mark
Two filament lamps are rated "230 V, 60 W" and "230 V, 100 W". They are connected in series to a 230 V mains supply. Assume the resistance of each lamp stays constant. Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
From R = V2/P, the 60 W lamp has R ≈ 880 Ω and the 100 W lamp has R ≈ 530 Ω. In series they carry the same current, so P = I2R is larger for the lamp with the larger resistance, which is the 60 W lamp. The pd across each lamp is less than 230 V, so each dissipates less than its rated power (about 23 W and 14 W). Option A applies the rating to a situation it does not describe. Option B ignores how the supply pd is shared. Option D overlooks that the resistances are different.
✓ 1
Answer C
Answer: C
Common error: A power rating applies only at the rated pd. Use it to find the resistance first, then analyse the actual circuit.
45IBP2H-017-00-07Solar cells as energy sourcesPaper 1AHard1 mark
A cell is tested by connecting it to different loads. When the current is 0.50 A the terminal pd is 2.7 V. When the current is 1.5 A the terminal pd is 2.1 V. What are the emf ε and internal resistance r of the cell?
Show mark scheme
Marking point
Mark
Notes
The terminal pd is V = ε − Ir. Subtracting the two equations gives 0.6 = 1.0 × r, so r = 0.60 Ω. Then ε = 2.7 + 0.50 × 0.60 = 3.0 V. Option A takes the first terminal pd as the emf, but the emf is the terminal pd at zero current. Option B subtracts Ir instead of adding it. Option C uses the reciprocal of the gradient for r.
✓ 1
Answer D
Answer: D
Exam tip: On a graph of V against I, the intercept on the V axis is ε and the gradient is −r. Extrapolate to I = 0 to find the emf.
46B-1B-20Ohmic behaviour and the heating effectPaper 1BEasy7 marks
A coil of thin copper wire is placed in a large water bath kept at 20 °C, and the potential difference V across it is measured for a range of currents I. The experiment is repeated with the coil in still air at 20 °C, each reading being taken once the coil has reached a steady temperature. The table and graph show the results; the line is the line of best fit for the water bath. The resistance of copper varies with temperature θ as R = R20(1 + 3.9 × 10−3 K−1 × (θ − 20 °C)).
I / A
0.20
0.40
0.60
0.80
1.00
1.20
V / V (water bath)
1.48
3.01
4.51
5.99
7.48
9.00
V / V (air)
1.51
3.07
4.68
6.45
8.42
10.59
Navy: coil in a water bath, with the line of best fit. Red: coil in air (drawn to scale).
(a)
(i)
Determine the resistance of the coil at 20 °C.
(2)
(b)
(i)
Determine the temperature of the coil in air when the current is 1.20 A.
(2)
(c)
(i)
Determine the rate of energy transfer from the coil to the air per kelvin of temperature difference at 1.20 A.
(2)
(d)
(i)
Explain, with reference to the lattice ions and the free electrons, why the resistance of the coil in air increases at large currents.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient of the water-bath line from well-separated points, e.g. 9.00/1.20
✓ 1
A straight line through the origin: the coil is ohmic at constant temperature.
R20 = 7.5 Ω
✓ 1
Accept 7.4–7.6 Ω.
Part (b)(i)
R = 10.59/1.20 = 8.83 Ω
✓ 1
θ − 20 = (8.83/7.5 − 1)/3.9 × 10−3 = 45 K, so θ ≈ 65 °C
✓ 1
Allow ECF from (a). Accept 60–70 °C.
Part (c)(i)
Power = IV = 10.59 × 1.20 = 12.7 W, all transferred to the air in the steady state
✓ 1
Rate per kelvin = 12.7/45 = 0.28 W K−1
✓ 1
Allow ECF from (b). Accept 0.25–0.32 W K−1.
Part (d)(i)
The coil is hotter, so the lattice ions vibrate with larger amplitude and the free electrons collide with them more frequently, reducing the drift for a given p.d.
✓ 1
Answers: (a)(i) 7.5 Ω · (b)(i) ≈ 65 °C · (c)(i) 0.28 W K−1(the remaining parts are explanations — see the table above)
A student measures the resistance R of different lengths L of a constantan wire with an ohmmeter connected to two crocodile clips. The mean diameter of the wire, measured with a micrometer at several points, is 0.32 mm with an uncertainty of ±0.01 mm. The graph shows R against L with the line of best fit; the line does not pass through the origin because of the resistance of the clips and leads.
L / m
0.200
0.400
0.600
0.800
1.000
R / Ω
1.62
2.84
4.05
5.26
6.50
R against L with the line of best fit (drawn to scale)
(a)
(i)
Determine the gradient of the line. Give a unit for your answer.
(2)
(b)
(i)
Determine the resistivity of constantan.
(2)
(c)
(i)
The percentage uncertainty in the gradient is 2 %. Determine the absolute uncertainty in your answer to (b).
(2)
(d)
(i)
A current of 0.150 A flows in the wire. Determine the electric field strength inside the wire.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (6.49 − 0.40)/1.000
✓ 1
The intercept (≈ 0.40 Ω, a systematic offset) does not affect the gradient.
= 6.1 Ω m−1
✓ 1
Accept 5.9–6.3. Unit required.
Part (b)(i)
A = πd²/4 = π(0.32 × 10−3)²/4 = 8.04 × 10−8 m²
✓ 1
ρ = gradient × A = 6.1 × 8.04 × 10−8 = 4.9 × 10−7 Ω m
Allow ECF from (b). Absolute uncertainty to 1 s.f.
Part (d)(i)
p.d. per metre of wire = I × gradient, so E = V/d = 0.150 × 6.1 = 0.91 V m−1
✓ 1
Allow ECF from (a). Using 6.50 Ω (which includes the clips) gives 0.98 V m−1: [0].
Answers: (a)(i) 6.1 Ω m−1 · (b)(i) 4.9 × 10−7 Ω m · (c)(i) ±0.4 × 10−7 Ω m · (d)(i) 0.91 V m−1(the remaining parts are explanations — see the table above)
48B-1B-23Charge and energy from a chemical cellPaper 1BEasy7 marks
A student investigates how much charge and energy a fully charged rechargeable cell can deliver. The cell is connected to an electronic load that keeps the current constant at 0.500 A, measured to ±1 %. A data logger records the terminal potential difference V of the cell at intervals until V falls to 1.00 V, when the load switches off (the cut-off).
The table gives the readings and the graph shows them joined by straight lines. The cut-off is marked by the dashed line.
t / h
0.00
0.25
0.50
1.00
1.50
2.00
2.50
3.00
3.50
3.75
4.00
4.25
V / V
1.38
1.30
1.27
1.25
1.24
1.23
1.21
1.19
1.15
1.11
1.05
0.93
Terminal potential difference against time during the discharge (drawn to scale)
(a)
(i)
Determine the time, in hours, at which the cut-off is reached.
(1)
(b)
(i)
Calculate the total charge that passes through the load.
(1)
(c)
(i)
Estimate the energy transferred by the cell to the load.
(2)
(d)
(i)
The time of the cut-off can be judged to ±0.05 h. Determine the absolute uncertainty in the charge.
(2)
(e)
(i)
The manufacturer states that the cell delivers 2.40 A h when it is discharged at 0.20 A to 1.00 V. Suggest why the charge found by the student is smaller than this.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Reading where the graph crosses 1.00 V: t ≈ 4.10 h
✓ 1
Accept 4.05–4.15 h. Linear interpolation between 4.00 h and 4.25 h gives 4.10 h.
Part (b)(i)
q = It = 0.500 × 4.10 × 3600 = 7.38 × 103 C
✓ 1
Allow ECF from (a). The time must be converted to seconds; 2.05 (A h) scores [0] unless converted.
Part (c)(i)
Energy = I × (area under the V–t graph), or mean pd × charge; the mean pd from the graph is about 1.21 V
✓ 1
Accept a mean pd of 1.20–1.24 V, or the area under the graph as 4.9–5.1 V h. Using 1.2 V (nominal) is acceptable.
Energy ≈ 1.21 × 7.38 × 103 = 9.0 × 103 J
✓ 1
Allow ECF from (b). Accept 8.6–9.4 × 103 J. Award [1 max] for 7.4 × 103 J (energy = Vq with the cut-off pd of 1.00 V).
Allow ECF from (a). Percentage (or fractional) uncertainties must be added.
Absolute uncertainty = 0.022 × 7.38 × 103 ≈ ±160 C (≈ ±2 × 102 C)
✓ 1
Allow ECF from (b). Accept ±150 C to ±200 C.
Part (e)(i)
At the larger current the pd across the internal resistance (Ir) is larger, so the terminal pd reaches the 1.00 V cut-off earlier (more of the energy is dissipated inside the cell)
✓ 1
Reference to the internal resistance or the "lost volts" is required; "the cell is faulty" or "energy is wasted" alone scores [0].
Answers: (a)(i) 4.10 h · (b)(i) 7.38 × 103 C · (c)(i) 9.0 × 103 J · (d)(i) ±160 C (the remaining parts are explanations — see the table above)
49B-1B-05Efficiency of an electric motorPaper 1BMedium7 marks
A small dc motor, connected to a 6.00 V supply, winds a thread onto an axle and lifts a load of mass 0.300 kg at constant speed through a vertical height of 0.800 m. The current is 0.62 A throughout. The student times the lift five times with a hand-held stopwatch (table). The uncertainties are: height ±0.005 m, current ±0.01 A, potential difference ±0.01 V. When the axle is held so that it cannot turn, the current from the same supply is 1.50 A.
Trial
1
2
3
4
5
t / s
2.38
2.52
2.45
2.41
2.49
(a)
(i)
Determine the efficiency of the motor.
(2)
(b)
(i)
Determine the absolute uncertainty in your answer to (a).
(2)
(c)
(i)
The resistance of the motor coil can be found from the current when the axle is held. Determine the rate at which energy is transferred to forms other than the work done on the load and the heating of the coil.
(2)
(d)
(i)
Using your answer to (b), suggest how the student could reduce the uncertainty in the efficiency using the same stopwatch.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Mean time = 2.45 s; output power = mgh/t = 0.300 × 9.81 × 0.800/2.45 = 0.961 W
✓ 1
Input power = 6.00 × 0.62 = 3.72 W, so η = 0.26 (26 %)
✓ 1
Accept 0.26.
Part (b)(i)
Δt = half the range = 0.07 s, i.e. 2.9 %; total = 2.9 + 0.6 + 1.6 + 0.2 = 5.3 %
✓ 1
Fractional uncertainties of t, h, I and V add.
Δη = 0.053 × 0.26 = ±0.01 (η = 0.26 ± 0.01)
✓ 1
Allow ECF from (a). Absolute uncertainty to 1 s.f.
Part (c)(i)
Coil resistance = 6.00/1.50 = 4.0 Ω; heating of coil = I²R = 0.62² × 4.0 = 1.54 W
✓ 1
When the axle cannot turn there is no back emf, so R = V/I.
Remaining power = 3.72 − 0.961 − 1.54 = 1.2 W (friction in the bearings and gears, air resistance, sound)
✓ 1
Allow ECF from (a). Accept 1.1–1.3 W.
Part (d)(i)
The timing contributes most (≈ 3 % of ≈ 5.3 %); lift the load through a much greater height (e.g. 2.0 m) so that the same ±0.07 s is a smaller fraction of a longer time
✓ 1
Allow ECF from (b). "Repeat more times" alone scores [0]: it does not reduce the reaction-time spread of each reading as effectively and does not use the analysis in (b).
Answers: (a)(i) 0.26 · (b)(i) ±0.01 · (c)(i) 1.2 W (the remaining parts are explanations — see the table above)
50B-1B-22emf and internal resistance of a lemon cellPaper 1BMedium7 marks
A zinc strip and a copper strip are pushed into a lemon to make a chemical cell of emf ε and internal resistance r. The cell is connected in series with a resistance box of resistance R and a microammeter. The current I is recorded for several values of R. The student assumes the microammeter is ideal, so that 1/I = R/ε + r/ε, and plots 1/I against R (graph). The microammeter in fact has a constant resistance of 0.10 kΩ.
R / kΩ
0.50
1.00
1.50
2.00
2.50
3.00
I / μA
765
539
421
341
289
250
(1/I) / mA−1
1.31
1.86
2.38
2.93
3.46
4.00
1/I against R with the line of best fit (drawn to scale)
(a)
(i)
Determine ε from the graph.
(2)
(b)
(i)
Determine the value of r given by the student's analysis.
(2)
(c)
(i)
Explain how the resistance of the microammeter affects your answers to (a) and (b), and determine a corrected value of r.
(2)
(d)
(i)
A student suggests that a single lemon cell could light an LED that needs a current of 20 mA. Use your answers to (a) and (c) to comment on this suggestion.
Any valid pair of well-separated points on the line.
ε = 1/gradient = 0.93 V
✓ 1
Accept 0.91–0.95 V.
Part (b)(i)
Intercept on the 1/I axis ≈ 0.78 mA−1 = r/ε
✓ 1
Accept 0.74–0.82 mA−1.
r = 0.78 × 0.93 = 0.72 kΩ
✓ 1
Allow ECF from (a). Accept 0.68–0.76 kΩ; or intercept/gradient.
Part (c)(i)
The meter resistance is in series with R, so the intercept gives (r + 0.10 kΩ)/ε: the value in (b) is too large, while the gradient, and so ε, is unaffected
✓ 1
r = 0.72 − 0.10 = 0.62 kΩ
✓ 1
Allow ECF from (b).
Part (d)(i)
The largest possible current is ε/r = 0.93/620 Ω ≈ 1.5 mA, far less than 20 mA, so the suggestion is wrong
✓ 1
Allow ECF from (a) and (c).
Answers: (a)(i) 0.93 V · (b)(i) 0.72 kΩ · (c)(i) 0.62 kΩ (the remaining parts are explanations — see the table above)
A uniform resistance wire of length 1.000 m and resistance 10.0 Ω is connected across a cell of emf 3.00 V and unknown internal resistance r. A sliding contact on the wire is attached to a trolley on a straight track inclined at 10.0° to the horizontal. An ideal voltmeter measures the potential difference V between one end of the wire and the contact, at distance x from that end. The student first calibrates the sensor by placing the contact at known positions (table and graph). The trolley is then released from rest with the contact at x = 0.100 m.
x / m
0.100
0.300
0.500
0.700
0.900
V / V
0.290
0.851
1.434
1.996
2.573
Calibration of the slide-wire sensor: V against x with the line of best fit (drawn to scale)
(a)
(i)
Determine the gradient of the calibration graph. Give a unit for your answer.
(2)
(b)
(i)
Determine r.
(2)
(c)
(i)
At 0.80 s after release the voltmeter reads 1.703 V. Determine the acceleration of the trolley.
(2)
(ii)
Compare your answer to (c)(i) with the acceleration expected for a trolley moving without friction, and suggest a cause of any difference.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. 2.573/0.900 = 2.86
✓ 1
Accept 2.83–2.88.
Unit V m−1
✓ 1
Part (b)(i)
Current = ε/(10.0 + r); gradient = current × 10.0 Ω m−1, so 10.0 + r = 3.00 × 10.0/2.86 = 10.5 Ω
✓ 1
The wire has 10.0 Ω per metre.
r = 0.5 Ω
✓ 1
Allow ECF from (a). Accept 0.4–0.6 Ω.
Part (c)(i)
x = 1.703/2.86 = 0.596 m, so the displacement is 0.596 − 0.100 = 0.496 m
✓ 1
Allow ECF from (a). The starting position must be subtracted.
a = 2s/t² = 2 × 0.496/0.80² = 1.55 m s−2
✓ 1
Accept 1.50–1.60 m s−2.
Part (c)(ii)
Frictionless: g sin 10.0° = 1.70 m s−2; the measured value is about 9 % smaller, caused by friction at the sliding contact (and in the wheels)
✓ 1
Allow ECF from (c)(i).
Answers: (a)(i) 2.86 V m−1 · (b)(i) 0.5 Ω · (c)(i) 1.55 m s−2(the remaining parts are explanations — see the table above)
52B-2-36Resistivity and heating stripsPaper 2Easy12 marks
The rear window of a car is demisted by 8 identical strips of silver paste printed on the glass. The strips are connected in parallel between two conducting busbars, which are connected to the car's 12.0 V supply of negligible internal resistance. Neglect the resistance of the busbars.
Each strip is 1.10 m long, 1.00 mm wide and 8.0 μm thick. Resistivity of the paste = 4.0 × 10−8 Ω m.
(a)
(i)
Show that the resistance of one strip is about 5.5 Ω.
(2)
(ii)
Calculate the total resistance of the demister.
(1)
(iii)
Calculate the total power dissipated in the demister.
(1)
(b)
(i)
One strip is scratched and breaks. State and explain the effect on the power dissipated in each of the other strips and on the total power.
(2)
(c)
(i)
On a cold morning the window is covered by a layer of frost (ice) of area 0.90 m² and thickness 0.12 mm at −6.0 °C. 40 % of the power of the demister is transferred to the frost. Determine the time taken to melt all the frost. Density of ice = 917 kg m−3; specific heat capacity of ice = 2.1 × 103 J kg−1 K−1; specific latent heat of fusion of ice = 3.34 × 105 J kg−1.
(3)
(ii)
Suggest why the frost takes longer to melt when the car is moving.
(1)
(d)
(i)
The engine converts the chemical energy of petrol to electrical energy with an overall efficiency of 20 %. The energy density of petrol is 4.6 × 107 J kg−1. Determine the mass of petrol used to run the demister for 10 minutes.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Cross-sectional area = 1.00 × 10−3 × 8.0 × 10−6 = 8.0 × 10−9 m²
✓ 1
R = ρL/A = 4.0 × 10−8 × 1.10/8.0 × 10−9 = 5.50 Ω
✓ 1
Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
R = 5.50/8 = 0.69 Ω
✓ 1
Allow ECF from (a)(i).
Part (a)(iii)
P = V2/R = 12.02/0.6875 = 209 W
✓ 1
Allow ECF from (a)(ii). ALT: 8 × 12.0²/5.50.
Part (b)(i)
The strips are in parallel, so the pd across each remaining strip is still 12.0 V and its current and power (26 W) are unchanged
✓ 1
An explanation in terms of the unchanged pd is required.
The total power falls by the power of one strip, to 7/8 of its value (183 W)
✓ 1
Allow ECF from (a)(iii).
Part (c)(i)
Mass of frost = 917 × 0.90 × 0.12 × 10−3 = 0.0990 kg
Both the warming to 0 °C and the melting must be included; award this mark for 3.31 × 104 J if the warming is omitted, then ECF.
Time = 3.43 × 104/(0.40 × 209) = 410 s (about 7 minutes)
✓ 1
Allow ECF from (a)(iii). Accept 400–420 s.
Part (c)(ii)
Air moving over the outside of the window removes energy from the glass and frost faster (by convection), so a smaller fraction of the power goes into the frost
✓ 1
Reference to increased energy loss to the air is required.
Part (d)(i)
Chemical energy needed = 209 × 600/0.20 = 6.3 × 105 J
✓ 1
Allow ECF from (a)(iii).
Mass = 6.3 × 105/4.6 × 107 = 0.014 kg (about 14 g)
✓ 1
Award [1 max] for 5.5 × 10−4 kg (efficiency multiplied instead of divided).
Answers: (a)(i) 5.50 Ω · (a)(ii) 0.69 Ω · (a)(iii) 209 W · (c)(i) 410 s · (d)(i) 0.014 kg (the remaining parts are explanations — see the table above)
54IBP2H-017-00-09emf and internal resistance of a batteryPaper 2Easy5 marks
The battery of an electric bicycle has an emf of 36 V and an internal resistance of 0.40 Ω. When the bicycle climbs a hill, the motor draws a current of 12 A.
(a)
Calculate the terminal potential difference of the battery.
(2)
(b)
Calculate the rate at which energy is dissipated inside the battery.
(2)
(c)
Calculate the percentage of the energy supplied by the battery that is delivered to the motor.
55B-2-15A solar panel as a source of emfPaper 2Medium11 marks
A solar module consists of 36 solar cells connected in series. The graph shows the current–voltage (I–V) characteristic of the module when it is illuminated normally by sunlight of intensity 1000 W m−2. The area of the module is 0.36 m².
I–V characteristic of the solar module (drawn to scale)
(a)
(i)
State the emf of the module.
(1)
(b)
(i)
Determine the maximum power that the module can deliver.
(3)
(c)
(i)
Calculate the resistance of the load that receives this maximum power.
(1)
(d)
(i)
Determine the efficiency of the module at maximum power.
(2)
(e)
(i)
A chemical cell can be modelled as a source of constant emf ε in series with a constant internal resistance r. Explain, with reference to the graph, why the module cannot be modelled in this way.
(2)
(f)
(i)
Each photon absorbed in a cell can release at most one electron. When the module is short-circuited the current is 3.0 A. Determine the minimum number of photons that must be absorbed per second by one cell of the module.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
21.6 V (the pd when the current is zero)
✓ 1
Accept 21.4–21.8 V.
Part (b)(i)
VI evaluated for at least two points near the bend of the curve
✓ 1
E.g. 17 V × 2.8 A = 48 W; 19 V × 2.5 A = 48 W.
e.g. 18 V × 2.7 A = 49 W
✓ 1
Maximum power ≈ 49 W
✓ 1
Accept 47–52 W.
Part (c)(i)
R = V/I = 18/2.7 = 6.7 Ω
✓ 1
Allow ECF from (b). Accept 6.2–7.2 Ω.
Part (d)(i)
Incident power = 1000 × 0.36 = 360 W
✓ 1
Efficiency = 49/360 = 0.14
✓ 1
Allow ECF from (b).
Part (e)(i)
For constant ε and r, V = ε − Ir, so the I–V graph would be a straight line (of gradient −1/r)
✓ 1
The graph is curved: almost horizontal at low pd and steep near the emf, so the effective internal resistance changes with the current
✓ 1
Part (f)(i)
Electrons flowing per second = 3.0/1.60 × 10−19 = 1.9 × 1019 s−1
✓ 1
The cells are in series, so the same charge flows through every cell: each cell must absorb at least 1.9 × 1019 photons per second
✓ 1
Award [1 max] for dividing or multiplying by 36.
Answers: (a)(i) 21.6 V · (b)(i) ≈ 49 W · (c)(i) 6.7 Ω · (d)(i) 0.14 · (f)(i) 1.9 × 1019 s−1(the remaining parts are explanations — see the table above)
56B-2-30Resistivity and power in a supply leadPaper 2Medium13 marks
An electric kettle is rated 2.40 kW when the potential difference across it is 230 V. It is used at the end of an extension lead 15.0 m long. The lead contains two copper conductors, one carrying the current to the kettle and one carrying it back; each has a cross-sectional area of 1.00 mm². The resistivity of copper is 1.7 × 10−8 Ω m. Treat the supply as a steady potential difference of 230 V at the socket and the resistance of the kettle as constant.
(a)
(i)
Calculate the resistance of the kettle and the current in it when the pd across it is 230 V.
(2)
(b)
(i)
Show that the resistance of the lead is about 0.5 Ω.
(2)
(c)
(i)
Determine the power of the kettle when it is used with the lead.
(3)
(d)
(i)
Determine the electric field strength inside one copper conductor and the magnitude of the electric force on a conduction electron in it.
(2)
(e)
(i)
The lead is replaced by one of the same length whose conductors have 1.5 times the diameter. Assuming that the current is unchanged, deduce the ratio (power dissipated in the new lead)/(power dissipated in the original lead).
(2)
(f)
(i)
The original lead is used while it is still tightly wound on its drum. Explain why it can become dangerously hot, although it does not when it is unwound.
(2)
Show mark scheme
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Mark
Notes
Part (a)(i)
R = V²/P = 230²/2400 = 22.04 Ω
✓ 1
I = P/V = 2400/230 = 10.4 A
✓ 1
Part (b)(i)
The current flows along both conductors, so the length of copper is 2 × 15.0 = 30.0 m
✓ 1
This hidden step must be seen.
R = ρL/A = 1.7 × 10−8 × 30.0/1.00 × 10−6 = 0.51 Ω
✓ 1
Full substitution or an answer to at least 2 s.f. is required. A value of 0.26 Ω (one conductor) scores [1 max].
Part (c)(i)
Total resistance = 22.04 + 0.51 = 22.55 Ω
✓ 1
Allow ECF from (a) and (b).
I = 230/22.55 = 10.2 A
✓ 1
P = I²R = 10.2² × 22.04 = 2.29 kW
✓ 1
Accept 2.28–2.30 kW (2.35 kW if one conductor only was used in (b)). Award [1 max] for 2.40 kW minus the power in the lead.
Part (d)(i)
pd along one conductor = 10.2 × 0.255 = 2.60 V, so E = 2.60/15.0 = 0.17 V m−1
✓ 1
Allow ECF from (b) and (c). The field is uniform along the wire, so E = V/L.
F = eE = 1.60 × 10−19 × 0.17 = 2.8 × 10−20 N
✓ 1
Allow ECF from the field strength.
Part (e)(i)
R ∝ A−1 ∝ d−2, so the resistance of the new lead is 1/1.5² of the original
✓ 1
Area ∝ diameter² must be used.
P = I²R with I unchanged, so the ratio = 1/2.25 = 0.44
✓ 1
Award [1 max] for 1/1.5 = 0.67.
Part (f)(i)
The rate of thermal energy production per unit length (I²R/L) is the same in both cases
✓ 1
On the drum each turn is surrounded by other warm turns, so little surface is exposed to the air: convection (and radiation) remove energy much more slowly, and the temperature rises until the losses balance the production, which may melt the insulation
✓ 1
"The heat cannot escape" alone scores 0; a reference to the reduced area for convection or radiation is required.
Answers: (a)(i) 22.04 Ω; 10.4 A · (b)(i) 0.51 Ω · (c)(i) 2.29 kW · (d)(i) 0.17 V m−1; 2.8 × 10−20 N · (e)(i) 0.44 (the remaining parts are explanations — see the table above)
57B-2-33Resistance of transmission cablesPaper 2Medium11 marks
A tram is supplied from a substation that maintains a constant potential difference of 600 V between an overhead copper contact wire and the steel rails. The current flows from the substation along the contact wire to the tram and returns along the rails. The contact wire has a cross-sectional area of 1.50 × 10−4 m²; the resistivity of copper is 1.7 × 10−8 Ω m. The rails have a resistance of 0.015 Ω per kilometre.
(a)
(i)
Show that the resistance of 1.0 km of the contact wire is about 0.11 Ω.
(1)
(b)
The tram is 2.0 km from the substation and draws a current of 400 A.
(i)
Determine the total resistance of the supply circuit between the substation and the tram.
(2)
(ii)
Calculate the potential difference between the tram's current collector and the rails.
(1)
(iii)
Determine the fraction of the power supplied by the substation that is dissipated in the supply circuit.
(2)
(c)
(i)
The tram, of mass 3.2 × 104 kg, climbs a straight slope at constant speed while drawing 400 A at the position in (b). The slope rises 5.0 m for every 100 m travelled along it, and the total resistive force on the tram is 2.0 × 103 N. 85 % of the electrical power delivered to the tram is converted to mechanical work by its motors. Determine the speed of the tram.
(3)
(d)
(i)
The tram draws a constant current I. Show that the pd at the tram at a distance x from the substation is V0 − Ikx, where V0 = 600 V and k is the resistance per unit length of the supply circuit (wire and rails together). The tram's equipment needs at least 450 V. Deduce the greatest distance from the substation at which the tram can draw 400 A.
Adding the return path through 2.0 km of rails: 0.227 + 0.030 = 0.257 Ω
✓ 1
The rail return is the hidden step; 0.227 Ω scores [1].
Part (b)(ii)
pd lost in the supply circuit = 400 × 0.257 = 103 V; pd at the tram = 600 − 103 = 497 V
✓ 1
Allow ECF from (b)(i).
Part (b)(iii)
Power dissipated = 400² × 0.257 = 4.11 × 104 W; power supplied = 600 × 400 = 2.40 × 105 W
✓ 1
Allow ECF from (b)(i).
Fraction = 0.17 (17 %)
✓ 1
Accept 0.17. ALT: (600 − 497)/600.
Part (c)(i)
Useful power = 0.85 × 497 × 400 = 1.69 × 105 W
✓ 1
Allow ECF from (b)(ii).
Driving force = mg sin θ + Fr = 3.2 × 104 × 9.81 × 0.050 + 2.0 × 103 = 1.77 × 104 N
✓ 1
Constant speed: zero resultant force along the slope.
v = P/F = 9.6 m s−1
✓ 1
Accept 9.4–9.7 m s−1. Award [2 max] for 11 m s−1 (resistive force omitted).
Part (d)(i)
Resistance of the circuit = kx, so the lost pd = Ikx and the pd at the tram = V0 − Ikx
✓ 1
x = (600 − 450)/(400 × 0.128 Ω km−1) = 2.9 km
✓ 1
Allow ECF from (a). Accept 2.9 km.
Answers: (a)(i) 0.113 Ω · (b)(i) 0.257 Ω · (b)(ii) 497 V · (b)(iii) 0.17 · (c)(i) 9.6 m s−1 · (d)(i) 2.9 km (the remaining parts are explanations — see the table above)
58IBP2H-017-01-10I–V characteristic of a light-emitting diodePaper 2Medium7 marks
The table gives measurements of the current I in a red LED used as a bicycle rear light for different potential differences V across it.
V / V: 1.60, 1.70, 1.80, 1.90, 2.00 I / mA: 0.0, 1.0, 4.0, 10.0, 20.0
Data from the table
(a)
Explain, using the data, whether the LED is an ohmic conductor.
(2)
(b)
Calculate the resistance of the LED when V = 2.00 V.
(1)
(c)
The LED is to operate at 20.0 mA from a 5.0 V battery of negligible internal resistance, with a fixed resistor in series. Use the table to calculate the resistance of the fixed resistor.
(2)
(d)
Calculate the power dissipated in the fixed resistor.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Valid quantitative use of the data, e.g. V/I calculated at two voltages (1700 Ω at 1.70 V and 100 Ω at 2.00 V), OR equal 0.10 V steps give unequal increases in current, OR I = 0 at a non-zero V ; so the current is not proportional to V / the resistance is not constant, and the LED is non-ohmic
✓ 2
Part (b)
R = 2.00/0.0200 = 100 Ω
✓ 1
Part (c)
pd across the resistor = 5.0 − 2.00 = 3.0 V ; R = 3.0/0.0200 = 150 Ω
✓ 2
Part (d)
P = VI = 3.0 × 0.0200, or P = I2R = 0.02002 × 150 ; P = 0.060 W . ECF from (c)
✓ 2
Exam tip: To show that a component is non-ohmic, back up your answer with numbers from the data, such as two calculated V/I values, rather than just describing the trend.
A welding machine is connected by a copper cable of length 5.0 m and cross-sectional area 25 mm2. The cable carries a current of 150 A. Number density of free electrons in copper = 8.5×1028 m−3 Resistivity of copper = 1.7×10−8 Ω m
(a)
Calculate the drift speed of the free electrons in the cable.
(2)
(b)
Calculate the time taken for an electron to drift the full length of the cable.
(1)
(c)
The welding arc starts almost immediately the machine is switched on. Explain this, given your answer to (b).
(2)
(d)
Calculate the power dissipated in the cable.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = I/(nAe) = 150/(8.5×1028 × 25×10−6 × 1.6×10−19) ; v = 4.4×10−4 m s−1
✓ 2
Part (b)
t = 5.0/4.41×10−4 = 1.1×104 s (about 3 hours) . ECF from (a)
✓ 1
Part (c)
The electric field is established throughout the circuit almost instantly, at close to the speed of light ; free electrons everywhere in the circuit, including at the welding tip, start drifting at the same time, so energy is transferred without any individual electron travelling the whole length
✓ 2
Part (d)
R = ρL/A = 1.7×10−8 × 5.0/25×10−6 = 3.4×10−3 Ω ; P = I2R = 1502 × 3.4×10−3 = 77 W
✓ 2
Common error: 1 mm² = 1×10⁻⁶ m², not 1×10⁻³ m². Square the conversion factor when converting areas.
60IBP2H-017-01-13Origin of resistance and the heating effectPaper 2Medium8 marks
A hot-wire foam cutter used by a model-maker uses a nichrome wire of length 0.60 m and diameter 0.40 mm. The resistivity of nichrome is 1.1×10−6 Ω m.
(a)
Show that the resistance of the wire is about 5 Ω.
(2)
(b)
Outline, in terms of free electrons and lattice ions, how electrical energy is transferred to internal energy in the wire.
(2)
(c)
The wire is replaced by a nichrome wire of the same length but half the diameter. The pd is kept the same. Determine the factor by which the power changes.
(2)
(d)
The original wire must dissipate 15 W to cut the foam cleanly. Calculate the pd required.
Free electrons are accelerated by the electric field and gain kinetic energy ; they collide with the lattice ions and transfer energy to them, increasing the ions' vibrational energy, so the internal energy and temperature of the wire rise
✓ 2
Part (c)
Area ∝ d2, so the area is ¼ and the resistance is 4 times larger ; P = V2/R, so the power becomes ¼ of its original value
✓ 2
Part (d)
V = √(PR) = √(15 × 5.25) ; V = 8.9 V . Accept 8.7 V if R = 5 Ω is used; ECF from (a).
✓ 2
Common error: Using the diameter instead of the radius in A = πr² makes the area four times too large.
61IBP2H-017-01-15Power in series and parallelPaper 2Medium7 marks
A spacecraft battery-warming system runs from a 28 V supply. It contains two identical heating elements. The elements can be switched to be in parallel (high setting) or in series (low setting). On the high setting the total power is 20 W.
(a)
Calculate the resistance of each element.
(2)
(b)
Calculate the power on the low setting.
(2)
(c)
The resistance of the elements increases as they warm up. Explain why.
(2)
(d)
At operating temperature, the resistance of each element is 8.0% greater than the value in (a). Calculate the power on the high setting at this temperature.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Total resistance on high setting = V2/P = 282/20 = 39.2 Ω ; two identical elements in parallel, so each element = 2 × 39.2 = 78.4 Ω ≈ 78 Ω
✓ 2
Part (b)
Series resistance = 2 × 78.4 = 156.8 Ω (≈157 Ω) ; P = 282/156.8 = 5.0 W . Accept the alternative argument: the series resistance is 4 × the parallel resistance, so P = 20/4 = 5.0 W. ECF from (a)
✓ 2
Part (c)
As temperature rises, the lattice ions vibrate with greater amplitude ; the free electrons collide with the ions more frequently, so the drift speed for a given pd falls, the current falls and the resistance rises
✓ 2
Part (d)
Each element = 1.08 × 78.4 = 84.7 Ω, so the parallel resistance = 42.3 Ω, and P = 282/42.3 = 18.5 W . Accept 20/1.08 = 18.5 W. ECF from (a)
✓ 1
Exam tip: At constant pd, P = V²/R, so an increase in resistance reduces the power. Use the form of the power equation that matches what is held constant.
An aircraft windscreen is kept free of ice by a nichrome heating wire. The wire is connected directly across the aircraft's 28 V DC supply. At its operating temperature it must dissipate 140 W. Resistivity of nichrome at operating temperature = 1.1 × 10−6 Ω m.
(a)
Show that the resistance of the wire at its operating temperature is about 6 Ω.
(2)
(b)
The wire has a diameter of 0.40 mm. Calculate the length of wire needed.
(3)
(c)
Suggest why nichrome is used for the heater rather than copper.
(2)
(d)
When the heater is first switched on, the wire is cold and its resistivity is slightly lower than the value given. Explain how the power dissipated by the wire changes as it warms up to its operating temperature.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Uses R = V2/P, or I = P/V = 5.0 A then R = V/I
✓ 1
R = 282/140 = 5.6 Ω
✓ 1
Part (b)
A = π(0.20 × 10−3)2 = 1.26 × 10−7 m2
✓ 1
L = RA/ρ
✓ 1
L = 5.6 × 1.26 × 10−7/1.1 × 10−6 = 0.64 m
✓ 1
Allow error carried forward from the resistance found in the first part.
—
Part (c)
Copper has a much lower resistivity than nichrome, about 60 times lower
✓ 1
So a copper wire with a resistance of 5.6 Ω would need to be impractically long or extremely thin and fragile, OR a copper wire of the same dimensions would have a far smaller resistance, so it would draw a much larger current and dissipate far too much power
✓ 1
For either mark, accept instead that nichrome resists oxidation or has a high melting point at high temperature (maximum 2 marks in total).
—
Part (d)
The resistivity, and so the resistance, of the wire increases as it warms up
✓ 1
The pd is fixed at 28 V, so P = V2/R decreases: the power starts slightly above 140 W and falls to 140 W
✓ 1
Exam tip: In A = πr², use the radius rather than the diameter, and convert millimetres to metres before squaring.
63IBP2H-017-02-12Charge and energy from a chemical cellPaper 2Medium7 marks
A cordless drill is powered by a rechargeable battery. When the drill draws a current of 2.0 A, the terminal potential difference of the battery is 11.6 V. When the drill draws 10.0 A, the terminal potential difference is 10.0 V.
(a)
Explain why the terminal potential difference is smaller when the current is larger.
(2)
(b)
Determine the internal resistance of the battery.
(2)
(c)
Determine the emf of the battery.
(1)
(d)
Calculate the percentage of the energy transferred by the battery that is delivered to the drill when the current is 10.0 A.
(2)
Show mark scheme
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Mark
Notes
Part (a)
Part of the emf is dropped across the internal resistance; this pd is Ir
✓ 1
Ir increases with current, so V = ε − Ir decreases
✓ 1
Part (b)
Writes 11.6 = ε − 2.0r and 10.0 = ε − 10.0r, or equivalent
64IBP2H-017-02-13Conductors and insulatorsPaper 2Medium7 marks
A copper track on a printed circuit board has a rectangular cross-section 1.0 mm wide and 35 μm thick. The track carries a steady current of 2.0 A to an LED. Number density of free electrons in copper = 8.5 × 1028 m−3.
(a)
Calculate the drift speed of the free electrons in the track.
(3)
(b)
The LED lights almost instantly when the circuit is switched on. Explain why, with reference to your answer to (a).
(2)
(c)
The board material around the track has a resistivity of about 1013 Ω m. Explain, in terms of charge carriers, why the board is an insulator while the track is a conductor.
(2)
Show mark scheme
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Notes
Part (a)
A = 1.0 × 10−3 × 35 × 10−6 = 3.5 × 10−8 m2
✓ 1
v = I/(nAq)
✓ 1
v = 2.0/(8.5 × 1028 × 3.5 × 10−8 × 1.60 × 10−19) = 4.2 × 10−3 m s−1
✓ 1
Part (b)
The drift speed is only a few mm s−1, so an electron starting at the switch would take a long time (many seconds or more) to reach the LED; the LED does not wait for it
✓ 1
The electric field is set up throughout the circuit almost instantly, at a large fraction of the speed of light, so free electrons already present everywhere, including in the LED, begin drifting at the same time
✓ 1
Part (c)
Copper has a very large number density of free (delocalised) electrons
✓ 1
In the board material, the electrons are bound to atoms, so the density of mobile carriers is negligible and the same pd drives almost no current
✓ 1
Common error: A micrometre (μm) is 10⁻⁶ m. Using 10⁻³ gives an area, and therefore a drift speed, that is wrong by a factor of 1000.
65IBP2H-017-02-16A solar panel as a source of emfPaper 2Medium9 marks
A small solar cell charges a fitness tracker. The cell is illuminated with light of intensity 1000 W m−2. The potential difference V across the cell and the current I are measured for different load resistances.
V / V: 0.00, 0.20, 0.40, 0.45, 0.50, 0.55, 0.60 I / mA: 120, 118, 112, 105, 90, 60, 0
The area of the cell is 3.0 cm2.
Data from the table
(a)
State (i) the open-circuit voltage and (ii) the short-circuit current of the cell.
(2)
(b)
Determine, using the data, the maximum power output of the cell.
(2)
(c)
Calculate the load resistance that gives this maximum power.
(1)
(d)
Calculate the efficiency of the cell at maximum power.
(2)
(e)
Explain, using the data, why the solar cell cannot be modelled as a source of fixed emf with a fixed internal resistance.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
(i) 0.60 V
✓ 1
(ii) 120 mA
✓ 1
Part (b)
Calculates VI for several rows, e.g. 44.8 mW, 47.3 mW and 45.0 mW
✓ 1
Maximum power ≈ 47 mW, at 0.45 V
✓ 1
Part (c)
R = 0.45/0.105 = 4.3 Ω
✓ 1
Allow follow-through from the maximum-power point chosen in the previous part.
—
Part (d)
Incident power = 1000 × 3.0 × 10−4 = 0.30 W
✓ 1
Efficiency = 0.04725/0.30 = 0.158 ≈ 16%
✓ 1
Allow follow-through from the maximum power found earlier.
—
Part (e)
For a fixed emf and a fixed internal resistance, V = ε − Ir, so the V–I relationship would be linear
✓ 1
Here the current falls by only 8 mA between 0 and 0.40 V, but by 90 mA between 0.50 V and 0.60 V. The relationship is therefore non-linear and the effective internal resistance is not constant
✓ 1
Exam tip: Convert cm² to m² by multiplying by 10⁻⁴ before you calculate the incident power.
66IBP2H-017-00-10emf and internal resistancePaper 2Medium9 marks
A mechanic tests a car battery by connecting it to a series of load resistors and recording the current I and the terminal potential difference V.
I / A: 20, 40, 60, 80 V / V: 12.2, 11.8, 11.4, 11.0
Data from the table
(a)
Outline why the terminal potential difference decreases as the current increases.
(2)
(b)
Determine the emf and the internal resistance of the battery.
(3)
(c)
When the engine is started, the starter motor draws a current of 150 A. Calculate the terminal potential difference during starting.
(2)
(d)
Explain, with a calculation, why the battery becomes warm while the engine is being started.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The battery has internal resistance, so there is a pd (Ir) across it inside the battery
✓ 1
This 'lost volts' increases with current, and the terminal pd is V = ε − Ir
✓ 1
Part (b)
r = −gradient = (12.2 − 11.0)/(80 − 20)
✓ 1
r = 0.020 Ω
✓ 1
ε = 12.2 + 20 × 0.020 = 12.6 V
✓ 1
Part (c)
V = 12.6 − 150 × 0.020
✓ 1
V = 9.6 V; ECF from (b)
✓ 1
Part (d)
Power dissipated in the internal resistance P = I2r = 1502 × 0.020 = 450 W; ECF from (b)
✓ 1
This energy is transferred as internal energy (thermal energy) inside the battery, so its temperature rises
✓ 1
Exam tip: In a data question, use data points that are far apart when calculating the gradient. Then check that the other points fit the same straight line.
67IBP2H-017-00-11A thermistor thermometerPaper 2Medium8 marks
A neonatal incubator has an over-temperature alarm. Its sensor is a potential divider made of a fixed resistor R in series with an NTC thermistor, connected across a 9.0 V supply of negligible internal resistance. The output voltage Vout is taken across the thermistor. The alarm sounds when Vout falls to 3.0 V. At 37 °C the thermistor has a resistance of 2.4 kΩ.
(a)
Calculate the value of R needed for the alarm to sound at 37 °C. Assume the alarm circuit draws no current.
(2)
(b)
Explain how Vout changes as the temperature rises above 37 °C.
(2)
(c)
The alarm circuit actually has an input resistance of 12 kΩ, connected across the thermistor. Determine Vout at 37 °C.
(3)
(d)
State the effect of the alarm circuit's input resistance on the temperature at which the alarm sounds.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
pd across R = 9.0 − 3.0 = 6.0 V, so R/2.4 kΩ = 6.0/3.0
✓ 1
R = 4.8 kΩ
✓ 1
Part (b)
The resistance of the NTC thermistor decreases as its temperature rises
✓ 1
The thermistor therefore takes a smaller share of the supply pd, so Vout decreases below 3.0 V and the alarm sounds
Vout is already below 3.0 V at 37 °C, so the alarm sounds at a temperature lower than 37 °C
✓ 1
Common error: A load connected across part of a potential divider is in parallel with that part. This lowers the output, so recalculate the divider with the combined resistance.
68IBP2H-017-00-13Resistivity and heating stripsPaper 2Medium8 marks
The rear-window demister of a car consists of six identical heating strips, each of resistance 4.8 Ω, connected in parallel across a 12 V supply. Assume the supply has negligible internal resistance and the strip resistances are constant.
(a)
Calculate the total resistance of the demister.
(1)
(b)
Calculate the total power of the demister.
(2)
(c)
One strip breaks. State and explain the effect on the total power.
(2)
(d)
Calculate the total power if the six strips were connected in series across the same supply, and hence explain why they are connected in parallel.
(3)
Show mark scheme
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Mark
Notes
Part (a)
R = 4.8/6 = 0.80 Ω
✓ 1
Part (b)
P = V2/R = 122/0.80
✓ 1
P = 180 W; allow ECF from (a)
✓ 1
Part (c)
Each remaining strip still has the full 12 V across it, so it still dissipates 122/4.8 = 30 W
✓ 1
The total power decreases, by 30 W, to 5 × 30 = 150 W
✓ 1
Part (d)
Series resistance R = 6 × 4.8 = 28.8 Ω
✓ 1
P = 122/28.8 = 5.0 W
✓ 1
Any one valid reason, e.g. 5.0 W is far too little power to clear the window, whereas in parallel each strip has the full 12 V and the total is 180 W; OR in series one broken strip would stop the whole heater, whereas in parallel the other strips keep working
✓ 1
Exam tip: With a fixed supply pd, use P = V²/R. A lower total resistance gives more power, and connecting the strips in parallel gives the lowest resistance.
69IBP2H-017-00-15Ohmic and non-ohmic conductorsPaper 2Medium7 marks
A caver's headlamp uses two identical cells, each of emf 1.5 V and internal resistance 0.30 Ω. The lamp has a constant resistance of 2.4 Ω.
(a)
The cells are connected in series. Calculate the power dissipated in the lamp.
(3)
(b)
The cells are instead connected in parallel. Calculate the power dissipated in the lamp.
(3)
(c)
Outline one advantage of the parallel arrangement for a caver.
(1)
Show mark scheme
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Notes
Part (a)
Total emf = 3.0 V and total internal resistance = 0.60 Ω
✓ 1
I = 3.0/(2.4 + 0.60) = 1.0 A
✓ 1
P = I2R = 1.02 × 2.4 = 2.4 W
✓ 1
Part (b)
Total emf = 1.5 V and total internal resistance = 0.15 Ω
✓ 1
I = 1.5/(2.4 + 0.15) = 0.588 A
✓ 1
P = 0.5882 × 2.4 = 0.83 W
✓ 1
Part (c)
Each cell supplies only half the current, so the cells last longer (or the lamp keeps working if one cell fails)
✓ 1
Exam tip: For identical cells, series connection adds both the emfs and the internal resistances. Parallel connection keeps the emf the same and divides the internal resistance by the number of cells.
70B-2-04I–V characteristic of a light-emitting diodePaper 2Hard11 marks
The graph shows the current–voltage (I–V) characteristic of a red light-emitting diode (LED). The LED emits light of wavelength 640 nm.
I–V characteristic of the LED (drawn to scale)
(a)
(i)
Explain, with reference to the graph, why the LED is a non-ohmic component.
(1)
(b)
(i)
Determine the resistance of the LED when the current in it is 20 mA.
(2)
(c)
(i)
The LED is connected in series with a resistor X to a battery of emf 5.0 V and negligible internal resistance so that the current is 20 mA. Calculate the resistance of X.
(1)
(d)
(i)
The battery and X are replaced by a cell of emf 3.0 V and negligible internal resistance in series with a 60 Ω resistor. By drawing a suitable line on the graph, determine the current in the LED.
(3)
(e)
(i)
Calculate, in eV, the energy of a photon emitted by the LED, and compare it with the pd across the LED when it conducts.
(2)
(f)
(i)
At a current of 20 mA, 25 % of the electrical power supplied to the LED is emitted as light. Determine the number of photons emitted per second.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The graph is not a straight line through the origin: the current is not proportional to the pd (V/I is not constant)
✓ 1
"The graph is curved" alone is not enough; the proportionality idea is needed.
Part (b)(i)
V read at 20 mA = 2.00 V
✓ 1
Accept 1.99–2.01 V.
R = 2.00/0.020 = 100 Ω
✓ 1
Accept 99–101 Ω. Do not accept the inverse gradient of the curve (about 3 Ω).
Part (c)(i)
pd across X = 5.0 − 2.00 = 3.0 V; RX = 3.0/0.020 = 150 Ω
✓ 1
Allow ECF from (b).
Part (d)(i)
VLED = 3.0 − 60I, e.g. the line through (1.80 V, 20 mA) and (2.10 V, 15 mA)
✓ 1
Any two correct points.
Straight line drawn correctly on the graph
✓ 1
Intersection with the curve: I ≈ 17 mA
✓ 1
Accept 16–18 mA. Award [1 max] for a trial-and-error answer without the line if it is within range.
71B-2-14Potential divider with a light-dependent resistorPaper 2Hard12 marks
A light-dependent resistor (LDR) is used to switch on a garden lamp automatically. The graph shows how the resistance R of the LDR varies with the illuminance E of the light falling on it.
The LDR is connected in series with a fixed resistor F across a 6.0 V supply of negligible internal resistance. A switching unit connected across the LDR turns the lamp on when the pd across the LDR rises to 4.0 V.
Resistance R of the LDR against illuminance E (drawn to scale)
(a)
(i)
Outline why the resistance of the LDR decreases as the illuminance increases.
(2)
(b)
(i)
The lamp is to switch on when the illuminance falls to 20 lx. Determine the resistance of F, assuming that the switching unit draws no current.
(3)
(c)
(i)
Deduce, without calculation, whether the resistance of F should be increased or decreased so that the lamp switches on only when it is darker.
(2)
(d)
(i)
As the battery that supplies the circuit runs down, its emf falls from 6.0 V to 5.4 V (internal resistance still negligible). The switching unit still turns the lamp on when the pd across the LDR rises to 4.0 V. Using your answer to (b), determine the illuminance at which the lamp now switches on, and state the consequence for the user.
(3)
(e)
(i)
In the LDR material a photon needs an energy of at least 2.4 eV to release a charge carrier. Determine the longest wavelength of light that can reduce the resistance of the LDR, and hence suggest why the LDR hardly responds to red light of wavelength 650 nm.
(2)
Show mark scheme
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Notes
Part (a)(i)
Absorbed light (photons) releases more charge carriers (electrons) in the material
✓ 1
More mobile charge carriers give a greater current for the same pd, so the resistance V/I is smaller
✓ 1
Part (b)(i)
Resistance of the LDR at 20 lx read from the graph = 28.7 kΩ
✓ 1
Accept 27–30 kΩ.
pd across F = 6.0 − 4.0 = 2.0 V, so RF = RLDR × 2.0/4.0
✓ 1
Or the current 4.0/28.7 kΩ = 0.139 mA.
RF = 14.4 kΩ
✓ 1
Accept 13.5–15 kΩ.
Part (c)(i)
The lamp switches on when RLDR = 2RF (pd shared in the ratio 4.0 : 2.0); darker means a larger RLDR at the switching point
✓ 1
Allow ECF from (b).
So RF must be increased
✓ 1
Dependent on a correct reason.
Part (d)(i)
pd across F = 5.4 − 4.0 = 1.4 V, so the current = 1.4/14.4 kΩ = 9.7 × 10−5 A
✓ 1
Allow ECF from (b).
RLDR = 4.0/9.7 × 10−5 = 41 kΩ (or 14.4 × 4.0/1.4)
✓ 1
Accept 40–42 kΩ. Allow ECF from (b).
From the graph E ≈ 13 lx: the lamp now switches on only when it is considerably darker (13 lx instead of 20 lx)
✓ 1
Accept 12–14 lx. The consequence must follow from the value read.
72B-2-31emf and internal resistancePaper 2Hard12 marks
A voltmeter of very high resistance is connected across the terminals of a car battery. At t = 2 s the two headlamps are switched on. At t = 5 s the starter motor is switched on, in parallel with the headlamps, and it turns the engine until t = 6.5 s. The graph shows the reading V of the voltmeter against time t.
Each headlamp is rated 60 W at 12.0 V. Unless stated otherwise, assume the resistance of a headlamp is constant.
Terminal pd V of the battery against time t (drawn to scale)
(a)
(i)
State the emf of the battery, and explain your answer.
(1)
(ii)
Show that the internal resistance of the battery is about 0.025 Ω.
(2)
(b)
(i)
Determine the current in the starter motor while it turns the engine.
(3)
(ii)
Show that (power of the lamps while the starter turns)/(power of the lamps before) = (V2/V1)², where V1 and V2 are the terminal pds, and evaluate the ratio.
(2)
(iii)
In fact the filaments of the lamps cool when the pd falls. Suggest the effect of this on your answer to (b)(i).
(2)
(c)
(i)
The current to and from the starter motor flows in two long straight parallel cables 1.5 cm apart. Determine the force per unit length between the cables while the starter turns, and state whether the cables attract or repel each other.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
12.50 V: before t = 2 s there is (almost) no current, so there is no pd across the internal resistance and the terminal pd equals the emf
✓ 1
Accept 12.4–12.6 V. Value and reason both needed.
Part (a)(ii)
Combined resistance of the two lamps = ½ × 12.0²/60 = 1.20 Ω, so I = 12.25/1.20 = 10.2 A
✓ 1
The two lamps are in parallel.
r = (12.50 − 12.25)/10.2 = 0.0245 Ω
✓ 1
Allow ECF from (a)(i). Answer to at least 2 s.f. required.
Part (b)(i)
Total current from the battery = (12.50 − 9.50)/0.0245 = 122 A
✓ 1
Allow ECF from (a)(ii). The lost pd is read from the graph (accept 9.4–9.6 V).
Current in the lamps = 9.50/1.20 = 7.92 A
✓ 1
The lamps now have only 9.50 V across them.
Starter current = 122 − 7.92 = 115 A
✓ 1
Accept 95–145 A (small differences read from the graph). Award [2 max] for 122 A (lamp current ignored).
Part (b)(ii)
P = V²/R with R the same in both cases, so the ratio is V2²/V1²
✓ 1
(9.50/12.25)² = 0.60
✓ 1
Accept 0.59–0.61.
Part (b)(iii)
A cooler metal filament has a lower resistance (smaller lattice vibrations), so the lamp current at 9.50 V is larger than 7.92 A
✓ 1
The total current is fixed by the lost pd, so the starter current is smaller than calculated in (b)(i)
✓ 1
Allow ECF from (b)(i). Dependent on the first marking point.
Repel, because the currents are in opposite directions
✓ 1
Answers: (a)(i) 12.50 V · (a)(ii) 0.0245 Ω · (b)(i) 115 A · (b)(ii) 0.60 · (c)(i) 0.18 N m−1; repel (the remaining parts are explanations — see the table above)
73B-2-32A thermistor thermometerPaper 2Hard14 marks
A thermistor is used as the sensor in an electronic thermometer. The circuit is shown. In the thermistor material the number of free charge carriers increases with temperature. The table gives the resistance RT of the thermistor at different temperatures θ.
θ / °C
0
10
20
30
40
50
RT / kΩ
33.6
20.2
12.5
8.04
5.30
3.59
The supply has negligible internal resistance.
(a)
The thermistor.
(i)
Explain why the resistance of the thermistor decreases as its temperature increases, even though its lattice vibrations increase.
(1)
(b)
Using the thermometer (ideal voltmeter).
(i)
Show that the voltmeter reading is 4.0 V at 20 °C.
(2)
(ii)
The voltmeter reads 5.0 V. Determine the temperature of the thermistor.
(2)
(c)
A non-ideal voltmeter.
(i)
The thermistor is at 30 °C. The voltmeter is replaced by one with a constant resistance of 40 kΩ. Determine the new reading.
(2)
(ii)
A user who assumes the voltmeter is ideal converts this reading to a temperature. Deduce the temperature obtained, and comment on the result.
(3)
(d)
Self-heating.
(i)
Show that, at 30 °C with an ideal voltmeter, the power dissipated in the thermistor is about 2 mW.
(2)
(ii)
The temperature of the thermistor rises by 1.0 K for every 1.5 mW dissipated in it. Estimate the error this causes and suggest one change to the circuit that would reduce it.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
The increase in the number of mobile charge carriers outweighs the effect of greater lattice vibration/more frequent collisions, so the current for a given pd increases and the resistance falls
✓ 1
Part (b)(i)
Current = 9.0/(12.5 + 10.0) kΩ = 0.40 mA
✓ 1
Or pd shared in the ratio 10.0 : 12.5.
Reading = 0.40 mA × 10.0 kΩ = 4.0 V
✓ 1
Part (b)(ii)
pd across thermistor = 4.0 V, so RT = 4.0/(5.0/10.0) = 8.0 kΩ
✓ 1
θ ≈ 30 °C from the table
✓ 1
Part (c)(i)
Parallel combination of 10.0 kΩ and 40 kΩ = 8.0 kΩ
The thermometer reads about 5 K too low: the meter resistance is not much greater than 10 kΩ, so it lowers the resistance of that section and the reading; a voltmeter of much higher resistance is needed
✓ 1
Part (d)(i)
I = 9.0/(10.0 + 8.04) × 103 = 4.99 × 10−4 A
✓ 1
P = I²RT = (4.99 × 10−4)² × 8.04 × 103 = 2.0 × 10−3 W
✓ 1
Part (d)(ii)
The thermistor is about 2.0/1.5 ≈ 1.3 K warmer than its surroundings, so the thermometer reads about 1.3 K high
✓ 1
Accept 1–1.5 K.
Reduce the current, e.g. use a lower supply pd (at 3.0 V the power falls by a factor of 9, to about 0.2 mW) or larger resistances
✓ 1
Accept any change that reduces the current through the thermistor, with a reason.
Answers: (b)(ii) 30 °C · (c)(i) 4.5 V · (c)(ii) ≈ 25 °C · (d)(ii) ≈ 1.3 K (the remaining parts are explanations — see the table above)
74B-2-34An electric-vehicle battery packPaper 2Hard12 marks
The battery pack of an electric car consists of 96 identical chemical cell modules connected in series. Each module has an emf of 3.70 V and an internal resistance of 1.50 mΩ.
(a)
The pack.
(i)
Calculate the emf and the internal resistance of the pack.
(2)
(b)
Hard acceleration.
(i)
During hard acceleration the current drawn from the pack is 300 A. Calculate the potential difference across the terminals of the pack.
(1)
(ii)
Determine the percentage of the energy transferred by the cells that is dissipated inside the pack.
(2)
(c)
Fast charging.
(i)
The pack is charged with a current of 250 A. Determine the potential difference that the charger must provide across the terminals of the pack, explaining why it is greater than the emf.
(2)
(ii)
The pack has a mass of 450 kg and a mean specific heat capacity of 1.0 × 103 J kg−1 K−1. Estimate the temperature rise of the pack during 20 minutes of fast charging if no cooling is provided.
(2)
(d)
A solar roof?
(i)
The fully charged pack stores 1.9 × 108 J. A solar panel of area 2.0 m² and efficiency 20 % could be fitted to the roof of the car. The maximum intensity of sunlight is about 1.0 × 103 W m−2. Discuss whether solar cells could replace chemical cells, or charging from the mains, as the energy source for the car.
(3)
Show mark scheme
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Mark
Notes
Part (a)(i)
emf = 96 × 3.70 = 355 V
✓ 1
Accept 355.2 V.
Internal resistance = 96 × 1.50 × 10−3 = 0.144 Ω
✓ 1
Part (b)(i)
V = ε − Ir = 355.2 − 300 × 0.144 = 312 V
✓ 1
ECF from (a).
Part (b)(ii)
Fraction = I²r/εI = Ir/ε = 43.2/355.2
✓ 1
Or 13.0 kW out of 106.6 kW.
= 12 %
✓ 1
Accept 12.2 %.
Part (c)(i)
The charger must drive current against the emf of the pack and also supply the pd across the internal resistance: V = ε + Ir
ΔT = 1.08 × 107/(450 × 1.0 × 103) = 24 K, so cooling is needed
✓ 1
Accept 24 K.
Part (d)(i)
Maximum solar power = 1.0 × 103 × 2.0 × 0.20 = 400 W
✓ 1
Time to store 1.9 × 108 J = 4.8 × 105 s ≈ 130 hours of full sunshine, far longer in practice (night, cloud, angle of the Sun)
✓ 1
Conclusion: the solar roof cannot replace mains charging, and a chemical store is still needed because the car needs about 100 kW, far more than 400 W, and must run when there is no sunlight; the solar roof could only add a small amount of range
✓ 1
Award for a conclusion supported by the power comparison and the need for storage.
Answers: (a)(i) 355 V; 0.144 Ω · (b)(i) 312 V · (b)(ii) 12 % · (c)(i) 391 V · (c)(ii) 24 K (the remaining parts are explanations — see the table above)
75B-2-39emf and internal resistance of a batteryPaper 2Hard12 marks
A camera drone of mass 1.60 kg is powered by a battery of four identical lithium-ion cells connected in series. Each cell has an emf of 4.10 V and an internal resistance of 12 mΩ. While the drone hovers at rest in still air, the current from the battery is 18.0 A.
(a)
(i)
Calculate the potential difference across the terminals of the battery while the drone hovers.
(1)
(ii)
Determine the power delivered to the motors and the effective resistance of the motors while the drone hovers.
(2)
(b)
(i)
The battery can deliver a charge of 5.0 A h. Estimate the longest time, in minutes, for which the drone can hover.
(1)
(c)
(i)
To hover, the rotors push air vertically downwards. The air above the rotors is at rest and it leaves the rotors with speed u. Show that the mass of air pushed down per unit time is mg/u, where m is the mass of the drone.
(2)
(ii)
u = 10.0 m s−1. Calculate the mass of air pushed down per second.
(1)
(iii)
Determine the efficiency with which the electrical power delivered to the motors is converted into kinetic energy of the air.
(2)
(d)
(i)
In cold weather the internal resistance of each cell rises to 30 mΩ; the emf is unchanged. Determine the current needed for the motors to receive the same power as in (a)(ii).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
V = ε − Ir = 16.4 − 18.0 × 0.048 = 15.5 V
✓ 1
The emfs and the internal resistances of the four cells both add.
Part (a)(ii)
Power delivered to the motors = VI = 15.5 × 18.0 = 280 W
✓ 1
Allow ECF from (a)(i). Accept 279–280 W. Award [0] for 295 W (εI, the power transferred by the cells).
Effective resistance = V/I = 15.5/18.0 = 0.863 Ω
✓ 1
Allow ECF from (a)(i). Accept 0.86 Ω.
Part (b)(i)
t = 5.0/18.0 = 0.278 h = 17 min
✓ 1
Accept 16–17 min.
Part (c)(i)
The force exerted on the air is its rate of change of momentum: (mass of air per unit time) × u
✓ 1
Newton's second law in the form F = Δp/Δt.
By Newton's third law the air pushes up on the drone with an equal force, which balances the weight mg when hovering, so mass per unit time = mg/u
✓ 1
Both the third-law pair and the equilibrium condition are required.
Part (c)(ii)
1.60 × 9.81/10.0 = 1.57 kg s−1
✓ 1
Allow ECF from (c)(i).
Part (c)(iii)
Kinetic energy given to the air per second = ½ × 1.57 × 10.02 = 78.5 W
✓ 1
Allow ECF from (c)(ii).
Efficiency = 78.5/280 = 0.28
✓ 1
Allow ECF from (a)(ii). Accept 0.275–0.285. Award [1 max] for 0.266 (power transferred by the cells, 295 W, used).
Part (d)(i)
Power to the motors = εI − I2r with r = 4 × 0.030 = 0.120 Ω: 279.6 = 16.4I − 0.120I2
✓ 1
Allow ECF from (a)(ii). The new internal resistance of the whole battery must be used.
0.120I2 − 16.4I + 279.6 = 0
✓ 1
I = 20.0 A (the other root, 117 A, would put most of the emf across the internal resistance and is rejected)
✓ 1
Accept 19.9–20.1 A. Award [1 max] for 18.0 A (current assumed unchanged).
Answers: (a)(i) 15.5 V · (a)(ii) 280 W; 0.863 Ω · (b)(i) 17 min · (c)(ii) 1.57 kg s−1 · (c)(iii) 0.28 · (d)(i) 20.0 A (the remaining parts are explanations — see the table above)
76B-2-42A hot-filament electron sourcePaper 2Hard18 marks
The electron source of an electron-beam instrument is a straight tungsten wire of length 30.0 mm and diameter 0.125 mm, in a vacuum. The wire is heated by a current from a supply of negligible internal resistance until electrons escape from its surface. The graph shows how the resistivity ρ of tungsten varies with temperature T.
Resistivity of tungsten at 293 K = 5.6 × 10−8 Ω m; emissivity of the hot tungsten surface = 0.30. Neglect the energy conducted away through the ends of the wire.
Resistivity ρ of tungsten against temperature T (drawn to scale)
(a)
(i)
Show that the resistance of the wire at 293 K is about 0.14 Ω.
(2)
(ii)
Outline why the resistivity of tungsten increases with temperature.
(1)
(b)
In operation the wire is kept at a steady temperature of 2500 K.
(i)
Show that the power radiated by the wire is about 8 W.
(2)
(ii)
Use the graph to determine the resistance of the wire at 2500 K.
(2)
(iii)
Determine the current in the wire and the pd across it at 2500 K.
(2)
(c)
(i)
The supply pd is fixed at the value found in (b)(iii). Deduce the ratio (power dissipated in the wire at the instant it is switched on at 293 K)/(power dissipated at 2500 K), and suggest one consequence for the supply.
(2)
(ii)
The density of tungsten is 1.93 × 104 kg m−3 and its mean specific heat capacity between 293 K and 2500 K is 150 J kg−1 K−1. Estimate the shortest possible time for the wire to reach 2500 K after it is switched on, and explain why the actual time is longer.
(3)
(d)
(i)
Calculate, in eV, the energy of a photon at the peak wavelength of the radiation emitted by the wire at 2500 K.
(2)
(e)
Electrons leave the wire with negligible speed and are accelerated through a potential difference of 2.00 kV.
(i)
Determine the speed of the electrons.
(1)
(ii)
The electrons then enter a region of uniform magnetic field of flux density 1.50 mT directed perpendicular to their velocity. Determine the radius of their path.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
A = π(0.125 × 10−3/2)² = 1.23 × 10−8 m²
✓ 1
Use of the diameter as the radius scores 0 for this mark.
At a higher temperature the metal ions vibrate with greater amplitude, so the conduction electrons collide with them more often (their drift is impeded more)
✓ 1
"The electrons collide more" without reference to lattice vibrations scores 0.
Part (b)(i)
Surface area = πdL = π × 0.125 × 10−3 × 0.030 = 1.18 × 10−5 m²
✓ 1
The curved surface area is the hidden step.
P = 0.30 × 5.67 × 10−8 × 1.18 × 10−5 × 2500⁴ = 7.83 W
In the steady state the electrical power equals the power radiated: I²R = P, I = √(7.83/1.78) = 2.09 A
✓ 1
Allow ECF from (b)(i) and (b)(ii). The energy balance must be stated or used.
V = IR = 2.09 × 1.78 = 3.74 V
✓ 1
Or V = √(PR).
Part (c)(i)
P = V²/R with V the same, so the ratio = R2500/R293 = ρ2500/ρ293 = 7.3/0.56 = 13
✓ 1
Allow ECF from (a)(i) and (b)(ii). The dimensions cancel; an inverted ratio scores 0.
The initial current is about 13 times the operating current (≈ 27 A), so the supply must be able to deliver a large surge current (or the current must be limited at switch-on)
✓ 1
Part (c)(ii)
Mass = 1.93 × 104 × 1.23 × 10−8 × 0.030 = 7.1 × 10−6 kg
✓ 1
Allow ECF from (a)(i).
Energy needed = 7.1 × 10−6 × 150 × 2207 = 2.4 J
✓ 1
Shortest time = 2.4/(13 × 7.8) = 0.02 s; actually longer because the power falls as the resistance rises and some energy is radiated during the warm-up
✓ 1
Allow ECF from (b)(i) and (c)(i). Accept 0.02–0.03 s. Both the value and the reason are needed.
Part (d)(i)
λmax = 2.9 × 10−3/2500 = 1.2 × 10−6 m
✓ 1
E = hc/λ = 1.1 eV
✓ 1
The answer must be in eV.
Part (e)(i)
eV = ½mev², so v = √(2 × 1.60 × 10−19 × 2000/9.11 × 10−31); v = 2.65 × 107 m s−1
✓ 1
This is about 0.09c, so the non-relativistic expression is acceptable.
Part (e)(ii)
evB = mev²/r, so r = mev/eB; r = 9.11 × 10−31 × 2.65 × 107/(1.60 × 10−19 × 1.50 × 10−3) = 0.10 m
✓ 1
Allow ECF from (e)(i).
Answers: (a)(i) 0.137 Ω · (b)(i) 7.83 W · (b)(ii) 1.78 Ω · (b)(iii) 2.09 A; 3.74 V · (c)(i) 13 · (c)(ii) 0.02 s · (d)(i) 1.1 eV · (e)(i) 2.65 × 107 m s−1 · (e)(ii) 0.10 m (the remaining parts are explanations — see the table above)
77IBP2H-017-01-12Resistivity and heating stripsPaper 2Hard8 marks
A car rear-window demister consists of 10 identical heating strips connected in parallel. Each strip has a resistance of 14 Ω. The demister is connected to a battery of emf 12.0 V. The combined internal resistance of the battery and leads is 0.10 Ω.
(a)
Calculate the total resistance of the 10 strips.
(1)
(b)
Calculate the current from the battery.
(2)
(c)
Calculate the power dissipated in one strip.
(2)
(d)
One strip breaks and becomes an open circuit. Determine the total power now dissipated in the remaining strips, and state, with a reason, whether the power in each remaining strip increases or decreases.
(3)
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Part (a)
R = 14/10 = 1.4 Ω
✓ 1
Part (b)
I = ε/(R + r) = 12.0/(1.4 + 0.10) ; I = 8.0 A . ECF from (a)
✓ 2
Part (c)
Current per strip = 0.80 A, or terminal pd = 11.2 V ; P = 0.802 × 14 = 9.0 W . ECF from (b)
✓ 2
Part (d)
New parallel resistance = 14/9 = 1.56 Ω, so I = 12.0/1.656 = 7.25 A ; total power in strips = I2R = 7.252 × 1.556 = 82 W ; terminal pd rises from 11.2 V to 11.3 V because less is lost across the internal resistance, so the power in each remaining strip increases slightly
✓ 3
Exam tip: When internal resistance is present, a change to the external circuit changes the terminal pd. Do not assume each branch still has the full emf.
78IBP2H-017-01-14A thermistor thermometerPaper 2Hard8 marks
A biomedical technician checks a sensor circuit. Two 20 kΩ resistors are connected in series across a 9.0 V battery of negligible internal resistance. A voltmeter of resistance 100 kΩ is connected across one of the resistors.
(a)
State the pd across each resistor before the voltmeter is connected.
(1)
(b)
Calculate the reading on the voltmeter.
(2)
(c)
Explain why the reading is less than your answer to (a).
(2)
(d)
Determine the minimum voltmeter resistance needed for the reading to be within 1.0% of the value in (a).
(3)
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Part (a)
4.5 V
✓ 1
Part (b)
Parallel combination = (1/20 + 1/100)−1 = 16.7 kΩ ; V = 9.0 × 16.7/(16.7 + 20) = 4.1 V
✓ 2
Part (c)
The voltmeter in parallel reduces the resistance of that section of the circuit ; the pd divides in proportion to resistance, so that section now takes a smaller share of the 9.0 V and the measured pd is lower
✓ 2
Part (d)
Reading must be at least 0.99 × 4.5 = 4.455 V, so Rp/(Rp + 20 kΩ) ≥ 0.495 ; Rp ≥ (0.495/0.505) × 20 kΩ = 19.604 kΩ, keeping this value unrounded ; 1/RV = 1/19.604 − 1/20 (in kΩ−1), so RV = 9.9×105 Ω; accept 9.8×105 Ω to 9.9×105 Ω where it comes from a rounded Rp
✓ 3
Common error: When a later step subtracts two nearly equal quantities, rounding the intermediate value early can change the final answer noticeably. Carry extra significant figures until the last step.
79IBP2H-017-01-16Ohmic and non-ohmic conductorsPaper 2Hard8 marks
The table shows data for the filament lamp in a stage lighting unit.
V / V: 0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0 I / A: 0, 0.20, 0.30, 0.37, 0.43, 0.48, 0.52
Data from the table
(a)
Calculate the resistance of the lamp at 1.0 V and at 6.0 V.
(2)
(b)
Explain the change in resistance you found in (a).
(2)
(c)
The lamp is connected in series with a 10 Ω resistor across a 6.0 V supply of negligible internal resistance. Determine the current in the circuit. Assume the lamp's characteristic is linear between adjacent data points.
(3)
(d)
Calculate the power dissipated in the lamp.
(1)
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Notes
Part (a)
At 1.0 V: R = 1.0/0.20 = 5.0 Ω ; at 6.0 V: R = 6.0/0.52 = 11.5 Ω ≈ 12 Ω
✓ 2
Part (b)
A larger current dissipates more power, so the filament temperature rises ; lattice ions vibrate with greater amplitude, increasing the rate of electron–ion collisions, so the resistance increases
✓ 2
Part (c)
Condition: Vlamp + 10I = 6.0 V. At 0.30 A the total pd is 2.0 + 3.0 = 5.0 V, and at 0.37 A it is 3.0 + 3.7 = 6.7 V, so I lies between 0.30 A and 0.37 A ; between these points Vlamp = 2.0 + (I − 0.30)/0.07, so 2.0 + (I − 0.30)/0.07 + 10I = 6.0 ; 1.7I = 0.58, I = 0.34 A
✓ 3
Part (d)
ECF from (c). Vlamp = 6.0 − 10 × 0.341 = 2.59 V, so P = 2.59 × 0.341 = 0.88 W
✓ 1
Common error: A non-ohmic lamp has no single fixed resistance. Using 12 Ω (its value at 6 V) in the series calculation gives the wrong current. Use the data and the condition that the pds sum to the emf.
80IBP2H-017-02-10Measuring resistivityPaper 2Hard9 marks
A student tests constantan wire for use in a strain gauge. The student measures the resistance R of different lengths L of the wire. Each length is connected using crocodile clips.
L / m: 0.20, 0.40, 0.60, 0.80, 1.00 R / Ω: 1.5, 2.9, 4.3, 5.6, 7.0
The diameter of the wire is measured as 0.30 ± 0.01 mm.
Data from the table
(a)
A graph of R against L is a straight line that does not pass through the origin. Outline what this suggests about the measurements.
(2)
(b)
Determine the gradient of the graph of R against L.
(2)
(c)
Calculate the percentage uncertainty in the cross-sectional area of the wire.
(2)
(d)
Using your answers to the previous two parts, determine the resistivity of constantan and its absolute uncertainty. Ignore the uncertainty in the gradient.
(3)
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Part (a)
There is a non-zero resistance (about 0.1–0.2 Ω) when L = 0, so there is a systematic (zero) error
✓ 1
This is likely due to the resistance of the clips, contacts or connecting leads
✓ 1
Part (b)
Uses two widely separated points on the line, e.g. (0.20, 1.5) and (1.00, 7.0), or a best-fit line
✓ 1
Gradient = 6.9 Ω m−1 (accept 6.8–7.0 Ω m−1)
✓ 1
Part (c)
Percentage uncertainty in d = 0.01/0.30 × 100 = 3.3%
✓ 1
A ∝ d2, so the percentage uncertainty in A = 2 × 3.3% = 6.7%
✓ 1
Part (d)
A = π(0.15 × 10−3)2 = 7.07 × 10−8 m2
✓ 1
Gradient = ρ/A, so ρ = gradient × A = 6.9 × 7.07 × 10−8 = 4.9 × 10−7 Ω m (accept 4.8–5.0 × 10−7 Ω m, consistent with the gradient found)
✓ 1
Uncertainty = 6.7% of ρ ≈ ±0.3 × 10−7 Ω m, so ρ = (4.9 ± 0.3) × 10−7 Ω m
✓ 1
Allow error carried forward from the gradient and percentage-uncertainty parts.
—
Exam tip: Using the gradient, rather than a single R/L value, removes the effect of a constant contact resistance. This is why the graphical method is more reliable here.
81IBP2H-017-02-11Solar cells as energy sourcesPaper 2Hard9 marks
A Mars lander is powered by a solar array of total area 4.0 m2 and efficiency 28%. The intensity of sunlight at the Earth's distance from the Sun is 1360 W m−2. Mars orbits at 1.52 times the Earth–Sun distance.
(a)
Show that the intensity of sunlight at the distance of Mars is about 590 W m−2.
(2)
(b)
Ignoring any absorption by the Martian atmosphere, calculate the maximum electrical power output of the array when sunlight is incident normally on it.
(2)
(c)
Suggest two reasons why the actual power output on the Martian surface is less than the value you calculated for the maximum power.
(2)
(d)
The array delivers the maximum power at a terminal potential difference of 32 V. Calculate the current supplied.
(1)
(e)
The array is built from many parallel strings, each made of cells connected in series. Explain this arrangement.
(2)
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Part (a)
Intensity ∝ 1/d2, so I = 1360/1.522
✓ 1
I = 589 W m−2 ≈ 590 W m−2
✓ 1
Part (b)
P = efficiency × intensity × area = 0.28 × 589 × 4.0
✓ 1
P = 660 W
✓ 1
Part (c)
First valid reason
✓ 1
Second valid reason
✓ 1
Acceptable reasons include: dust settling on the panels blocks light; the atmosphere and dust storms absorb or scatter sunlight; sunlight is not incident normally for most of the day, or there is no sunlight at night; the efficiency of the cells changes with temperature.
—
Part (d)
I = P/V = 660/32 = 21 A (allow error carried forward from the maximum power)
✓ 1
Part (e)
Cells in series add their emfs, so each string provides the required 32 V
✓ 1
Strings in parallel add their currents to give the total current; if a cell fails, only its own string stops supplying current and the rest keep working
✓ 1
Common error: Intensity follows an inverse-square law. Divide by 1.52², not by 1.52.
82IBP2H-017-02-15Resistance of transmission cablesPaper 2Hard8 marks
An overhead power cable consists of a steel core surrounded by aluminium strands. The steel and the aluminium act as two conductors in parallel along the whole length of the cable. Steel: cross-sectional area 50 mm2, resistivity 1.4 × 10−7 Ω m. Aluminium: total cross-sectional area 300 mm2, resistivity 2.8 × 10−8 Ω m.
(a)
Calculate the resistance of a 1.0 km length of (i) the steel core and (ii) the aluminium strands.
(2)
(b)
Determine the resistance of 1.0 km of the cable.
(2)
(c)
Show that about 3% of the current is carried by the steel core.
(2)
(d)
The cable carries a current of 400 A. Calculate the power dissipated in each kilometre of cable.
83IBP2H-017-00-14Circuit diagrams and metersPaper 2Hard9 marks
A technician measures the resistance of a strain-gauge sensor using an ammeter of resistance 5.0 Ω and a voltmeter of resistance 2000 Ω. The resistance is calculated as (voltmeter reading)/(ammeter reading). The true resistance of the sensor is 100 Ω. Arrangement A: the ammeter is in series with the sensor, and the voltmeter is connected across both the sensor and the ammeter together. Arrangement B: the voltmeter is connected directly across the sensor only, and the ammeter is in series with the sensor–voltmeter combination.
(a)
State the resistance of an ideal ammeter and of an ideal voltmeter, and give a reason for each.
(2)
(b)
Calculate the resistance that the technician obtains using arrangement A.
(2)
(c)
Calculate the resistance that the technician obtains using arrangement B.
(2)
(d)
The sensor is replaced by one with a true resistance of 10 Ω. Determine which arrangement now gives the more accurate value, and calculate the value obtained using that arrangement.
(3)
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Notes
Part (a)
Ideal ammeter: zero resistance, so that it causes no pd and does not change the current it measures
✓ 1
Ideal voltmeter: infinite resistance, so that it draws no current from the circuit
✓ 1
Part (b)
The voltmeter measures the pd across the sensor and ammeter in series, so the measured value is R + RA
✓ 1
= 100 + 5.0 = 105 Ω
✓ 1
Part (c)
The ammeter measures the total current through the sensor and voltmeter in parallel, so the measured value is RRV/(R + RV)
✓ 1
= (100 × 2000)/2100 = 95 Ω
✓ 1
Part (d)
Arrangement A would give 15 Ω, which is a 50% error
✓ 1
Arrangement B gives (10 × 2000)/2010 = 9.95 Ω
✓ 1
The error with B is only 0.5%, so B is better for low resistances
✓ 1
Common error: Real meters are not ideal. Model the ammeter as a small series resistance and the voltmeter as a large parallel resistance, then work out which quantity each one actually measures.
A potentiometer of total resistance 200 Ω is connected across a 12 V battery of negligible internal resistance. It supplies the pd for an athletics timing-gate sensor of resistance 300 Ω. The sensor is connected between the sliding contact and the negative end of the potentiometer. Let x be the resistance of the section of the potentiometer between the slider and the negative end.
(a)
The sensor is disconnected and the slider is at the midpoint. State the pd between the slider and the negative end.
(1)
(b)
The sensor is reconnected with the slider still at the midpoint. Calculate the pd across the sensor.
(3)
(c)
Calculate the power supplied by the battery in (b).
(2)
(d)
Determine the value of x for which the sensor has a pd of 6.0 V across it.
(3)
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Part (a)
6.0 V
✓ 1
Part (b)
Lower section in parallel with the sensor: (100 × 300)/400 = 75 Ω
✓ 1
Total resistance = 75 + 100 = 175 Ω
✓ 1
V = 12 × 75/175 = 5.1 V
✓ 1
Part (c)
P = V2/Rtotal = 122/175
✓ 1
P = 0.82 W; ECF from (b)
✓ 1
Part (d)
The pd across the sensor is half the supply pd, so 300x/(300 + x) = 200 − x
✓ 1
This rearranges to x2 + 400x − 60000 = 0
✓ 1
x = 116 Ω, taking the positive root
✓ 1
Common error: The unloaded output of a potential divider is not the loaded output. Combine the load in parallel with its section before applying the ratio rule.
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