B.5 Current and Circuits Questions

1B-1A-09

Current, charge and potential difference·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

In a time t, N electrons, each of charge e, pass through a resistor. The energy transferred in the resistor in this time is E.

What is the resistance of the resistor?

Show mark scheme

Marking point	Mark	Notes
Step 1Current: I = Δq/Δt = Ne/t. Potential difference: V = W/q = E/(Ne).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2R = V/I = [E/(Ne)] × [t/(Ne)] = Et/(N²e²).	—
Step 3Check with R = P/I²: (E/t)/(Ne/t)² gives the same result.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis takes the current to be the charge Ne (not divided by t) in R = P/I²: (E/t)/(Ne)².
BThis is the reciprocal of the resistance (I²/P).
CThis uses the energy E instead of the power E/t in R = P/I².
DCorrect: R = V/I with V = E/(Ne) and I = Ne/t.

Syllabus understandingB.5 — direct current as a flow of charge carriers as given by I = Δq/Δt; the electric potential difference as the work done per unit charge as given by V = W/q; electrical resistance R = V/I Command term: Deduce

2B-1A-10

Resistivity·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A uniform wire of resistance R is drawn out so that its length becomes three times its original length. Its volume and resistivity are unchanged.

What is the new resistance?

Show mark scheme

Marking point	Mark	Notes
Step 1Volume = AL is constant, so tripling L reduces the cross-sectional area A to one third.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2R = ρL/A: the length triples (×3) and the area falls to a third (another ×3).	—
Step 3New resistance = 3 × 3 × R = 9R.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis treats the resistance as inversely proportional to the length; a longer wire has a larger resistance.
BThis assumes the length and area changes cancel; in fact both changes increase the resistance.
CThis accounts for the length only, forgetting that the wire becomes thinner.
DCorrect: R ∝ L/A, with L tripled and A reduced to a third.

Syllabus understandingB.5 — resistivity as given by ρ = RA/L Command term: Determine

3B-1A-12

emf and internal resistance·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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

Marking point	Mark	Notes
Step 1The intercept on the V axis (at I = 0) is the emf: ε = 1.50 V.	—	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.	—
Step 3With zero external resistance ε = I(0 + r), so I = 1.50/1.25 = 1.2 A.	✓ 1	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.

Syllabus understandingB.5 — that electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r) Command term: Determine

4B-1A-16

Potential dividers·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark4 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
Step 1Without the load the divider gives 6.0 V; connecting the load changes the ratio.	—	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.	—
Step 3The divider is now 1.0 kΩ above 0.50 kΩ: V_out = 12 × 0.50/(1.0 + 0.50).	—
Step 4V_out = 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.
DThis is the pd across the upper resistor.

Syllabus understandingB.5 — the combinations of resistors in series and parallel circuits (series: V = V₁ + V₂ + …; parallel: 1/R_p = 1/R₁ + 1/R₂ + …); electrical resistance R as given by R = V/I Command term: Determine

5B-1A-21

Conductors and insulators·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

Copper is a good electrical conductor and polythene is an electrical insulator.

Which statement explains this difference?

Show mark scheme

Marking point	Mark	Notes
Step 1A current needs mobile charge carriers.	—	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.	—
Step 3In polythene the electrons are held in covalent bonds, so there are almost no mobile charge carriers and the resistivity is enormous.	✓ 1	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.

Syllabus understandingB.5 — the properties of electrical conductors and insulators in terms of mobility of charge carriers Command term: Identify

6B-1A-26

Power in series and parallel·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two identical heating elements are connected first in series and then in parallel across the same ideal supply.

What is the ratio (total power in parallel)/(total power in series)?

Show mark scheme

Marking point	Mark	Notes
Step 1With each element of resistance R: series total 2R, parallel total R/2.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2P = V²/R_total, so P_series = V²/2R and P_parallel = 2V²/R.	—
Step 3Ratio = (2V²/R)/(V²/2R) = 4.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis is the inverse ratio.
BThis compares the total series power with the power in one parallel element, the wrong way up: (V²/2R)/(V²/R) = 1/2.
CHalf of the correct value — the total resistance changes by a factor of 4, not 2.
DCorrect: the total resistance is four times smaller in parallel, so at fixed V the power is four times greater.

Syllabus understandingB.5 — the combinations of resistors in series and parallel circuits; electrical power P dissipated by a resistor as given by P = IV = I²R = V²/R Command term: Determine

7B-1A-28

Series–parallel circuit·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The circuit shown contains a 12 V cell of negligible internal resistance.

What is the reading of the ammeter?

The cell has negligible internal resistance.

Show mark scheme

Marking point	Mark	Notes
Step 1Parallel pair: 1/R_p = 1/6.0 + 1/3.0 = 1/2.0, so R_p = 2.0 Ω.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Total resistance = 4.0 + 2.0 = 6.0 Ω.	—
Step 3I = 12/6.0 = 2.0 A.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis is 12/13 — adding all three resistances in series.
BThis is 12/9 — ignoring the 4 Ω resistor and adding the 6 Ω and 3 Ω in series instead of in parallel.
CCorrect: 12/(4.0 + 2.0) = 2.0 A.
DThis is 12/4.0, ignoring the parallel pair altogether.

Syllabus understandingB.5 — the combinations of resistors in series and parallel circuits; electrical resistance R as given by R = V/I Command term: Determine

8B-1A-29

Solar cells as energy sources·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
Step 1Input power = intensity × area = 900 × 2.0 × 10⁻² = 18 W.	—	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).	—
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.	✓ 1	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 %.

Syllabus understandingB.5 — that cells provide a source of emf; chemical cells and solar cells as the energy source in circuits; electrical power P dissipated by a resistor as given by P = IV = I²R = V²/R; A.3 — efficiency η = P_output/P_input Command term: Determine

9B-1A-51

Circuit diagrams and meters·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

A student wants to determine the resistance of the resistor R from one pair of meter readings. The four circuits W, X, Y and Z are shown.

Which circuit allows the resistance of R to be determined?

The meters are ideal.

Show mark scheme

Marking point	Mark	Notes
Step 1The ammeter must be in series with R so that it carries the same current as R.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The voltmeter must be connected in parallel with R so that it measures the potential difference across R only.	—
Step 3Only in circuit X are both conditions met; then R = V/I.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AIn W the meters are the wrong way round: the ideal voltmeter in series stops the current, and the ammeter placed across R short-circuits it.
BCorrect: ammeter in series with R, voltmeter in parallel with R.
CIn Y the voltmeter is in series; its very large resistance reduces the current to almost zero and it reads the pd of the cell, not the pd across R.
DIn Z the ammeter is placed correctly, but the voltmeter is connected across the ammeter, which has (almost) zero pd across it, not across R.

Syllabus understandingB.5 — that circuit diagrams represent the arrangement of components in a circuit; electrical resistance R as given by R = V/I Command term: Identify

10B-1A-52

Ohmic and non-ohmic conductors·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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 V_lamp + V_resistor = 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.

Syllabus understandingB.5 — the ohmic and non-ohmic behaviour of electrical conductors; Ohm's law; electrical resistance R = V/I; the combinations of resistors in series circuits Command term: Determine

11B-1A-53

Origin of resistance and the heating effect·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksExplain

When there is a current in a metal resistor, the resistor becomes warm.

Which statement explains this heating effect?

Show mark scheme

Marking point	Mark	Notes
Step 1The pd sets up an electric field in the metal; the free electrons are accelerated by it and gain kinetic energy.	—	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.	—
Step 3In each collision energy is transferred to the ions, which vibrate with greater amplitude: the internal energy and temperature of the resistor rise.	✓ 1	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.

Syllabus understandingB.5 — electric resistance and its origin; the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors Command term: Explain

12B-1A-54

Chemical cells as energy sources·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
Step 1Charge: q = IΔt = 10.0 A × 3600 s = 3.6 × 10⁴ C.	—	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.	—
Step 3W = 3.6 × 10⁴ × 3.7 = 1.3 × 10⁵ J.	✓ 1	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.

Syllabus understandingB.5 — chemical cells as the energy source in circuits; direct current as given by I = Δq/Δt; potential difference as the work done per unit charge V = W/q Command term: Determine

13B-1A-55

Resistivity·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
Step 1A = 0.25 × 10⁻³ × 35 × 10⁻⁶ = 8.75 × 10⁻⁹ m².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2R = ρL/A = 1.7 × 10⁻⁸ × 0.080/8.75 × 10⁻⁹ = 0.155 Ω.	—
Step 3P = I²R, so I = √(P/R) = √(0.50/0.155) = 1.8 A.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis is √(PR) = 0.28, which is the pd across the track in volts, not the current.
BCorrect: R = 0.155 Ω and I = √(P/R) = 1.8 A.
CThis is P/R = 3.2, which is I²; the square root has not been taken.
DThis takes the cross-sectional area as (width)² = 6.25 × 10⁻⁸ m², as if the track were square, giving R = 0.022 Ω.

Syllabus understandingB.5 — resistivity as given by ρ = RA/L; electrical power P dissipated by a resistor as given by P = I²R Command term: Determine

14B-1A-56

Resistance of a non-ohmic component·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The graph shows how the current I in a component X varies with the potential difference V across it. The tangent to the curve at point P is drawn.

What is the resistance of X at point P?

The dashed line is the tangent to the curve at P.

Show mark scheme

Marking point	Mark	Notes
Step 1Resistance is defined as R = V/I at the operating point, whatever the shape of the graph.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At P: V = 4.0 V and I = 0.25 A.	—
Step 3R = 4.0/0.25 = 16 Ω.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis is I/V = 0.25/4.0, the ratio inverted.
BThis is the reciprocal of the gradient of the tangent at P, (6.0 − 2.0)/(0.50 − 0) = 8.0 Ω; resistance is V/I, not ΔV/ΔI.
CThis is ΔV/ΔI between the points at 2.0 V and 4.0 V: 2.0/(0.25 − 0.0625) = 11 Ω.
DCorrect: R = V/I = 4.0/0.25 = 16 Ω.

Syllabus understandingB.5 — electrical resistance R as given by R = V/I; the ohmic and non-ohmic behaviour of electrical conductors Command term: Determine

15B-1A-57

Potentiometers·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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

Show mark scheme

Marking point	Mark	Notes
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.	—	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.	—
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.	✓ 1	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.
DThe circuits have been interchanged.

Syllabus understandingB.5 — that resistors can have variable resistance (potentiometers); the combinations of resistors in series and parallel circuits; circuit diagrams Command term: Deduce

16B-1A-58

Light-dependent resistor sensor·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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 V_out must then be 4.0 V.

Which row gives the required value of R and the change in V_out as it becomes darker?

The cell has negligible internal resistance and the voltmeter is ideal.

Value of RChange in V_out as it becomes darker

Show mark scheme

Marking point	Mark	Notes
Step 1The pd across R must be 4.0 V, so the pd across the LDR is 6.0 − 4.0 = 2.0 V.	—	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.	—
Step 3As it becomes darker the resistance of the LDR increases, so the LDR takes a larger share of the 6.0 V and V_out across R decreases.	✓ 1	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 V_out falls.
DThe resistance of an LDR increases, not decreases, as the light intensity falls.

Syllabus understandingB.5 — that resistors can have variable resistance (light-dependent resistors); the combinations of resistors in series circuits Command term: Deduce

17B-1A-59

Non-ideal meters·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

The ammeter has negligible resistance.

Show mark scheme

Marking point	Mark	Notes
Step 1The ammeter measures the total current in R and the voltmeter together.	—	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.	—
Step 3R = 6.0/0.034 = 176 Ω ≈ 180 Ω.	✓ 1	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.

Syllabus understandingB.5 — non-ideal meters of constant resistance (Guidance); the combinations of resistors in parallel; R = V/I Command term: Determine

18B-1A-60

Internal resistance and lamp brightness·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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.

Syllabus understandingB.5 — electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r); electrical power P = I²R; combinations of resistors in parallel Command term: Determine

19B-1A-70

emf and internal resistance·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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.

Syllabus understandingB.5 — that electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r); electrical power P = I²R = V²/R; that resistors can have variable resistance Command term: Determine

20B-1A-71

Current, charge and potential difference·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Charge crossing the plane: Δq = (6.0 × 10¹⁸ + 1.0 × 10¹⁸) × 1.60 × 10⁻¹⁹ = 1.12 C.	—
Step 3I = Δq/Δt = 1.12/2.0 = 0.56 A.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

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.

Syllabus understandingB.5 — direct current (dc) I as a flow of charge carriers as given by I = Δq/Δt Command term: Determine

21B-1A-72

Ohmic and non-ohmic conductors·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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.

Syllabus understandingB.5 — the ohmic and non-ohmic behaviour of electrical conductors, including heating effects; the combinations of resistors in series circuits; electrical power P = IV Command term: Determine

22B-1A-125

Short-circuiting a component·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two identical lamps L1 and L2 are connected in series with an ammeter to a cell of emf 6.0 V and negligible internal resistance. A switch S is connected directly across L2, as shown. The resistance of each lamp may be assumed constant.

With S open the ammeter reads 0.20 A. What does the ammeter read when S is closed?

S is shown open. The cell has negligible internal resistance.

Show mark scheme

Marking point	Mark	Notes
Step 1With S open the circuit resistance is 2R, where R = 6.0/(2 × 0.20) = 15 Ω is the resistance of one lamp.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Closing S connects a path of (almost) zero resistance across L2: L2 is short-circuited, there is no pd across it and no current in it.	—
Step 3The circuit resistance is now that of L1 only, so the current doubles: I = 6.0/15 = 0.40 A.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis takes "short circuit" to mean that the whole circuit stops working; only L2 is bypassed, and L1 still completes the circuit.
BThis splits the original current equally between L2 and the switch, as if they were two equal parallel paths; the zero-resistance switch takes all the current, and the total resistance falls.
CThis assumes the cell always drives the same current; the current depends on the total resistance, which has halved.
DCorrect: with L2 bypassed the resistance halves, so the current doubles to 0.40 A (L1 now has the full 6.0 V across it).

Syllabus understandingB.5 — that circuit diagrams represent the arrangement of components in a circuit; the combinations of resistors in series circuits; electrical resistance R as given by R = V/I Command term: Determine

23B-1A-126

Charge carriers in conductors and insulators·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Three statements are made about charge carriers.

I. In a metal wire the charge carriers are free electrons, which drift in the direction opposite to the conventional current.
II. In a salt solution the current is carried by positive and negative ions moving in opposite directions.
III. An insulator has a very high resistivity because it contains very few charged particles.

Which statements are correct?

Show mark scheme

Marking point	Mark	Notes
Step 1I is correct: conventional current is the direction of flow of positive charge, so the negatively charged free electrons in a metal drift the opposite way.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2II is correct: dissolved ions are free to move; the positive ions drift with the conventional current and the negative ions against it, and both flows carry charge in the same direction.	—
Step 3III is wrong: an insulator contains as many electrons and nuclei as any other solid; its charges are bound to atoms or molecules, so it has very few mobile charge carriers.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis rejects II, as if only electrons could carry a current; ions in a solution are mobile charge carriers.
BCorrect: conduction depends on the presence of mobile charge carriers, whatever their sign.
CThis rejects I by taking the electrons to drift in the direction of the conventional current, and accepts the false statement III.
DThis accepts III, which confuses "very few charged particles" with "very few mobile charge carriers"; the charges in an insulator are present but bound.

Syllabus understandingB.5 — the properties of electrical conductors and insulators in terms of mobility of charge carriers; direct current (dc) I as a flow of charge carriers Command term: Deduce

24B-1A-127

emf and energy transfer in a cell·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A battery of emf 6.0 V is connected to a resistor. The potential difference across the terminals of the battery is 5.0 V.

What is the energy dissipated inside the battery while a charge of 3.0 C passes through it?

Show mark scheme

Marking point	Mark	Notes
Step 1The emf is the energy transferred from chemical energy per unit charge: W = 6.0 × 3.0 = 18 J in total.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The terminal pd is the energy delivered to the external circuit per unit charge: 5.0 × 3.0 = 15 J goes to the resistor.	—
Step 3The rest, (6.0 − 5.0) × 3.0 = 3.0 J, is dissipated in the internal resistance of the battery.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis divides the lost pd by the charge (1.0/3.0); energy is pd × charge, from V = W/q.
BCorrect: the pd across the internal resistance is 1.0 V, and 1.0 V × 3.0 C = 3.0 J.
CThis is the energy delivered to the resistor (terminal pd × charge), not the energy dissipated inside the battery.
DThis is the total energy transferred from chemical energy by the battery (emf × charge).

Syllabus understandingB.5 — that cells provide a source of emf; the electric potential difference as the work done per unit charge as given by V = W/q; that electric cells are characterized by their emf ε and internal resistance r Command term: Determine

25B-1A-128

Current in parallel branches·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

In the circuit shown, ammeter A1 reads 1.5 A. The ammeters are ideal.

What is the reading of ammeter A2?

The cell has negligible internal resistance.

Show mark scheme

Marking point	Mark	Notes
Step 1The two branches are in parallel, so they have the same pd: 6.0I₂ = 3.0I₃, so the 3.0 Ω branch carries twice the current of the 6.0 Ω branch.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The branch currents add to the current in A1: I₂ + 2I₂ = 1.5 A.	—
Step 3So A2 reads 0.50 A (and the 3.0 Ω resistor carries 1.0 A).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the current divides in the inverse ratio of the resistances, so one third of 1.5 A passes through the 6.0 Ω resistor.
BThis splits the current equally between the branches; equal currents would need equal resistances.
CThis gives the larger share to the larger resistance; the 1.0 A is the current in the 3.0 Ω resistor.
DThis treats A2 as being in series with the cell; the current in A1 divides at the junction.

Syllabus understandingB.5 — that circuit diagrams represent the arrangement of components in a circuit; the combinations of resistors in series and parallel circuits (parallel: I = I₁ + I₂ + …, V = V₁ = V₂ = …) Command term: Determine

26B-1A-129

Solar and chemical cells as energy sources·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

A weather station on a remote mountain is powered by a solar panel. A rechargeable chemical battery is also fitted.

Which statement gives a disadvantage of the solar panel, compared with the chemical battery, as the energy source for the station?

Show mark scheme

Marking point	Mark	Notes
Step 1A solar cell transfers energy from the light it absorbs into electrical energy; it has no store of energy of its own.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Its emf and the current it can deliver fall when the light intensity falls, and at night it supplies nothing; this is why the station also needs a battery.	—
Step 3A chemical battery stores energy, so it can supply the station at any time, but it must be recharged (here by the solar panel) or replaced when its chemical store is used up.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the solar panel depends on light, so a store of energy is needed for the night and for dull weather.
BThis is a disadvantage of the chemical battery, not of the solar panel: a solar panel has no chemical store to run down.
CA solar panel produces no greenhouse gases while it operates; the energy comes from sunlight, not from burning a fuel.
DThis describes a chemical cell running down as its store is used; the emf of a solar cell depends on the light, not on the charge already delivered.

Syllabus understandingB.5 — that cells provide a source of emf; chemical cells and solar cells as the energy source in circuits (Guidance: advantages and disadvantages of different sources of electrical energy) Command term: Identify

27B-1A-130

Resistivity and field in series wires·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two wires P and Q are made of the same metal. P has length L and diameter d. Q has length 2L and diameter 2d. The wires are connected in series to a cell.

What is (electric field strength inside P)/(electric field strength inside Q)?

Show mark scheme

Marking point	Mark	Notes
Step 1R = ρL/A with A ∝ d²: R_P ∝ L/d² and R_Q ∝ 2L/(4d²), so R_P = 2R_Q. In series the current is the same, so V_P = 2V_Q.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The field in a uniform wire is the pd per unit length: E_P = V_P/L and E_Q = V_Q/2L.	—
Step 3E_P/E_Q = (V_P/V_Q) × 2 = 2 × 2 = 4. (Equivalently E = ρI/A: the same current through a quarter of the area.)	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis is the correct ratio inverted: the thinner wire P has the larger field.
BThis assumes the field depends only on the material and the current; for the same current the field is inversely proportional to the cross-sectional area.
CThis is the ratio of the potential differences; the pd across Q is spread over twice the length.
DCorrect: P has twice the pd of Q over half the length, so its field is four times as large.

Syllabus understandingB.5 — resistivity as given by ρ = RA/L; the combinations of resistors in series circuits; D.2 — electric field strength E = V/d for a uniform field Command term: Deduce

28B-1A-131

Resistance of a ring·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A uniform wire of resistance R is bent into a circular ring. Leads are connected to the ring at points X and Y, which are a quarter of the circumference apart, as shown.

What is the resistance of the ring between X and Y?

The ring is made from a single uniform wire.

Show mark scheme

Marking point	Mark	Notes
Step 1The resistance of a uniform wire is proportional to its length, so the short arc has resistance R/4 and the long arc 3R/4.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Between X and Y the two arcs are in parallel: 1/R_XY = 4/R + 4/(3R) = 16/(3R).	—
Step 3R_XY = 3R/16 (smaller than the resistance of either arc).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the arcs of R/4 and 3R/4 in parallel give (R/4 × 3R/4)/R = 3R/16.
BThis is the resistance of the short arc alone; the long arc provides a second, parallel path, which lowers the resistance further.
CThis is the resistance of the long arc alone, ignoring the parallel path through the short arc.
DThis adds the two arcs in series; they are joined at both ends, so they are in parallel.

Syllabus understandingB.5 — the combinations of resistors in series and parallel circuits (parallel: 1/R_p = 1/R₁ + 1/R₂ + …); resistivity as given by ρ = RA/L Command term: Deduce

29B-1A-132

I–V characteristic of a thermistor·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The graph shows the I–V characteristic of component X, a thermistor that warms up as the current in it increases.

Which row describes how the resistance of X and the number of mobile charge carriers in X change as V increases?

I–V characteristic of component X (drawn to scale)

Resistance of XNumber of mobile charge carriers in X

Show mark scheme

Marking point	Mark	Notes
Step 1Resistance is V/I at each point. Reading the graph: 2.0 V/1.24 mA = 1.6 kΩ and 6.0 V/9.48 mA = 0.63 kΩ, so the resistance decreases.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The current warms the thermistor. In a thermistor material a rise in temperature releases more charge carriers that are free to move.	—
Step 3This increase in the number of mobile charge carriers outweighs the greater lattice vibrations, so the current rises faster than the pd: decreases / increases.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis reads the steepening graph as a rising resistance; on an I–V graph a steeper line means a smaller resistance (V/I falls).
BThis gets the resistance right but explains it as for a metal at constant carrier number; fewer collisions cannot be the cause, since the lattice vibrations increase with temperature.
CCorrect: V/I falls along the curve, because heating increases the number of mobile charge carriers.
DThis is the behaviour of a metal filament (resistance rises as the ions vibrate more, carrier number fixed), and it misreads the graph.

Syllabus understandingB.5 — that resistors can have variable resistance (Guidance: thermistors); the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors; the properties of electrical conductors and insulators in terms of mobility of charge carriers; electrical resistance R = V/I Command term: Deduce

30B-1A-133

Power of heaters in series·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Heater X is rated P and heater Y is rated 2P, both when the potential difference across them is V. The heaters are connected in series to a supply of constant potential difference V. The resistance of each heater is constant.

Which row gives the total power of the two heaters and the heater that dissipates the greater power?

Total powerHeater with the greater power

Show mark scheme

Marking point	Mark	Notes
Step 1From P = V²/R: R_X = V²/P and R_Y = V²/2P, so in series the total resistance is 3V²/2P.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Total power = V²/(3V²/2P) = 2P/3.	—
Step 3In series the current is the same, so by P = I²R the heater with the larger resistance, X (the lower-rated heater), dissipates more: 4P/9 in X and 2P/9 in Y.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis assumes the supply pd is shared equally (V/2 each), giving P/4 + 2P/4; the pd divides in the ratio of the resistances, 2 : 1.
BThe total is right, but the higher-rated heater has the smaller resistance and, with the same current, the smaller power.
CThis adds the rated powers, which is the result for the heaters in parallel, each with the full pd V.
DCorrect: the total resistance is 1.5 times that of X, so the total power is 2P/3, and the larger resistance X takes the larger share.

Syllabus understandingB.5 — electrical power P dissipated by a resistor as given by P = IV = I²R = V²/R; the combinations of resistors in series circuits Command term: Deduce

31B-1A-134

Heating effect in wires of different diameter·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two wires X and Y are made of the same metal and have the same length. The diameter of X is half the diameter of Y. The wires are connected in series to a supply and the current is switched on.

Energy transfer to the surroundings is negligible at first. What is (initial rate of rise of temperature of X)/(initial rate of rise of temperature of Y)?

Show mark scheme

Marking point	Mark	Notes
Step 1Same current in series, so the power I²R ∝ R ∝ 1/d²: X dissipates 4 times the power of Y.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Mass ∝ volume ∝ d² (same length and density): X has 1/4 of the mass of Y.	—
Step 3Rate of rise of temperature = power/(mc) ∝ (1/d²)/d² = 1/d⁴, so the ratio is 4 × 4 = 16.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis is the result for the wires in parallel (same pd, power ∝ 1/R ∝ d², mass ∝ d²); in series the current, not the pd, is common.
BThis is the ratio of the powers only; the thinner wire also has a quarter of the mass to heat.
CThis takes the cross-sectional area (and so the resistance) to depend linearly on the diameter, giving a power ratio of 2 instead of 4.
DCorrect: X has 4 times the power and a quarter of the mass, so its temperature rises 16 times as fast — why a thin wire in series can act as a fuse.

Syllabus understandingB.5 — the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors; electrical power P = I²R; resistivity ρ = RA/L; B.1 — quantitative analysis of thermal energy transfers Q = mcΔT Command term: Deduce

32B-1A-135

Current between two potential dividers·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

In the circuit shown, the cell has emf 12 V and negligible internal resistance, and the ammeter is ideal.

What is the reading of the ammeter?

The cell has negligible internal resistance; the ammeter is ideal.

Show mark scheme

Marking point	Mark	Notes
Step 1The ideal ammeter joins P and Q, so they are at the same potential. The two upper resistors are then in parallel (2.0 Ω ∥ 4.0 Ω = 1.33 Ω), and so are the two lower ones (1.33 Ω).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2These two equal combinations share the 12 V equally: the pd across each upper resistor and across each lower resistor is 6.0 V.	—
Step 3Current into P through the upper 2.0 Ω resistor = 6.0/2.0 = 3.0 A; current leaving P through the lower 4.0 Ω resistor = 6.0/4.0 = 1.5 A. The difference, 1.5 A, passes through the ammeter from P to Q.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis assumes the two branches are equivalent because each contains 2.0 Ω and 4.0 Ω; the resistors are in opposite orders, so without the ammeter P would be at 8.0 V and Q at 4.0 V (above the bottom wire).
BCorrect: the upper-left resistor delivers 3.0 A to P but the lower-left resistor takes only 1.5 A away, so 1.5 A crosses to Q.
CThis is the current in each 2.0 Ω resistor, not the difference of the currents at P.
DThis is the total current from the cell, 12 V/2.67 Ω.

Syllabus understandingB.5 — that circuit diagrams represent the arrangement of components in a circuit; the combinations of resistors in series and parallel circuits; electrical resistance R = V/I Command term: Determine

33B-1A-136

Resistance of a hollow conductor·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A metal tube of resistivity ρ and length L has an outer diameter 2d and an inner diameter d. A current flows along the tube, parallel to its axis.

What is the resistance of the tube?

Show mark scheme

Marking point	Mark	Notes
Step 1The current flows through the metal only, so the cross-sectional area is the area of the ring between the two circles.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Outer radius = d, inner radius = d/2: A = πd² − π(d/2)² = 3πd²/4.	—
Step 3R = ρL/A = 4ρL/(3πd²).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis uses the diameters as radii: π[(2d)² − d²] = 3πd², an area four times too large.
BThis uses the whole circular cross-section of outer radius d, ignoring the hole.
CCorrect: the conducting area is the ring of area 3πd²/4.
DThis uses the area of the hole, π(d/2)², as if the current flowed through it.

Syllabus understandingB.5 — resistivity as given by ρ = RA/L; electrical resistance and its origin Command term: Deduce

34B-1B-05

Efficiency of an electric motor·B.5 Current and circuits

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

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)

Syllabus understandingB.5 — electrical power P dissipated by a resistor as given by P = IV = I²R; A.3 — power and efficiency, gravitational potential energy; Tools — propagation of uncertainties, improving an experiment Command term: Determine

35B-1B-09

Potential dividers·B.5 Current and circuits

Paper 1BHard7 marks

Data-based question7 steps to full marksDetermine

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)

Syllabus understandingB.5 — that electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r); that resistors can have variable resistance (potentiometers); A.1 — the equations of motion for solving problems with uniformly accelerated motion; Tools — calibration graph, gradient with a unit Command term: Determine

36B-1B-20

Ohmic behaviour and the heating effect·B.5 Current and circuits

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

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)

Syllabus understandingB.5 — Ohm's law; the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors; electric resistance and its origin; P = IV; Tools — gradient of a line through the origin Command term: Determine

37B-1B-21

Measuring resistivity·B.5 Current and circuits

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

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	✓ 1	Allow ECF from (a). Accept 4.7–5.1 × 10⁻⁷ Ω m.
Part (c)(i)
% uncertainty = 2 × (0.01/0.32) × 100 + 2 = 6.25 + 2 = 8.25 %	✓ 1	The diameter term must be doubled.
Δρ = 0.0825 × 4.9 × 10⁻⁷ = ±0.4 × 10⁻⁷ Ω m	✓ 1	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)

Syllabus understandingB.5 — resistivity as given by ρ = RA/L; D.2 — electric field strength E = V/d for a uniform field; Tools — gradient with unit, systematic offset (intercept), propagation of uncertainties through a power Command term: Determine

38B-1B-22

emf and internal resistance of a lemon cell·B.5 Current and circuits

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

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.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient = (4.00 − 1.31)/(3.00 − 0.50) = 1.07 mA⁻¹ kΩ⁻¹ (= V⁻¹)	✓ 1	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)

Syllabus understandingB.5 — that cells provide a source of emf; chemical cells as the energy source in circuits; cells characterized by their emf ε and internal resistance r as given by ε = I(R + r); Tools — linearising a relationship, gradient and intercept, systematic error from a non-ideal meter Command term: Determine

39B-1B-23

Charge and energy from a chemical cell·B.5 Current and circuits

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

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).
Part (d)(i)
Percentage uncertainty = 1 % + (0.05/4.10) × 100 % = 1 % + 1.2 % = 2.2 %	✓ 1	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)

Syllabus understandingB.5 — direct current as a flow of charge carriers as given by I = Δq/Δt; the electric potential difference as the work done per unit charge as given by V = W/q; chemical cells as the energy source in circuits; electric cells characterized by their emf and internal resistance; Tools — interpolation, area under a graph, propagation of uncertainties Command term: Determine

40B-1B-38

Identical resistors in parallel·B.5 Current and circuits

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

A student has six resistors of the same nominal value R from one packet. She connects n of them in parallel between two crocodile clips and measures the resistance R_m of the arrangement with a digital ohmmeter, for n = 1 to 6. Each reading has an uncertainty of ±0.05 Ω. The leads and clips add a constant resistance r_c in series with the resistors.

The table gives the readings. The graph shows R_m against 1/n with the line of best fit.

n	1	2	3	4	5	6
R_m / Ω	10.79	5.82	4.13	3.29	2.77	2.46
1/n	1.000	0.500	0.333	0.250	0.200	0.167

R_m against 1/n with the line of best fit (drawn to scale)

(a)

(i)

The student first assumes that r_c is negligible, so that R_m = R/n. Use three readings from the table to test this assumption.

(2)

(b)

(i)

Explain why the graph of R_m against 1/n is a straight line that does not pass through the origin.

(1)

(c)

(i)

Determine R.

(2)

(d)

(i)

Determine r_c, and suggest how the student could correct each reading for it using the same ohmmeter.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
nR_m calculated for at least three values of n, e.g. 10.8 Ω, 12.4 Ω and 14.8 Ω (n = 1, 3, 6)	✓ 1	All six products: 10.8, 11.6, 12.4, 13.2, 13.8, 14.8 Ω.
nR_m is not constant: it increases by about 37 % with n, much more than the reading uncertainty allows (at most ±0.3 Ω), so the assumption is not supported	✓ 1	A conclusion supported by the trend is required. Two values only: [1 max].
Part (b)(i)
n identical resistors in parallel have resistance R/n, so R_m = R(1/n) + r_c: a straight line of gradient R and intercept r_c	✓ 1	Reference to the form y = mx + c is required.
Part (c)(i)
Gradient from two well-separated points on the line, e.g. (10.8 − 0.8)/(1.000 − 0)	✓ 1	Points from the table may be used if they lie on the line.
R = 10.0 Ω	✓ 1	Allow ECF from (b)(i) for the identification of the gradient as R. Accept 9.7–10.3 Ω.
Part (d)(i)
r_c = intercept on the R_m axis ≈ 0.8 Ω	✓ 1	Accept 0.7–0.9 Ω, read from the candidate's own line. Allow ECF from (b)(i) for the identification of the intercept as r_c.
Clip the two crocodile clips directly together, read the ohmmeter (a zero-error reading) and subtract this value from every reading	✓ 1	The correction must use the same leads and clips. "Use better leads" alone scores [0].

Answers: (c)(i) 10.0 Ω · (d)(i) ≈ 0.8 Ω (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — the combinations of resistors in series and parallel circuits (parallel: 1/R_p = 1/R₁ + 1/R₂ + …); electrical resistance R = V/I; Tools — testing a relationship with at least three data points, linearising a relationship, gradient and intercept, systematic error (zero offset) Command term: Determine

41B-1B-39

Melting current of fuse wire·B.5 Current and circuits

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A student investigates the current I at which tinned-copper fuse wires of different diameter d melt. Each wire is 5.0 cm long and is held horizontally between two terminal posts in still air. The current is increased slowly until the wire melts. d is measured with a micrometer to ±0.005 mm and I is read to ±0.1 A.

The student models the wire as follows: just before it melts, its temperature is the same for every wire, the resistivity ρ is the same for every wire, and the rate of energy transfer to the air per unit area of the curved surface is the same for every wire. Energy conducted to the posts is neglected.

d / mm	0.10	0.15	0.20	0.25	0.30	0.35
I / A	1.7	3.3	5.1	6.9	9.0	11.6

(a)

(i)

Show that the model predicts that I² ∝ d³.

(2)

(b)

(i)

Determine the percentage uncertainty in I²/d³ for the wire of diameter 0.20 mm.

(2)

(c)

(i)

Evaluate I²/d³ for three of the wires and deduce whether the data support the model.

(2)

(d)

(i)

Predict the diameter of fuse wire that melts at a current of 13 A.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Power dissipated = I²R = I²ρL/(πd²/4) ∝ I²/d² (for fixed length)	✓ 1	Use of R = ρL/A with A ∝ d² is required.
In the steady state this equals the rate of energy transfer, which is ∝ curved surface area πdL ∝ d; so I²/d² ∝ d, giving I² ∝ d³	✓ 1	The equality of input power and loss must be stated or used.
Part (b)(i)
%I = 0.1/5.1 = 2.0 %, %d = 0.005/0.20 = 2.5 %	✓ 1
% uncertainty = 2 × 2.0 + 3 × 2.5 = 11 %	✓ 1	Accept 11–12 %. Award [1 max] if the powers are not applied (4.5 %).
Part (c)(i)
Values for at least three wires, e.g. 0.10 → 2890, 0.15 → 3230, 0.20 → 3250, 0.25 → 3050, 0.30 → 3000, 0.35 → 3140 (A² mm⁻³)	✓ 1	Units are not required. Any three wires spread over the range.
The values differ from their mean (≈ 3100 A² mm⁻³) by at most about 7 %, which is within the ≈ 11 % uncertainty, so the data support the model	✓ 1	Allow ECF from (b). The conclusion must compare the variation with the uncertainty.
Part (d)(i)
d = (13²/3100)^1/3 = 0.38 mm	✓ 1	Allow ECF from (c). Accept 0.37–0.39 mm.

Answers: (b)(i) 11 % · (d)(i) 0.38 mm (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors; resistivity as given by ρ = RA/L; electrical power P = I²R; B.1 — thermal energy transfer from a surface; Tools — propagation of uncertainties through a power, testing a relationship with at least three data points Command term: Determine

42B-1B-40

Linearising a thermistor characteristic·B.5 Current and circuits

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A student places a thermistor in a water bath and measures its resistance R with an ohmmeter at different temperatures θ. Because the thermistor takes time to reach the temperature of the water, each temperature is uncertain by ±1 K; the uncertainty in R is negligible.

Over this range the resistance is expected to follow R = R_∞e^B/T, where T is the kelvin temperature and R_∞ and B are constants. The graph shows ln(R/kΩ) against 1/T with uncertainty bars and the line of best fit.

θ / °C	10	20	30	40	50	60	70
R / kΩ	20.0	12.6	8.00	5.39	3.63	2.53	1.83
(1/T) / 10⁻³ K⁻¹	3.532	3.411	3.299	?	3.095	3.002	2.914
ln(R / kΩ)	2.996	2.534	2.079	?	1.289	0.928	0.604

ln(R / kΩ) against 1/T with uncertainty bars and the line of best fit (drawn to scale)

(a)

(i)

Calculate the two missing values in the table for θ = 40 °C.

(1)

(b)

(i)

Determine B. Give a unit for your answer.

(2)

(c)

(i)

Determine the absolute uncertainty in B by drawing lines of maximum and minimum gradient on the graph.

(2)

(d)

(i)

Determine, in °C, the temperature at which the resistance of the thermistor is 3.0 kΩ.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1/T = 1/313.15 K = 3.193 × 10⁻³ K⁻¹ and ln(5.39) = 1.685	✓ 1	Both values required. Accept 3.19 and 1.68. Use of 313 K is acceptable.
Part (b)(i)
ln R = ln R_∞ + B(1/T), so B is the gradient; gradient from two well-separated points on the line = 3.89 per 10⁻³ K⁻¹	✓ 1	Accept 3.7–4.1 per 10⁻³ K⁻¹. The power of ten of the axis must be used.
B = 3900 K	✓ 1	Accept 3700–4100 K. Unit K required. Award [1 max] for 3.89 K (factor 10³ omitted).
Part (c)(i)
Lines of maximum and minimum gradient drawn through all the uncertainty bars; gradients ≈ 4.01 and 3.75 per 10⁻³ K⁻¹	✓ 1	Lines must pass through every bar. Accept gradients within ±0.05 of these.
ΔB = (max − min)/2 ≈ ±130 K	✓ 1	Allow ECF from (b). Accept ±50 K to ±250 K, given to 1 or 2 s.f. Award [1 max] if the factor 10³ is omitted consistently with (b).
Part (d)(i)
ln 3.0 = 1.10; from the line (or the equation of the line), 1/T = 3.04 × 10⁻³ K⁻¹	✓ 1	Allow ECF from (b). Accept 3.04–3.07 × 10⁻³ K⁻¹.
T = 329 K, so θ = 55 °C	✓ 1	Accept 53–56 °C. Using 3.0 instead of ln 3.0 scores [0].

Answers: (a)(i) 3.193; 1.685 · (b)(i) 3900 K · (c)(i) ±130 K · (d)(i) 55 °C (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that resistors can have variable resistance (Guidance: thermistors); electrical resistance R = V/I; Tools — linearising a relationship using logarithms, gradient with a unit, uncertainty bars and lines of maximum and minimum gradient, interpolation Command term: Determine

43B-2-04

I–V characteristic of a light-emitting diode·B.5 Current and circuits

Paper 2Hard11 marks

Short answer & extended response11 steps to full marksDetermine

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; R_X = 3.0/0.020 = 150 Ω	✓ 1	Allow ECF from (b).
Part (d)(i)
V_LED = 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.
Part (e)(i)
E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/640 × 10⁻⁹ = 3.11 × 10⁻¹⁹ J = 1.94 eV	✓ 1	Answer in eV required.
About equal to the ≈ 2.0 V across the LED: each electron passing through transfers ≈ 2 eV, enough to produce one photon	✓ 1	Allow ECF from (b) for the pd.
Part (f)(i)
Light power = 0.25 × 2.00 × 0.020 = 0.010 W	✓ 1	Allow ECF from (b).
Number per second = 0.010/3.11 × 10⁻¹⁹ = 3.2 × 10¹⁶ s⁻¹	✓ 1	Allow ECF from (e).

Answers: (b)(i) 100 Ω · (c)(i) 150 Ω · (d)(i) 17 mA · (e)(i) 1.94 eV · (f)(i) 3.2 × 10¹⁶ s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — Ohm's law and the ohmic and non-ohmic behaviour of electrical conductors; electrical resistance R = V/I; electric cells characterized by ε = I(R + r); electrical power P = IV; E.1 — that photons are emitted with energy E = hf Command term: Determine

44B-2-14

Potential divider with a light-dependent resistor·B.5 Current and circuits

Paper 2Hard12 marks

Short answer & extended response12 steps to full marksDetermine

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

Marking point	Mark	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 R_F = R_LDR × 2.0/4.0	✓ 1	Or the current 4.0/28.7 kΩ = 0.139 mA.
R_F = 14.4 kΩ	✓ 1	Accept 13.5–15 kΩ.
Part (c)(i)
The lamp switches on when R_LDR = 2R_F (pd shared in the ratio 4.0 : 2.0); darker means a larger R_LDR at the switching point	✓ 1	Allow ECF from (b).
So R_F 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).
R_LDR = 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.
Part (e)(i)
λ = hc/E = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/(2.4 × 1.60 × 10⁻¹⁹) = 5.2 × 10⁻⁷ m	✓ 1
650 nm is longer than this, so each red photon (about 1.9 eV) has too little energy to release a charge carrier	✓ 1

Answers: (b)(i) 14.4 kΩ · (d)(i) ≈ 13 lx · (e)(i) 5.2 × 10⁻⁷ m (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that resistors can have variable resistance (Guidance: light-dependent resistors); the properties of electrical conductors and insulators in terms of mobility of charge carriers; electrical resistance R = V/I; the combinations of resistors in series and parallel circuits; E.2 — that photons of a certain frequency, known as the threshold frequency, are required to release photoelectrons from the metal (applied here to the release of charge carriers in the LDR); E.1 — E = hf Command term: Determine

45B-2-15

A solar panel as a source of emf·B.5 Current and circuits

Paper 2Medium11 marks

Short answer & extended response11 steps to full marksDetermine

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)

Syllabus understandingB.5 — chemical cells and solar cells as the energy source in circuits; that electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r); electrical power P = IV; direct current I as a flow of charge carriers as given by I = Δq/Δt; the combinations of resistors in series and parallel circuits (series: I = I₁ = I₂); A.3 — efficiency; E.2 — the photoelectric effect as evidence of the particle nature of light Command term: Determine

46B-2-30

Resistivity and power in a supply lead·B.5 Current and circuits

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDetermine

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

Marking point	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)

Syllabus understandingB.5 — resistivity as given by ρ = RA/L; combinations of resistors in series; electrical power P = IV = I²R = V²/R; the heating effect of resistors; D.2 — electric field strength as E = F/q and the uniform field E = V/d; B.1 — convection and radiation as mechanisms of thermal energy transfer Command term: Determine

47B-2-31

emf and internal resistance·B.5 Current and circuits

Paper 2Hard12 marks

Short answer & extended response12 steps to full marksDetermine

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.
Part (c)(i)
F/L = μ₀I₁I₂/2πr = 4π × 10⁻⁷ × 115²/(2π × 0.015) = 0.18 N m⁻¹	✓ 1	Allow ECF from (b)(i).
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)

Syllabus understandingB.5 — that electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r); the combinations of resistors in series and parallel circuits; electrical power P = IV = I²R = V²/R; electric resistance and its origin; D.3 — the force per unit length between parallel wires as given by F/L = μ₀I₁I₂/2πr Command term: Determine

48B-2-32

A thermistor thermometer·B.5 Current and circuits

Paper 2Hard14 marks

Short answer & extended response9 steps to full marksDeduce

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 R_T of the thermistor at different temperatures θ.

θ / °C	0	10	20	30	40	50
R_T / 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 R_T = 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	✓ 1
Reading = 9.0 × 8.0/(8.0 + 8.04) = 4.5 V	✓ 1	Accept 4.49 V.
Part (c)(ii)
Assuming an ideal meter: R_T = 10.0 × (9.0 − 4.49)/4.49 = 10.1 kΩ	✓ 1	Accept 10.0–10.1 kΩ; ECF from (c)(i).
From the table θ ≈ 25 °C	✓ 1	Accept 24–26 °C.
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 = I²R_T = (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)

Syllabus understandingB.5 — that resistors can have variable resistance (Guidance: thermistors); the properties of electrical conductors and insulators in terms of mobility of charge carriers; the combinations of resistors in series and parallel circuits; the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors; electrical power P = I²R (Guidance: in cases where non-ideal meters are used, the resistance will be constant) Command term: Deduce

49B-2-33

Resistance of transmission cables·B.5 Current and circuits

Paper 2Medium11 marks

Short answer & extended response11 steps to full marksDetermine

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.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
R = ρL/A = 1.7 × 10⁻⁸ × 1000/1.50 × 10⁻⁴ = 0.113 Ω	✓ 1	An answer to at least 3 s.f. is required.
Part (b)(i)
Contact wire: 2.0 × 0.113 = 0.227 Ω	✓ 1	Allow ECF from (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 + F_r = 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)

Syllabus understandingB.5 — resistivity as given by ρ = RA/L; combinations of resistors in series; potential difference; electrical power P = IV = I²R; A.2 — forces on a slope in translational equilibrium; A.3 — power as P = Fv and efficiency Command term: Determine

50B-2-34

An electric-vehicle battery pack·B.5 Current and circuits

Paper 2Hard12 marks

Short answer & extended response9 steps to full marksDiscuss

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

Marking point	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	✓ 1
V = 355.2 + 250 × 0.144 = 391 V	✓ 1
Part (c)(ii)
Thermal energy = I²rt = 250² × 0.144 × 1200 = 1.08 × 10⁷ J	✓ 1
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)

Syllabus understandingB.5 — that cells provide a source of emf; chemical cells and solar cells as the energy source in circuits (advantages and disadvantages of different sources of electrical energy); ε = I(R + r); P = I²R; B.1 — specific heat capacity Q = mcΔT Command term: Discuss

51B-2-36

Resistivity and heating strips·B.5 Current and circuits

Paper 2Easy12 marks

Short answer & extended response12 steps to full marksDetermine

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	✓ 1
Energy = 0.0990 × (2.1 × 10³ × 6.0 + 3.34 × 10⁵) = 3.43 × 10⁴ J	✓ 1	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)

Syllabus understandingB.5 — resistivity as given by ρ = RA/L; the combinations of resistors in parallel circuits; electrical power P = IV = I²R = V²/R; the heating effect of resistors; B.1 — Q = mcΔT and Q = mL; convection; A.3 — efficiency and the energy density of fuel sources Command term: Determine

52B-2-39

emf and internal resistance of a battery·B.5 Current and circuits

Paper 2Hard12 marks

Short answer & extended response12 steps to full marksDetermine

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)

Syllabus understandingB.5 — that electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r); combinations of cells/resistors in series; electrical resistance R = V/I; electrical power P = IV = I²R; direct current I = Δq/Δt; A.2 — Newton's three laws of motion and F = Δp/Δt; A.3 — kinetic energy, power and efficiency Command term: Determine

53B-2-60

Heating pads in a jacket·B.5 Current and circuits

Paper 2Easy12 marks

Short answer & extended response12 steps to full marksDetermine

A heated jacket contains two identical heating pads, one in the back and one in the front. Each pad is rated 5.0 W when the potential difference across it is 5.0 V. The jacket is powered by a power bank whose output potential difference is a constant 5.0 V. The resistance of each pad is constant.

(a)

(i)

Calculate the resistance of one pad.

(1)

(b)

(i)

Draw a circuit diagram of the power bank, the two pads and two switches, so that the jacket can be switched off, run with the back pad only, or run with both pads each working at its rated power.

(2)

(ii)

Calculate the current from the power bank when both pads are switched on.

(1)

(c)

(i)

When fully charged, the power bank stores 18.5 W h of energy, and 85 % of this can be delivered to the jacket. Determine the time, in hours, for which both pads can operate.

(3)

(ii)

Calculate the charge that passes through the power bank in this time.

(1)

(d)

(i)

The back pad is damaged so that half of the parallel carbon fibres in its heating element break. The remaining fibres are undamaged. State and explain the effect on the power of the back pad when both pads are switched on.

(2)

(e)

(i)

The front pad and the fabric in contact with it have a total mass of 0.12 kg and a mean specific heat capacity of 1.3 × 10³ J kg⁻¹ K⁻¹. Estimate the initial rate of rise of their temperature when the front pad is switched on, and state one assumption you made.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
R = V²/P = 5.0²/5.0 = 5.0 Ω	✓ 1
Part (b)(i)
Both pads connected in parallel across the power bank	✓ 1	Correct circuit symbols are expected; the power bank may be drawn as a cell or battery.
One switch in the main circuit (controlling both pads) and the second switch in series with the front pad only, in its own branch	✓ 1	Do not award if a switch is placed so that it short-circuits the power bank or a pad.
Part (b)(ii)
I = 2 × 5.0/5.0 = 2.0 A	✓ 1	Allow ECF from (a). ALT: 10 W/5.0 V.
Part (c)(i)
Energy stored = 18.5 × 3600 = 6.66 × 10⁴ J	✓ 1	Allow the calculation in W h throughout.
Energy delivered = 0.85 × 6.66 × 10⁴ = 5.66 × 10⁴ J	✓ 1	Award [1 max] overall for 18.5/10 = 1.85 h (efficiency ignored).
t = 5.66 × 10⁴/10 = 5.66 × 10³ s = 1.6 h	✓ 1	Allow ECF from (b)(ii). Accept 1.6 h. ALT: 0.85 × 18.5/10.
Part (c)(ii)
q = It = 2.0 × 5.66 × 10³ = 1.1 × 10⁴ C	✓ 1	Allow ECF from (b)(ii) and (c)(i). The time must be in seconds.
Part (d)(i)
The cross-sectional area of conductor halves, so by R = ρL/A its resistance doubles (to 10 Ω)	✓ 1	Allow ECF from (a).
The pad is in parallel with the power bank, so the pd across it is still 5.0 V; P = V²/R halves, to 2.5 W	✓ 1	Use of P = I²R with an unchanged current scores [0] for this mark.
Part (e)(i)
Rate = P/(mc) = 5.0/(0.12 × 1.3 × 10³) = 0.032 K s⁻¹	✓ 1	Accept 0.032 K s⁻¹ (1.9 K per minute).
Assumption: no energy is transferred to the surroundings (the body or the air) at first / the temperature is uniform throughout the pad and fabric	✓ 1	Any one sensible assumption.

Answers: (a)(i) 5.0 Ω · (b)(ii) 2.0 A · (c)(i) 1.6 h · (c)(ii) 1.1 × 10⁴ C · (d)(i) 2.5 W · (e)(i) 0.032 K s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that circuit diagrams represent the arrangement of components in a circuit; the combinations of resistors in parallel circuits; electrical power P = IV = I²R = V²/R; direct current I = Δq/Δt; resistivity ρ = RA/L; A.3 — efficiency; B.1 — Q = mcΔT Command term: Determine

54B-2-61

A pencil line as a resistor·B.5 Current and circuits

Paper 2Easy12 marks

Short answer & extended response12 steps to full marksDetermine

A student draws a uniform line AB with a soft pencil on a sheet of paper. The line is 12.0 cm long and 5.0 mm wide, and the layer of graphite is 2.0 μm thick. An ohmmeter connected between A and B reads 3.6 kΩ.

(a)

(i)

Outline, in terms of charge carriers, why the graphite line conducts while the paper is an insulator.

(2)

(ii)

Determine the resistivity of the pencil line.

(2)

(b)

Ends A and B are connected to a battery of emf 9.0 V and negligible internal resistance. An ideal voltmeter is connected between A and a probe that touches the line at a distance x from A.

(i)

Calculate the current in the line.

(1)

(ii)

Calculate the reading of the voltmeter when x = 4.0 cm.

(1)

(iii)

The student draws over the half of the line nearer to B again, so that the thickness of graphite in that half doubles. The other half is unchanged. Determine the new reading of the voltmeter when x = 4.0 cm.

(3)

(c)

The line is restored to its original condition. A second, identical line is drawn beside it, and the two lines are joined at both ends by thick metal strips of negligible resistance. The 9.0 V battery is connected across the strips.

(i)

Determine the current from the battery.

(2)

(ii)

State and explain whether the voltmeter reading at x = 4.0 cm on one of the lines differs from your answer to (b)(ii).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Graphite has electrons that are free to move through the material (mobile charge carriers)	✓ 1
In paper almost all electrons are bound to their atoms or molecules, so there are (almost) no mobile charge carriers to form a current	✓ 1	"Paper has no electrons" or "paper has no charges" scores [0] for this mark.
Part (a)(ii)
A = 5.0 × 10⁻³ × 2.0 × 10⁻⁶ = 1.0 × 10⁻⁸ m²	✓ 1	The cross-section is a rectangle, width × thickness.
ρ = RA/L = 3.6 × 10³ × 1.0 × 10⁻⁸/0.120 = 3.0 × 10⁻⁴ Ω m	✓ 1
Part (b)(i)
I = 9.0/3.6 × 10³ = 2.5 × 10⁻³ A	✓ 1
Part (b)(ii)
The line is uniform, so the pd is proportional to the length: 9.0 × 4.0/12.0 = 3.0 V	✓ 1	ALT: 2.5 × 10⁻³ × (3.6 kΩ × 4.0/12.0). Allow ECF from (b)(i).
Part (b)(iii)
Half nearer A: 1.8 kΩ (unchanged); half nearer B: thickness doubled, so 1.8/2 = 0.90 kΩ	✓ 1	Resistance ∝ 1/thickness.
Total resistance = 2.7 kΩ, so I = 9.0/2.7 × 10³ = 3.33 × 10⁻³ A	✓ 1
Reading = 3.33 × 10⁻³ × 1.2 × 10³ = 4.0 V	✓ 1	Allow ECF from the new total resistance. Award [1 max] for 3.0 V (no change).
Part (c)(i)
Two equal resistances in parallel: R = 3.6/2 = 1.8 kΩ	✓ 1
I = 9.0/1.8 × 10³ = 5.0 × 10⁻³ A	✓ 1	Award [1 max] for 1.25 × 10⁻³ A (lines treated as in series).
Part (c)(ii)
No change (still 3.0 V): each line has the full 9.0 V across it, because the lines are in parallel with the battery, and each line is still uniform	✓ 1	Allow ECF from (b)(ii). The reason must refer to the unchanged pd across each line.

Answers: (a)(ii) 3.0 × 10⁻⁴ Ω m · (b)(i) 2.5 × 10⁻³ A · (b)(ii) 3.0 V · (b)(iii) 4.0 V · (c)(i) 5.0 × 10⁻³ A (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — the properties of electrical conductors and insulators in terms of mobility of charge carriers; resistivity as given by ρ = RA/L; that resistors can have variable resistance (Guidance: potentiometers); the combinations of resistors in series and parallel circuits; electrical resistance R = V/I Command term: Determine

55B-2-62

Potential along two wires in series·B.5 Current and circuits

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDetermine

A wire X of length 1.20 m and a wire Y of length 0.80 m are joined end to end at J. Both wires have a diameter of 0.46 mm. The free end A of X and the free end C of Y are connected to a cell of emf 1.50 V and an ideal ammeter in series; the ammeter reads 0.150 A.

An ideal voltmeter measures the potential V at a point on the wires, at distance x from A, relative to C. The graph shows the results.

Potential V relative to C against distance x from A (drawn to scale)

(a)

(i)

State the potential difference across the terminals of the cell.

(1)

(ii)

Determine the internal resistance of the cell.

(2)

(b)

(i)

Determine the resistance of Y.

(1)

(ii)

Determine the resistivity of the material of Y.

(2)

(c)

(i)

Deduce, using only the gradients of the graph, the ratio (resistivity of Y)/(resistivity of X).

(2)

(ii)

State the magnitude of the electric field strength inside Y.

(1)

(d)

(i)

Determine, in eV, the energy transferred in Y by one conduction electron as it passes through the whole length of Y.

(1)

(e)

(i)

Y is replaced by a wire of the same material and length but half the diameter. Determine the new potential at J.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1.35 V: the potential of A relative to C, read at x = 0 (the ideal ammeter has no pd across it)	✓ 1	Accept 1.34–1.36 V.
Part (a)(ii)
pd across the internal resistance = 1.50 − 1.35 = 0.15 V	✓ 1	Allow ECF from (a)(i).
r = 0.15/0.150 = 1.0 Ω	✓ 1	Accept 0.9–1.1 Ω.
Part (b)(i)
pd across Y = potential at J = 0.81 V, so R_Y = 0.81/0.150 = 5.4 Ω	✓ 1	Accept 0.80–0.82 V and 5.3–5.5 Ω.
Part (b)(ii)
A = π(0.23 × 10⁻³)² = 1.66 × 10⁻⁷ m²	✓ 1	Use of the diameter as the radius scores [0] for this mark.
ρ = RA/L = 5.4 × 1.66 × 10⁻⁷/0.80 = 1.1 × 10⁻⁶ Ω m	✓ 1	Allow ECF from (b)(i). Accept 1.1 × 10⁻⁶ Ω m.
Part (c)(i)
The gradient magnitude is the pd per unit length, IR/L = Iρ/A; I and A are the same for both wires	✓ 1
Gradients ≈ 0.45 V m⁻¹ (X) and 1.0 V m⁻¹ (Y), so the ratio = 2.25	✓ 1	Accept 2.1–2.4.
Part (c)(ii)
E = 1.0 V m⁻¹ (the magnitude of the potential gradient in Y)	✓ 1	Allow ECF from (c)(i). Accept 0.95–1.05 V m⁻¹.
Part (d)(i)
W = qV with V = 0.81 V: energy = 0.81 eV	✓ 1	Allow ECF from (b)(i). The answer must be in eV; 1.3 × 10⁻¹⁹ J is not accepted without conversion.
Part (e)(i)
New resistance of Y = 4 × 5.4 = 21.6 Ω	✓ 1	Allow ECF from (b)(i). R ∝ 1/d².
I = 1.50/(3.6 + 21.6 + 1.0) = 0.0573 A	✓ 1	Allow ECF from (a)(ii). The internal resistance must be included.
Potential at J = 0.0573 × 21.6 = 1.24 V	✓ 1	Accept 1.22–1.25 V. Award [2 max] for 1.29 V (internal resistance ignored).

Answers: (a)(i) 1.35 V · (a)(ii) 1.0 Ω · (b)(i) 5.4 Ω · (b)(ii) 1.1 × 10⁻⁶ Ω m · (c)(i) 2.25 · (c)(ii) 1.0 V m⁻¹ · (d)(i) 0.81 eV · (e)(i) 1.24 V (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r); resistivity as given by ρ = RA/L; the combinations of resistors in series circuits; the electric potential difference as the work done per unit charge V = W/q; D.2 — the electric field strength as the potential gradient E = −ΔV/Δr; work done in eV Command term: Determine

56B-2-63

Hot-wire anemometer·B.5 Current and circuits

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksDetermine

A hot-wire anemometer measures the speed u of air. Its sensor is a platinum wire of length 2.0 mm and diameter 10 μm, held between two supports in the air stream. The resistivity of platinum at 20 °C is 1.06 × 10⁻⁷ Ω m. The air is at 20 °C.

The resistance of the wire at temperature θ is R = R₀[1 + α(θ − 20 °C)], where R₀ is its resistance at 20 °C and α = 3.9 × 10⁻³ K⁻¹. The rate at which energy is transferred from the wire to the air is k(θ − 20 °C), where k = k₀ + k₁√u with k₀ = 6.0 × 10⁻⁵ W K⁻¹ and k₁ = 4.5 × 10⁻⁵ W K⁻¹ m^−1/2 s^1/2.

(a)

(i)

Show that R₀ is about 2.7 Ω.

(2)

(b)

In use, an electronic circuit adjusts the current so that the wire stays at 220 °C.

(i)

Determine the current in the wire when the air speed is 5.0 m s⁻¹.

(3)

(ii)

Explain why a larger current is needed when the air speed increases.

(2)

(c)

(i)

The current is 0.100 A. Determine the air speed.

(2)

(d)

A simpler design keeps the current constant at the value found in (b)(i) instead of keeping the temperature constant. Let Δθ = θ − 20 °C.

(i)

Show that, in the steady state, Δθ = I²R₀/(k − αI²R₀).

(2)

(ii)

Determine the temperature of the wire when the air speed is 20 m s⁻¹.

(2)

(iii)

Deduce what happens to the wire in this design if the air flow stops.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
A = π(5.0 × 10⁻⁶)² = 7.85 × 10⁻¹¹ m²	✓ 1
R₀ = 1.06 × 10⁻⁷ × 2.0 × 10⁻³/7.85 × 10⁻¹¹ = 2.70 Ω	✓ 1	An answer to at least 3 s.f. or full substitution is required.
Part (b)(i)
Resistance at 220 °C = 2.70 × (1 + 3.9 × 10⁻³ × 200) = 4.80 Ω	✓ 1	Allow ECF from (a).
k = 6.0 × 10⁻⁵ + 4.5 × 10⁻⁵ × √5.0 = 1.61 × 10⁻⁴ W K⁻¹; rate of energy transfer = 1.61 × 10⁻⁴ × 200 = 0.0321 W	✓ 1
Steady temperature: I²R = 0.0321 W, so I = √(0.0321/4.80) = 0.0818 A	✓ 1	Accept 0.081–0.082 A. The energy balance must be used.
Part (b)(ii)
Faster air carries energy away from the wire more rapidly (forced convection: k increases with u), so the wire would cool	✓ 1
To keep the temperature, and so the resistance, constant, the electrical power I²R must increase to match the larger loss, so I must increase	✓ 1	Reference to the power balance at constant R is required.
Part (c)(i)
k = I²R/200 = 0.100² × 4.80/200 = 2.40 × 10⁻⁴ W K⁻¹	✓ 1	Allow ECF from (b)(i). The resistance at 220 °C must be used.
√u = (2.40 × 10⁻⁴ − 6.0 × 10⁻⁵)/4.5 × 10⁻⁵ = 4.01, so u = 16 m s⁻¹	✓ 1	Accept 15–17 m s⁻¹.
Part (d)(i)
Energy balance: I²R₀(1 + αΔθ) = kΔθ	✓ 1
I²R₀ = Δθ(k − αI²R₀), which rearranges to the given expression	✓ 1	The collection of the Δθ terms must be seen.
Part (d)(ii)
k = 2.61 × 10⁻⁴ W K⁻¹; I²R₀ = 1.80 × 10⁻² W and αI²R₀ = 7.04 × 10⁻⁵ W K⁻¹	✓ 1	Allow ECF from (a)(i) and (b)(i).
Δθ = 95 K, so θ ≈ 115 °C	✓ 1	Accept 105–120 °C.
Part (d)(iii)
With u = 0, k = k₀ = 6.0 × 10⁻⁵ W K⁻¹, which is less than αI²R₀ = 7.04 × 10⁻⁵ W K⁻¹, so there is no steady temperature: the resistance and the power rise together and the temperature rises until the wire burns out	✓ 1	Allow ECF from (d)(ii). The comparison of the two terms (or a negative denominator) is required.

Answers: (a)(i) 2.70 Ω · (b)(i) 0.0818 A · (c)(i) 16 m s⁻¹ · (d)(ii) 115 °C (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors; electric resistance and its origin; resistivity as given by ρ = RA/L; electrical power P = I²R; B.1 — thermal energy transfer by convection Command term: Determine

57B-2-64

Current in an ionisation smoke detector·B.5 Current and circuits

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksDetermine

In an ionisation smoke detector, two parallel metal plates with air between them are connected in series with a resistor of resistance 1.0 × 10¹⁰ Ω and a battery of emf 9.0 V and negligible internal resistance. A sealed americium-241 source of activity 3.7 × 10⁴ Bq emits alpha particles of kinetic energy 5.5 MeV.

On average the alpha particles deposit 20 % of their kinetic energy in the air between the plates, and 34 eV is needed to produce one ion pair (a positive ion and a negative ion). When all the ions are collected by the plates, each ion pair causes a charge e to flow round the circuit. This happens whenever the pd across the plates is at least 3 V.

(a)

(i)

Explain why there is almost no current between the plates without the source, but a current with the source.

(2)

(b)

(i)

Calculate the number of ion pairs produced by an alpha particle that transfers all of its kinetic energy to air.

(1)

(ii)

Determine the current in the circuit.

(2)

(iii)

Determine the pd across the plates, and confirm that all the ions are collected.

(2)

(c)

(i)

When smoke enters the space between the plates, the ions attach to smoke particles and the current falls. The alarm sounds when the pd across the plates rises to 7.5 V. Determine the smallest percentage decrease in the current that sounds the alarm.

(2)

(ii)

Suggest why the current falls when the ions attach to smoke particles.

(1)

(d)

(i)

The half-life of americium-241 is 432 years. Determine the percentage decrease in the current, in clear air, after 10 years.

(2)

(ii)

Comment on whether the detector needs to be adjusted during this time.

(1)

(iii)

Suggest why an alpha source, rather than a beta source, is used.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Air molecules are neutral (electrons bound), so without the source there are almost no mobile charge carriers in the air: air is an insulator	✓ 1
The alpha particles ionise the air, producing mobile positive and negative ions, which the electric field between the plates moves to the plates, forming a current	✓ 1	Both the creation of mobile carriers and their motion in the field are required.
Part (b)(i)
5.5 × 10⁶/34 = 1.6 × 10⁵	✓ 1
Part (b)(ii)
Ion pairs produced per second = 3.7 × 10⁴ × 0.20 × 1.62 × 10⁵ = 1.2 × 10⁹ s⁻¹	✓ 1	Allow ECF from (b)(i). The 20 % must be applied.
I = 1.20 × 10⁹ × 1.60 × 10⁻¹⁹ = 1.9 × 10⁻¹⁰ A	✓ 1	Award [1 max] for 3.8 × 10⁻¹⁰ A (charge 2e per pair).
Part (b)(iii)
pd across the resistor = 1.92 × 10⁻¹⁰ × 1.0 × 10¹⁰ = 1.92 V	✓ 1	Allow ECF from (b)(ii).
pd across the plates = 9.0 − 1.92 = 7.08 V, which is more than 3 V, so all the ions are collected	✓ 1	The comparison with 3 V is required.
Part (c)(i)
pd across the resistor at the alarm = 9.0 − 7.5 = 1.5 V, so the current = 1.5/1.0 × 10¹⁰ = 1.5 × 10⁻¹⁰ A	✓ 1
Decrease = (1.92 − 1.50)/1.92 = 22 % (currents in units of 10⁻¹⁰ A)	✓ 1	Allow ECF from (b)(ii). Accept 21–23 %.
Part (c)(ii)
Smoke particles are much more massive, so the charged particles move much more slowly in the field (lower mobility) and many recombine before reaching the plates, so less charge reaches the plates per second	✓ 1	"Smoke blocks the current" scores [0].
Part (d)(i)
Fraction remaining = (1/2)^10/432 = 0.9841 (or λ = ln 2/432 = 1.60 × 10⁻³ y⁻¹ and e^−10λ)	✓ 1	The current is proportional to the activity.
Decrease = 1.6 %	✓ 1	Accept 1.5–1.7 %.
Part (d)(ii)
No: a 1.6 % fall is much smaller than the 22 % fall needed to sound the alarm, so it will not cause a false alarm	✓ 1	Allow ECF from (c)(i) and (d)(i).
Part (d)(iii)
Alpha particles are strongly ionising, producing many ion pairs (a large current) in a small volume, and their short range in air means they do not leave the detector	✓ 1	Either idea, linked to the detector.

Answers: (b)(i) 1.6 × 10⁵ · (b)(ii) 1.9 × 10⁻¹⁰ A · (b)(iii) 7.08 V · (c)(i) 22 % · (d)(i) 1.6 % (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — direct current (dc) I as a flow of charge carriers as given by I = Δq/Δt; the properties of electrical conductors and insulators in terms of mobility of charge carriers; the combinations of resistors in series circuits; electrical resistance R = V/I; E.3 — the properties of alpha radiation (ionising ability and range); activity and half-life Command term: Determine

58B-2-67

A hot-filament electron source·B.5 Current and circuits

Paper 2Hard18 marks

Short answer & extended response18 steps to full marksDetermine

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.
R = ρL/A = 5.6 × 10⁻⁸ × 0.030/1.23 × 10⁻⁸ = 0.137 Ω	✓ 1	Answer to at least 3 s.f. required.
Part (a)(ii)
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	✓ 1	Answer to at least 2 s.f. required.
Part (b)(ii)
ρ at 2500 K read from the graph = 7.3 × 10⁻⁷ Ω m	✓ 1	Accept 7.1–7.5 × 10⁻⁷ Ω m.
R = 7.3 × 10⁻⁷ × 0.030/1.23 × 10⁻⁸ = 1.78 Ω	✓ 1	Allow ECF from (a)(i). ALT: 0.137 × 7.3/0.56. Accept 1.7–1.85 Ω.
Part (b)(iii)
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)
λ_max = 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 = ½m_ev², 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 = m_ev²/r, so r = m_ev/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)

Syllabus understandingB.5 — resistivity ρ = RA/L; electrical power P = IV = I²R = V²/R; the origin of resistance; B.1 — the Stefan–Boltzmann law with emissivity; Q = mcΔT; Wien's displacement law; E.1 — photon energy E = hf; D.2 — the work done on a charge, qV; D.3 — the motion of a charged particle in a uniform magnetic field, F = qvB sin θ Command term: Determine

B.5 Current and circuits: IB Physics HL exam-style questions

32 practice questions on B.5 Current and circuits

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