B.5 Current and Circuits Questions

1B-1A-20

Current & charge·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

In a time of 8.0 s, a charge of 24 C flows through a heating element and 288 J of electrical energy is transferred in the element.

Which row gives the current in the element and the potential difference across it?

CurrentPotential difference

Show mark scheme

Marking point	Mark	Notes
Step 1I = Δq/Δt = 24 / 8.0 = 3.0 A.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2V = W/q = 288 / 24 = 12 V.	✓ 1	Answer D

Answer: D · 2 stages of work, one mark

Every option, and why

AThe current has been calculated as Δt/Δq. Current is charge per unit time.
BTwo errors: Δt/Δq for the current, and energy per unit time (the power, 36 W) for the potential difference.
C36 is the energy transferred per second — the power in watts. Potential difference is energy per unit charge.
DCorrect: I = 24/8.0 = 3.0 A and V = 288/24 = 12 V.

Syllabus understandingB.5 — direct current (dc) I as a flow of charge carriers as given by I = Δq/Δt; that the electric potential difference V is the work done per unit charge on moving a positive charge between two points along the path of the current as given by V = W/q Command term: Determine

2B-1A-21

Resistivity·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

Wires X and Y are made of the same metal and are at the same temperature. Y is twice as long as X and has twice the diameter of X. The two wires are connected in series to a battery.

R is the resistance of a wire and P is the power dissipated in it. Which row gives R_Y/R_X and P_Y/P_X?

R_Y / R_XP_Y / P_X

Show mark scheme

Marking point	Mark	Notes
Step 1R = ρL/A with A ∝ d²: R_Y/R_X = 2 / 2² = 1/2.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2In series the wires carry the same current, so P = I²R ∝ R: P_Y/P_X = 1/2.	✓ 1	Answer D

Answer: D · 2 stages of work, one mark

Every option, and why

AThe change in cross-section has been ignored, so only the doubled length has been counted.
BThe area has been taken as proportional to the diameter instead of the diameter squared, so the two factors of 2 cancel.
CP = V²/R has been used as if the wires had the same potential difference. In series they carry the same current, so the smaller resistance dissipates less.
DCorrect: R ∝ L/d² gives 2/4 = 1/2, and with equal currents P = I²R gives the same ratio.

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

3B-1A-22

emf & internal resistance·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

A cell is connected to a variable resistor. The graph shows the variation of the potential difference V across the terminals of the cell with the current I in the cell.

Which row gives the emf and the internal resistance of the cell?

Graph drawn to scale

emfInternal resistance

Show mark scheme

Marking point	Mark	Notes
Step 1V = ε − Ir, so the intercept on the V axis (where I = 0) is the emf: ε = 6.0 V.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2The gradient is −r: r = 6.0 / 3.0 = 2.0 Ω.	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

A3.0 is the intercept on the current axis — the current in amperes when the terminals are short-circuited — not the emf.
BThe reciprocal of the gradient (ΔI/ΔV) has been taken. The internal resistance is −ΔV/ΔI.
CCorrect: the intercept on the V axis gives ε = 6.0 V and the magnitude of the gradient gives r = 2.0 Ω.
DV/I at a single point (2.0 V / 2.0 A) gives the resistance of the external circuit at that current, not the internal resistance.

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

4B-1B-07

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

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

A student investigates how the current I in a 12 V filament lamp depends on the potential difference V across it. The voltmeter has a resolution of 0.01 V and the ammeter a resolution of 0.001 A.

The student suggests that I = kVⁿ, where k and n are constants, and plots lg I against lg V. The graph shows six of the seven points.

V / V	I / A	lg(V / V)	lg(I / A)
1.00	0.160	0.000	−0.796
2.00	0.238	0.301	−0.623
3.50	0.332	0.544	−0.479
5.00	0.406
7.00	0.497	0.845	−0.304
9.50	0.589	0.978	−0.230
12.00	0.677	1.079	−0.169

Graph drawn to scale

(a)

Calculate the percentage uncertainty in the current when V = 1.00 V.

(1)

(b)

Calculate the missing values in the table for V = 5.00 V and plot the point on the graph.

(1)

(c)

Draw the line of best fit for the data.

(1)

(d)

Determine n.

(2)

(e)

Determine k.

(2)

(f)

Explain how your value of n shows that the filament is a non-ohmic conductor, and why it behaves in this way.

(2)

(g)

Another student claims that the current is proportional to √V. Comment on this claim.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.001 / 0.160 × 100 = 0.63 %	✓ 1	Accept 0.6 % or 0.625 %.
Part (b)
0.699 AND −0.391 AND plotted correctly	✓ 1	Accept 2 to 4 decimal places.
Part (c)
single straight line of best fit through the points	✓ 1
Part (d)
lg I = n lg V + lg k, so n = gradient	✓ 1	Must see the gradient identified with n.
n = 0.58	✓ 1	Accept 0.56–0.60.
Part (e)
lg k = intercept on the lg I axis «at lg V = 0» ≈ −0.80	✓ 1	Or substitute a point on the line.
k = 10^−0.80 = 0.16 «A V^−0.58»	✓ 1	Accept 0.15–0.17. Allow ECF.
Part (f)
for an ohmic conductor I ∝ V, so n would be 1; n < 1 means V/I increases with V	✓ 1
a larger current raises the temperature of the filament; the lattice ions vibrate more so the electrons collide more often and the resistance increases	✓ 1	OWTTE.
Part (g)
the claim predicts n = 0.5 but n ≈ 0.58; the points lie so close to the line that a gradient of 0.5 cannot be drawn through them, so the claim is not supported	✓ 1	Allow ECF from (d). Accept a comparison with the candidate’s range of acceptable gradients.

Answers: (a) 0.63 % · (b) 0.699, −0.391 · (d) 0.58 · (e) 0.16 (the remaining parts are explanations — see the table above)

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

5B-1B-08

emf & internal resistance·B.5 Current and circuits

Paper 1BMedium10 marks

Data-based question10 steps to full marksDetermine

A student investigates a used 1.5 V zinc–carbon cell. The cell is connected in series with an ammeter, a rheostat and a switch, and a voltmeter is connected across the terminals of the cell. For each setting of the rheostat the switch is closed only long enough to read the meters.

The uncertainties are ±0.02 V in V and ±0.01 A in I, as shown by the error bars.

I / A	0.10	0.20	0.30	0.40	0.50	0.60	0.70
V / V	1.41	1.30	1.23	1.15	1.04	0.98	0.88

Graph drawn to scale

(a)

Outline why the terminal p.d. decreases as the current increases.

(1)

(b)

Draw the line of best fit for the data.

(1)

(c)

Determine the emf ε of the cell.

(1)

(d)

Determine the internal resistance r of the cell.

(2)

(e)

By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in r, and write r with its uncertainty.

(3)

(f)

The manufacturer states that a new cell of this type has an internal resistance of 0.20 Ω. Comment on the student’s result in the light of your answer to (e).

(1)

(g)

Explain why the switch was closed only briefly for each reading.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
V = ε − Ir: more energy per unit charge is transferred in the internal resistance as I increases	✓ 1	OWTTE.
Part (b)
single straight line through all the error bars	✓ 1
Part (c)
ε = intercept on the V axis = 1.49 V	✓ 1	Accept 1.47–1.51 V.
Part (d)
r = −gradient, from a large triangle	✓ 1
r = 0.86 Ω	✓ 1	Accept 0.82–0.90 Ω.
Part (e)
steepest and shallowest lines through all error bars «gradients ≈ −0.95 and −0.79 Ω»	✓ 1
Δr = ½(max − min) ≈ 0.08 Ω	✓ 1	Accept 0.05–0.12 Ω. Allow ECF.
r = (0.86 ± 0.08) Ω	✓ 1	MP3 is for matching the precision of value and uncertainty. Accept (0.9 ± 0.1) Ω.
Part (f)
0.20 Ω is well outside the range of r, so the internal resistance of the cell has increased as it has been used	✓ 1	Allow ECF from (e).
Part (g)
to avoid heating/discharging the cell «and the rheostat» during the experiment, which would change ε and r	✓ 1	OWTTE.

Answers: (c) 1.49 V · (d) 0.86 Ω · (e) (0.86 ± 0.08) Ω (the remaining parts are explanations — see the table above)

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

6B-1B-09

Resistivity·B.5 Current and circuits

Paper 1BMedium11 marks

Data-based question11 steps to full marksDetermine

A student determines the resistivity ρ of constantan. The diameter d of the wire is measured with a digital micrometer (resolution 0.001 mm) at five places. The wire is then taped to a metre rule and its resistance R is measured with an ohmmeter connected by crocodile clips at different lengths L from one end. Each value of R has an uncertainty of ±0.05 Ω.

Reading	1	2	3	4	5
d / mm	0.315	0.318	0.312	0.316	0.314

L / m	R / Ω
0.200	1.54
0.350	2.44
0.500	3.44
0.650	4.41
0.800	5.31
0.950	6.29

Graph drawn to scale

(a)

Calculate the mean diameter of the wire and its absolute uncertainty.

(2)

(b)

Draw the line of best fit for the data.

(1)

(c)

Determine ρ using the gradient of your line.

(3)

(d)

The uncertainty in the gradient is ±2 %. Determine the absolute uncertainty in ρ.

(2)

(e)

The literature value for constantan is 4.9 × 10⁻⁷ Ω m. Comment on the student’s result.

(1)

(f)

The line does not pass through the origin. Suggest a reason for this and explain why it does not affect the value of ρ found in (c).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
mean d = 0.315 mm	✓ 1
uncertainty = ½ × range = ±0.003 mm	✓ 1	Do not accept ±0.001 mm: the spread of the readings is larger than the resolution.
Part (b)
single straight line through all the error bars, not forced through the origin	✓ 1
Part (c)
gradient = 6.35 Ω m⁻¹	✓ 1	Accept 6.16–6.54 Ω m⁻¹.
A = πd²/4 = 7.79 × 10⁻⁸ m²	✓ 1	Allow ECF from (a).
ρ = gradient × A = 4.95 × 10⁻⁷ Ω m	✓ 1	Allow ECF. Unit needed.
Part (d)
Δρ/ρ = 2 % + 2 × 0.95 % = 3.9 %	✓ 1	The diameter term is doubled because A ∝ d².
Δρ = ±2 × 10⁻⁸ Ω m	✓ 1	Allow ECF. Accept ±0.19 × 10⁻⁷ Ω m.
Part (e)
4.9 × 10⁻⁷ lies within the range 4.75–5.14 × 10⁻⁷ Ω m, so the result is consistent with the literature value	✓ 1	Allow ECF from (d).
Part (f)
the resistance of the leads/crocodile clips/contacts is included in every reading	✓ 1	OWTTE.
this adds the same resistance to every value, shifting the line up without changing its gradient, and only the gradient is used for ρ	✓ 1

Answers: (a) (0.315 ± 0.003) mm · (c) 4.95 × 10⁻⁷ Ω m · (d) ±2 × 10⁻⁸ Ω m (the remaining parts are explanations — see the table above)

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

7B-1B-10

Thermistors, LDRs & dividers·B.5 Current and circuits

Paper 1BHard11 marks

Data-based question11 steps to full marksDetermine

The resistance R of a negative temperature coefficient (NTC) thermistor is measured with an ohmmeter while the thermistor is in a stirred water bath at temperature θ. The manufacturer states that R = R₀ e^B/T, where T is the absolute temperature and R₀ and B are constants.

The student plots ln(R/kΩ) against 1/T. The graph shows six of the seven points.

θ / °C	R / kΩ	(1/T) / 10⁻³ K⁻¹	ln(R / kΩ)
10.0	20.3	3.534	3.011
20.0	12.5	3.413	2.526
30.0	8.06	3.300	2.087
40.0	5.30
50.0	3.57	3.096	1.273
60.0	2.50	3.003	0.916
70.0	1.75	2.915	0.560

Graph drawn to scale

(a)

Calculate the two missing values in the table for θ = 40.0 °C and plot the point on the graph.

(2)

(b)

Draw the line of best fit for the data.

(1)

(c)

Determine B, giving its unit.

(2)

(d)

The uncertainty in the gradient is ±3 %. The manufacturer quotes B = 3950 K ± 1 %. Comment on the agreement.

(1)

(e)

The thermistor is connected in series with a fixed 4.70 kΩ resistor across a 5.00 V supply of negligible internal resistance. Calculate the p.d. across the fixed resistor when the thermistor is at 40.0 °C.

(2)

(f)

State and explain how the p.d. across the fixed resistor changes as the temperature rises.

(2)

(g)

The ohmmeter passes a small current through the thermistor. Suggest how this could cause a systematic error in R.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
1/T = 1/313 K = 3.195 × 10⁻³ K⁻¹	✓ 1	Temperature must be converted to kelvin. Accept 3.19 × 10⁻³.
ln(R/kΩ) = ln 5.30 = 1.668 AND point plotted	✓ 1	Accept 1.67.
Part (b)
single straight line of best fit through the points	✓ 1
Part (c)
ln R = ln R₀ + B(1/T) so B = gradient, allowing for the 10⁻³ in the axis	✓ 1
B = 3.96 × 10³ K	✓ 1	Accept 3.8 × 10³–4.1 × 10³ K. Unit needed. Award [1] max for 3.9 «power of ten missing».
Part (d)
the student’s range (3840–4070 K) overlaps the manufacturer’s range (3910–3990 K) so the values agree	✓ 1	Allow ECF from (c).
Part (e)
V = 5.00 × 4.70 / (4.70 + 5.30)	✓ 1	Uses the table value of R at 40 °C.
= 2.35 V	✓ 1	Accept 2.3–2.4 V.
Part (f)
the resistance of the thermistor decreases «NTC»	✓ 1
so the fixed resistor takes a larger share of the supply p.d.: the p.d. across it increases	✓ 1
Part (g)
the current heats the thermistor above the bath temperature, so R is lower than the value at the recorded temperature	✓ 1	OWTTE.

Answers: (a) 3.195 × 10⁻³ K⁻¹, 1.668 · (c) 3.96 × 10³ K · (e) 2.35 V (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that resistors can have variable resistance; variable resistors will be limited to thermistors, light-dependent resistors (LDR) and potentiometers; the combinations of resistors in series circuits Command term: Determine

8B-2-09

Conductors, insulators & resistance·B.5 Current and circuits

Paper 2Easy11 marks

Short answer & extended response11 steps to full marksDeduce

Before a new cable is installed, an electrician tests its insulation. The cable has a copper core surrounded by a layer of PVC plastic. A metal foil is wrapped tightly around the outside of the PVC. An insulation tester applies a direct potential difference V between the copper core and the foil and measures the very small current I that passes through the PVC.

The table gives the results for two cables. Cable 2 has a damaged section into which water has entered. The graph shows the results for cable 1.

V / V	I / μA (cable 1)	I / μA (cable 2)
100	0.40	0.50
250	1.00	1.40
500	2.00	3.40
1000	4.00	11.0

Graph drawn to scale

(a)

State Ohm's law.

(1)

(b)

Deduce, using the data, whether the PVC of cable 1 obeys Ohm's law.

(2)

(c)

A cable is acceptable only if the resistance of its insulation is at least 1.0 × 10⁸ Ω at every test potential difference. Determine whether cable 2 is acceptable.

(2)

(d)

Draw, on the graph, the line of best fit for cable 1. Sketch, on the same axes, the graph of I against V for cable 2.

(2)

(e)

Outline, in terms of charge carriers, why the PVC is an electrical insulator whereas the copper core is a good electrical conductor.

(2)

(f)

Calculate the number of electrons that flow each second along the wire connecting the tester to the copper core of cable 1 when V = 1000 V.

(1)

(g)

Suggest why the tester applies a large potential difference, such as 1000 V, rather than the 1.5 V of a single cell.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
the current in a conductor is (directly) proportional to the potential difference across it, provided the temperature «and other physical conditions» stay constant	✓ 1	Both proportionality and the condition needed.
Part (b)
V/I = 2.5 × 10⁸ Ω for every reading OR doubling V doubles I	✓ 1	At least two readings must be tested.
V/I is constant «I ∝ V», so the PVC of cable 1 obeys Ohm's law «is ohmic»	✓ 1	MP2 only scores if MP1 scores.
Part (c)
at 1000 V: R = 1000/(11.0 × 10⁻⁶) = 9.1 × 10⁷ Ω	✓ 1	Values at the other p.d.s: 2.0 × 10⁸, 1.8 × 10⁸ and 1.5 × 10⁸ Ω.
9.1 × 10⁷ Ω is less than 1.0 × 10⁸ Ω, so cable 2 is not acceptable	✓ 1	MP2 needs a comparison with a calculated value at 1000 V and a consistent conclusion.
Part (d)
cable 1: a single ruled straight line through the origin and through all four points	✓ 1
cable 2: a curve through the origin, lying above the line for cable 1, whose gradient increases with V and that passes close to (1000 V, 11.0 μA)	✓ 1	Accept the four points of cable 2 plotted and joined by a smooth curve.
Part (e)
copper contains a very large number of free «delocalised» electrons that can drift through the metal when a potential difference is applied	✓ 1
in PVC the electrons are bound to their atoms/molecules, so there are almost no mobile charge carriers «and the current is tiny»	✓ 1	OWTTE. Do not accept 'PVC has no electrons'.
Part (f)
N = I/e = 4.00 × 10⁻⁶/1.60 × 10⁻¹⁹ = 2.5 × 10¹³ «s⁻¹»	✓ 1
Part (g)
at 1.5 V the current would be only about 6 × 10⁻⁹ A, too small to be measured accurately «so a large V gives a measurable current / tests the insulation under the conditions of use»	✓ 1	Accept a reference to the very large resistance requiring a large p.d. for a measurable current.

Answers: (b) 2.5 × 10⁸ Ω, ohmic · (c) 9.1 × 10⁷ Ω — not acceptable · (f) 2.5 × 10¹³ s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — Ohm's law; the ohmic and non-ohmic behaviour of electrical conductors; the properties of electrical conductors and insulators in terms of mobility of charge carriers; electrical resistance R as given by R = V/I; direct current (dc) I as a flow of charge carriers as given by I = Δq/Δt Command term: Deduce

9B-2-10

Thermistors, LDRs & dividers·B.5 Current and circuits

Paper 2Hard12 marks

Short answer & extended response12 steps to full marksDetermine

The heated mirror on a car door contains a thin heating element made of a positive temperature coefficient (PTC) thermistor material, whose resistance increases as its temperature rises. The element is connected directly across the 12.0 V battery of the car, which has negligible internal resistance.

The graph shows how the power P dissipated in the element varies with the temperature θ of the mirror. The mirror and its element have a total mass of 0.15 kg and a mean specific heat capacity of 750 J kg⁻¹ K⁻¹.

Graph drawn to scale

(a)

(i)

Calculate, using the graph, the resistance of the element at 0 °C.

(1)

(ii)

Describe how the resistance of the element varies as θ increases from 0 °C to 70 °C.

(2)

(b)

On a frosty morning the air and the mirror are at −10 °C when the heater is switched on. Estimate the initial rate of increase of the temperature of the mirror.

(2)

(c)

The mirror transfers energy to the surrounding air at a rate of 0.50 W for every kelvin by which the mirror is hotter than the air.

(i)

Draw, on the graph, a line to show the rate of energy transfer from the mirror to the air when the air is at −10 °C.

(2)

(ii)

Hence determine the temperature at which the mirror reaches thermal equilibrium.

(1)

(iii)

On a mild day the air is at 20 °C. Determine the new equilibrium temperature of the mirror.

(2)

(d)

The manufacturer could instead have used a fixed resistor of resistance equal to that of the PTC element at 0 °C. Explain, with reference to your answers to (c), one advantage of the PTC element.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
R = V²/P = 12.0²/34.6 = 4.16 Ω	✓ 1	Accept 4.1–4.2 Ω «P read as 34.3–35.1 W».
Part (a)(ii)
from 0 °C to about 40 °C the resistance increases only slowly «from about 4.2 Ω to about 5.3 Ω, as P falls slowly»	✓ 1
above about 40–45 °C it increases very rapidly, to about 79 Ω at 70 °C «P falls steeply»	✓ 1	Accept any value from 60 to 100 Ω at 70 °C. Accept 'about 20 times larger'.
Part (b)
P at −10 °C = 36 W; the mirror is at the temperature of the air, so «initially» no energy is transferred to the air	✓ 1	Accept 35–36 W.
Δθ/Δt = P/mc = 36/(0.15 × 750) = 0.32 K s⁻¹	✓ 1	Accept 0.31–0.32 K s⁻¹. Award [2] for CNA.
Part (c)(i)
straight line starting at (−10 °C, 0)	✓ 1
with gradient 0.50 W K⁻¹, e.g. passing through (30 °C, 20 W) and (50 °C, 30 W)	✓ 1	± half a small square.
Part (c)(ii)
intersection of the line with the curve: θ ≈ 42 °C	✓ 1	Accept 40–45 °C. Allow ECF from the candidate's line.
Part (c)(iii)
a parallel line «same gradient» starting at (20 °C, 0) drawn or used	✓ 1
intersection at θ ≈ 52 °C	✓ 1	Accept 50–55 °C. ALTERNATIVE: trial values from the graph that balance P and 0.50(θ − 20).
Part (d)
with the PTC element a 30 K rise in air temperature raises the equilibrium temperature by only about 10 K, because as the mirror warms above about 40 °C its resistance rises sharply and P = V²/R falls	✓ 1
a fixed resistor would dissipate about 35 W at all temperatures, giving equilibrium temperatures of about 59 °C and 89 °C: the PTC mirror cannot overheat «and wastes less energy on mild days»	✓ 1	OWTTE. Accept a qualitative comparison for MP2 if the fixed resistor's temperature rise ∝ air temperature is clearly explained.

Answers: (a)(i) 4.16 Ω · (b) 0.32 K s⁻¹ · (c)(ii) ≈ 42 °C · (c)(iii) ≈ 52 °C (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that resistors can have variable resistance (thermistors); electrical resistance R as given by R = V/I; electrical power P dissipated by a resistor as given by P = IV = I²R = V²/R; the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors (synoptic link: B.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c as given by Q = mcΔT; that temperature difference determines the direction of the resultant thermal energy transfer between bodies) Command term: Determine

10B-2-11

Resistivity·B.5 Current and circuits

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDetermine

An engineer designs the heating element of a 230 V, 1.00 kW electric heater. The element is to be made from nichrome wire of diameter 0.40 mm. At the working temperature of the element the resistivity of nichrome is 1.10 × 10⁻⁶ Ω m.

The heater contains two identical elements of this kind. A switch connects to the supply either one element alone, or the two elements in series, or the two elements in parallel. Assume that the resistance of each element is the same on every setting.

(a)

Outline, in terms of charge carriers, the origin of the electrical resistance of a metal.

(2)

(b)

(i)

Show that the resistance of the element at its working temperature must be about 53 Ω.

(1)

(ii)

Determine the length of nichrome wire needed.

(3)

(c)

(i)

Determine the three power outputs that the heater can provide.

(3)

(ii)

The heater is to be used with an extension cable rated for a maximum current of 6.0 A. Determine the highest setting that can be used safely.

(2)

(iii)

The engineer considers using nichrome wire of twice the diameter. State and explain how the length of wire needed would change.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
free electrons (charge carriers) drift through the metal under the potential difference	✓ 1
they collide with the (vibrating) lattice ions, transferring energy to them, which opposes their flow and heats the metal	✓ 1	OWTTE.
Part (b)(i)
R = V²/P = 230² / 1000 = 52.9 Ω	✓ 1	Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
A = π(0.20 × 10⁻³)² = 1.26 × 10⁻⁷ m²	✓ 1	Award [0] for this mark if the diameter is used as the radius.
L = RA/ρ = 52.9 × 1.26 × 10⁻⁷ / 1.10 × 10⁻⁶	✓ 1
L = 6.0 m	✓ 1	Accept 6.0–6.1 m. Award [3] for CNA. A diameter–radius error gives 24 m: award [2].
Part (c)(i)
series: 2 × 52.9 = 106 Ω AND parallel: 52.9/2 = 26.5 Ω	✓ 1	ALTERNATIVE: series — the same p.d. across twice the resistance halves the power; parallel — each element has the full 230 V, so the power doubles.
series: P = 230²/106 = 500 W	✓ 1	Allow ECF from (b)(i).
parallel: P = 230²/26.5 = 2.00 × 10³ W; with one element alone 1.00 × 10³ W	✓ 1	Award [3] for 500 W, 1.00 kW and 2.00 kW with some working.
Part (c)(ii)
I = P/V: 2.00 kW setting 8.70 A; 1.00 kW setting 4.35 A	✓ 1	Allow ECF from (c)(i). Accept I = V/R with the resistances from (c)(i).
8.70 A > 6.0 A but 4.35 A < 6.0 A, so the highest safe setting is one element alone «1.00 kW»	✓ 1	MP2 requires a comparison of calculated currents with 6.0 A.
Part (c)(iii)
the length must be four times as great	✓ 1
the cross-sectional area ∝ d² is four times as great, and R = ρL/A must stay the same, so L must also increase four times	✓ 1

Answers: (b)(i) 52.9 Ω · (b)(ii) 6.0 m · (c)(i) 500 W, 1.00 × 10³ W, 2.00 × 10³ W · (c)(ii) one element «1.00 kW» · (c)(iii) 4 times longer (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — electric resistance and its origin; resistivity as given by ρ = RA/L; 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

11B-2-12

Thermistors, LDRs & dividers·B.5 Current and circuits

Paper 2Easy10 marks

Short answer & extended response10 steps to full marksCalculate

A wind vane turns the slider S of a rotary potentiometer. The circular track of the potentiometer has a total resistance of 10.0 kΩ and is connected across a supply of emf 5.00 V and negligible internal resistance. The output is the potential difference between S and the end E of the track. The resistance between E and S is proportional to the angle θ through which the vane has turned from north; θ = 360° corresponds to the whole track.

Diagram NOT accurately drawn

Axes for (b)

(a)

(i)

Calculate the current in the track.

(1)

(ii)

Show that the output potential difference is about 1.7 V when θ = 120°.

(1)

(iii)

Calculate the power dissipated in the track.

(1)

(b)

Draw, on the axes, a graph of the output potential difference against θ for 0 ≤ θ ≤ 360°.

(2)

(c)

The output is connected to a data logger of very large resistance that accepts a maximum input of 4.00 V. A fixed resistor is connected in series between the supply and the upper end of the track.

(i)

Calculate the resistance of the fixed resistor needed for the output to be 4.00 V when θ = 360°.

(2)

(ii)

Calculate the output potential difference when θ = 270°.

(1)

(iii)

Explain why the output potential difference is still proportional to θ.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
I = V/R = 5.00 / 10000 = 5.0 × 10⁻⁴ A	✓ 1
Part (a)(ii)
V = 5.00 × 120/360 = 1.67 V	✓ 1	Must see the fraction of the track OR the answer to 3 s.f.
Part (a)(iii)
P = V²/R = 5.00²/10000 = 2.5 × 10⁻³ W	✓ 1	Accept IV with ECF from (a)(i).
Part (b)
straight line through the origin	✓ 1
reaching 5.00 V at 360° (passing through about 1.7 V at 120°)	✓ 1	Ruled line needed.
Part (c)(i)
current in the track = 4.00/(10.0 × 10³) = 4.0 × 10⁻⁴ A AND pd across the fixed resistor = 5.00 − 4.00 = 1.00 V	✓ 1	ALTERNATIVE: potential-divider ratio R/10.0 kΩ = 1.00/4.00.
R = 1.00/(4.0 × 10⁻⁴) = 2.5 × 10³ Ω	✓ 1	Award [2] for CNA.
Part (c)(ii)
4.00 × 270/360 = 3.00 V	✓ 1	Accept 4.0 × 10⁻⁴ × 7.5 × 10³. Allow ECF from (c)(i).
Part (c)(iii)
the logger draws negligible current, so the current in the track is the same for every position of S	✓ 1
the resistance between E and S is proportional to θ, so V = IR_ES is proportional to θ	✓ 1	OWTTE.

Answers: (a)(i) 5.0 × 10⁻⁴ A · (a)(ii) 1.67 V · (a)(iii) 2.5 × 10⁻³ W · (c)(i) 2.5 kΩ · (c)(ii) 3.00 V (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that resistors can have variable resistance (variable resistors limited to thermistors, LDRs and potentiometers); the combinations of resistors in series and parallel circuits; electrical resistance R as given by R = V/I Command term: Calculate

12B-2-17

Current & charge·B.5 Current and circuits

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksCalculate

A student coats a steel key with copper. The key and a copper plate are dipped into copper sulfate solution and connected in the circuit shown. The supply has an emf of 6.0 V and negligible internal resistance, and the meters are ideal.

In the solution, charge is carried by positive copper ions Cu²⁺, each of charge +2e, and by negative sulfate ions. Each copper ion that reaches the key gains two electrons and is deposited on the key as a copper atom. The current is kept constant at 0.250 A for 40.0 minutes.

Molar mass of copper = 0.0635 kg mol⁻¹. Density of copper = 8.96 × 10³ kg m⁻³. Surface area of the key = 18.0 cm².

Diagram NOT accurately drawn

(a)

Outline how charge flows in the connecting wires and how it flows in the solution.

(2)

(b)

(i)

Calculate the charge that flows through the circuit.

(1)

(ii)

Show that about 1.9 × 10²¹ copper ions are deposited on the key.

(1)

(iii)

Calculate the mass of copper deposited on the key.

(2)

(iv)

Determine the thickness of the copper coating, assuming that it is uniform.

(2)

(c)

(i)

The voltmeter reads 1.5 V. Calculate the resistance of the fixed resistor R.

(2)

(ii)

Calculate the energy transferred in the solution and electrodes during the coating.

(1)

(iii)

The student now coats a larger key, of twice the surface area, with a coating of the same thickness in the same time. The voltmeter reading stays the same. Determine the resistance that the fixed resistor must now have.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
in the wires: free electrons drift «from the negative terminal of the supply towards the key and from the copper plate to the positive terminal»	✓ 1
in the solution: positive «copper» ions move towards the key and negative «sulfate» ions move in the opposite direction	✓ 1	OWTTE. Do not accept 'electrons flow through the solution'.
Part (b)(i)
q = It = 0.250 × 2400 = 600 C	✓ 1
Part (b)(ii)
N = q/2e = 600/(2 × 1.60 × 10⁻¹⁹) = 1.88 × 10²¹	✓ 1	Must see full substitution OR answer to at least 3 s.f. The factor 2 must be seen.
Part (b)(iii)
n = N/N_A = 1.88 × 10²¹/6.02 × 10²³ = 3.11 × 10⁻³ mol	✓ 1
m = nM = 3.11 × 10⁻³ × 0.0635 = 2.0 × 10⁻⁴ kg «0.20 g»	✓ 1	Allow ECF from (b)(ii). Award [1 max] for 4.0 × 10⁻⁴ kg «factor 2 omitted». Award [2] for CNA.
Part (b)(iv)
volume = m/ρ = 1.98 × 10⁻⁴/8.96 × 10³ = 2.21 × 10⁻⁸ m³	✓ 1	Allow ECF from (b)(iii).
thickness = volume/area = 2.21 × 10⁻⁸/1.8 × 10⁻³ = 1.2 × 10⁻⁵ m «12 μm»	✓ 1	Award [1 max] if the area is not converted to m².
Part (c)(i)
p.d. across R = 6.0 − 1.5 = 4.5 V	✓ 1
R = 4.5/0.250 = 18 Ω	✓ 1	Award [1 max] for 24 Ω «6.0 V used». Award [2] for CNA.
Part (c)(ii)
W = qV = 600 × 1.5 = 900 J	✓ 1	Allow ECF from (b)(i).
Part (c)(iii)
twice the mass must be deposited in the same time, so the current must double to 0.50 A	✓ 1
R = 4.5/0.50 = 9.0 Ω	✓ 1	Allow ECF from (c)(i). Award [1 max] for 36 Ω «resistance doubled».

Answers: (b)(i) 600 C · (b)(iii) 2.0 × 10⁻⁴ kg · (b)(iv) 1.2 × 10⁻⁵ m · (c)(i) 18 Ω · (c)(ii) 900 J · (c)(iii) 9.0 Ω (the remaining parts are explanations — see the table above)

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; that the electric potential difference V is the work done per unit charge on moving a positive charge between two points along the path of the current as given by V = W/q; the combinations of resistors in series and parallel circuits (linked: B.3 — the amount of substance n as given by n = N/N_A; B.1 — density ρ as given by ρ = m/V) Command term: Calculate

13B-2-18

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

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksExplain

A 230 V, 60 W filament lamp is replaced by a 230 V LED lamp of power 8.5 W that gives the same light output. When the filament lamp is cold, its resistance is 70 Ω.

Axes for (b)

(a)

(i)

Show that the resistance of the filament at its normal working temperature is about 880 Ω.

(1)

(ii)

Explain why the resistance of the filament at its working temperature is much greater than when it is cold.

(3)

(b)

Sketch, on the axes, the variation of the current I in the filament lamp with the potential difference V across it, for both positive and negative values of V.

(2)

(c)

The lamp is switched on when the filament is cold. Determine the ratio of the current at the instant of switching on to the current in normal operation.

(2)

(d)

(i)

The lamp is used for 5.0 hours every day for a year of 365 days, and electrical energy costs €0.30 per kW h. Calculate the money saved in a year by using the LED lamp instead of the filament lamp.

(3)

(ii)

Deduce how many times more efficient the LED lamp is than the filament lamp, and outline what happens to most of the energy supplied to the filament lamp.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
R = V²/P = 230² / 60 = 882 Ω	✓ 1	Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
the current heats the filament to a very high temperature	✓ 1
the lattice ions vibrate with a greater amplitude	✓ 1
so the conduction electrons collide with them more often, which increases the resistance	✓ 1	OWTTE.
Part (b)
curve through the origin whose gradient decreases as V increases (current increases less than in proportion)	✓ 1
the curve for negative V is the same shape, rotated by 180° about the origin (third quadrant)	✓ 1
Part (c)
switch-on current = 230/70 = 3.29 A; normal current = 60/230 = 0.261 A	✓ 1	ALTERNATIVE: ratio = R_hot/R_cold = 882/70.
ratio = 12.6	✓ 1	Accept 12–13. Award [2] for CNA.
Part (d)(i)
power saved = 60 − 8.5 = 51.5 W	✓ 1
energy saved = 0.0515 kW × 1825 h = 94 kW h	✓ 1
saving = 94 × 0.30 = €28	✓ 1	Accept €28.
Part (d)(ii)
same useful output, so efficiency ratio = 60/8.5 ≈ 7	✓ 1	Accept 7.1.
most of the energy supplied to the filament lamp increases the internal energy of the filament and is transferred to the surroundings as thermal energy (infrared radiation)	✓ 1

Answers: (a)(i) 882 Ω · (c) 12.6 · (d)(i) €28 · (d)(ii) about 7 times (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors; electric resistance and its origin; electrical resistance R as given by R = V/I; electrical power P dissipated by a resistor as given by P = IV = I²R = V²/R Command term: Explain

14B-1A-41

Conductors, insulators & resistance·B.5 Current and circuits

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

Why is copper a much better electrical conductor than glass?

Show mark scheme

Marking point	Mark	Notes
Step 1Conduction depends on the availability and mobility of charge carriers. Copper has about 10²⁹ free electrons per cubic metre that can drift through the ionic lattice; in glass essentially every electron is bound.	✓ 1	Answer C

Answer: C · 1 stage of work, one mark

Every option, and why

AAtomic size is not what determines conduction. What matters is whether charge carriers are free to move.
BCopper is in fact much denser than glass, and density is not what governs conductivity.
CCorrect: in a metal the outer electrons are delocalised and drift readily under a potential difference. In glass the electrons are all bound to particular atoms.
DA metal is a lattice of positive ions surrounded by free electrons. It is the mobility of those electrons, not the absence of ions, that makes it conduct.

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

15B-1A-42

Cells & sources of emf·B.5 Current and circuits

Paper 1AMedium1 mark

Multiple choice · 1 mark1 step to full marksIdentify

Chemical cells and solar cells are both used as sources of emf in circuits.

Which statement is correct?

Show mark scheme

Marking point	Mark	Notes
Step 1A solar cell transfers the energy of incident electromagnetic radiation directly to electrical energy, so its output depends on the intensity of the radiation. A chemical cell transfers chemical energy to electrical energy while it drives a current.	✓ 1	Answer B

Answer: B · 1 stage of work, one mark

Every option, and why

AEmf is the energy transferred per unit charge (J C⁻¹ = V), not per unit current.
BCorrect: a solar cell's source of energy is the incident radiation, so less intense light gives a smaller output.
CThis is the reverse transfer, which happens while a rechargeable cell is being charged. A cell driving a current transfers chemical energy to electrical energy.
DA solar cell stores no energy; with no light falling on it there is no output. Energy for the night needs a separate storage cell — one disadvantage of solar cells.

Syllabus understandingB.5 — that cells provide a source of emf; chemical cells and solar cells as the energy source in circuits Command term: Identify

16B-1A-43

Thermistors, LDRs & dividers·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDeduce

A thermistor is connected in series with a fixed resistor R, an ideal ammeter and a cell of negligible internal resistance. An ideal voltmeter is connected across the thermistor. The resistance of the thermistor decreases as its temperature increases.

The temperature of the thermistor is increased. Which row gives the change in the ammeter reading and the change in the voltmeter reading?

Diagram NOT accurately drawn

Ammeter readingVoltmeter reading

Show mark scheme

Marking point	Mark	Notes
Step 1The thermistor's resistance falls, so the total resistance falls and the current I = ε / (R + R_T) increases.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2The pd across R, IR, increases. The two pds add up to the constant emf, so the pd across the thermistor decreases.	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

ATwo errors: the thermistor's resistance has been taken to rise with temperature, like a metal wire, and V = IR has then been applied to the thermistor using the fall in current alone.
BThe thermistor's resistance has been taken to rise with temperature, like a metal wire. Here it falls, so the current rises and the thermistor takes a smaller share of the emf.
CCorrect: the total resistance falls, so the current rises; R then takes a larger share of the emf and the thermistor a smaller one.
DV = IR has been applied to the thermistor using the rise in current alone, ignoring the fall in its resistance. The pd across R rises, so the pd across the thermistor falls.

Syllabus understandingB.5 — that resistors can have variable resistance (thermistors, light-dependent resistors and potentiometers); the combinations of resistors in series and parallel circuits; that circuit diagrams represent the arrangement of components in a circuit Command term: Deduce

17B-1A-44

Series & parallel circuits·B.5 Current and circuits

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

Resistor X, of resistance R, is connected in series with a parallel combination of resistor Y, of resistance R, and resistor Z, of resistance 2R. The supply has emf V and negligible internal resistance.

What is the current in Z?

Diagram NOT accurately drawn

Show mark scheme

Marking point	Mark	Notes
Step 1Y and Z in parallel: (1/R + 1/(2R))⁻¹ = 2R/3. Total resistance = R + 2R/3 = 5R/3, so the supply current is 3V / (5R).	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2The pd across the parallel pair is 3V/(5R) × 2R/3 = 2V/5, so the current in Z is (2V/5) / 2R = V / (5R).	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: Z, with twice Y's resistance, takes one third of the supply current: ⅓ × 3V/(5R) = V/(5R).
BThe supply current has been split equally between Y and Z. Parallel branches share the current in inverse proportion to their resistances.
CThe larger share of the current has been given to the larger resistance (⅔ × 3V/(5R)). More current flows through the branch with less resistance.
DThe whole supply emf has been taken across Z. Part of the emf is across X, so the pd across Z is only 2V/5.

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

18B-1B-17

Resistivity·B.5 Current and circuits

Paper 1BMedium9 marks

Data-based question9 steps to full marksDetermine

A manufacturer’s data table gives the resistance per unit length R/L at 20 °C of nichrome heating wire for several standard diameters d. A student uses these secondary data to test whether R/L is inversely proportional to the cross-sectional area of the wire, and plots lg(R/L) against lg d. The graph shows six of the seven points.

d / mm	(R/L) / Ω m⁻¹	lg(d / mm)	lg(R/L / Ω m⁻¹)
0.200	34.8	−0.699	1.542
0.250	21.8	−0.602	1.338
0.315	14.0	−0.502	1.146
0.400	8.54
0.500	5.56	−0.301	0.745
0.630	3.42	−0.201	0.534
0.800	2.16	−0.097	0.334

Graph drawn to scale

(a)

Calculate the missing values in the table for d = 0.400 mm and plot the point on the graph.

(1)

(b)

Draw the line of best fit for the data.

(1)

(c)

Determine the gradient of your line.

(2)

(d)

Explain how your answer to (c) supports the hypothesis that R/L is inversely proportional to the cross-sectional area.

(2)

(e)

Using the graph, determine the resistivity of nichrome.

(2)

(f)

The wire in an electric heater operates at about 600 °C. Suggest why its resistance is not correctly predicted by the table.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
−0.398 AND 0.931 AND plotted correctly	✓ 1	Accept 2 to 4 decimal places.
Part (b)
single straight line of best fit through the points	✓ 1
Part (c)
gradient from a large triangle «negative sign shown»	✓ 1
gradient = −2.00	✓ 1	Accept −2.06 to −1.94.
Part (d)
A = πd²/4 so R/L ∝ 1/A means R/L ∝ d⁻²	✓ 1
so the graph of lg(R/L) against lg d should be a straight line of gradient −2, as found	✓ 1	Allow ECF from (c).
Part (e)
at lg d = 0 «d = 1.00 mm», lg(R/L) ≈ 0.14 so R / L ≈ 1.38 Ω m⁻¹	✓ 1	Any point read from the line may be used.
ρ = (R/L) × πd²/4 = 1.1 × 10⁻⁶ Ω m	✓ 1	Accept 1.0–1.2 × 10⁻⁶ Ω m. Allow ECF.
Part (f)
the resistivity «resistance» increases with temperature «greater lattice vibration», so the resistance is larger than the 20 °C value	✓ 1	OWTTE.

Answers: (a) −0.398, 0.931 · (c) −2.00 · (e) 1.1 × 10⁻⁶ Ω m (the remaining parts are explanations — see the table above)

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

19B-1B-18

emf & internal resistance·B.5 Current and circuits

Paper 1BHard10 marks

Data-based question10 steps to full marksDetermine

A zinc strip and a copper strip are pushed into a lemon to make a cell. The cell is connected in series with a resistance box of resistance R and a digital microammeter. No voltmeter is used. The current I is recorded for several values of R.

The graph shows 1/I against R for five of the six readings.

R / kΩ	I / μA	(1/I) / mA⁻¹
1.0	278.1	3.60
2.0	211.9	4.72
3.0	174.9	5.72
5.0	126.5
7.0	100.7	9.93
10.0	75.4	13.26

Graph drawn to scale

(a)

Show that, for a cell of emf ε and internal resistance r, 1/I = R/ε + r/ε.

(1)

(b)

Calculate the missing value of 1/I and plot the point on the graph.

(1)

(c)

Draw the line of best fit for the data.

(1)

(d)

Determine ε.

(2)

(e)

Determine r, assuming the microammeter is ideal.

(2)

(f)

The microammeter has a resistance of 0.10 kΩ. Explain how this affects your answer to (e) and state a corrected value of r.

(2)

(g)

The current from the lemon cell slowly decreases while the circuit is connected. Suggest one change to the procedure that would reduce the effect of this on the results.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
ε = I(R + r) ⇒ 1/I = (R + r)/ε = R/ε + r/ε	✓ 1	Must see the starting equation and the division by εI.
Part (b)
1000 / 126.5 = 7.91 «mA⁻¹» AND plotted	✓ 1	Accept 7.9.
Part (c)
single straight line of best fit through the points	✓ 1
Part (d)
gradient = 1/ε ≈ 1.07 «mA⁻¹ kΩ⁻¹ =» V⁻¹	✓ 1	Accept 1.04–1.10.
ε = 0.94 V	✓ 1	Accept 0.91–0.97 V. Allow ECF.
Part (e)
intercept = r/ε ≈ 2.54 mA⁻¹	✓ 1	Accept 2.39–2.69.
r = intercept × ε = 2.4 kΩ	✓ 1	Allow ECF from (d) and their intercept.
Part (f)
the total resistance in the circuit is R + r + R_A, so the intercept gives r + R_A and (e) overestimates r «a systematic error»	✓ 1
r = 2.4 − 0.1 = 2.3 kΩ	✓ 1	Allow ECF from (e).
Part (g)
disconnect the circuit between readings / take each reading quickly after connecting / take readings in a random or reversed order and average	✓ 1	OWTTE.

Answers: (b) 7.91 mA⁻¹ · (d) 0.94 V · (e) 2.4 kΩ · (f) 2.3 kΩ (the remaining parts are explanations — see the table above)

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

20B-1B-19

Thermistors, LDRs & dividers·B.5 Current and circuits

Paper 1BEasy9 marks

Data-based question9 steps to full marksDetermine

A light-dependent resistor (LDR) is placed next to the sensor of a light meter in a dark box lit by an adjustable lamp. For each illuminance E the resistance R of the LDR is measured three times with a digital ohmmeter.

The graph shows the mean resistance for six of the seven illuminances.

E / lx	R₁ / kΩ	R₂ / kΩ	R₃ / kΩ	mean R / kΩ
50	13.4	13.0	13.3	13.2
100	8.2	7.8	8.1
200	5.0	4.8	4.9	4.9
300	3.6	3.7	3.6	3.6
400	2.8	3.0	2.9	2.9
600	2.2	2.2	2.1	2.2
800	1.8	1.7	1.8	1.8

Graph drawn to scale

(a)

State the resolution of the ohmmeter.

(1)

(b)

Calculate the mean resistance for E = 100 lx and its absolute uncertainty.

(2)

(c)

Plot your mean from (b) on the graph and draw the curve of best fit.

(2)

(d)

Using your curve, determine R when E = 250 lx.

(1)

(e)

The LDR is connected in series with a 10.0 kΩ resistor across a 6.0 V supply of negligible internal resistance. Calculate the p.d. across the LDR when E = 250 lx.

(2)

(f)

Suggest one improvement to the range or distribution of the measurements.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
0.1 kΩ	✓ 1	Accept ±0.1 kΩ.
Part (b)
mean = 8.0 kΩ	✓ 1	Accept 8.03.
uncertainty = ½ × range = ±0.2 kΩ	✓ 1	Do not accept ±0.1 kΩ «the resolution»: the spread of the repeats is larger.
Part (c)
point plotted correctly	✓ 1	Allow ECF from (b).
smooth curve through the points, steep at low E and levelling off at high E; not dot-to-dot	✓ 1
Part (d)
≈ 4.1 kΩ	✓ 1	Accept 3.8–4.5 kΩ.
Part (e)
V = 6.0 × R / (R + 10.0)	✓ 1
= 1.74 V	✓ 1	Allow ECF from (d). Accept 1.6–1.9 V.
Part (f)
take more readings at low illuminance «below 200 lx» where R changes most rapidly / extend the range below 50 lx	✓ 1	Do not accept “repeat the readings” alone.

Answers: (b) (8.0 ± 0.2) kΩ · (d) ≈ 4.1 kΩ · (e) 1.74 V (the remaining parts are explanations — see the table above)

21B-2-30

Chemical & solar cells·B.5 Current and circuits

Paper 2Easy10 marks

Short answer & extended response10 steps to full marksDetermine

A portable power bank contains lithium-ion cells with a combined emf of 3.7 V. It is labelled “10 000 mA h”, which means that when fully charged it can deliver a current of 10 000 mA for 1 hour.

(a)

State what is meant by the emf of a cell.

(1)

(b)

(i)

Show that the charge that the fully charged power bank can deliver is about 3.6 × 10⁴ C.

(1)

(ii)

Calculate the energy stored in the fully charged power bank.

(2)

(c)

The power bank is recharged by a portable solar panel of area 0.060 m² and efficiency 18 %, in sunlight of intensity 650 W m⁻². Of the electrical energy produced by the panel, 85 % ends up stored in the power bank.

(i)

Calculate the electrical power output of the solar panel.

(1)

(ii)

Determine whether 5.0 hours of this sunlight is enough to charge the power bank fully from empty.

(3)

(d)

Later in the day the Sun's rays make an angle of 30° with the plane of the panel. The intensity of the sunlight, measured perpendicular to the rays, is still 650 W m⁻². Determine the electrical power output of the panel now.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
the energy transferred «from chemical to electrical energy» per unit charge passing through the cell	✓ 1	Accept “work done per unit charge by the cell”. Do not accept “the voltage of the cell”.
Part (b)(i)
q = It = 10.0 × 3600 = 3.6 × 10⁴ C	✓ 1	Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
W = qε = 3.6 × 10⁴ × 3.7	✓ 1
W = 1.3 × 10⁵ J	✓ 1	Award [2] for CNA.
Part (c)(i)
P = 0.060 × 650 × 0.18 = 7.0 W	✓ 1
Part (c)(ii)
energy produced in 5.0 h = 7.0 × 18 000 = 1.26 × 10⁵ J	✓ 1	Allow ECF from (c)(i).
energy stored = 0.85 × 1.26 × 10⁵ = 1.07 × 10⁵ J	✓ 1
1.07 × 10⁵ J is less than 1.33 × 10⁵ J, so it is not enough	✓ 1	ALTERNATIVE: time needed = 1.33 × 10⁵/(0.85 × 7.0) = 2.2 × 10⁴ s ✓✓ ≈ 6.2 h, longer than 5.0 h ✓. MP3 requires a comparison. Allow ECF from (b)(ii).
Part (d)
area of the panel projected perpendicular to the rays = 0.060 × sin 30° = 0.030 m²	✓ 1	OWTTE: the panel intercepts half as much radiation.
P = 0.030 × 650 × 0.18 = 3.5 W	✓ 1	Award [1 max] for 6.1 W «cos 30° used».

Answers: (b)(i) 3.6 × 10⁴ C · (b)(ii) 1.3 × 10⁵ J · (c)(i) 7.0 W · (c)(ii) 1.07 × 10⁵ J stored: no · (d) 3.5 W (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that cells provide a source of emf; chemical cells and solar cells as the energy source in circuits; direct current (dc) I as a flow of charge carriers as given by I = Δq/Δt; that the electric potential difference V is the work done per unit charge on moving a positive charge between two points along the path of the current as given by V = W/q (linked: B.2 — that the incoming radiative power is dependent on the projected surface along the direction of the path of the rays) Command term: Determine

22B-2-31

Series & parallel circuits·B.5 Current and circuits

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDraw

Four identical filament lamps are each rated “6.0 V, 3.0 W”. They are to be connected to a battery of emf 12 V and negligible internal resistance so that every lamp operates at its rated values. The figure shows the battery.

Diagram NOT accurately drawn

(a)

(i)

Calculate the current in a lamp when it operates at its rated values.

(1)

(ii)

Calculate the resistance of a lamp when it operates at its rated values.

(1)

(b)

Draw, on the figure, a circuit diagram to show how the four lamps can be connected to the battery so that each lamp operates at its rated values.

(2)

(c)

(i)

Calculate the current in the battery.

(1)

(ii)

Show that the total resistance of the four lamps is equal to the resistance of one lamp.

(1)

(d)

In the arrangement used, lamps P and Q are in series in one branch, and lamps R and S are in series in a second branch connected in parallel with the first. A fifth identical lamp T is connected between the junction of P and Q and the junction of R and S. State and explain whether lamp T lights.

(2)

(e)

Lamp T is removed. Lamp P is then replaced by a lamp of resistance 6.0 Ω. Assume that the resistance of every lamp stays constant.

(i)

Determine whether lamp Q now operates above its rated power.

(3)

(ii)

Outline why the brightness of lamp R does not change.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
I = P/V = 3.0/6.0 = 0.50 A	✓ 1
Part (a)(ii)
R = V/I = 6.0/0.50 = 12 Ω	✓ 1	Accept R = V²/P. Allow ECF from (a)(i).
Part (b)
two lamps in series in each of two branches	✓ 1	Correct lamp symbols needed.
the two branches connected in parallel across the battery	✓ 1	ALTERNATIVE: two pairs of lamps in parallel, the two pairs connected in series ✓✓. Award [0] for all four lamps in series or all four in parallel.
Part (c)(i)
I = 2 × 0.50 = 1.0 A OR I = 4 × 3.0/12 = 1.0 A	✓ 1
Part (c)(ii)
each branch is 24 Ω and two equal branches in parallel give 12 Ω OR 12 V/1.0 A = 12 Ω	✓ 1
Part (d)
T does not light	✓ 1	No mark for a bare statement without reasoning. MP1 only scores if MP2 scores.
each junction is midway between two equal resistances, so both junctions are at the same potential «6.0 V above the negative terminal»; there is no potential difference across T and so no current in it	✓ 1	OWTTE.
Part (e)(i)
current in the branch = 12/(6.0 + 12) = 0.67 A	✓ 1	Allow ECF from (a)(ii).
power in Q = 0.67² × 12 = 5.3 W	✓ 1
5.3 W is greater than 3.0 W, so Q operates above its rated power «and may fail»	✓ 1	ALTERNATIVE: pd across Q = 0.67 × 12 = 8.0 V ✓✓, greater than 6.0 V ✓. MP3 requires a comparison of calculated values.
Part (e)(ii)
R and S form a separate branch connected directly across the battery, which has negligible internal resistance, so the pd across R is still 6.0 V «and its current is unchanged»	✓ 1	OWTTE.

Answers: (a)(i) 0.50 A · (a)(ii) 12 Ω · (c)(i) 1.0 A · (e)(i) 5.3 W > 3.0 W: yes (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that circuit diagrams represent the arrangement of components in a circuit; the combinations of resistors in series and parallel circuits; electrical power P dissipated by a resistor as given by P = IV = I²R = V²/R; that the electric potential difference V is the work done per unit charge on moving a positive charge between two points along the path of the current as given by V = W/q Command term: Draw

23B-2-32

emf & internal resistance·B.5 Current and circuits

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksDetermine

A car battery has an emf of 12.6 V. When the driver operates the starter motor with the headlamps switched off, the battery supplies a current of 150 A to the motor and the potential difference between the battery terminals falls to 9.9 V.

The headlamps are connected directly across the battery terminals. Each headlamp has a power of 60 W when the terminal potential difference is 12.6 V.

(a)

(i)

Show that the internal resistance of the battery is about 0.02 Ω.

(1)

(ii)

Calculate the rate at which thermal energy is generated inside the battery while the starter motor operates.

(2)

(b)

Deduce the fraction of the energy transferred from chemical energy by the battery that is delivered to the starter motor.

(2)

(c)

The headlamps are now switched on while the starter motor operates. Assume that the headlamps have a constant resistance and that the terminal potential difference stays at 9.9 V.

(i)

Determine the percentage reduction in the power of a headlamp while the starter motor operates.

(3)

(ii)

Suggest why the actual reduction in the power of a headlamp is smaller than your answer to (c)(i).

(2)

(d)

On a very cold morning the internal resistance of the battery rises to 0.030 Ω but its emf is unchanged. The headlamps are switched off. Treat the starter motor as a fixed resistance. The motor needs a potential difference of at least 9.5 V across it to turn the engine.

(i)

Determine whether the motor can turn the engine on this morning.

(3)

(ii)

Suggest why car batteries are designed to have a very small internal resistance.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
r = (12.6 − 9.9)/150 = 0.018 Ω	✓ 1	Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
P = I²r = 150² × 0.018 OR P = I(ε − V) = 150 × 2.7	✓ 1
P = 405 W «4.1 × 10² W»	✓ 1	Accept 4.0–4.5 × 10² W «450 W from r = 0.02 Ω». Award [2] for CNA.
Part (b)
power from chemical energy = εI = 12.6 × 150 = 1890 W AND power delivered to the motor = VI = 9.9 × 150 = 1485 W	✓ 1	ALTERNATIVE: fraction = VI/εI = V/ε ✓.
fraction = 1485/1890 = 0.79	✓ 1	Accept 79 %. Award [2] for CNA.
Part (c)(i)
R = V²/P = 12.6²/60 = 2.65 Ω	✓ 1
P = 9.9²/2.65 = 37 W	✓ 1
reduction = (60 − 37)/60 ≈ 38 %	✓ 1	ALTERNATIVE: P ∝ V², so P₂/P₁ = (9.9/12.6)² = 0.617 ✓✓, reduction 38 % ✓. Award [1 max] for 21 % «P ∝ V used».
Part (c)(ii)
at the lower potential difference the filament is cooler	✓ 1
so its resistance is lower and P = V²/R is larger than predicted «the filament is non-ohmic»	✓ 1
Part (d)(i)
resistance of motor = 9.9/150 = 0.066 Ω	✓ 1
I = 12.6/(0.066 + 0.030) = 131 A	✓ 1	Allow ECF.
V = 131 × 0.066 = 8.7 V, less than 9.5 V, so the motor cannot turn the engine	✓ 1	MP3 requires a comparison with 9.5 V.
Part (d)(ii)
so that the very large current needed by the starter motor causes only a small drop in terminal potential difference «and little energy is wasted in the battery»	✓ 1	OWTTE.

Answers: (a)(i) 0.018 Ω · (a)(ii) 405 W · (b) 0.79 · (c)(i) 38 % · (d)(i) 8.7 V: no (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — that electric cells are characterized by their emf ε and internal resistance r as given by ε = I(R + r); 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; the ohmic and non-ohmic behaviour of electrical conductors, including the heating effect of resistors Command term: Determine

24B-2-33

Circuit diagrams & meters·B.5 Current and circuits

Paper 2Hard12 marks

Short answer & extended response12 steps to full marksDetermine

A student determines the resistance of a resistor X using the circuit shown. The battery has an emf of 3.0 V and negligible internal resistance. The voltmeter is ideal. The ammeter has a resistance of 0.50 Ω. The student calculates the resistance of X as (voltmeter reading)/(ammeter reading). The resistance of X is 4.0 Ω.

Diagram NOT accurately drawn

(a)

State what is meant by an ideal ammeter.

(1)

(b)

(i)

Calculate the reading on the ammeter.

(2)

(ii)

State the reading on the voltmeter.

(1)

(iii)

Show that the student's value for the resistance of X is about 4.5 Ω.

(1)

(c)

Explain why the student's value is larger than the resistance of X.

(2)

(d)

The student uses the same circuit to measure other resistors. Determine the smallest resistance that can be measured in this way with an error of no more than 1 %.

(3)

(e)

Suggest how the student should change the circuit so that the value obtained equals the resistance of X, and explain why.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
an ammeter with zero resistance «so that there is no potential difference across it and it does not change the current»	✓ 1	OWTTE.
Part (b)(i)
total resistance = 4.0 + 0.50 = 4.5 Ω	✓ 1
I = 3.0/4.5 = 0.67 A	✓ 1	Award [1 max] for 0.75 A «ammeter resistance ignored».
Part (b)(ii)
3.0 V «it is connected across the battery terminals and the internal resistance is negligible»	✓ 1
Part (b)(iii)
R = 3.0/0.667 = 4.50 Ω	✓ 1	Must see full substitution OR answer to at least 3 s.f. Allow ECF from (b)(i).
Part (c)
the voltmeter measures the potential difference across X AND the ammeter	✓ 1
there is a potential difference IR_A across the ammeter, so V/I = R_X + R_A = 4.5 Ω	✓ 1	OWTTE.
Part (d)
the measured value is always larger than the true value by the ammeter resistance, 0.50 Ω	✓ 1
fractional error = 0.50/R ≤ 0.01	✓ 1
R ≥ 50 Ω	✓ 1	Award [3] for CNA. Award [2 max] for a correct check of a single value «e.g. 100 Ω gives 0.5 %» without finding the smallest resistance.
Part (e)
connect the voltmeter directly across X only «between the junction of the ammeter and X and the other end of X»	✓ 1
the voltmeter then reads only the pd across X, and as it is ideal it takes no current, so the ammeter still reads the current in X and V/I = R_X	✓ 1	OWTTE.

Answers: (b)(i) 0.67 A · (b)(ii) 3.0 V · (b)(iii) 4.50 Ω · (d) 50 Ω (the remaining parts are explanations — see the table above)

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 as given by R = V/I; unless otherwise stated, ammeters and voltmeters will be considered as ideal; in cases where non-ideal meters are used, the resistance will be constant Command term: Determine

25B-2-34

Electrical power & energy·B.5 Current and circuits

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDetermine

An electric car is charged from a direct-current (dc) fast charger that delivers a power of 50 kW. The charging cable is 4.0 m long and contains two copper conductors, one carrying the current to the car and one returning it. Each conductor has a cross-sectional area of 35 mm². The resistivity of copper is 1.7 × 10⁻⁸ Ω m.

The battery of car X charges at a potential difference of 400 V; the battery of car Y charges at 800 V.

(a)

(i)

Show that the total resistance of the two conductors is about 4 mΩ.

(1)

(ii)

Calculate the current in the cable while car X is charging.

(1)

(iii)

Calculate the power dissipated in the cable while car X is charging.

(2)

(iv)

Calculate the charge that flows through the cable in 30 minutes while car X is charging.

(1)

(b)

Deduce, without calculating the current, the power dissipated in the cable while car Y is charging at 50 kW.

(2)

(c)

The cable may safely dissipate no more than 50 W per metre of its length. A new charger delivers 150 kW. Determine whether this cable can be used safely with the new charger for car X and for car Y.

(4)

(d)

Explain why, for the same power loss, car Y can be charged at a given power through a thinner cable than car X.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
R = ρL/A = 1.7 × 10⁻⁸ × 8.0/3.5 × 10⁻⁵ = 3.9 × 10⁻³ Ω	✓ 1	Must see L = 8.0 m and A in m². Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
I = P/V = 50 000/400 = 125 A	✓ 1
Part (a)(iii)
P = I²R = 125² × 3.9 × 10⁻³	✓ 1	Allow ECF from (a)(i) and (a)(ii).
P = 61 W	✓ 1	Accept 59–63 W.
Part (a)(iv)
q = It = 125 × 1800 = 2.25 × 10⁵ C	✓ 1	Accept 2.3 × 10⁵ C.
Part (b)
the same power at twice the potential difference needs half the current	✓ 1
P = I²R, so the power dissipated is a quarter: about 15 W	✓ 1	Allow ECF from (a)(iii).
Part (c)
currents at 150 kW: X 375 A; Y 188 A	✓ 1
power dissipated with car X = 375² × 3.9 × 10⁻³ = 5.5 × 10² W	✓ 1	Allow ECF from (a)(i).
power dissipated with car Y = 188² × 3.9 × 10⁻³ = 1.4 × 10² W	✓ 1
greatest safe power = 4.0 × 50 = 200 W, so the cable is not safe for car X but is safe for car Y	✓ 1	ALTERNATIVE for MP4: per metre, X 137 W m⁻¹ and Y 34 W m⁻¹, compared with 50 W m⁻¹. MP4 requires both conclusions supported by values.
Part (d)
at twice the potential difference the current is halved, so for the same resistance the power loss I²R is a quarter	✓ 1
R = ρL/A, so the cable resistance can be four times larger, i.e. the cross-sectional area can be a quarter, for the same power loss	✓ 1

Answers: (a)(i) 3.9 × 10⁻³ Ω · (a)(ii) 125 A · (a)(iii) 61 W · (a)(iv) 2.25 × 10⁵ C · (b) 15 W · (c) X: 550 W > 200 W, no; Y: 140 W < 200 W, yes (the remaining parts are explanations — see the table above)

Syllabus understandingB.5 — electrical power P dissipated by a resistor as given by P = IV = I²R = V²/R; resistivity as given by ρ = RA/L; direct current (dc) I as a flow of charge carriers as given by I = Δq/Δt Command term: Determine

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

25 practice questions on B.5 Current and circuits

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