Circuit symbols, charge, current, potential difference and resistance (RP3) – GCSE Physics exam questions with mark scheme

1Grade 1–3Foundation3 marks

The diagram shows a circuit used to investigate a lamp.

(a) The ammeter reads 0.25 A and the voltmeter reads 3.0 V.

Calculate the resistance of the lamp. [2]

(b) Describe how the voltmeter is connected in the circuit. [1]

(a) $R = \dfrac{3.0}{0.25}$ allow $3.0 = 0.25 \times R$(1)
(a) 12 Ω an answer of 12 (Ω) scores 2 marks(1)
(b) in parallel (with / across the lamp) allow connected across the lamp / to either side of the lamp(1)
Answer: (a) 12 Ω (b) in parallel with the lamp
Examiner tip: Rearrange $V = IR$ to $R = V/I$; $0.25 \div 3.0$ is the common wrong way round.

2Grade 4–53 marks

A current of 0.40 A flows through a resistor for 5.0 minutes.

Calculate the charge that flows through the resistor. [3]

$t = 300$ s(1)
$Q = 0.40 \times 300$(1)
120 C allow ecf for $Q = 0.40 \times 5.0 = 2.0$ C for 1 mark only(1)
Answer: 120 C
Examiner tip: Time must be in seconds for $Q = It$ — 5.0 minutes is 300 s.

3Grade 6–7Higher6 marks

A student investigates how the resistance of a wire depends on its length. She connects a length of wire to a battery, an ammeter and a voltmeter.

Length (cm)	20	40	60	80	100
Resistance (Ω)	1.0	2.1	2.9	4.0	5.0

(a) Describe how she uses the meter readings to find the resistance of the wire. [1]

(b) Give one control variable in this investigation. [1]

Explain why. [2]

(d) Describe the relationship between resistance and length shown by the results. Give a reason for your answer. [2]

(a) divide the voltmeter reading by the ammeter reading / $R = V/I$(1)
(b) any one from: thickness / diameter / cross-sectional area of the wire; material / type of wire; temperature (of wire) ignore current / pd / voltage (these change as length changes)(1)
(c) (otherwise) the wire would heat up(1)
(c) (and) the resistance increases with temperature, which would affect the readings / temperature would not be controlled(1)
(d) resistance is (directly) proportional to length do not accept 'resistance increases as length increases' / 'positive correlation' / 'linear' alone(1)
(d) supported by data, e.g. doubling the length (approximately) doubles the resistance (20 cm → 1.0 Ω, 40 cm → 2.1 Ω; 40 cm → 2.1 Ω, 80 cm → 4.0 Ω) or R/L is (approximately) constant at 0.05 Ω/cm or a graph of R against L would be a straight line through the origin(1)
Answer: (a) $R=V/I$ (b) e.g. wire thickness (c) wire heats up, increasing its resistance (d) directly proportional — e.g. R/L ≈ 0.05 Ω/cm for every reading (20 cm → 1.0 Ω, 100 cm → 5.0 Ω)
Examiner tip: 'Proportional' needs evidence from the data — quote values showing that R ÷ L stays (about) the same, or that doubling L doubles R.

4Grade 1–3Foundation4 marks

The diagram shows the circuit symbols for four components, A, B, C and D.

(a) Name components A, B and C. [3]

(b) Component D is a variable resistor.

Give one reason for including a variable resistor in a circuit. [1]

(a) A: fuse(1)
(a) B: light dependent resistor allow LDR do not accept light emitting diode / LED(1)
(a) C: diode do not accept LED / light emitting diode(1)
(b) to change / control / adjust the current (in the circuit) allow to change the resistance (of the circuit) / to change the potential difference across a component allow to change the brightness of a lamp / speed of a motor(1)
Answer: (a) A fuse; B light dependent resistor (LDR); C diode (b) to vary the current in the circuit
Examiner tip: LDR arrows point in (light falling on it) and a plain diode has no arrows — calling either an LED loses the mark.

5Grade 1–3Foundation4 marks

(a) Complete the sentence.

Electric current is the rate of flow of ____________ . [1]

(b) Name the unit of electric charge. [1]

(c) A student connects a lamp, a switch and an ammeter in a complete loop of wire. There is no cell or battery in the loop. The switch is closed.

Give the reason why the ammeter reads zero. [1]

(d) The student adds a cell to the loop but opens the switch.

Explain why there is still no current in the lamp. [1]

(a) (electric) charge do not accept electrons(1)
(b) coulomb allow C(1)
(c) there is no source of potential difference (to make the charge flow) allow no source of pd / voltage / no power supply ignore no energy alone(1)
(d) the circuit is not (a) closed / complete (circuit) allow there is a gap in the circuit / break in the circuit(1)
Answer: (a) charge (b) coulomb, C (c) no source of potential difference (d) circuit is not closed / complete
Examiner tip: Charge only flows when there is both a closed circuit and a source of potential difference — name the missing one, not just 'no power'.

6Grade 4–53 marks

A resistor on a circuit board has a resistance of 470 Ω.

The current in the resistor is 20 mA.

Calculate the potential difference across the resistor. [3]

$I = 0.020$ A allow 20 mA = 0.02 A(1)
$V = 0.020 \times 470$(1)
9.4 (V) an answer of 9.4 (V) scores 3 marks an answer of 9400 (V) scores 2 marks(1)
Answer: 9.4 V
Examiner tip: Convert milliamps to amps before multiplying — leaving 20 gives 9400 V and drops a mark.

7Grade 6–7Higher4 marks

A small heating element is used to stop the lens of a security camera misting up.

When the heating element is switched on, a charge of 72 C flows through it in 4.0 minutes.

The potential difference across the heating element is 6.0 V.

Calculate the resistance of the heating element. [4]

$t = 240$ s(1)
$I = \dfrac{72}{240} = 0.30$ (A) allow $72 = I \times 240$(1)
$R = \dfrac{6.0}{0.30}$ allow ecf from their current(1)
20 (Ω) an answer of 20 (Ω) scores 4 marks allow ecf: using $t = 4.0$ gives $I = 18$ A and $R = 0.33$ Ω for 3 marks(1)
Answer: 20 Ω
Examiner tip: Find the current first with $Q = It$ (time in seconds), then use $V = IR$ — you can't jump straight from charge to resistance.

8Grade 6–76 marks

A student has three identical resistors.

She wants to find out how the total resistance of the resistors depends on whether they are connected in series or in parallel.

Describe a method she could use.

You should include the measurements she should take and how she should use them to find each total resistance. [6]

connect a power supply / battery, a switch and an ammeter in series with the resistor(s)(1)
connect a voltmeter in parallel / across the resistor or combination of resistors (being tested)(1)
first test one resistor on its own(1)
then connect two resistors in series and then three resistors in series(1)
then connect two resistors in parallel and then three resistors in parallel(1)
for each arrangement, record the ammeter reading and the voltmeter reading(1)
calculate total resistance using $R = V \div I$(1)
repeat the readings and calculate a mean for each arrangement(1)
use a low pd / switch off between readings so the resistors do not heat up (and change resistance)(1)
compare the totals: in series the total is the sum of the individual resistances; in parallel the total is less than the resistance of one resistor(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically ordered method that could be followed to produce valid results. Correct placement of both meters, testing of both series and parallel arrangements, the readings taken and how total resistance is calculated are all included.
Level 2	3–4	A method with the main steps present but some detail missing or out of order, e.g. meters described but only one type of arrangement tested, or no clear use of $R = V/I$.
Level 1	1–2	Some relevant isolated steps, e.g. 'use an ammeter and a voltmeter' or 'put the resistors in series then in parallel', without a coherent method.
0	0	No relevant content.

Answer: Circuit: supply, switch, ammeter in series; voltmeter across the resistor combination. Measure V and I for one resistor, then 2 and 3 in series, then 2 and 3 in parallel; calculate $R = V/I$ each time; repeat and average; keep pd low / switch off between readings to avoid heating; compare totals.
Examiner tip: Top-band answers say exactly where each meter goes and how $V$ and $I$ become a resistance — 'measure the resistance' with no meters or calculation stays in Level 1.

9Grade 4–54 marks

A student wants to find the resistance of a resistor. She draws the circuit shown in the diagram.

(a) The circuit has a mistake.

Describe the mistake and how the circuit should be corrected. [3]

(b) The student corrects the circuit. She then replaces the resistor with one of greater resistance. The potential difference across the resistor stays the same.

State what happens to the reading on the ammeter. [1]

(a) the ammeter and the voltmeter are in the wrong places / have been swapped(1)
(a) the ammeter should be (connected) in series with the resistor(1)
(a) the voltmeter should be (connected) in parallel / across the resistor allow a correctly drawn corrected circuit for the last two marks(1)
(b) (reading) decreases allow goes down / gets smaller(1)
Answer: (a) Meters swapped: ammeter should be in series with the resistor, voltmeter in parallel across it. (b) Ammeter reading decreases.
Examiner tip: Say where each meter should go — 'the meters are the wrong way round' on its own only earns the first mark.

10Grade 8–9Higher5 marks

A student investigated how the resistance of a wire depended on its length. She used crocodile clips to connect different lengths of the wire into a circuit and calculated the resistance from meter readings.

The graph shows her results.

(a) The line of best fit does not pass through the origin.

Suggest one reason why. [1]

(b) Determine the gradient of the line. Give the unit. [2]

Use your answer to part (b) to calculate the resistance of a 1.5 m length of the wire. [2]

(a) resistance of the connecting leads / crocodile clips (was included in each reading) allow zero error (in a meter) / systematic error allow contact resistance at the clips ignore random error / human error(1)
(b) gradient = 0.030 allow 0.029 – 0.031 allow 3.0 (0.029 – 0.031 × 100) if length converted to m working e.g. $\dfrac{3.4 - 0.4}{100 - 0}$ from a large triangle(1)
(b) Ω/cm allow Ω/m if gradient given as 3.0 unit must be consistent with the value given(1)
(c) $R = 0.030 \times 150$ allow ecf from (b) allow $3.0 \times 1.5$(1)
(c) 4.5 (Ω) do not accept 4.9 (Ω) (intercept added)(1)
Answer: (a) resistance of the leads/clips (zero error) (b) 0.030 Ω/cm (c) 4.5 Ω
Examiner tip: The gradient is unaffected by a zero error, so use gradient × length — don't add the intercept, and match 150 cm with Ω/cm.

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