Series and parallel circuits – GCSE Physics exam questions with mark scheme

1Grade 4–54 marks

The diagram shows two resistors connected in series to a 6.0 V battery.

(a) Calculate the total resistance of the circuit. [1]

(b) Calculate the current in the circuit. [2]

(a) 12 Ω(1)
(b) $I = \dfrac{6.0}{12}$ allow ecf from (a)(1)
(b) 0.50 A(1)
(c) 4.0 V allow ecf: their (b) × 8.0(1)
Answer: (a) 12 Ω (b) 0.50 A (c) 4.0 V
Examiner tip: In series the current is the same everywhere, so use the 0.50 A with the 8.0 Ω resistor in (c).

2Grade 6–7Higher5 marks

The diagram shows a 6.0 Ω resistor and a 12 Ω resistor connected in parallel to a 12 V battery.

(a) Calculate the current drawn from the battery. [3]

(b) Explain why the total resistance of two resistors in parallel is less than the resistance of the smallest individual resistor. [2]

(a) pd across each resistor is 12 V or $I = 12/6.0 = 2.0$ A alternative method: allow $\dfrac{1}{R} = \dfrac{1}{6.0} + \dfrac{1}{12}$ for this mark, $R = 4.0$ Ω for the second mark, then $I = 12/4.0$(1)
(a) $I = 12/12 = 1.0$ A(1)
(a) 3.0 A an answer of 3 (A) scores 3 marks(1)
(b) adding a resistor in parallel provides an extra path (for the charge / current)(1)
(b) so the total current is greater than the current through either resistor on its own (for the same pd), so the total resistance ($R = V/I$) is smaller allow so more charge flows per second (for the same pd)(1)
Answer: (a) 3.0 A (b) extra path, so total current is larger for the same pd
Examiner tip: Each parallel branch has the full battery pd across it — do not split the 12 V between the branches.

3Grade 4–54 marks

A light-emitting diode (LED) is connected in series with a 220 Ω resistor and a 9.0 V battery.

The resistor stops the current in the LED from becoming too large.

The current in the circuit is 30 mA.

(a) Calculate the potential difference across the resistor. [3]

(b) Calculate the potential difference across the LED. [1]

(a) 30 mA = 0.030 A allow conversion shown anywhere, e.g. 30 × 10⁻³(1)
(a) $V = 0.030 \times 220$(1)
(a) 6.6 (V) an answer of 6.6 scores 3 marks; an answer of 6600 scores 2 marks(1)
(b) 2.4 V allow ecf: 9.0 − their (a), provided their (a) is less than 9.0 V(1)
Answer: (a) 6.6 V (b) 2.4 V
Examiner tip: Convert milliamps to amps before using V = IR — leaving 30 in the equation gives 6600 V and loses the conversion mark.

4Grade 8–9Higher6 marks

The diagram shows a circuit containing a resistor X and two identical lamps.

The ammeter reads 0.80 A. The voltmeter reads 4.0 V.

(a) Calculate the resistance of resistor X. [2]

(b) Calculate the resistance of one of the lamps. [4]

(a) $R = \dfrac{4.0}{0.80}$(1)
(a) 5.0 (Ω)(1)
(b) pd across (each) lamp = 12 − 4.0 = 8.0 V(1)
(b) current in each lamp = 0.80 ÷ 2 = 0.40 A(1)
(b) $R = \dfrac{8.0}{0.40}$ allow ecf from incorrect pd or current(1)
(b) 20 (Ω) an answer of 20 scores 4 marks; an answer of 10 (Ω) scores 2 marks (8.0/0.80, both lamps treated as one); allow alternative: combined resistance of lamps = 8.0/0.80 = 10 Ω (2 marks), one lamp = 2 × 10 = 20 Ω (2 marks)(1)
Answer: (a) 5.0 Ω (b) 20 Ω
Examiner tip: The 0.80 A splits equally between the two identical lamps, but both lamps have the same 8.0 V across them — divide the current, not the pd.

5Grade 6–7Higher6 marks

The graph shows how the current in two resistors, P and Q, depends on the potential difference across each resistor.

(a) Use the graph to determine the resistance of resistor P. [2]

(b) Resistors P and Q are connected in series to a 6.0 V supply.

Use the graph to determine the current in the circuit.

Explain how you used the graph. [3]

(c) Which resistor has the larger potential difference across it when they are connected in series? Give a reason. [1]

(a) correct pair of values read from line P, e.g. 4.0 V and 0.40 A any correct point; allow $R = V/I$ with correct substitution(1)
(a) 10 (Ω)(1)
(b) the current is the same in P and Q (because they are in series)(1)
(b) find the current at which the pd across P plus the pd across Q equals 6.0 V e.g. at 0.20 A, P has 2.0 V and Q has 4.0 V(1)
(b) 0.20 A allow alternative method: R_Q = 20 Ω from graph, total R = 10 + 20 = 30 Ω (1 mark), I = 6.0/30 (1 mark), 0.20 A (1 mark)(1)
(c) Q and it has the larger resistance (and the same current) allow Q has 4.0 V and P has 2.0 V(1)
Answer: (a) 10 Ω (b) 0.20 A — same current in each; pds 2.0 V + 4.0 V = 6.0 V (c) Q, larger resistance
Examiner tip: Reading 6.0 V off each line separately and adding the currents (0.6 + 0.3 A) treats the resistors as if they were in parallel — in series it is the pds that add, not the currents.

6Grade 6–76 marks

A garden designer has six identical lamps and a 12 V battery.

Each lamp is designed to work at its normal brightness with a potential difference of 12 V across it.

The designer could connect the six lamps in series or in parallel with the battery.

Compare connecting the lamps in series with connecting them in parallel.

In your answer you should refer to current, potential difference and resistance. [6]

Series: the current is the same in every lamp(1)
Series: the 12 V is shared between the six lamps, so each lamp has only 2 V across it(1)
Series: lamps would be much dimmer than normal / would not work at normal brightness(1)
Series: total resistance is the sum of the six resistances, so it is large and the current from the battery is small(1)
Series: if one lamp breaks (or is removed) the circuit is broken and all the lamps go out(1)
Series: the lamps cannot be switched on and off separately(1)
Parallel: each lamp has the full 12 V across it, so each lamp is at normal brightness(1)
Parallel: the current from the battery is the sum of the six branch currents, so it is larger(1)
Parallel: total resistance is less than the resistance of one lamp (adding lamps adds extra paths)(1)
Parallel: if one lamp breaks the others stay on; each lamp could have its own switch(1)
Comparison: parallel draws a larger current, so the battery transfers energy faster / goes flat sooner(1)
Comparison: parallel is the better arrangement for lamps rated at 12 V; series would need lamps rated at 2 V(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed and coherent comparison of both arrangements. Ideas about current, potential difference and resistance are used and linked logically to consequences such as lamp brightness, what happens if a lamp fails and the current drawn from the battery.
Level 2	3–4	Relevant points are made about both arrangements, with some linking to consequences, but the comparison is incomplete or some links are missing or unclear.
Level 1	1–2	Simple, relevant statements about series and/or parallel circuits. Statements may be about one arrangement only and are not linked to consequences.
No marks	0	No relevant content.

Answer: Series: same current, pd shared (2 V each), lamps dim, high total R, one fails → all go out. Parallel: 12 V across each lamp, normal brightness, branch currents add so larger battery current, lower total R, others stay on if one fails; battery runs down faster.
Examiner tip: A ‘compare’ question needs linked statements about BOTH arrangements — listing series facts only, however detailed, caps the answer at Level 1–2.

7Grade 1–3Foundation5 marks

The diagram shows a circuit containing two different lamps, P and Q, connected in series to a 4.5 V battery.

Ammeter A₁ reads 0.25 A.

(a) Give the readings on ammeters A₂ and A₃. [1]

(b) The voltmeter reads 1.8 V.

Determine the potential difference across lamp Q. [1]

less than the difference between the sum of

The total resistance of two components in series is ____________ the resistances of the two components. [1]

(d) The filament in lamp P breaks.

What happens to lamp Q? Give a reason for your answer. [2]

(a) A₂ = 0.25 A and A₃ = 0.25 A both needed for the mark; allow ‘0.25 A for both’ / ‘the same as A₁’ only if 0.25 is stated(1)
(b) 2.7 V allow 2.7 with no unit; allow 4.5 − 1.8(1)
(c) the sum of(1)
(d) lamp Q goes out / switches off do not accept 'goes dim' or any answer saying Q stays on or gets brighter(1)
(d) the circuit is broken / incomplete so there is no current (in Q) allow there is only one path for the current; allow the current stops; reason only credited if Q goes out(1)
Answer: (a) 0.25 A and 0.25 A (b) 2.7 V (c) the sum of (d) Q goes out — the series circuit is broken so there is no current
Examiner tip: In a series circuit the current is the same everywhere, but the battery pd is shared — students often write 0.25 A ÷ 2 for the ammeters or forget to subtract for lamp Q.

8Grade 4–5Foundation5 marks

The diagram shows a lamp and a resistor R connected in parallel to a 6.0 V battery.

Ammeter A₁ reads 0.30 A. Ammeter A₂ reads 0.45 A.

(a) Determine the reading on ammeter A₃. [1]

(b) Give the potential difference across resistor R. [1]

Use the equation:

$$\text{potential difference} = \text{current} \times \text{resistance}$$

Give your answer to 2 significant figures. [3]

(a) 0.75 A(1)
(b) 6.0 V allow 6 V; the battery pd is across each branch(1)
(c) $6.0 = 0.45 \times R$ allow ecf from (b) for the pd used(1)
(c) $R = \dfrac{6.0}{0.45}$(1)
(c) 13 (Ω) an answer of 13 scores 3 marks; 13.3 or 13.33 (not given to 2 s.f.) scores 2 marks(1)
Answer: (a) 0.75 A (b) 6.0 V (c) 13 Ω
Examiner tip: In parallel each branch gets the full battery pd — using 3.0 V (half of 6.0 V) in part (c) is the most common error.

9Grade 6–7Higher7 marks

The diagram shows a light-dependent resistor (LDR) connected in series with a 4.0 kΩ resistor and a 6.0 V battery.

When the room is dark, the voltmeter reads 1.2 V.

(a) Calculate the current in the circuit. [2]

(b) Calculate the resistance of the LDR when the room is dark. [2]

Explain what happens to the reading on the voltmeter. [3]

(a) $1.2 = I \times 4000$ allow $I = 1.2/4000$; allow use of 4.0 if final answer is given as 0.30 mA(1)
(a) 0.0003(0) A allow 3.0 × 10⁻⁴ A or 0.30 mA; 0.3 with no unit scores 1 mark only(1)
(b) pd across LDR = 6.0 − 1.2 = 4.8 V(1)
(b) $R = 4.8 / 0.0003 = 16\,000$ Ω allow 16 kΩ; allow ecf from (a); allow ratio method $4.8/1.2 \times 4.0$ kΩ for both marks(1)
(c) (voltmeter) reading increases(1)
(c) resistance of the LDR decreases (as light intensity increases)(1)
(c) so the current in the circuit increases and pd across the fixed resistor ($V = IR$) increases allow the LDR now takes a smaller share of the 6.0 V, so the resistor takes a larger share(1)
Answer: (a) 3.0 × 10⁻⁴ A (b) 16 000 Ω (16 kΩ) (c) reading increases — LDR resistance falls, current rises, so pd across the fixed resistor rises
Examiner tip: The voltmeter is across the fixed resistor, not the LDR — subtract 1.2 V from 6.0 V to get the pd across the LDR before finding its resistance.

10Grade 4–56 marks

A student investigated how the total resistance of a combination of identical resistors depends on how they are connected.

For each arrangement she measured the potential difference across the combination and the current through it.

The table shows her results.

Arrangement	Potential difference in V	Current in A	Total resistance in Ω
1 resistor	3.0	0.30	10.0
2 in series	3.0	0.15	20.0
3 in series	3.0	0.10	30.0
2 in parallel	2.9	0.58	5.0
3 in parallel	2.8	0.84	X

(a) Calculate value X. Give your answer to 2 significant figures. [2]

(b) Describe the pattern shown by the results for the resistors connected in series. [1]

Give one piece of evidence from the table that supports her conclusion. [1]

(d) While taking one reading, the student noticed that the ammeter reading slowly decreased.

Suggest why the ammeter reading decreased and how the student could reduce this effect. [2]

(a) $R = \dfrac{2.8}{0.84}$(1)
(a) 3.3 (Ω) an answer of 3.3 scores 2 marks; 3.33 or 3.333 (not given to 2 s.f.) scores 1 mark(1)
(b) total resistance increases by 10 Ω for each extra resistor allow total resistance = number of resistors × 10 Ω; allow total resistance is the sum of the individual resistances; ‘resistance increases’ alone is insufficient(1)
(c) 2 resistors in parallel have a total resistance of 5.0 Ω, which is less than the 10.0 Ω of one resistor allow 3 in parallel (3.3 Ω or their X) less than 2 in parallel (5.0 Ω) / less than 10 Ω; must quote values(1)
(d) the resistor(s) / wires heated up, so the resistance increased allow temperature of resistor increased(1)
(d) switch the circuit off between readings allow take readings quickly; allow use a lower potential difference / smaller current(1)
Answer: (a) 3.3 Ω (b) total R rises by 10 Ω per resistor (sum of resistances) (c) e.g. 2 in parallel = 5.0 Ω < 10.0 Ω for one (d) resistor heats up so R increases; switch off between readings
Examiner tip: When asked for evidence from a table, quote the actual numbers — ‘the resistance is lower in parallel’ without values usually scores zero.

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