A Level · Physics 9702 · Fields & motion

Capacitance: A Level Physics questions with answers

Free Capacitance questions for A Level Physics (9702), part of Fields & motion. Each question has the answer and the marking points an examiner would look for.

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Capacitance practice questions

Question 1Structured · 7 marksA* challenge

A 1000 μF capacitor is charged to 6 V and then discharged through a 22 kΩ resistor.

(a)

Calculate the initial charge and the energy stored.

[2]
(b)

Calculate the time constant.

[1]
(c)

Calculate the potential difference across the capacitor 5 s after discharge begins.

[2]
(d)

Calculate the time for the charge to fall to half its initial value.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)6.00 × 10−3 C; 0.0180 J
Q = CV = 6.00 × 10−3 C A1
E = ½CV2 = 0.0180 J A1
2
(b)22.0 s
τ = RC = 22.0 s B1
1
(c)4.78 V
V = V0e−t/RC C1
= 4.78 V A1
2
(d)15.2 s
t = RC ln 2 C1
= 15.2 s A1
2
Question 2Structured · 10 marks

Capacitors of 2 μF and 12 μF are connected in series to a 24 V supply.

(a)

Explain why the two capacitors store the same charge.

[2]
(b)

Calculate the combined capacitance.

[2]
(c)

Use your answer to (b) to calculate the charge on each capacitor and the p.d. across each.

[3]
(d)

Calculate the total energy stored.

[2]
(e)

State one use of a capacitor.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)same current flows through series circuit; charge on plates equal and opposite
in series the same charging current flows for the same time B1
each plate pair gains equal and opposite charge B1
2
(b)1.71 μF
1/C = 1/C₁ + 1/C₂ C1
1.71 μF A1
2
(c)4.11 × 10−5 C; 20.6 V and 3.43 V
Q = CV = 4.11 × 10−5 C C1
V₁ = Q/C₁ = 20.6 V A1
V₂ = 3.43 V A1
3
(d)4.94 × 10−4 J
W = ½CV2 C1
4.94 × 10−4 J A1
2
(e)smoothing / timing / flash / backup
smoothing a rectified supply / timing circuits / camera flash / backup power B1
1
Question 3Structured · 10 marks

A 470 μF capacitor is charged through a 4.7 kΩ resistor from a 9.0 V supply, starting uncharged.

(a)

Sketch graphs of current and of capacitor p.d. against time during charging.

Fig. 3.1
[2]
(b)

Calculate the initial current.

[1]
(c)

Calculate the p.d. across the capacitor after one time constant.

[2]
(d)

Calculate the time for the p.d. to reach 90% of the supply p.d.

[3]
(e)

Explain why the current decreases as the capacitor charges.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)current decays exponentially from V/R; p.d. rises to V exponentially
current: exponential decay from maximum B1
p.d.: exponential rise approaching 9.0 V B1
2
(b)1.91 mA
I = V/R = 1.91 mA B1
1
(c)5.69 V
V = V₀(1 − e−1) C1
5.69 V A1
2
(d)5.09 s
0.90 = 1 − e−t/RC C1
t = RC ln 10 C1
5.09 s A1
3
(e)p.d. across capacitor opposes supply; p.d. across resistor falls
the capacitor p.d. increases and opposes the supply B1
so the p.d. across the resistor, and hence the current, decreases B1
2
Question 4Structured · 3 marks

A 2012 μF capacitor charged to 6.0 V discharges through a 33 kΩ resistor. Calculate the potential difference after 73 s.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
2 V
τ = RC C1
V = V0e−t/RC C1
2 A1
3
Question 5Structured · 9 marks

A capacitor of capacitance 220 μF is charged to 6.0 V and then discharged through a resistor of resistance 22 kΩ.

(a)

Define capacitance.

[1]
(b)

Calculate the initial charge and energy stored.

[3]
(c)

Calculate the time constant of the circuit.

[1]
(d)

Use your answer to (c) to calculate the p.d. across the capacitor 15 s after discharge begins.

[2]
(e)

A second, identical capacitor is connected in series with the first. State the total capacitance and explain the effect on the time constant.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)charge stored per unit p.d.
charge per unit potential difference (C = Q/V) B1
1
(b)1.32 × 10−3 C; 3.96 × 10−3 J
Q = CV = 1.32 × 10−3 C A1
W = ½CV2 C1
3.96 × 10−3 J A1
3
(c)4.84 s
τ = RC = 4.84 s B1
1
(d)0.271 V
V = V0 e−t/RC C1
0.271 V A1
2
(e)110 μF; time constant halves
110 μF B1
RC halves so discharge is faster B1
2
Question 6Structured · 4 marks

Capacitors can be combined in series and parallel.

(a)

Derive the formula for the combined capacitance of two capacitors in parallel.

[2]
(b)

Calculate the combined capacitance of 20 μF and 30 μF in series.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)same V, charges add
same p.d. across both B1
total charge Q = Q₁ + Q₂ so C = C₁ + C₂ B1
2
(b)12 μF
1/C = 1/20 + 1/30 C1
12 μF A1
2

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in