A 5.0 microF capacitor charged to 12 V discharges through a 2.0 Mohm resistor. Calculate the potential difference after 8.0 s.
[4 marks]
Total for this question: 4
All equation deep-dives
Capacitor discharge
Fields and capacitors
AQA 7408
In the data booklet
ORIGINAL
Q = Q0 e−t / (R C)
Derived rearrangement
V = V0 e−t / (R C)Derived using Q = C V for a fixed capacitor.The booklet prints the charge form. The voltage form used in the applications follows from Q = C V for a fixed capacitor.
Board and spec code confirmed against AQA 7408 · registry checked 2026-07-11How this checking works
Know the equation
| Symbol | Quantity | Unit |
|---|---|---|
| Q | charge after time t | C |
| Q0 | initial charge | C |
| V | potential difference after time t (derived form) | V |
| V0 | initial potential difference (derived form) | V |
| t | time | s |
| R | resistance | ohm |
| C | capacitance | F |
Rearrangements
- t = -R C ln(Q / Q0)
- R = -t / (C ln(Q / Q0))
- C = -t / (R ln(Q / Q0))
- ln(Q) = ln(Q0) - t / (R C)
Apply it — mark your own working
Work each one out on paper first, then reveal the mark scheme and tick the marks you actually earned. That is exactly how you should mark past papers.
Q1
Do the calculation on paper first — then mark it.
Q2
The potential difference across a capacitor falls from 9.0 V to 2.0 V in 24 s while it discharges through a 1.5 Mohm resistor. Calculate the capacitance.
[4 marks]
Total for this question: 4
Do the calculation on paper first — then mark it.
Where the marks get lost
- Entering e−t / (R C) instead of putting the whole -t/(R C) in the exponent.
- Using log base 10 when rearranging the exponential; the inverse of e is ln.
- Leaving R in Mohm or C in microF. Their product is in seconds only after SI conversion.
Exam tip: Calculate the time constant R C on its own line. After one time constant, Q is about 0.37 Q0; for a fixed capacitor, V is likewise about 0.37 V0.
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