Capacitor discharge
Q = Q0 e−t / (R C)
The booklet prints the charge form. The voltage form used in the applications follows from Q = C V for a fixed capacitor.
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.
A 5.0 microF capacitor charged to 12 V discharges through a 2.0 Mohm resistor. Calculate the potential difference after 8.0 s.
Do the calculation on paper first — then mark it.
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.
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.
Still losing marks on the calculations?
I'll go through your working line by line and show you exactly where the marks are — your first lesson is free.