Charging and discharging capacitorsEdexcel International A Level Physics: Revision notes
Section 1
Charging and discharging curves
When a capacitor discharges through a resistor, the current and the charge on the capacitor fall exponentially: rapidly at first, then more slowly, never quite reaching zero. The p.d. across the capacitor follows the same curve.
When a capacitor charges through a resistor from a supply, the current is largest at the start (I0 = V/R, as the capacitor has no p.d.) and decreases exponentially to zero, while the charge and capacitor p.d. rise exponentially towards their final values.
The reason is that the p.d. across the resistor is the supply p.d. minus the capacitor p.d., and it falls as the capacitor charges.
The charging current is not constant. It is greatest at the start and falls to zero as the capacitor p.d. approaches the supply p.d.
Section 2
The time constant
The time constant τ = RC (in seconds, with R in Ω and C in F) is a measure of how quickly a capacitor charges or discharges. It is the time for the charge (or current or p.d.) to fall to 1/e (about 37%) of its initial value when discharging, and for it to rise to about 63% of its final value when charging.
A larger R or larger C gives a longer time constant, so a slower change.
Worked example: 2200 μF and 4.7 kΩ give RC = 4.7 × 10³ × 2200 × 10⁻⁶ = 10 s.
Section 3
Equations for exponential discharge
The charge, current and p.d. all decay with the same time constant:
Q = Q0e^(−t/RC), I = I0e^(−t/RC), V = V0e^(−t/RC)
Taking natural logs gives straight-line forms:
ln Q = ln Q0 − t/RC, similarly for I and V.
A graph of ln V against t is a straight line with gradient −1/RC and intercept ln V0. To find the time for the p.d. to fall to V: t = RC ln(V0/V).
Worked example: with V0 = 9.0 V and RC = 10.3 s, after 20 s, V = 9.0 × e^(−20/10.3) = 1.3 V.
Check your calculator: ln and e^x are inverses. Use brackets for the whole index −t/RC.
Section 4
Core Practical 11: discharge of a capacitor
Charge a capacitor fully from a supply, then switch it to discharge through a resistor, with a data logger or oscilloscope recording the p.d. across the capacitor.
- Record V against t at regular intervals.
- Plot V against t to see the exponential curve.
- Plot ln V against t. A straight line shows exponential decay, and its gradient is −1/RC.
You can also find RC directly as the time for V to fall to 37% of V0. Use a large RC so the readings are not too fast to record, and repeat the readings. Capacitors typically have tolerances of 10% or more, so the measured RC may differ from the calculated value.
Must know
- RC is the time constant, in seconds
- Q, I and V decay as e^(−t/RC) when discharging
- ln V = ln V0 − t/RC; the ln V–t gradient is −1/RC
- After one time constant the charge has fallen to 37%
- The charging current starts at V/R and falls to zero
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Charging and discharging capacitors
- A 2200 μF capacitor is charged to a p.d. of 9.0 V and is then discharged through a 4.7 kΩ resistor.Calculate the p.d. across the capacitor 20 s after the discharge begins.2 marks
- A 100 μF capacitor, initially uncharged, is charged through a 47 kΩ resistor using a 6.0 V supply that has negligible internal resistance.Explain why the current in the circuit decreases as the capacitor charges.2 marks
- In Core Practical 11 a student records the p.d. V across a capacitor as it discharges through a resistor, using a data logger. She plots ln V against the time t in seconds (V in volts). Her line of best fit is straight, with a gradient of −0.050 s⁻¹ and an intercept of 2.20 on the ln V axis.Use the equation for exponential discharge to explain the shape of her graph, and determine the time constant of the circuit and the initial p.d. across the capacitor.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).