Charging, discharging and the time constant

Discharging

A charged capacitor connected across a resistor R drives a current I = V / R. As charge flows off, V falls, so I falls too: the rate of loss of charge is proportional to the charge left. This gives exponential decay:

Q = Q_0e^-t/RC, I = I_0e^-t/RC, V = V_0e^-t/RC

The time constant

The time constant τ = RC (in seconds, when R is in Ω and C in F).

• After one time constant, Q, V and I have fallen to 1/e, about 37%, of their starting values.
• The time to halve is t_1/2 = RC ln 2 ≈ 0.69RC.
• After about 5RC the capacitor is effectively fully discharged.

A bigger R or a bigger C means a slower discharge.

Charging

Charging through a resistor from a supply of e.m.f. ε: the current starts at ε / R and decays exponentially with the same time constant. The p.d. and charge rise quickly at first, then level off at V = ε and Q = Cε. The graphs of V and Q against t are 'upside-down' exponential curves.

Logarithmic form

Taking natural logs: ln V = ln V_0 - t / RC (and the same for Q and I). A graph of ln V against t is a straight line with gradient -1/RC and intercept ln V_0. This is how RC is found from experimental data.

Worked example

A 100 µF capacitor charged to 12 V discharges through a 47 kΩ resistor. Find the time constant and the p.d. after 3.0 s.

1. RC = 47 x 10³ x 100 x 10^-6 = 4.7 s
2. V = 12e^-3.0/4.7 = 12 x 0.528

Answer: RC = 4.7 s; V = 6.3 V

Check yourself

1. Find the time constant for a 220 µF capacitor and a 10 kΩ resistor, in s.
   Show answer

   2.2 s. RC = 10 x 10³ x 220 x 10^-6 = 2.2 s.

2. A capacitor charged to 9.0 V discharges with RC = 2.0 s. Find the p.d. after 1.5 s, in V, to 3 s.f.
   Show answer

   4.25 V. V = 9.0e^-1.5/2.0 = 9.0 x 0.472 = 4.25 V.

3. What percentage of the initial charge remains after one time constant? Give 3 s.f.
   Show answer

   36.8 %. Q / Q_0 = e^-1 = 0.368, which is 36.8%.

4. A circuit has RC = 4.7 s. How long does the p.d. take to halve, in s, to 3 s.f.?
   Show answer

   3.26 s. t_1/2 = RC ln 2 = 4.7 x 0.693 = 3.26 s.

5. A capacitor charged to 12 V starts to discharge through a 47 kΩ resistor. Find the initial current in mA, to 3 s.f.
   Show answer

   0.255 mA. I_0 = V_0 / R = 12 / 47000 = 2.55 x 10^-4 A = 0.255 mA.

6. A graph of ln V against t has a gradient of -0.25 s^-1. The resistor is 10 kΩ. Find C in µF.
   Show answer

   400 µF. Gradient = -1/RC, so RC = 4.0 s and C = 4.0 / 10000 = 4.0 x 10^-4 F = 400 µF.

7. What is the shape of a graph of ln V against t for a discharging capacitor?
   • A curve that levels off
   • A straight line through the origin with gradient RC
   • A horizontal line
   • A straight line with a negative gradient of -1/RC
   Show answer

   A straight line with a negative gradient of -1/RC. ln V = ln V_0 - t / RC is in the form y = c + mx.

8. The resistor in a discharge circuit is replaced by one with a larger resistance. What happens?
   • The capacitor discharges more quickly
   • The initial current increases
   • The capacitor discharges more slowly
   • The capacitor stores more charge
   Show answer

   The capacitor discharges more slowly. A larger R gives a larger RC and a smaller current, so discharge is slower.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 4 (WPH14) question paper, June 2024: list of data, formulae and relationships

Last updated 7 November 2026. Spotted a mistake? Tell us.