Charging & Discharging Capacitors
Exponential filling and emptying, paced by CR.
Lesson at a glanceShow the key points
- Charging: q = ℰC(1 − e^(−t/CR)) (32.16); current (ℰ/R)e^(−t/CR).
- Discharging: q = Qe^(−t/CR) (32.17).
- Time constant τ = CR: 63 % after one τ; nearly complete after 4–5τ.
A capacitor joined to a battery through a resistor doesn’t fill instantly. The charge creeps up along an exponential curve whose pace is set by one number, CR.
Charging
A capacitor C, a cell ℰ and a resistance R (the wires plus the cell’s internal resistance) are joined in series at t = 0. At time t the capacitor holds q and the current is i = dq/dt. Round the loop:
The product CR has the dimensions of time; it is the time constant τ. The final charge ℰC is reached only as t → ∞, but after one time constant q = ℰC(1 − 1/e) = 0.63 ℰC.
A 47 µF capacitor charges through 10 kΩ from a 9 V cell. Find the time constant and the time to reach 90 % of full charge.
Discharging
A capacitor holding Q is joined across a resistance R at t = 0. Now the charge falls, so the current is i = −dq/dt, and the loop law gives q/C − Ri = 0, i.e. R dq/dt = −q/C:
After one time constant 37 % (1/e) of the charge is left, so 63 % of the discharge is done. Again it is complete only at t → ∞, but four or five time constants leave less than 2 %.
A 200 µF capacitor holding 1.0 mC is discharged through 5.0 kΩ. How much charge remains after 3.0 s?
- 1Charging: q = ℰC(1 − e^(−t/CR)) (32.16); current ℰ/R e^(−t/CR).
- 2Discharging: q = Q e^(−t/CR) (32.17).
- 3Time constant τ = CR: 63 % done after one τ; ~99 % after about 5τ.
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Keep exploring
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