Capacitor charge and dischargeEdexcel A-Level Physics: Revision notes

When a charged capacitor discharges through a resistor $R$, the current is largest at the start and falls as the capacitor loses charge. The rate of discharge is proportional to the charge remaining, so the charge, current and potential difference all decay exponentially:

$Q = Q_0 e^{-t/RC} \qquad I = I_0 e^{-t/RC} \qquad V = V_0 e^{-t/RC}$

where $Q_0$, $I_0$ and $V_0$ are the values at $t = 0$, and $I_0 = V_0/R$. The graphs are curves that fall steeply at first and approach zero but never reach it.

The time constant $\tau = RC$ is the time for the charge, current or potential difference to fall to $1/e$ (about 37%) of its initial value. Its unit is the second, since $\Omega \times \text{F} = \text{s}$.

A large $R$ or $C$ gives a long time constant and a slow discharge. After a time $\tau$ the value is $0.37$ of the start, after $2\tau$ it is $0.14$ and after $5\tau$ it is less than 1%. The time for the quantity to halve is $0.69RC$.

When a capacitor charges through a resistor the charge and the p.d. across the capacitor rise from zero, steeply at first, and level off as they approach the supply voltage. The current in the circuit is largest at the start and falls exponentially to zero as the capacitor fills, so it follows the same $I = I_0 e^{-t/RC}$ form. The time constant for charging is also $RC$.

When the capacitor discharges, charge, current and p.d. all fall exponentially from their starting values.

Taking natural logarithms of $V = V_0 e^{-t/RC}$ gives

$\ln V = \ln V_0 - \frac{t}{RC}$

so a graph of $\ln V$ against $t$ is a straight line with gradient $-1/RC$ and intercept $\ln V_0$. The same applies to $\ln Q$ and $\ln I$. A straight line confirms that the decay is exponential, and the gradient gives the time constant.

Core practical: investigate the p.d. across a capacitor while it charges and discharges. Connect a supply, switch, resistor and capacitor in series with a data logger or voltmeter across the capacitor, and record $V$ at regular time intervals. The voltmeter must have a very high resistance so that it does not discharge the capacitor itself. Discharge the capacitor fully between runs.

A 470 μF capacitor charged to 12 V discharges through 10 kΩ.

Time constant: $RC = 10\times10^{3}\times 470\times10^{-6} = 4.7$ s.

Initial current: $I_0 = V_0/R = 12/10\,000 = 1.2$ mA.

P.d. after 6.0 s: $V = 12\,e^{-6.0/4.7} = 12 \times 0.279 = 3.3$ V.

Time to reach 2.0 V: $t = RC\ln(V_0/V) = 4.7\ln 6 = 8.4$ s.