Capacitor charge and dischargeAQA A-Level Physics: Revision notes

When a capacitor discharges through a resistor, charge Q, p.d. V and current I all fall exponentially: the graph starts steeply and flattens, with each quantity falling by the same fraction in equal times.

When a capacitor charges through a resistor from a supply, Q and V rise exponentially towards a maximum (Q₀ and V₀), while the current falls exponentially from V₀/R towards zero.

On the graphs:

• the gradient of a Q–t graph is the current I;
• the area under an I–t graph is the charge Q that has flowed.

The time constant is

$\tau = RC$

with R in ohms and C in farads giving seconds. During discharge it is the time for Q, V or I to fall to 1/e (about 37%) of its initial value. During charging it is the time for Q or V to rise to about 63% of its final value.

A larger R or C makes the exponential change slower.

Worked example. R = 47 kΩ, C = 220 µF: $RC = 47 \times 10^3 \times 220 \times 10^{-6} = 10.3$ s.

For a capacitor discharging through a resistor:

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

where $I_0 = V_0/R$ is the initial current. All three have the same form because Q = CV and I = V/R.

Worked example. C = 470 µF, R = 10 kΩ, V₀ = 9.0 V. RC = 4.7 s. At t = 10 s: $V = 9.0\,e^{-10/4.7} = 1.1$ V. The initial current is 9.0 / 10 000 = 9.0 × 10⁻⁴ A, and at t = 10 s it is $9.0 \times 10^{-4} \times e^{-10/4.7} = 1.1 \times 10^{-4}$ A.

For a capacitor charging from a supply of p.d. V₀ through a resistor R:

$Q = Q_0\left(1 - e^{-t/RC}\right), \quad V = V_0\left(1 - e^{-t/RC}\right), \quad I = I_0 e^{-t/RC}$

where Q₀ = CV₀ and I₀ = V₀/R. The p.d. across the resistor is V₀ − V, which falls as the capacitor charges.

Worked example. C = 1000 µF, R = 20 kΩ, V₀ = 6.0 V, so RC = 20 s. At t = 30 s: $V = 6.0(1 - e^{-1.5}) = 6.0 \times 0.777 = 4.7$ V.

The time for Q, V or I to fall to half its value is $T_{1/2}$. Setting $V = V_0/2$ in $V = V_0 e^{-t/RC}$ gives $e^{-T/RC} = \tfrac{1}{2}$, so

$T_{1/2} = RC \ln 2 = 0.69\,RC$

Worked example. RC = 10 s gives $T_{1/2} = 6.9$ s. The p.d. falls to 1/4 after two half-lives (13.8 s).

Method. Charge a capacitor, then discharge it through a resistor. Record V at regular time intervals using a voltmeter (a data logger or a stopwatch). Use a high-resistance voltmeter so that it does not discharge the capacitor, and a long time constant so timing errors are a small fraction of each reading.

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

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

A graph of ln V against t is a straight line with gradient −1/RC and intercept ln V₀. The time constant is RC = −1/gradient, and C = RC/R. A straight line confirms the exponential law, and a best-fit line uses all the data.