Discharging and Energy in RC Circuits

Disconnect the battery from a charged capacitor and connect the capacitor across a resistor instead. Charge flows from one plate, through the resistor, to the other, and the capacitor empties. The equation is even simpler than charging, and its solution has a striking property: a fixed "half-life".

Why it works

Around the loop the capacitor's voltage drives the current through the resistor: $\dfrac{q}{C} = IR$. The current drains the charge, $I = -\dfrac{dq}{dt}$, so $$\frac{dq}{dt} = -\frac{q}{RC}.$$ The charge drains at a rate proportional to how much is left. Whenever "rate is proportional to amount", the solution is an exponential, $q = Q_0e^{-t/RC}$, the same law as radioactive decay.

Key idea

$$q = Q_0e^{-t/RC}, \qquad V = V_0e^{-t/RC}, \qquad I = \frac{V_0}{R}e^{-t/RC}.$$ The charge halves every $t_{1/2} = RC\ln 2 \approx 0.693\,RC$, however much there is. All of the stored energy, $\tfrac{1}{2}CV_0^2$, ends up as heat in the resistor.

Discharging from two different starting voltages: both curves reach half their start at the same moment, $t_{1/2} = RC\ln 2$. The dotted curve is the stored energy, which halves twice as fast.

Worked examples

Worked example: Charge left after a while

A $100\ \mu\text{F}$ capacitor charged to $50\ \text{V}$ discharges through $10\ \text{k}\Omega$. How much charge is left after $2.0\ \text{s}$, and what is the current then?

$\tau = RC = (10^4)(10^{-4}) = 1.0\ \text{s}$ and $Q_0 = CV_0 = 5000\ \mu\text{C}$.
$q = 5000\,e^{-2.0} = 677\ \text{μC}$.
$I = \dfrac{V_0}{R}e^{-2.0} = (5.0\ \text{mA})(0.135) = 0.677\ \text{mA}$.

Answer: $q \approx 677\ \text{μC}$

Worked example: The half time

How long does it take the charge to halve, and then to fall to a quarter?

$e^{-t/RC} = \tfrac{1}{2}$ gives $t_{1/2} = RC\ln 2 = 0.693\ \text{s}$.
A quarter is half of half: another $0.693\ \text{s}$, so $1.39\ \text{s}$ in all.

Answer: $t_{1/2} \approx 0.693\ \text{s}$

Worked example: All the energy becomes heat

How much energy does the resistor turn into heat over the whole discharge?

$$\int_0^\infty I^2R\,dt = \frac{V_0^2}{R}\int_0^\infty e^{-2t/RC}\,dt = \frac{V_0^2}{R}\cdot\frac{RC}{2} = \tfrac{1}{2}CV_0^2$$
$\tfrac{1}{2}(100 \times 10^{-6})(50)^2 = 0.125\ \text{J}$: exactly the energy the capacitor started with.

Answer: $0.125\ \text{J}$

This gives a neat way to measure $RC$ in a lab. Take the natural log of the voltage: $\ln V = \ln V_0 - t/RC$. Plotted against time, $\ln V$ falls along a straight line with slope $-1/RC$. Two readings are enough in principle: $RC = \dfrac{t_2 - t_1}{\ln(V_1/V_2)}$, and the starting voltage cancels out.

Common mistake

"A capacitor with more charge takes longer to lose half of it." More charge means a higher voltage, which drives a bigger current, so it drains faster in exact proportion. The half time is $RC\ln 2$, with no $Q_0$ in it. Check with the visual: two capacitors starting at different voltages reach half their starting values together.

Discharging and Energy in RC Circuits

Worked examples

Practice

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Sources and further reading