Put a resistor and a capacitor in a loop with a battery and the current changes with time: large at first, then fading away as the capacitor fills. The loop rule explains it, and the answer is the exponential decay you know from algebra, with the capacitance setting the clock.
Why it works
The loop rule says $\varepsilon = IR + \dfrac{q}{C}$ at every moment. At the start $q = 0$, so $I = \varepsilon/R$: the empty capacitor acts like a plain wire. As charge builds, the capacitor's voltage $q/C$ pushes back, so less voltage is left for the resistor and the current drops. The charging rate is proportional to how far the capacitor still has to go, so the gap shrinks by the same fraction in each equal time: exponential decay, with time constant $\tau = RC$.
Key idea
$\tau = RC$. Charging: $V_C = \varepsilon\left(1 - e^{-t/RC}\right)$, $I = \dfrac{\varepsilon}{R}e^{-t/RC}$. Discharging: $V_C = V_0e^{-t/RC}$. At $t = 0$ an empty capacitor acts like a wire; after many $\tau$ a full capacitor blocks steady current.
| Time | $\tau$ | $2\tau$ | $3\tau$ | $5\tau$ |
|---|---|---|---|---|
| Fraction charged, $1 - e^{-t/\tau}$ | 63% | 86% | 95% | 99.3% |
| Current left, $e^{-t/\tau}$ | 37% | 14% | 5% | 0.7% |
Worked examples
Worked example: Time constant and first current
A $10\,\text{k}\Omega$ resistor, a $100\,\mu\text{F}$ capacitor (empty) and a $9.0\,\text{V}$ battery are switched into a loop. Find $\tau$ and the starting current.
- $\tau = RC = (10 \times 10^{3}\,\Omega)(100 \times 10^{-6}\,\text{F}) = 1.0\,\text{s}$.
- At $t = 0$ the capacitor acts like a wire: $I_0 = \dfrac{9.0\,\text{V}}{10{,}000\,\Omega} = 0.90\,\text{mA}$.
Answer: $1.0\,\text{s}$; $0.90\,\text{mA}$
Worked example: Partly charged
In the same circuit, what is the capacitor's voltage after $2.0\,\text{s}$?
- $t = 2\tau$, so $V_C = 9.0\left(1 - e^{-2}\right) = 9.0(1 - 0.135)$.
- $V_C = 7.78\,\text{V}$, about $86\%$ of the way.
Answer: $7.78\,\text{V}$
Worked example: Early and late
A $12\,\text{V}$ battery drives $R_1 = 4\,\Omega$ in series with $R_2 = 8\,\Omega$, and an empty capacitor sits in parallel with $R_2$. Find the current through $R_1$ just after closing the switch and a long time later.
- Just after: the empty capacitor acts like a wire across $R_2$, so only $R_1$ limits the current: $I = 12/4 = 3.0\,\text{A}$.
- Long after: the full capacitor passes no current, so $R_1$ and $R_2$ are in series: $I = 12/12 = 1.0\,\text{A}$.
Answer: $3.0\,\text{A}$, then $1.0\,\text{A}$
Common mistake
Thinking an empty capacitor blocks current the moment the switch closes. It does the opposite: with no charge it has no voltage, so at first it acts like a plain wire and the current is largest. Only once it is full does it act like a break. Check: use $\varepsilon = IR + q/C$ at the moment in question.