Young & Freedman · Chapter 26.4
Capacitors, resistors, time constants, and exponential dynamics — built from the ground up.
Before any equations, here is the physical picture. A capacitor stores charge on two parallel plates separated by a gap — it's like a rechargeable tank of charge. A resistor limits how fast charge can flow in or out — it's like a narrow pipe.
When you open a valve (close the switch), water flows fast at first — the pressure is high. But as the tank fills, pressure drops and flow slows. Eventually flow stops entirely. That exponential slowdown is the RC circuit's signature behavior.
The crucial insight: the driving force decreases as the capacitor fills. This self-limiting nature is exactly why the math produces an exponential, not a linear, curve.
An RC circuit in its simplest charging form has a battery (EMF $\mathcal{E}$), a resistor $R$, a capacitor $C$, and a switch, all in series.
Once the switch is closed, we apply Kirchhoff's Voltage Law (KVL): the sum of all voltage drops around the loop equals zero.
Going around the loop, we encounter three voltage contributions:
This is the master equation. Everything follows from it.
Now we make one substitution to turn this into a differential equation. The current $I$ is simply the rate at which charge accumulates on the capacitor:
Substituting into KVL:
Rearranging to isolate the derivative:
This is a first-order linear ODE. Notice: the rate of change of charge depends on how much charge is already there. That coupling between $Q$ and $dQ/dt$ is the reason the solution is exponential.
Initial condition: at $t = 0$, the capacitor is uncharged, so $Q(0) = 0$.
We need to solve:
From $Q(t)$, we can immediately get the current by differentiating:
where $I_0 = \mathcal{E}/R$ is the initial current (the maximum, at $t=0$), limited only by the resistor since the capacitor starts with zero voltage.
And the voltage across the capacitor:
The combination $RC$ appears so often we give it its own symbol:
$\tau$ has units of seconds (check: $[\Omega][\text{F}] = [\text{V/A}][\text{C/V}] = [\text{C/A}] = [\text{s}]$). It is the characteristic timescale of the circuit.
With $\tau$, the equations become beautifully clean:
At $t = \tau$:
So after one time constant, the capacitor has charged to 63.2% of its final value.
| Time elapsed | Charge Q / Cℰ | Current I / I₀ |
|---|---|---|
| $t = 0$ | $0\%$ | $100\%$ |
| $t = \tau$ | $63.2\%$ | $36.8\%$ |
| $t = 2\tau$ | $86.5\%$ | $13.5\%$ |
| $t = 3\tau$ | $95.0\%$ | $5.0\%$ |
| $t = 5\tau$ | $99.3\%$ | $0.7\%$ |
| $t \to \infty$ | $100\%$ | $0\%$ |
In practice, the circuit is considered "fully charged" after about $5\tau$.
Now we disconnect the battery (or the battery was never present). The capacitor, already holding charge $Q_0 = C\mathcal{E}$, now drives current through the resistor to discharge.
KVL with no battery: There's no EMF source; the only voltage source is the capacitor itself. The voltage it supplies is $Q/C$, and the resistor drops $IR$:
Since charge is leaving the capacitor, $I = -dQ/dt$ (the minus sign: as $Q$ decreases, positive current flows). Substituting:
The current during discharge (positive, flowing in the conventional direction):
And the voltage across the capacitor:
where $V_0 = Q_0/C$ is the initial voltage. Notice discharge is a pure exponential decay — the same $\tau = RC$ governs both charging and discharging.
A charged capacitor stores energy in its electric field. The energy stored when it holds charge $Q$ at voltage $V = Q/C$ is:
During charging from 0 to $Q_f = C\mathcal{E}$, the battery delivers total energy:
But the energy stored in the capacitor is only:
Exactly half the energy the battery delivers is lost to heat in the resistor. You can verify this by integrating the power dissipated in $R$:
This result is independent of R — no matter how large or small the resistor, always exactly half the battery energy is dissipated. (With a smaller $R$, current is larger but the process finishes faster; it balances out.)
Adjust the sliders to explore how $R$, $C$, and $\mathcal{E}$ affect the charging and discharging curves. The time constant $\tau = RC$ updates live.
| Quantity | Charging | Discharging |
|---|---|---|
| $Q(t)$ | $C\mathcal{E}(1 - e^{-t/\tau})$ | $Q_0\,e^{-t/\tau}$ |
| $V_C(t)$ | $\mathcal{E}(1 - e^{-t/\tau})$ | $V_0\,e^{-t/\tau}$ |
| $I(t)$ | $\dfrac{\mathcal{E}}{R}e^{-t/\tau}$ | $\dfrac{V_0}{R}e^{-t/\tau}$ |
| At $t = 0$ | $Q=0$, $I=\mathcal{E}/R$ (max) | $Q=Q_0$, $I=V_0/R$ (max) |
| At $t \to \infty$ | $Q=C\mathcal{E}$, $I=0$ | $Q=0$, $I=0$ |
| Time constant | $\tau = RC$ | |
| ODE source | KVL around the loop | |
| Math structure | First-order linear ODE → exponential solution | |
An RC circuit charges (or discharges) exponentially because KVL produces a differential equation of the form $dQ/dt \propto -Q$, whose solution is always $e^{-t/\tau}$ with the time constant $\tau = RC$ setting the speed.