RC Circuits

An RC circuit is simply a resistor (the R) and a capacitor (the C) wired together, so the capacitor fills or drains through the resistor like a bucket through a narrow pipe: fast at first, then slower and slower as it gets full. It's how circuits make delays and timers, and how they smooth out jittery signals. The equation $R\frac{dQ}{dt} + \frac{Q}{C} = \mathcal{E}$ looks like a lot, but it's just three voltages: $R\frac{dQ}{dt}$ (the resistor's share, bigger when charge flows faster), $\frac{Q}{C}$ (the capacitor pushing back, harder the fuller it gets) and $\mathcal{E}$ (the battery's push). The first two have to split the battery's voltage at every instant, and that's the whole equation. It's not as bad as it looks; bear with it.

A camera flash whines while it charges. Inside, a battery is filling a capacitor through a resistance, and the pitch shifts as the current tapers off. When the ready light comes on, the capacitor is nearly full. Defibrillators, blinking LEDs, the debounce on a keyboard switch and the timing of a pacemaker all run on the same slow fill and drain.

The equation, from Kirchhoff. Last lesson every circuit was steady, and KVL gave plain algebra. A capacitor changes that, because its voltage depends on how much charge has piled up so far. Close a switch on a battery $\mathcal{E}$, a resistor $R$ and an uncharged capacitor $C$ in series. Walk the loop: up through the battery ($+\mathcal{E}$), across the resistor ($-IR$), across the capacitor ($-Q/C$). The current is the rate charge piles onto the plates, $I = dQ/dt$, so

Now KVL hands us a differential equation instead of algebra.

Solving it. It's a linear first-order equation, the same kind that gave the electron's drift speed in the first lesson. Divide by $R$: $Q' + \frac{1}{RC}Q = \frac{\mathcal{E}}{R}$. The integrating factor is $e^{t/RC}$, and

Starting empty, $Q(0) = 0$ gives $K = -C\mathcal{E}$, so $Q = C\mathcal{E}\left(1 - e^{-t/RC}\right)$. Differentiate for the current: $I = \frac{\mathcal{E}}{R}e^{-t/RC}$.

Reading the solution. Think about why the current tapers off. Every bit of charge on the plates adds to the capacitor's voltage $Q/C$, which pushes back against the battery. The resistor only feels what's left over, $\mathcal{E} - Q/C$, so the fuller the capacitor gets, the less current flows, and the slower it fills. A rate that shrinks in proportion to what's left is exactly what an exponential does.

• At $t = 0$ the capacitor is empty and has no voltage, so the whole $\mathcal{E}$ sits across the resistor: $I = \mathcal{E}/R$. An uncharged capacitor acts like a wire.
• As $t \to \infty$ the capacitor reaches $Q = C\mathcal{E}$, its voltage matches the battery's, and the current stops. A full capacitor acts like a break.
• In between, everything moves by the same exponential with time constant $\tau = RC$. Ohms times farads is seconds: $\frac{\text{V}}{\text{A}}\cdot\frac{\text{C}}{\text{V}} = \frac{\text{C}}{\text{C/s}}$. After $\tau$ the capacitor is $1 - e^{-1} = 63\%$ charged, after $3\tau$ it's $95\%$, and after $5\tau$ it's over $99\%$, which engineers call done. The initial rate, $C\mathcal{E}/\tau$, would finish the job in exactly one $\tau$ if it kept going.

Discharging. Remove the battery and connect the charged capacitor across $R$: now $RQ' + Q/C = 0$, which separates to $Q = Q_0e^{-t/RC}$. The charge halves every $\tau\ln 2 \approx 0.69\tau$.

Every first-order circuit looks like this. Any voltage or current in a circuit with one capacitor (or, later, one inductor) moves as

so you only need three things: the starting value (the capacitor's voltage can't jump), the final value (treat the capacitor as a break) and $\tau = R_{\text{th}}C$, where $R_{\text{th}}$ is the resistance the capacitor sees with the sources turned off. You never have to solve the differential equation again. If this shape looks familiar, it should: it's the same "rate in minus rate out" picture as a mixing tank in Modeling with First Order Equations.

Where does the energy go? The battery pushes a total charge $C\mathcal{E}$ through a voltage $\mathcal{E}$, doing work $C\mathcal{E}^2$. The capacitor ends up storing $\frac12C\mathcal{E}^2$ (Capacitance and Stored Energy). The other half heats the resistor:

That's an improper integral, and notice its answer doesn't depend on $R$ at all. A smaller resistor charges faster with a bigger current and burns exactly the same energy.

A common mistake. At the moment a switch flips, it's tempting to think the capacitor's voltage jumps to its new value along with everything else. It can't: a sudden jump in $Q/C$ would take an infinite current. The capacitor's current can jump, but its voltage always starts where it left off. Two more slips to watch for: writing $\tau = R/C$ (check the units), and using the wrong $R$ when several resistors surround the capacitor.

Next, an inductor plays the same game with current instead of charge (Inductance and RL Circuits), and putting both together gives oscillations (LC and RLC Oscillations).

RC time constants appear wherever a circuit has to wait: the debounce filter that keeps one keypress from registering as five, the power-on reset that holds a microcontroller still until its supply settles, camera flashes and defibrillators that charge slowly and dump fast, and the low-pass filters that turn a microcontroller's on-off signal into a smooth analog voltage. Even a chip's speed is RC-limited: every wire and transistor gate is a tiny resistor charging a tiny capacitor, and $\tau$ sets how fast a bit can flip.