RC Circuits & Transients

A camera flash takes a couple of seconds to charge and then dumps everything it has in about a millisecond. Same capacitor, same energy, two wildly different timescales. What changed was the resistance the charge had to squeeze through on its way in and on its way out.

That is the whole subject. In Kirchhoff's Laws you solved networks where every current was fixed the moment the switch closed and stayed there forever. Put a capacitor in and that stops being true: the circuit has a memory, and the currents drift from one set of values to another over a time you can predict.

The capacitor's one rule

A capacitor stores charge $Q$ on its plates and, as a result, holds a voltage across itself:

$V_C = \frac{Q}{C}$

$C$ is the capacitance in farads, and a farad is a coulomb per volt. A microfarad, $1\ \mu\mathrm{F} = 10^{-6}\ \mathrm{F}$, is the size you will actually meet.

The useful way to read that equation is backwards. Charge is what you can move, and moving charge takes current, and current takes time. So $V_C$ cannot jump. To change the capacitor's voltage by $\Delta V$ you have to deliver $\Delta Q = C\,\Delta V$ of charge, and if the current is finite that takes a finite time. Every other voltage in the circuit can change instantly when you flip a switch. The capacitor's cannot.

That single fact is worth more than the formulas that follow. Hold onto it.

Why the response is an exponential

Take the simplest loop: a battery of emf $\mathcal{E}$, a resistor $R$, a capacitor $C$, and a switch you close at $t = 0$ with the capacitor empty. Go around the loop the way you learned:

$\mathcal{E} = IR + \frac{Q}{C}$

The current in the loop is charge arriving at the capacitor, so $I = dQ/dt$. Substitute:

$\mathcal{E} = R\frac{dQ}{dt} + \frac{Q}{C}$

Rearrange to isolate the rate:

$\frac{dQ}{dt} = \frac{1}{RC}\left(C\mathcal{E} - Q\right)$

Look at what that says before you solve it. The final charge, once nothing is changing, is $Q_\infty = C\mathcal{E}$. The bracket is how far the capacitor still has to go. So the rate of change is proportional to the remaining gap: far from done, fast; nearly done, slow. A quantity whose rate of approach is proportional to its remaining distance is an exponential approach, always. Nothing else has that property.

Solving it (separate and integrate, $\int dQ/(C\mathcal{E}-Q) = \int dt/RC$) gives

$Q(t) = C\mathcal{E}\left(1 - e^{-t/RC}\right), \qquad V_C(t) = \mathcal{E}\left(1 - e^{-t/RC}\right)$

and differentiating once returns the current:

$I(t) = \frac{\mathcal{E}}{R}e^{-t/RC}$

At $t = 0$ the capacitor has no voltage, so the resistor takes the whole emf and the current is the largest it will ever be, $\mathcal{E}/R$. As charge piles up, the capacitor's voltage eats into the emf, less is left for the resistor, and the current dies.

For a capacitor with initial charge $Q_0$ discharging through $R$ with no battery in the loop, the same algebra with $\mathcal{E} = 0$ gives

$Q(t) = Q_0 e^{-t/RC}, \qquad V_C(t) = V_0 e^{-t/RC}, \qquad I(t) = \frac{V_0}{R}e^{-t/RC}$

The current decays identically in both cases. Only the charge differs, because charging climbs to something and discharging falls to nothing.

The time constant does the thinking

The product

$\tau = RC$

has units of seconds. Check it: ohms times farads is (volts per amp) times (coulombs per volt), which is coulombs per amp, which is seconds. It is the only timescale in the problem, and it is the number to compute first in any transient question.

Why the product, rather than a ratio? Bigger $R$ means less current for the same voltage, so charge arrives more slowly. Bigger $C$ means more charge is needed to reach a given voltage. Both stretch the process, so both belong upstairs.

Numbers worth memorising. After one $\tau$ the gap to the final value has fallen to $e^{-1} = 0.368$ of what it was, so a charging capacitor is 63 percent of the way there. After three, $e^{-3} = 0.0498$, about 5 percent remains. After five, under 1 percent.

Run the logic backwards to solve for a time. To reach a fraction $f$ of the final value while charging, you need $1 - e^{-t/\tau} = f$, so

$t = -\tau\ln(1 - f)$

To fall to a fraction $f$ while discharging, $t = -\tau\ln f = \tau\ln(1/f)$. The half-life of a discharge is $t_{1/2} = \tau\ln 2 \approx 0.693\,\tau$.

Worked example 1. A $22\ \mu\mathrm{F}$ capacitor is charged and then discharged through a $10\ \mathrm{k\Omega}$ resistor. How long until its voltage is half the starting value?

First the time constant:

$\tau = RC = (10 \times 10^{3}\ \Omega)(22 \times 10^{-6}\ \mathrm{F}) = 0.22\ \mathrm{s}$

Then the half-life:

$t = \tau\ln 2 = (0.22)(0.6931) = 0.1525\ \mathrm{s}$

Notice the starting voltage never entered. Every exponential decay halves in the same time no matter where you start it, which is why the question could leave that number out.

The two limits answer most questions

Before you reach for $e^{-t/\tau}$, check whether the question only asks about $t = 0^+$ or $t \to \infty$. It usually does, and those two cases need no calculus at all, just two resistive circuits you already know how to solve.

Immediately after the switch closes, the capacitor's voltage is still whatever it was a moment earlier. If it started empty, that voltage is zero, and a component with zero volts across it is a plain wire. Redraw the circuit with the capacitor replaced by a short and solve.

After a long time, nothing is changing, so $dQ/dt = 0$, so no current flows in the capacitor's branch. A branch carrying no current is a break. Redraw with the capacitor replaced by an open gap and solve.

Everything in between slides exponentially from the first answer to the second.

Worked example 2. A $12\ \mathrm{V}$ battery feeds node $A$ through $R_1 = 6.0\ \mathrm{k\Omega}$. A resistor $R_2 = 3.0\ \mathrm{k\Omega}$ runs from $A$ to ground, and an uncharged $100\ \mu\mathrm{F}$ capacitor also runs from $A$ to ground. Find the battery current just after the switch closes, the capacitor voltage long after, and the time constant.

Just after. The capacitor is empty, so node $A$ is pinned to ground through a wire. $R_2$ is shorted out and irrelevant. The battery sees only $R_1$:

$I(0^+) = \frac{12\ \mathrm{V}}{6.0 \times 10^{3}\ \Omega} = 2.0 \times 10^{-3}\ \mathrm{A}$

Long after. The capacitor is a gap, so the current runs straight through $R_1$ and $R_2$ in series. That is a voltage divider:

$V_A = 12\,\frac{3.0}{6.0 + 3.0} = 4.0\ \mathrm{V}$

and the capacitor, sitting from $A$ to ground, holds that same 4.0 V.

The time constant. This is where people slip. The resistance in $\tau$ is not $R_1$, and it is not $R_2$. It is the resistance the capacitor sees looking back into the circuit from its own two terminals, with the sources killed: batteries replaced by wires, since an ideal battery holds its terminals at a fixed difference and so offers no resistance.

Kill the 12 V source and the top of $R_1$ is connected straight to ground. Now $R_1$ and $R_2$ both run from node $A$ to ground: they are in parallel.

$R_{\text{th}} = \frac{R_1R_2}{R_1 + R_2} = \frac{(6.0)(3.0)}{9.0} = 2.0\ \mathrm{k\Omega}$

$\tau = R_{\text{th}}C = (2.0 \times 10^{3})(100 \times 10^{-6}) = 0.20\ \mathrm{s}$

So $V_C(t) = 4.0\left(1 - e^{-t/0.20}\right)$ volts, and you got every piece of it from two resistive sketches plus one parallel combination.

Finding the R the capacitor sees

The recipe generalises, and it is the only technique you need for messy networks.

1. Pull the capacitor out, leaving two terminals.
2. Kill every source: short the ideal batteries, open the ideal current sources.
3. Compute the resistance between those two terminals with series and parallel rules.
4. That resistance times $C$ is $\tau$. Put the capacitor back and find the final voltage from the ordinary steady-state circuit.

A worked case from the problem bank. A 12 V battery drives two parallel branches: $R_1 = 1.0\ \mathrm{k\Omega}$ down to node $A$ then $R_2 = 1.0\ \mathrm{k\Omega}$ to the negative terminal, and $R_3 = 2.0\ \mathrm{k\Omega}$ down to node $B$ then $R_4 = 3.0\ \mathrm{k\Omega}$ to the negative terminal. A capacitor bridges $A$ and $B$.

In the final state the capacitor carries no current, so each branch is just a divider on its own. $V_A = 12(1.0/2.0) = 6.0\ \mathrm{V}$ and $V_B = 12(3.0/5.0) = 7.2\ \mathrm{V}$, so the capacitor settles at $|V_A - V_B| = 1.2\ \mathrm{V}$.

For $\tau$, short the battery. Then $R_1$ and $R_2$ both run from $A$ to the same node, giving $0.5\ \mathrm{k\Omega}$, and $R_3$ and $R_4$ both run from $B$ to that node, giving $(2)(3)/5 = 1.2\ \mathrm{k\Omega}$. Walking from terminal $A$ to terminal $B$ you pass through both in series:

$R_{\text{th}} = 0.5 + 1.2 = 1.7\ \mathrm{k\Omega}$

With $C = 47\ \mu\mathrm{F}$, $\tau = (1.7\times10^{3})(47\times10^{-6}) = 0.0799\ \mathrm{s}$.

Capacitors in the same loop combine before you do any of this. Two capacitors in series store the same charge and their voltages add, so $1/C_{\text{eq}} = 1/C_1 + 1/C_2$; in parallel they share a voltage and their charges add, so $C_{\text{eq}} = C_1 + C_2$. A $10\ \mu\mathrm{F}$ and a $15\ \mu\mathrm{F}$ in series give $(10)(15)/25 = 6.0\ \mu\mathrm{F}$, and with a $100\ \mathrm{k\Omega}$ resistor that is $\tau = 0.60\ \mathrm{s}$. Note that this is the opposite of the resistor rules, which catches people out constantly.

Energy, heat, and the factor of one half

A charged capacitor stores

$U = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}$

A camera flash with $120\ \mu\mathrm{F}$ at $350\ \mathrm{V}$ holds $\tfrac12(120\times10^{-6})(350)^2 = 7.35\ \mathrm{J}$.

Now charge that capacitor from a battery through a resistor and count the energy. The battery pushes a total charge $Q = C\mathcal{E}$ around the loop at emf $\mathcal{E}$, so it does work $Q\mathcal{E} = C\mathcal{E}^2$. The capacitor ends up with $\tfrac12 C\mathcal{E}^2$. The missing half went into the resistor:

$W_R = C\mathcal{E}^2 - \tfrac{1}{2}C\mathcal{E}^2 = \tfrac{1}{2}C\mathcal{E}^2$

Exactly half the battery's work is wasted as heat, whatever $R$ is. Making the resistor smaller makes the charging faster and the current bigger, and those two effects cancel perfectly. For $470\ \mu\mathrm{F}$ charged by $9.0\ \mathrm{V}$, that heat is $\tfrac12(470\times10^{-6})(81) = 0.0190\ \mathrm{J}$.

If you want the heat over part of the process, integrate the power. With $I = (\mathcal{E}/R)e^{-t/\tau}$,

$P_R = I^2R = \frac{\mathcal{E}^2}{R}e^{-2t/\tau}$

The instantaneous power falls twice as fast as the current, because it goes as the square. Integrating from 0 to a time $T$:

$W_R = \frac{\mathcal{E}^2}{R}\cdot\frac{\tau}{2}\left(1 - e^{-2T/\tau}\right) = \tfrac{1}{2}C\mathcal{E}^2\left(1 - e^{-2T/\tau}\right)$

using $\tau = RC$ to cancel the $R$. The same expression with $\mathcal{E}$ replaced by $V_0$ gives the energy delivered by a discharge in its first $T$ seconds.

Worked example 3. A $200\ \mu\mathrm{F}$ capacitor is charged from zero through a resistor by a $50\ \mathrm{V}$ battery. How much energy is dissipated in the resistor between $t = 0$ and $t = \tau$?

Put $T = \tau$ in the result above, so the exponent is $-2$:

$W_R = \tfrac{1}{2}(200\times10^{-6})(50)^2\left(1 - e^{-2}\right)$

$\tfrac{1}{2}(200\times10^{-6})(2500) = 0.250\ \mathrm{J}, \qquad 1 - e^{-2} = 1 - 0.1353 = 0.8647$

$W_R = (0.250)(0.8647) = 0.216\ \mathrm{J}$

So 86 percent of the waste heat appears in the first time constant. The resistance never appeared, and it could not have: the answer depends only on how far along the process you are, and $\tau$ measures exactly that.

When two capacitors share charge

Connect a charged capacitor $C_1$ at $V_0$ through a resistor to an uncharged $C_2$ in a single loop. Current flows until the loop has no net driving voltage left.

For the time constant, treat the two capacitors as being in series around that loop (the same current passes through both, delivering the same charge to each) so

$\tau = R\,\frac{C_1C_2}{C_1 + C_2}$

For $C_1 = 12\ \mu\mathrm{F}$, $C_2 = 4.0\ \mu\mathrm{F}$, $R = 2.2\ \mathrm{k\Omega}$: the series combination is $(12)(4)/16 = 3.0\ \mu\mathrm{F}$, so $\tau = (2200)(3.0\times10^{-6}) = 0.0066\ \mathrm{s}$.

For the energy lost, do not chase the integral. Let $q$ be the charge that has moved when the current finally stops. At that moment the loop voltage is zero, so $V_0 - q/C_1 - q/C_2 = 0$, giving $q = C_s V_0$ where $C_s$ is that same series combination. The energy dissipated is the drop in stored energy, and grinding through the algebra leaves a clean result:

$W_R = \tfrac{1}{2}C_s\,(\Delta V_0)^2$

where $\Delta V_0$ is the initial voltage mismatch around the loop. Two $10\ \mu\mathrm{F}$ capacitors, one at 100 V and one empty, have $C_s = 5.0\ \mu\mathrm{F}$ and $\Delta V_0 = 100\ \mathrm{V}$, so $W_R = \tfrac12(5.0\times10^{-6})(10^4) = 0.025\ \mathrm{J}$. A quarter of the original 0.05 J is gone, and the resistance does not appear. Replace the wire with a better conductor and you lose exactly the same energy in a shorter, fiercer pulse.

Worked example 4. An uncharged $100\ \mu\mathrm{F}$ capacitor charges through a $10\ \mathrm{k\Omega}$ resistor from a $20\ \mathrm{V}$ battery. At $t = 1.0\ \mathrm{s}$ the battery is swapped for a plain wire, so the capacitor now discharges through the same resistor. Find $V_C$ at $t = 2.0\ \mathrm{s}$.

Break it at the switch-over. The time constant is the same in both phases:

$\tau = (10\times10^{3})(100\times10^{-6}) = 1.0\ \mathrm{s}$

Phase one, charging for one full time constant:

$V_C(1.0) = 20\left(1 - e^{-1.0/1.0}\right) = 20(1 - 0.3679) = 20(0.6321) = 12.64\ \mathrm{V}$

Phase two. The capacitor's voltage cannot jump, so 12.64 V is the starting value for the discharge. Reset the clock: one more second elapses, which is again one time constant.

$V_C(2.0) = 12.64\,e^{-1.0/1.0} = (12.64)(0.3679) = 4.65\ \mathrm{V}$

Two exponentials, each handled on its own, stitched together by the one thing that carries across the switch: the capacitor voltage. That is the pattern for every multi-stage transient. Find the voltage at the moment of the change, then start a fresh problem with that as the initial condition and whatever new $\tau$ and final value the new circuit has.

Recognising the technique

Reach for this machinery when you see any of these.

A capacitor plus the words "switch", "at $t=0$", "just after", or "a long time". That is a transient question, and the two-limit sketches come first. If the question only asks about one of those two moments, you are done without an exponential.

A number that is a fraction of a final value (90 percent charged, down to a tenth, within 1 percent) means you want $t = -\tau\ln(1-f)$ or $t = \tau\ln(1/f)$. Settling to 1 percent of final takes $-\ln(0.01) = 4.6$ time constants.

More than one resistor around a capacitor. Do not guess which one sets $\tau$. Kill the sources and find the resistance across the capacitor's terminals. If a question gives you a network and asks for a time constant, that is the whole point of the question.

The words energy, heat, or power. Then you are in $e^{-2t/\tau}$ territory, and often you can skip the integral entirely with $\tfrac12 CV^2$ before and after.

Two capacitors connected to each other. Conserve charge, find the final common state, and use $\tfrac12C_s\Delta V^2$ for the heat. The resistor's value will not matter, and a question that omits it is telling you so.

A capacitor turns an algebra problem into the same algebra problem plus a clock. Solve the circuit at the start, solve it at the end, and $\tau$ fills in the middle. Next you will meet networks where even the steady-state algebra fights back (bridges, ladders, and symmetry tricks) in Tricky Networks.