Charging a Capacitor: the RC Circuit

A capacitor in a loop

Until now every circuit was steady: currents that never changed. A capacitor changes that. The chapter Capacitance and Dielectrics showed that its voltage is \(q/C\), so its voltage grows as charge arrives. Charge a capacitor from a battery through a resistor, and the current must change as the capacitor fills. Such a circuit is an RC circuit, and its changing current is a transient.

Everything about the transient is set by one product, \(RC\). An ohm is a volt per ampere and a farad is a coulomb per volt. So an ohm times a farad is a coulomb per ampere, which is a second. The product \(RC\) is a time, the time constant \(\tau\).

A \(500\,\mu\text{F}\) capacitor charged through \(2\,\text{k}\Omega\) has \(\tau = 2 \times 10^3 \times 500 \times 10^{-6}\) \(= 1\,\text{s}\). The flash's \(1000\,\mu\text{F}\) capacitor, charged through about \(300\,\Omega\), has \(\tau = 0.3\,\text{s}\). A large resistance slows the flow of charge, and a large capacitance needs more charge. Both make the transient longer.

The loop equation

Close the switch at \(t = 0\) on a battery \(\mathcal E\), a resistor \(R\) and an empty capacitor \(C\). At a later instant the capacitor holds charge \(q\) and the current is \(i\). Walk round the loop with the current. The battery gives \(+\mathcal E\), the resistor \(-iR\), and the capacitor, entered at its positive plate, \(-q/C\). The current is the rate at which charge arrives, \(i = dq/dt\). So \[ \mathcal E - \dfrac{q}{C} - R\dfrac{dq}{dt} = 0 \]

The current is the slope of this curve. Differentiating \(e^{-t/\tau}\) brings down a factor \(-1/\tau\), and that minus sign meets the minus in front of the exponential. The two cancel, so the current is positive and falls away: \(i = C\mathcal E/\tau \times e^{-t/\tau}\) \(= (\mathcal E/R)e^{-t/\tau}\). Put \(q\) and \(i\) back in the loop equation: \(\mathcal E - \mathcal E(1 - e^{-t/\tau}) - \mathcal Ee^{-t/\tau}\) \(= 0\). The solution checks.

Watch Fig 5.1. At the first instant the capacitor is empty, so the whole \(12\,\text{V}\) is across the resistor and \(i = 12/2000 = 6\,\text{mA}\). As charge gathers, the capacitor's voltage rises and opposes the battery, so less is left to drive the current. The current dies away, and the charge creeps up towards \(C\mathcal E = 500\,\mu\text{F} \times 12\,\text{V} = 6\,\text{mC}\).

The tangent at \(t = 0\) in Fig 5.1 has slope \(\mathcal E/R\). Kept up, that first rate would fill the capacitor to \(C\mathcal E\) in a time \(C\mathcal E/(\mathcal E/R) = RC\). So \(\tau\) is the time the capacitor would take if the current did not fall. It does fall, so at \(t = \tau\) only \(1 - 1/e = 63\,\%\) is in. After \(2\tau\) it is \(86\,\%\), after \(3\tau\) it is \(95\,\%\).

The capacitor is half full when \(e^{-t/\tau} = 1/2\), at \(t = \tau\ln 2 = 0.69\tau\). It reaches \(99\,\%\) when \(e^{-t/\tau} = 0.01\), at \(t = \tau\ln 100 = 4.6\tau\). In theory it is never quite full. In practice, five time constants are "full" for every purpose.

The two voltages always add to the emf

At every instant the loop rule says \(V_R + V_C = \mathcal E\). The two trade places. With \(10\,\text{V}\) charging a capacitor, the resistor has \(10\,\text{V}\) and the capacitor \(0\) at \(t = 0\). At \(t = \tau\), the capacitor has \(10 \times 0.63 = 6.3\,\text{V}\) and the resistor \(3.7\,\text{V}\). After a long time the capacitor has all \(10\,\text{V}\) and the resistor none.

That pays the promise of the flash. With \(\tau\) about \(0.3\,\text{s}\), the capacitor is \(90\,\%\) full in about \(0.7\,\text{s}\) and \(99\,\%\) full only after about \(1.4\,\text{s}\). The last few percent come slowly, because the current dies as the capacitor fills. So the maker lights the ready lamp early, and the flash is nearly as bright.

A wire at first, a gap at the end

Many questions ask only about two instants. Just after the switch closes, the capacitor has no charge, so its voltage is zero. A component with zero voltage across it, whatever current it carries, behaves as a plain wire. After a long time the capacitor is full and no more current enters it. A component that carries no current, whatever voltage it holds, behaves as a gap.

In Fig 5.2, just after the switch closes, the capacitor shorts the \(4\,\Omega\) resistor. The battery sees only \(2\,\Omega\) and drives \(6\,\text{A}\), all of it into the capacitor. After a long time the capacitor branch is cut, and the two resistors are in series: \(12/6 = 2\,\text{A}\). The capacitor sits across the \(4\,\Omega\), so it ends at \(2 \times 4 = 8\,\text{V}\), holding \(3\,\mu\text{F} \times 8\,\text{V} = 24\,\mu\text{C}\).

The first current does not depend on \(C\): at \(t = 0\) the capacitor has no voltage, so it plays no part yet. The final charge does not depend on \(R\) in series with it: when the current has stopped, no resistor has any voltage across it. The resistance and the capacitance together decide only how long the change takes.

Half the energy is always lost

The battery moves the whole charge \(C\mathcal E\) through its full emf, so it does work \(W = C\mathcal E \times \mathcal E = C\mathcal E^2\). The capacitor ends with \(\tfrac12 C\mathcal E^2\), from the chapter Capacitance and Dielectrics. The rest, the other half, must have become heat in the resistor. For the \(500\,\mu\text{F}\) capacitor on \(12\,\text{V}\), the battery gives \(72\,\text{mJ}\), the capacitor keeps \(36\,\text{mJ}\) and the resistor takes \(36\,\text{mJ}\).

A small resistance lets a large current flow for a short time; a large one, a small current for a long time. The integral shows the two effects cancel exactly. The chapter Capacitance and Dielectrics promised this: whatever the resistance, charging from a battery wastes half the battery's work.