<em>RC</em> Circuits

When you use a flash camera, it takes a few seconds to charge the capacitor that powers the flash. The light flash discharges the capacitor in a tiny fraction of a second. Why does charging take longer than discharging? This question and a number of other phenomena that involve charging and discharging capacitors are discussed in this module.

RC Circuits

An \(\mathbf{\text{RC}}\) circuit is one containing a resistor \(\phantom{\rule{0.25em}{0ex}}R\) and a capacitor \(C\). The capacitor is an electrical component that stores electric charge.

This figure shows a simple \(\text{RC}\) circuit that employs a DC (direct current) voltage source. The capacitor is initially uncharged. As soon as the switch is closed, current flows to and from the initially uncharged capacitor. As charge increases on the capacitor plates, there is increasing opposition to the flow of charge by the repulsion of like charges on each plate.

In terms of voltage, this is because voltage across the capacitor is given by \({V}_{\text{c}}=Q/C\), where \(Q\) is the amount of charge stored on each plate and \(C\) is the capacitance. This voltage opposes the battery, growing from zero to the maximum emf when fully charged. The current thus decreases from its initial value of \({I}_{0}=\cfrac{\text{emf}}{R}\) to zero as the voltage on the capacitor reaches the same value as the emf. When there is no current, there is no \(\text{IR}\) drop, and so the voltage on the capacitor must then equal the emf of the voltage source. This can also be explained with Kirchhoff’s second rule (the loop rule), discussed in Kirchhoff’s Rules, which says that the algebraic sum of changes in potential around any closed loop must be zero.

The initial current is \({I}_{0}=\cfrac{\text{emf}}{R}\), because all of the \(\text{IR}\) drop is in the resistance. Therefore, the smaller the resistance, the faster a given capacitor will be charged. Note that the internal resistance of the voltage source is included in \(R\), as are the resistances of the capacitor and the connecting wires. In the flash camera scenario above, when the batteries powering the camera begin to wear out, their internal resistance rises, reducing the current and lengthening the time it takes to get ready for the next flash.

(a) An \(\text{RC}\) circuit with an initially uncharged capacitor. Current flows in the direction shown (opposite of electron flow) as soon as the switch is closed. Mutual repulsion of like charges in the capacitor progressively slows the flow as the capacitor is charged, stopping the current when the capacitor is fully charged and \(Q=C\cdot \text{emf}\). (b) A graph of voltage across the capacitor versus time, with the switch closing at time \(t=0\). (Note that in the two parts of the figure, the capital script E stands for emf, \(q\) stands for the charge stored on the capacitor, and \(\tau \) is the \(\mathrm{RC}\) time constant.)

Voltage on the capacitor is initially zero and rises rapidly at first, since the initial current is a maximum. This figure (b) shows a graph of capacitor voltage versus time (\(t\)) starting when the switch is closed at \(t=0\). The voltage approaches emf asymptotically, since the closer it gets to emf the less current flows. The equation for voltage versus time when charging a capacitor \(C\) through a resistor \(R\), derived using calculus, is

\(V=\text{emf}(1-{e}^{-t/\text{RC}}) (\text{charging}),\)

where \(V\) is the voltage across the capacitor, emf is equal to the emf of the DC voltage source, and the exponential e = 2.718 … is the base of the natural logarithm. Note that the units of \(\text{RC}\) are seconds. We define

\(\tau =\text{RC},\)

where \(\tau \) (the Greek letter tau) is called the time constant for an \(\text{RC}\) circuit. As noted before, a small resistance \(R\) allows the capacitor to charge faster. This is reasonable, since a larger current flows through a smaller resistance. It is also reasonable that the smaller the capacitor \(C\), the less time needed to charge it. Both factors are contained in \(\tau =\text{RC}\).

More quantitatively, consider what happens when \(t=\tau =\text{RC}\). Then the voltage on the capacitor is

\(V=\text{emf}(1-{e}^{-1})=\text{emf}(1-0\text{.}\text{368})=0\text{.}\text{632}\cdot \text{emf}.\)

This means that in the time \(\tau =\text{RC}\), the voltage rises to 0.632 of its final value. The voltage will rise 0.632 of the remainder in the next time \(\tau \). It is a characteristic of the exponential function that the final value is never reached, but 0.632 of the remainder to that value is achieved in every time, \(\tau \). In just a few multiples of the time constant \(\tau \), then, the final value is very nearly achieved, as the graph in this figure (b) illustrates.