DC Circuits Containing Resistors and Capacitors

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 RC circuit is one containing a resistor R and a capacitor C. The capacitor is an electrical component that stores electric charge.

Figure 21.38 shows a simple 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 $V_{c}=Q/C$, 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 $I_{o}=\frac{emf}{R}$ to zero as the voltage on the capacitor reaches the same value as the emf. When there is no current, there is no $IR$ 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 $I_{o}=\frac{emf}{R}$, because all of the $IR$ 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.

Figure 21.38 (a) An $RC$ 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 $Q=C$. (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 $r$ is the $RC$ time constant.)

Voltage on the capacitor is initially zero and rises rapidly at first, since the initial current is a maximum. Figure 21.38(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

$V=emf(1-e^{\frac{-t}{RC}})(charging)$ [Equation 21.77]

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 $RC$ are seconds. We define

$\tau =RC$ [Equation 21.78]

where $\tau$ (the Greek letter tau) is called the time constant for an $RC$ 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 $\tau =RC$.

More quantitatively, consider what happens when $t=\tau =RC$. Then the voltage on the capacitor is

$V=emf(1-e^{-1})=emf(1-0.368)=0.632\cdot emf$ [Equation 21.79]

This means that in the time $\tau =RC$, 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 Figure 21.38(b) illustrates.

Discharging a Capacitor

Discharging a capacitor through a resistor proceeds in a similar fashion, as Figure 21.39 illustrates. Initially, the current is $I_{0}=\frac{V_{0}}{R}$, driven by the initial voltage $V_{0}$ on the capacitor. As the voltage decreases, the current and hence the rate of discharge decreases, implying another exponential formula for $V$. Using calculus, the voltage $V$ on a capacitor $C$ being discharged through a resistor $R$ is found to be

$V=V_{0}e^{-t/RC} (discharging)$ [Equation 21.80]

Figure 21.39 (a) Closing the switch discharges the capacitor $C$ through the resistor $R$. Mutual repulsion of like charges on each plate drives the current. (b) A graph of voltage across the capacitor versus time, with $V=V_{0}$ at $\tau =0$. The voltage decreases exponentially, falling a fixed fraction of the way to zero in each subsequent time constant $\tau$.

The graph in Figure 21.39 (b) is an example of this exponential decay. Again, the time constant is $\tau =RC$. A small resistance $R$ allows the capacitor to discharge in a small time, since the current is larger. Similarly, a small capacitance requires less time to discharge, since less charge is stored. In the first time interval $\tau =RC$ after the switch is closed, the voltage falls to 0.368 of its initial value, since $V=V_{0}\cdot e^{-1}=0.368V_{0}$.

During each successive time $\tau$, the voltage falls to 0.368 of its preceding value. In a few multiples of $\tau$, the voltage becomes very close to zero, as indicated by the graph in Figure 21.39(b).

Now we can explain why the flash camera in our scenario takes so much longer to charge than discharge; the resistance while charging is significantly greater than while discharging. The internal resistance of the battery accounts for most of the resistance while charging. As the battery ages, the increasing internal resistance makes the charging process even slower. (You may have noticed this.)

The flash discharge is through a low-resistance ionized gas in the flash tube and proceeds very rapidly. Flash photographs, such as in Figure 21.40, can capture a brief instant of a rapid motion because the flash can be less than a microsecond in duration. Such flashes can be made extremely intense.

During World War II, nighttime reconnaissance photographs were made from the air with a single flash illuminating more than a square kilometer of enemy territory. The brevity of the flash eliminated blurring due to the surveillance aircraft’s motion. Today, an important use of intense flash lamps is to pump energy into a laser. The short intense flash can rapidly energize a laser and allow it to reemit the energy in another form.

Figure 21.40 This stop-motion photograph of a rufous hummingbird (Selasphorus rufus) feeding on a flower was obtained with an extremely brief and intense flash of light powered by the discharge of a capacitor through a gas. (credit: Dean E. Biggins, U.S. Fish and Wildlife Service)

Source: Rice University, https://openstax.org/books/college-physics/pages/21-6-dc-circuits-containing-resistors-and-capacitors
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