28.9 RC Circuits

A resistor circuit has a steady current. By adding a capacitor and a switch, we can make a circuit in which the current varies with time as the capacitor charges and discharges. Circuits with resistors and capacitors are called RC circuits. RC circuits are at the heart of timekeeping circuits in applications ranging from the intermittent windshield wipers on your car to computers and other digital electronics.

FIGURE 28.34a shows a charged capacitor, a switch, and a resistor. The capacitor has charge $Q_{0}$ and potential difference $Δ V_{0} = Q_{0} / C$ . There is no current, so the potential difference across the resistor is zero. Then, at $t = 0$ , the switch closes and the capacitor begins to discharge through the resistor.

The figure labeled (“a”) illustrates a series R C circuit before a switch closes.

The figure labeled (b) illustrates a series R C circuit after a switch closes.

Full Alternative Text

How long does the capacitor take to discharge? How does the current through the resistor vary as a function of time? To answer these questions, FIGURE 28.34b shows the circuit at a later instant when the capacitor is partially discharged.

Kirchhoff’s loop law is valid for any circuit, not just circuits with batteries. The loop law applied to the circuit of Figure 28.34b, going around the loop cw, is

Δ V_{cap} + Δ V_{res} = \frac{Q}{C} - I R = 0

(28.25)

Q and $I$ in this equation are the instantaneous values of the capacitor charge and the resistor current.

The current $I$ is the rate at which charge flows through the resistor: $I = d q / d t$ . But the charge flowing through the resistor is charge that was removed from the capacitor. That is, an infinitesimal charge dq flows through the resistor when the capacitor charge decreases by dQ. Thus $d q = - d Q$ , and the resistor current is related to the instantaneous capacitor charge by

I = - \frac{d Q}{d t}

(28.26)

Equation 28.26 tells us that $I$ is positive when $Q$ is decreasing $(d Q < 0)$ , as we would expect. The reasoning that has led to Equation 28.26 is rather subtle but very important.

A photograph of the rear flasher on a bike helmet flashing on and off. — The rear flasher on a bike helmet flashes on and off. The timing is controlled by an RC circuit.

If we substitute Equation 28.26 into Equation 28.25 and then divide by $R$ , the loop law for the RC circuit becomes

\frac{d Q}{d t} + \frac{Q}{R C} = 0

(28.27)

Equation 28.27 is a first-order differential equation for the capacitor charge $Q$ , but one that we can solve by direct integration. First, we rearrange Equation 28.27 to get all the charge terms on one side of the equation:

\frac{d Q}{Q} = - \frac{1}{R C} d t

The product RC is a constant for any particular circuit.

The capacitor charge was $Q_{0}$ at $t = 0$ when the switch was closed. We want to integrate from these starting conditions to charge $Q$ at a later time t. That is,

\int_{Q_{0}}^{Q} \frac{d Q}{Q} = - \frac{1}{R C} \int_{0}^{t} d t

(28.28)

Both are well-known integrals, giving

ln Q |_{Q_{0}}^{Q} = ln Q - {ln Q}_{0} = ln (\frac{Q}{Q_{0}}) = - \frac{t}{R C}

We can solve for the capacitor charge $Q$ by taking the exponential of both sides, then multiplying by $Q_{0}$ . Doing so gives

Q = Q_{0} e^{- t / R C}

(28.29)

Notice that $Q = Q_{0}$ at $t = 0$ , as expected.

The argument of an exponential function must be dimensionless, so the quantity RC must have dimensions of time. It is useful to define the time constant $τ$ to be

τ = R C

(28.30)

We can then write Equation 28.29 as

Q = Q_{0} e^{- t / τ}

(28.31)

The capacitor voltage, directly proportional to the charge, also decays exponentially as

Δ V_{C} = Δ V_{0} e^{- t / τ} (capacitor discharging)

(28.32)

The meaning of Equation 28.32 is easier to understand if we portray it graphically. FIGURE 28.35a shows the capacitor voltage as a function of time. The voltage decays exponentially, starting from $Δ V_{0}$ at $t = 0$ and asymptotically approaching zero as $t \to \infty.$ The time constant $τ$ is the time at which the voltage has decreased to $e^{- 1}$ (about 37%) of its initial value. At time $t = 2 τ$ , the voltage has decreased to $e^{- 2}$ (about 13%) of its initial value. A graph of the capacitor’s charge would have the same shape.

The figure labeled (“a”) illustrates a voltage versus time graph.

The figure labeled (b) illustrates a current versus time graph.

Full Alternative Text

NOTE The shape of the voltage graph is always the same, regardless of the specific value of the time constant $τ$ .

We find the resistor current by using Equation 28.26:

I = - \frac{d Q}{d t} = \frac{Q_{0}}{τ} e^{- t / τ} = \frac{Q_{0}}{R C} e^{- t / τ} = \frac{Δ V_{0}}{R} e^{- t / τ} = I_{0} e^{- t / τ}

(28.33)

where $I_{0} = Δ V_{0} / R$ is the initial current, immediately after the switch closes. FIGURE 28.35b is a graph of the resistor current versus t. You can see that the current undergoes the same exponential decay, with the same time constant, as the capacitor voltage.

NOTE There’s no specific time at which the capacitor has been completely discharged, because $Δ V$ approaches zero asymptotically, but the voltage and current have dropped to less than 1% of their initial values at $t = 5 τ$ . Thus $5 τ$ is a reasonable answer to the question “How long does it take to discharge a capacitor?”

EXAMPLE 28.12 Measuring capacitance

To determine the capacitance of an unmarked capacitor, you set up the circuit shown in FIGURE 28.36. After holding the switch in position a for several seconds, you suddenly flip it—at $t = 0$ s—to position b while monitoring the resistor voltage with a voltmeter. Your measurements are shown in the table. What is the capacitance? And what was the resistor current 5.0 s after the switch changed position?

The figure illustrates the circuit diagram of a parallel R C circuit for measuring capacitance.

FIGURE 28.36 Full Alternative Text

Time (s)	Voltage (V)
0.0	9.0
2.0	5.4
4.0	2.7
6.0	1.6
8.0	1.0

MODEL The battery charges the capacitor to 9.0 V. Then, when the switch is flipped to position b, the capacitor discharges through the $25, 000 Ω$ resistor with time constant $τ = R C$ .

SOLVE With the switch in position b, the resistor is in parallel with the capacitor and both have the same potential difference $Δ V_{R} = Δ V_{C} = Q / C$ at all times. The capacitor charge decays exponentially as

Q = Q_{0} e^{- t / τ}

Consequently, the resistor (and capacitor) voltage also decays exponentially:

Δ V_{R} = \frac{Q_{0}}{C} e^{- t / τ} = Δ V_{0} e^{- t / τ}

where $Δ V_{0} = 9.0 V$ is the potential difference at the instant the switch closes. To analyze exponential decays, we take the natural logarithm of both sides. This gives

ln (Δ V_{R}) = ln (Δ V_{0}) + ln (e^{- t / τ}) = ln (Δ V_{0}) - \frac{1}{τ} t

This result tells us that a graph of $ln (Δ V_{R})$ versus t—a semi-log graph—should be linear with y-intercept $ln (Δ V_{0})$ and slope $- 1 / τ$ . If this turns out to be true, we can determine $τ$ and hence $C$ from an experimental measurement of the slope.

FIGURE 28.37 is a graph of $ln (Δ V_{R})$ versus $t$ . It is, indeed, linear with a negative slope. From the y-intercept of the best-fit line, we find $Δ V_{0} = e^{2.20} = 9.0 V,$ as expected. This gives us confidence in our analysis. Using the slope, we find

The figure illustrates an integral of voltage delta V subscript R versus time graph.

FIGURE 28.37 Full Alternative Text

τ = - \frac{1}{slope} = - \frac{1}{- 0.28 s^{- 1}} = 3.6 s

With this, we can calculate

C = \frac{τ}{R} = \frac{3.6 s}{25, 000 Ω} = 1.4 \times 10^{- 4} F = 140 μ F

The initial current is $I_{0} = (9.0 V) / (25, 000 Ω) = 360 μ A$ . Current also decays exponentially with the same time constant, so the current after 5.0 s is

I = I_{0} e^{- t / τ} = (360 μ A) e^{- (5.0 s) / (3.6 s)} = 90 μ A

REVIEW The time constant of an exponential decay can be estimated as the time required to decay to one-third of the initial value. Looking at the data, we see that the voltage drops to one-third of the initial 9.0 V in just under 4 s. This is consistent with the more precise $τ = 3.6 s,$ so we have confidence in our results.

28.9 RC Circuits

Figure 28.34 Discharging a capacitor.

FIGURE 28.35 The decay curves of the capacitor voltage and the resistor current.

EXAMPLE 28.12 Measuring capacitance

FIGURE 28.36 An RC circuit for measuring capacitance.

FIGURE 28.37 A semi-log graph of the data.