28.9.1 Charging a Capacitor

FIGURE 28.38a shows a circuit that charges a capacitor. After the switch is closed, the battery’s charge escalator moves charge from the bottom electrode of the capacitor to the top electrode. The resistor, by limiting the current, slows the process but doesn’t stop it. The capacitor charges until $Δ V_{C} = ε;$ then the charging current ceases. The full charge of the capacitor is $Q_{0} = C (Δ V_{C})_{max} = C ε$ .

The figure labeled (“a”) illustrates the circuit diagram of a series R C circuit for charging a capacitor.

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

Full Alternative Text

The analysis is much like that of discharging a capacitor. As a homework problem, you can show that the capacitor voltage and the circuit current at time $t$ are

\begin{array}{l} Δ V = ε (1 - e^{- t / τ}) \\ \begin{matrix} I = I_{0} e^{- t / τ} & (capacitor charging) \end{matrix} \end{array}

(28.34)

where $I_{0} = ε / R$ and, again, $τ = R C$ . The “upside-down decay” of the capacitor voltage to $ε_{}$ is shown graphically in FIGURE 28.38b.

CHAPTER 28 CHALLENGE EXAMPLE Energy dissipated during a capacitor discharge

The switch in FIGURE 28.39 has been in position 1 for a long time. It is suddenly switched to position 2 for 1.0 s, then back to 1. How much energy is dissipated by the $5500 Ω$ resistor?

The figure illustrates the circuit diagram of two resistors, a capacitor, and a switch connected to a battery.

FIGURE 28.39 Full Alternative Text

MODEL With the switch in position 1, the capacitor charges through the $1200 Ω$ resistor with time constant $τ_{charge} = (1200 Ω) \times (2.0 \times 10^{- 4} F) = 0.24 s$ . Because the switch has been in position 1 for a “long time,” which we interpret as being much longer than 0.24 s, we will assume that the capacitor is fully charged to 50 V when the switch is changed to position 2. The capacitor then discharges through the $5500 Ω$ resistor until the switch is returned to position 1. Assume ideal wires.

SOLVE Let $t = 0 s$ be the time when the switch is moved from 1 to 2, initiating the discharge. The battery and $1200 Ω$ resistor are irrelevant during the discharge, so the circuit looks like that of Figure 28.34b. The time constant is $τ = (5500 Ω) (2.0 \times 10^{- 4} F) =$ $1 .1 s,$ so the capacitor voltage decreases from 50 V at $t = 0 s$ to

Δ V_{C} = (50 V) e^{- (1.0 s) / (1.1 s)} = 20 V

at $t = 1.0 s$ .

There are two ways to determine the energy dissipated in the resistor. We learned in Section 28.3 that a resistor dissipates energy at the rate $d E / d t = P_{R} = I^{2} R$ . The current decays exponentially as $I = I_{0} exp (- t / τ)$ , with $I_{0} = Δ V_{0} / R = 9.09 mA$ . We can find the energy that has been dissipated at time $t = T$ by integrating:

\begin{array}{l} Δ E & = & \int_{0}^{T} I^{2} R d t = I_{0}^{2} R \int_{0}^{T} e^{- 2 t / τ} d t = - \frac{1}{2} τ I_{0}^{2} R e^{- 2 t / τ} |_{0}^{T} \\ = & \frac{1}{2} τ I_{0}^{2} R (1 - e^{- 2 T / τ}) \end{array}

The 2 in the exponent appears because we squared the expression for $I$ . Evaluating for $T = 1.0 s,$ we find

Δ E = \frac{1}{2} (1.1 s) {(0.00909 A)}^{2} (5500 Ω) (1 - e^{- (2.0 s) / (1.1 s)}) = 0.21 J

Alternatively, we can use the known capacitor voltages at $t = 0 s$ and $t = 1.0 s$ and $U_{C} = \frac{1}{2} C (Δ V_{C})^{2}$ to calculate the energy stored in the capacitor at these times:

\begin{array}{l} U_{C} (t = 0.0 s) = \frac{1}{2} (2.0 \times 10^{- 4} F) {(50 V)}^{2} = 0.25 J \\ U_{C} (t = 1.0 s) = \frac{1}{2} (2.0 \times 10^{- 4} F) {(20 V)}^{2} = 0.04 J \end{array}

The capacitor has lost $Δ E = 0.21 J$ of energy, and this energy was dissipated by the current through the resistor.

REVIEW Not every problem can be solved in two ways, but doing so when it’s possible gives us great confidence in our result.

28.9.1 Charging a Capacitor

FIGURE 28.38 Charging a capacitor.

CHAPTER 28 CHALLENGE EXAMPLE Energy dissipated during a capacitor discharge

FIGURE 28.39 Circuit of a switched capacitor.