32.2 Capacitor Circuits

FIGURE 32.7a shows current $i_{C}$ charging a capacitor with capacitance $C$ . The instantaneous capacitor voltage is $v_{C} = q / C$ , where $\pm q$ is the charge on the two capacitor plates at this instant. It is useful to compare Figure 32.7a to Figure 32.3 for a resistor.

The figure labeled (“a”) illustrates a current charging a capacitor.

The figure labeled (b) illustrates a capacitor connected to an "A" C voltage source.

Full Alternative Text

FIGURE 32.7b, where capacitance $C$ is connected across an AC source of emf $ε$ , is the most basic capacitor circuit. The capacitor is in parallel with the source, so the capacitor voltage equals the emf: $v_{C} = ε = ε_{0} cos ω t$ . It will be useful to write

v_{C} = V_{C} \cos ω t

(32.6)

where $V_{C}$ is the peak or maximum voltage across the capacitor. You can see that $V_{C} = ε_{0}$ in this single-capacitor circuit.

To find the current to and from the capacitor, we first write the charge

q = C v_{C} = C V_{C} \cos ω t

(32.7)

The current is the rate at which charge flows through the wires, $i_{C} = d q / d t$ , thus

i_{C} = \frac{d q}{d t} = \frac{d}{d t} (C V_{C} \cos ω t) = - ω C V_{C} \sin ω t

(32.8)

We can most easily see the relationship between the capacitor voltage and current if we use the trigonometric identity $- sin (x) = cos (x + π / 2)$ to write

i_{C} = ω C V_{C} \cos (ω t + \frac{π}{2})

(32.9)

In contrast to a resistor, a capacitor’s current (Equation 32.9) and voltage (Equation 32.6) are not in phase. In FIGURE 32.8a, a graph of the instantaneous voltage $v_{C}$ and current $i_{C}$ , you can see that the current peaks one-quarter of a period before the voltage peaks. The phase angle of the current phasor on the phasor diagram of FIGURE 32.8b is $π / 2$ rad—a quarter of a circle—larger than the phase angle of the voltage phasor.

The figure illustrates a graphical representation of a capacitor current and voltage.

The figure illustrates a capacitor current and voltage in a phasor diagram.

Full Alternative Text

We can summarize this finding:

The AC current of a capacitor leads the capacitor voltage by $π / 2$ rad, or $90 °$ .

The current reaches its peak value $I_{C}$ at the instant the capacitor is fully discharged and $v_{C} = 0$ . The current is zero at the instant the capacitor is fully charged.

A simple harmonic oscillator provides a mechanical analogy of the $90 °$ phase difference between current and voltage. You learned in Chapter 15 that the position and velocity of a simple harmonic oscillator are

\begin{array}{l} x & = & A \cos ω t \\ v & = & \frac{d x}{d t} = - ω A \sin ω t = - v_{\max} \sin ω t = v_{\max} \cos (ω t + \frac{π}{2}) \end{array}

You can see that the velocity of an oscillator leads the position by $90 °$ in the same way that the capacitor current leads the voltage.

32.2 Capacitor Circuits

FIGURE 32.7 An AC capacitor circuit.

FIGURE 32.8 Graph and phasor diagrams of the capacitor current and voltage.