32.2.1 Capacitive Reactance

We can use Equation 32.9 to see that the peak current to and from a capacitor is $I_{C} = ω C V_{C}$ . This relationship between the peak voltage and peak current looks much like Ohm’s law for a resistor if we define the capacitive reactance $X_{C}$ to be

X_{C} \equiv \frac{1}{ω C}

(32.10)

With this definition,

I_{C} = \frac{V_{C}}{X_{C}} or V_{C} = I_{C} X_{C}

(32.11)

The units of reactance, like those of resistance, are ohms.

NOTE Reactance relates the peak voltage $V_{C}$ and current $I_{C}$ . But reactance differs from resistance in that it does not relate the instantaneous capacitor voltage and current because they are out of phase. That is, $v_{C} \neq i_{C} X_{C}$ .

A resistor’s resistance $R$ is independent of the emf frequency. In contrast, as seen in FIGURE 32.9, a capacitor’s reactance $X_{C}$ depends inversely on the frequency. The reactance becomes very large at low frequencies (i.e., the capacitor is a large impediment to current). This makes sense because $ω = 0$ would be a nonoscillating DC circuit, and we know that a steady DC current cannot pass through a capacitor. The reactance decreases as the frequency increases until, at very high frequencies, $X_{C} \approx 0$ and the capacitor begins to act like an ideal wire. This result has important consequences for how capacitors are used in many circuits.

The figure illustrates a capacitive reactance versus frequency graph.

FIGURE 32.9 Full Alternative Text

EXAMPLE 32.2 Capacitive reactance

What is the capacitive reactance of a $0.10 μ F$ capacitor at 100 Hz (an audio frequency) and at 100 MHz (an FM-radio frequency)?

SOLVE At 100 Hz,

X_{C} (at 100 Hz) = \frac{1}{ω C} = \frac{1}{2 π (100 Hz) (1.0 \times 10^{- 7} F)} = 16, 000 Ω

Increasing the frequency by a factor of $10^{6}$ decreases $X_{C}$ by a factor of $10^{6}$ , giving

X_{C} (at 100 MHz) = 0.016 Ω

REVIEW A capacitor with a substantial reactance at audio frequencies has virtually no reactance at FM-radio frequencies.

EXAMPLE 32.3 Capacitor current

A $10 μ F$ capacitor is connected to a 1000 Hz oscillator with a peak emf of 5.0 V. What is the peak current to the capacitor?

VISUALIZE Figure 32.7b showed the circuit diagram. It is a simple one-capacitor circuit.

SOLVE The capacitive reactance at $ω = 2 π f = 6280 rad / s$ is

X_{C} = \frac{1}{ω C} = \frac{1}{(6280 rad/s) (10 \times 10^{- 6} F)} = 16 Ω

The peak voltage across the capacitor is $V_{C} = ε_{0} = 5.0 V;$ hence the peak current is

I_{C} = \frac{V_{C}}{X_{C}} = \frac{5.0 V}{16 Ω} = 0.31 A

REVIEW Using reactance is just like using Ohm’s law, but don’t forget it applies to only the peak current and voltage, not the instantaneous values.