32.4 Inductor Circuits

FIGURE 32.13a shows the instantaneous current $i_{L}$ through an inductor. If the current is changing, the instantaneous inductor voltage is

v_{L} = L \frac{d i_{L}}{d t}

(32.16)

The figure labeled (“a”) illustrates the instantaneous current through an inductor.

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

Full Alternative Text

You learned in Chapter 30 that the potential decreases in the direction of the current if the current is increasing $(d i_{L} / d t > 0)$ and increases if the current is decreasing $(d i_{L} / d t < 0)$ .

FIGURE 32.13b is the simplest inductor circuit. The inductor $L$ is connected across the AC source, so the inductor voltage equals the emf: $v_{L} = ε = ε_{0} cos ω t$ . We can write

v_{L} = V_{L} \cos ω t

(32.17)

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

We can find the inductor current $i_{L}$ by integrating Equation 32.17. First, we use Equation 32.17 to write Equation 32.16 as

d i_{L} = \frac{v_{L}}{L} d t = \frac{V_{L}}{L} cos ω t d t

(32.18)

Integrating gives

\begin{array}{l} i_{L} & = & \frac{V_{L}}{L} \int cos ω t d t = \frac{V_{L}}{ω L} sin ω t = \frac{V_{L}}{ω L} cos (ω t - \frac{π}{2}) \\ = & I_{L} \cos (ω t - \frac{π}{2}) \end{array}

(32.19)

where $I_{L} = V_{L} / ω L$ is the peak or maximum inductor current.

NOTE Mathematically, Equation 32.19 could have an integration constant $i_{0}$ . An integration constant would represent a constant DC current through the inductor, but there is no DC source of potential in an AC circuit. Hence, on physical grounds, we set $i_{0} = 0$ for an AC circuit.

We define the inductive reactance, analogous to the capacitive reactance, to be

X_{L} \equiv ω L

(32.20)

Then the peak current $I_{L} = V_{L} / ω L$ and the peak voltage are related by

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

(32.21)

FIGURE 32.14 shows that the inductive reactance increases as the frequency increases. This makes sense. Faraday’s law tells us that the induced voltage across a coil increases as the time rate of change of $\vec{B}$ increases, and $\vec{B}$ is directly proportional to the inductor current. For a given peak current $I_{L}$ , $\vec{B}$ changes more rapidly at higher frequencies than at lower frequencies, and thus $V_{ L}$ is larger at higher frequencies than at lower frequencies.

The figure illustrates a plot of inductive reactance as a function of frequency.

FIGURE 32.14 Full Alternative Text

FIGURE 32.15a is a graph of the inductor voltage and current. You can see that the current peaks one-quarter of a period after the voltage peaks. The angle of the current phasor on the phasor diagram of FIGURE 32.15b is $π / 2$ rad less than the angle of the voltage phasor. We can summarize this finding:

The figure labeled (“a”) illustrates a graph of the inductor voltage and current versus time.

The figure labeled (b) illustrates a phasor diagram of the inductor voltage and current.

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EXAMPLE 32.5 Current and voltage of an inductor

A $25 μ H$ inductor is used in a circuit that oscillates at 100 kHz. The current through the inductor reaches a peak value of 20 mA at $t = 5.0 μ s$ . What is the peak inductor voltage, and when, closest to $t = 5.0 μ s,$ does it occur?

MODEL The inductor current lags the voltage by $90 °$ , or, equivalently, the voltage reaches its peak value one-quarter period before the current.

VISUALIZE The circuit looks like Figure 32.15b.

SOLVE The inductive reactance at $f = 100 kHz$ is

X_{L} = ω L = 2 π (1.0 \times 10^{5} Hz) (25 \times 10^{- 6} H) = 16 Ω

Thus the peak voltage is $V_{L} = I_{L} X_{L} = (20 mA) (16 Ω) = 320 mV$ . The voltage peak occurs one-quarter period before the current peaks, and we know that the current peaks at $t = 5.0 μ s$ . The period of a 100 kHz oscillation is $10.0 μ s,$ so the voltage peaks at

t = 5.0 μ s - \frac{10.0 μ s}{4} = 2.5 μ s

32.4 Inductor Circuits

FIGURE 32.13 Using an inductor in an AC circuit.

FIGURE 32.14 The inductive reactance as a function of frequency.

FIGURE 32.15 Graph and phasor diagrams of the inductor current and voltage.

EXAMPLE 32.5 Current and voltage of an inductor