AN INDUCTOR IN AN AC CIRCUIT
The current amplitude in a pure inductor in a radio receiver is to be 250 μA when the voltage amplitude is 3.60 V at a frequency of 1.60 MHz (at the upper end of the AM broadcast band). (a) What inductive reactance is needed? What inductance? (b) If the voltage amplitude is kept constant, what will be the current amplitude through this inductor at 16.0 MHz? At 160 kHz?
IDENTIFY and SET UP:
The circuit may have other elements, but in this example we don’t care: All they do is provide the inductor with an oscillating voltage, so the other elements are lumped into the ac source shown in Fig. 31.8a. We are given the current amplitude I and the voltage amplitude V. Our target variables in part (a) are the inductive reactance X_L at 1.60 MHz and the inductance L, which we find from Eqs. (31.13) and (31.12). Knowing L, we use these equations in part (b) to find X_L and I at any frequency.
X_{L}=\omega L (31.12)
V_{L}=I X_{L} (31.13)
EXECUTE:
(a) From Eq. (31.13),
X_{L}=\frac{V_{L}}{I}=\frac{3.60 \mathrm{~V}}{250 \times 10^{-6} \mathrm{~A}}=1.44 \times 10^{4} \Omega=14.4 \mathrm{k} \OmegaFrom Eq. (31.12), with \omega=2 \pi f,
L=\frac{X_{L}}{2 \pi f}=\frac{1.44 \times 10^{4} \Omega}{2 \pi\left(1.60 \times 10^{6} \mathrm{~Hz}\right)}=1.43 \times 10^{-3} \mathrm{H}=1.43 \mathrm{mH}(b) Combining Eqs. (31.12) and (31.13), we find I=V_{L} / X_{L}= V_{L} / \omega L=V_{L} / 2 \pi f L Thus the current amplitude is inversely proportional to the frequency f. Since I = 250 μA at f = 1.60 MHz, the current amplitudes at 16.0 MHz (10f) and 160 kHz = 0.160 MHz (f/10) will be, respectively, one-tenth as great (25.0 μA) and ten times as great (2500 μA = 2.50 mA).
EVALUATE: In general, the lower the frequency of an oscillating voltage applied across an inductor, the greater the amplitude of the resulting oscillating current.
