TUNING A RADIO
The series circuit in Fig. 31.20 is similar to some radio tuning circuits. It is connected to a variable-frequency ac source with an rms terminal voltage of 1.0 V. (a) Find the resonance frequency. At the resonance frequency, find (b) the inductive reactance X_L, the capacitive reactance X_C, and the impedance Z; (c) the rms current I_{rms}; (d) the rms voltage across each circuit element.
IDENTIFY and SET UP:
Figure 31.20 shows an L-R-C series circuit, with ideal meters inserted to measure the rms current and voltages, our target variables. Equation (31.32) gives the formula for the resonance angular frequency \omega_{0}, from which we find the resonance frequency f_0. We use Eqs. (31.12) and (31.18) to find X_L and X_C, which are equal at resonance; at resonance, from Eq. (31.23), we have Z = R. We use Eqs. (31.7), (31.13), and (31.19) to find the voltages across the circuit elements.
V_{R}=I R (31.7)
X_{L}=\omega L (31.12)
V_{L}=I X_{L} (31.13)
X_{C}=\frac{1}{\omega C} (31.18)
V_{C}=I X_{C} (31.19)
Z=\sqrt{R^{2}+[\omega L-(1 / \omega C)]^{2}} (31.23)
\omega_{0}=\frac{1}{\sqrt{L C}} (31.32)
EXECUTE:
(a) The values of \omega_{0} and f_{0} are
\begin{aligned}\omega_{0} &=\frac{1}{\sqrt{L C}}=\frac{1}{\sqrt{\left(0.40 \times 10^{-3}\mathrm{H}\right)\left(100 \times 10^{-12} \mathrm{~F}\right)}} \\&=5.0 \times 10^{6}\mathrm{rad} / \mathrm{s} \\f_{0} &=8.0 \times 10^{5} \mathrm{~Hz}=800 \mathrm{kHz}\end{aligned}This frequency is in the lower part of the AM radio band.
(b) At this frequency,
Since X_L = X_C at resonance as stated above, Z = R = 500 Ω.
(c) From Eq. (31.26) (V_{\mathrm{rms}}=I_{\mathrm{rms}} Z) the rms current at resonance is
(d) The rms potential difference across the resistor is
V_{R-\mathrm{rms}}=I_{\mathrm{rms}} R=(0.0020 \mathrm{~A})(500 \Omega)=1.0 \mathrm{~V}The rms potential differences across the inductor and capacitor are
\begin{aligned}V_{L-\mathrm{rms}} &=I_{\mathrm{rms}} X_{L}=(0.0020 \mathrm{~A})(2000\Omega)=4.0 \mathrm{~V} \\V_{C-\mathrm{rms}} &=I_{\mathrm{rms}} X_{C}=(0.0020\mathrm{~A})(2000 \Omega)=4.0 \mathrm{~V}\end{aligned}
EVALUATE: The potential differences across the inductor and the capacitor have equal rms values and amplitudes, but are 180° out of phase and so add to zero at each instant. Note also that at resonance, V_{R-rms} is equal to the source voltage V_{rms}, while in this example, V_{L-rms} and V_{C-rms} are both considerably larger than V_{rms}.