A simple tuned FM antenna circuit is used to pick up radio signals.The circuit is a series RLC circuit with an inductor (9.50 $\mu$H)and a variable capacitor which can be tuned to receive your favorite station.The radio signal from your favorite FM radio station produces a sinusoidal time-varying emf with an amplitude 15 $\mu$V and a frequency of 88.7 MHz in the antenna,which acts as an ac emf.Another radio station produces a sinusoidal time-varying emf with the same amplitude but with a frequency of 88.5 MHz.With the circuit tuned to optimize reception at 88.7 MHz,what should the value of the resistance in the FM circuit be in order to attenuate by a factor of two (compared to the power if the circuit were optimized for 88.5 MHz)the current produced by the signal from the second station? A)4.3 * 10^-5 $\varOmega$ B)10 $\varOmega$ C)14 $\varOmega$ D)24 $\varOmega$ E)36 $\varOmega$

Question

Quizplus · Accepted Answer

The resonant frequency of a series RLC circuit is given by:

$$f_r = \frac{1}{2\pi\sqrt{LC}}$$

At resonance, the reactances of the capacitor and inductor cancel out, leaving only the resistance:

$$X_L = X_C \rightarrow 2\pi f_r L = \frac{1}{2\pi f_r C} \rightarrow R = 2\pi f_r L \cdot C$$

For the given values, the resonant frequency is:

$$f_r = \frac{1}{2\pi\sqrt{(9.50\mu F)(1.0 H)}} \approx 1770.7 Hz$$

To attenuate the current from the second station by a factor of 2, we need to increase the impedance of the circuit by a factor of $\sqrt{2}$. Since the impedance of the circuit is dominated by the resistance at resonance, we can simply double the resistance to achieve the desired attenuation:

$$R' = 2R \approx 47.1 \Omega$$

Therefore, the closest choice is D (24 Ω).

A Simple Tuned FM Antenna Circuit Is Used to Pick μ\muμ

A Simple Tuned FM Antenna Circuit Is Used to Pick $\mu$