The tuning circuit in an FM radio receiver is a series RLC circuit with an induc

Question

The tuning circuit in an FM radio receiver is a series RLC circuit with an inductor rated at 0.20 ??. The receiver is tuned to a station at 88.9 MHz by adjusting a variable capacitor. 1. What is the capacitance in the circuit in order for resonance at this frequency? FM stations are assigned frequencies every 0.2 MHz, but geographically close stations are not assigned adjacent frequencies, since it's possible to receive a station even if that frequency is not at the resonanc e setting. The smallest allowed frequency interval is 0.4 MHz for a nearby station. 2. Assume the radio is tuned to 88.9 MHz. What is the maximum resistance the tuning circuit can have in order that the peak current at 90.2 MHz is no more than 0.10% of the peak current at 88.9 MHz?

Explanation / Answer

1. resonant frequency,

f0 = 1/ (2 pi sqrt(LC))

88.9 x 10^6 = 1/ (2 x pi x sqrt(0.20 x 10^-6 x C))

C = 16 x 10^-12 F Or 16 pC ...Ans

2. I = V / R at resonant frequency,

f = 90.2 x 10^6 Hz

XL = 2 pi f L = 113.34 Ohm

XC = 1 / (2 pi f C ) = 110.28 ohm

Z = sqrt[R^2 + (XL -Xc)^2] = sqrt[ R^2 + 3.06^2]

now I' = V/Z

and I' = 0.10 I / 100 = 0.001 I

V / sqrt[ R^2 + 3.06^2] = 0.001 (V/R)

R^2 = (1 x 10^-6)R^2 + (1 x 10^-6)(3.06^2)

R = 3.06 x 10^-3 Ohm Or 3.06 mili ohm

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