Concider a RLC circuit, which we\'ll use as a model for a radio. Radio waves wit

Question

Concider a RLC circuit, which we'll use as a model for a radio. Radio waves with different frequencies,f , hit the antenna and exert oscillating electric forces on the electrons in the antena. Thus they act, via the antenna, as individual batteries, each oscillating at their particular frequency f. Suppose that R=1200 ohms, then what should L & C be so that the largest current in the circuit at frequency f=89.9 MHz, but at the frequency f=90.3 MHz, the current is one tenth as large (this will insure that we get a good reception). The circuit loop goes antena to capacitor to resistor to inductor and back at antena.

Explanation / Answer

Rnet = R = 1200 ohms


2*pi*f*L = 1/(2*pi*f*C) at f = 89.9*10^6 Hz


L = (1/4*pi^2*f^2)*1/C

L = (1/(4*3.14*^2*89.9^2*10^12)*1/C

L = 3.137*10^-18/C


at f = 90.3*10^6 Hz

Rnet = sqrt( R^2 + (XL - XC)^2)

when current is 1/10 times, resistance becomes 10 times

Rnet = sqrt(R^2 + (XL-Xc)^2)

1200*10 = sqrt(1200^2 + (XL - XC)^2)

12000 = sqrt(1440000 + (XL - Xc)^2)

144000000 = 1440000 + (XL-XC)^2


(XL - XC)^2 = 142560000

XL - XC = sqrt(142560000)

XL - XC = 11940

2*pi*f*L - (1/(2*pi*f*C) = 11940

4*pi^2*f^2*L*C - 1 = 11940*2*pi*f*C


4*pi^2*90.3^2*10^12*3.137*10^-18*C/C - 1 = 11940*2*3.14*90.3**10^6C

1.008809 - 1 = 6.77*10^12*C

c = 0.008809/(4.77*10^12)

= 1.847*10^-15 F


L = 3.137*10^-18/(1.847*10^-15)

= 1.6984*10^-3 H

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