Consider the following RLC circuit, which we Consider the following RLC circuit,

Question

Consider the following RLC circuit, which we

Consider the following 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 antenna. Thus they act, via the antenna, as individual batteries, each oscillating at their particular frequency f. Suppose that R = 1200?, then what should L and C be so that we get the largest current in the circuit at frequency f = 89.9 MHz, but at the frequency f = 90.3 MHz, the current is only 1/10 as large (this will insure that we get a good reception)?

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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