Question

You've decided to build a radio to listen to your favourite FM radio station, which broadcasts at 93.90 MHz. For the tuner, you'll be using an RLC circuit, but the only inductor you happen to have on hand has a value of 0.510 μH. Unfortunately, there's another radio station — which you don't want to listen to — at a nearby frequency of 93.50 MHz. To prevent any interference from this 93.50 MHz radio station, you want the peak current from it to be no more than 0.370% of the peak current from your favourite 93.90 MHz station. What value of resistor should you use in your RLC circuit? (Assume the two stations have the same strength.) Hint:You'll have to first calculate the value of the capacitor so that the radio tunes to 93.90 MHz

Answer #1

Resonant frequency is given by

=>C=1/4pi^{2}f^{2}L
=1/4pi^{2}*(93.9*10^{6})^{2}*(0.51*10^{-6})

C=5.633*10^{-12} F

Peak Current at f=93.9MHz

I_{1}=V/R

at f=93.5 MHz

Given I_{2} = (0.37/100)I_{1} = 0.0037(V/R)

Inductive reactance

X_{L}=2pifL=2pi*93.5*10^{6}*0.51*10^{-6}
=299.6137 ohms

Capacitvie reactance

X_{C}=1/2pifC
=1/2pi*(93.5*10^{6})(5.633*10^{-12}) =302.182124
ohms

Impedance

Z=sqrt[R^{2}+(X_{L}-X_{C})^{2}]
=sqrt[R^{2}+(299.6137-302.182124)^{2}]

Z=sqrt[R^{2}+6.5968]

Peak current

I=V/Z =V/sqrt[R^{2}+6.5968]

=>0.0037(V/R) = V/sqrt[R^{2}+6.5968]

sqrt[R^{2}+6.5968] =270.27R

73046R^{2} = R^{2}+6.5968

R=9.503*10^{-3} ohms

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 minute ago

asked 23 minutes ago

asked 52 minutes ago

asked 54 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago