InterviewSolution
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InterviewSolution
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A radio can be turned over a frequency range from `500 kHz to 1.5 MHz`. If its L-C circuit has an effective inductance of `400 (mu)H`, what is the range of its variable capacitor. |
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Answer» We know that resonant frequency of a tuning circuit is given by `f_(0)=(1)/(2 pi sqrt(LC))`. From here we can find C, which is given as `C=(1)/(4 pi^(2)f_(0)^(2)L)` We are given the range of frequency and we have to find the range of variable capacitor. By putting the extreme value of frequency, we can find the range of capacitor. By Putting `F_(0)=500KHz`, we get `(C_1)=(1)/(4 pi^(2)(500xx10^(3))^(2)xx400xx10^(-6))=253pF` `C_(2)=(1)/(4 pi^(2)(1.5xx10^(6))^(2)xx400xx10^(-6))=28 pF` so the range of variable capacitor comes as , 28 PF-253 pF`. |
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