Question

A constant voltage at a frequency of `1 MHz` is applied to an inductor in series with variable capacitor, when capacitor is `500 pF`, the current has its maximum value, while it is reduced to half when capacitance is `600 pF`. Find
The inductance `L`
A. `0.05 mH`
B. `0.5 mH`
C. `0.005 mH`
D. `5 mH`

Answer 1

Correct Answer - A
`omega_(0)^(2) = (1)/(LC) , L = (10^(3))/(500 xx 4 pi^(2)) mH = 0.05 mH`