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
Resistance `(R )`
A. `30 Omega`
B. `20 Omega`
C. `40 Omega`
D. `50 Omega`

Answer 1

Correct Answer - A
`i_(0) = (E)/(R )` and `(i_(0))/(2) = (E)/(sqrt(R^(2) + (X_(L) - X_(C ))^(2)))`
Solving `sqrt(3) R = (10^(3))/(6 pi) implies R ~= 30 Omega`