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A constant voltage at a frequency of 1 M...

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`

Text Solution

Verified by Experts

The correct Answer is:
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`
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