The tuning circuit of a radio receiver has a resistance of `50Omega` , an inductor of 10 mH and a variable capacitor. A 1 MHz radio wave produces a potential difference of 0.1 mV. The values of the capacitor to produce resonance is `("Take "pi_(2) = 10)`

Options

A. `2.5 pF`
B. `5.0 pF`
C. `25 pF`
D. `50 pF`

Answer

A

Explanation

`L = 10 mHz = 10^(-2) Hz`
`f = 1 MHz = 10^(6) Hz`
`f = (1)/(2pi sqrt(LC)) rightarrow f^(2) = (1)/(4pi^(2)LC)`
`rightarrow C = (1)/(4pi^(2)f^(2)L) = (1)/(4 xx 10 xx 10^(-2) xx 10^(12)) = (10^(-12))/(4) = 2.5 pF`