A 200 km long telegraph wire has a capacitance of `0.025 mu F//km`. It carries an alternating current of 50 KHz. What should be the value of an inductance required to be connected in series, so that the impedance is minimum. (take `pi^(2)=10`)

Options

A. `1 mu H`
B. `2 mu H`
C. `5 mu H`
D. `8 mu H`

Answer

B

Explanation

Let L be the required inductance. The impedance is minimum, if `X_(L) =X_(C)`
or `2 pi f L = 1/(2 pi fC) or L = 1/(4pi^(2)f^(2)C)`
Total Capacitance =`0.025 xx 200 = 5 mu F = 5 xx 10^(-6) F`
`:. L = 1/(4xx10xx(50xx10^(3))^(2)xx5xx10^(-6))`
`=(1)/(40 xx 2500 xx 5) = 2 xx 10^(-6) = 2 muH`.