A telephone wire of length `200 km` has a capacitance of `0.01 4 (muF)/(km)`. If it carries an `AC` of frequency `5kHz` what should be the value of an inductor required to be connected in series so that impedence of the circuit is minimum ?

Options

A. `0.35 mH`
B. `35 mH`
C. `3.5 mH`
D. Zero

Answer

A

Explanation

Capacitance of wire
`C=0.014xx10^(-6)xx200=2.8xx10^(-6)F=2.8 muF`
For impedence of the circuit to be minimum `X_(L)=X_(C)`
`implies 2pivL=1/(2pivC)`
`implies L=1/(4pi^(2)v^(2)V)=1/(4(3.14)^(2)xx(5xx10^(3))^(2)xx2.8xx10^(-6))`
`=0.35xx10^(-3)H=0.35 mH`