Question

A 200 km telephone wire has capacity of `0.014 muF km^(-1)`. If it carries an alternating current oif frequency 50kHz, what should be the value of an inductance required to be connected in series so that impedance is minimumn ?

Answer 1

Capacitance , `C = 0.014 xx 200muF = 2.8 xx 10^(-16)F`
Frequency, `f = 50 k Hz = 50 xx 10^(3) cycl es s^(-1)`
The impedance will be minimum at resonant frequency , when
applied frequency will be same as resonant frequency `(f_(0) = f)`.
The resonant frequency , `f_(0) = 1/(2pisqrt (LC)) implies L = 1/(4pi^(2) f_(0)^(2) C)`
` L = 1/(4 xx 22/7 xx 22/7 xx (50 xx 10^(3))^(2) xx (2.8 xx 10^(-6))) = 0.36 xx 10^(-5) H`