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=0.014xx200 mu F = 2.8 xx 10^(-6)F`
Frequency ` f = 50 kC//s = 50 xx 10^(3)` cycles/s.
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) / ( 2pi sqrt(LC))rArr L = (1) / ( 4 pi^(2) f_(0)^(2)C)`
`L=(1)/(4xx(22)/(7)xx(22)/(7) xx (50xx10^(3))^(2)xx(2.8xx10^(-6)))=0.36xx10^(-3)H`