Question

A $200km$ telephone wire has capacity of $0.14km$.If it carries an alternating current of frequency $50kHz,$ what should be the value of an inductance required to be connected in series so that impedance is minimum?

Answer 1

Capacitance of wire

$C=0.014\times 10-6\times 200=2.8\times 10-6F=2.8\mu F$

For impedance of the circuit to be minimum

$XL=XC$

$2\pi \nu L=1/2\pi \nu C$

$L=1/4\pi 2\nu 2C=1/\left[4\right(3.14{)}^2\times (5\times {10}^3){”}^2\times 2.8\times {10}^{-}6]$

$=0.35\times {10}^{-}3H=0.35mH$