Question

A $200$km long telegraph wire has a capacitance of $0.014$ $\mu F/km$. If it carries an alternating current of $50\times {10}^3$Hz, what should be the value of an inductance required to be connected in series so that impedance is minimum?

• A. $0.48\times {10}^{-2}$mH
• B. $0.36\times {10}^{-2}$mH
• C. $0.52\times {10}^{-2}$mH
• D. $0.49\times {10}^{-2}$mH

Answer 1

The correct option is B $0.36\times {10}^{-2}$mH

Given, $C^{\prime }=0.014\mu F/km$
$f=50\times {10}^3$Hz
Total capacitance,
$C=C^{\prime }\times 200$
$=0.014\mu F\times 200=2.8\mu F$
For impedance to be minimum
$X_L=X_C,$ i.e., $\omega L=\frac{1}{\omega C}$
$L=\frac{1}{{\omega }^{2}C}$
$=\frac{1}{4{\pi }^2{l}^{2}C}$
$=\frac{1}{4{\pi }^2\times (50\times {10}^3{)}^2\times 2.8\times {10}^{-6}}$
$=0.36\times {10}^{-3}H=0.36\times {10}^{-2}$mH.