Question

A 100 km telegraph wire has a capacity of $0.02\mu F/km$ if it carries an alternating current of frequency $5kHz.$ The value of an inductance required to be connected in series so that the impedance is minimum is

• A. $50.7mH$
• B. $5.07mH$
• C. $0.507mH$
• D. $507mH$

Answer 1

Answer: (C)

$0.507mH$

Impedance minimum = 0

So, $X_L=X_C$

$WL=\frac{1}{WC}$

$L=\frac{1}{W^{2}C}$

$=\frac{1}{(2\pi f{)}^{2}C}$ $(\because w=2\pi f)$

$=\frac{1}{(2\pi 5\times {10}^3{)}^2\times 0.02\times {10}^{-6}\times 100}$
$=\frac{1}{4\times {\pi }^2\times 25\times {10}^6\times 0.02\times {10}^{-6}\times 100}$
$=0.507mH$