Question

A telephone wire of length 200 km has a capacitance of 0.014$\mu$F per km. If it carries an ac of frequency 2.5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

• A. 0.35 mH
• B. 35 mH
• C. 1.4 mH
• D. Zero

Answer 1

The correct option is C 1.4 mH

Capacitance of wire
$C=0.014\times {10}^{-6}\times 200=2.8\times {10}^{-6}F=2.8\mu$F
For impedance of the circuit to be minimum $X_L=X_c\Rightarrow 2\pi vL=\frac{1}{2\pi vC}$

$\Rightarrow L=\frac{1}{4{\pi }^2{v}^{2}C}=\frac{1}{4(3.14{)}^2\times (2.5\times {10}^3{)}^2\times 2.8\times {10}^{-6}}$
$=1.4\times {10}^{-3}$H = 1.4mH