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 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. 3.5 mH
• D. Zero

Answer 1

The correct option is A 0.35 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 $X_L=X_C\Rightarrow 2\pi vL=\frac{1}{2\pi vC}$
$\Rightarrow L=\frac{1}{4{\pi }^2{v}^{2}C}=\frac{1}{4(3.14{)}^2\times (5\times {10}^3{)}^2\times 2.8\times {10}^{-6}}=0.35\times {10}^{-3}H=0.35mH$