Q. A telephone wire of length 200 km has a capacitance of 0.014 $\mu F$ per km. If it carries an AC 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) 3.5 mH
C) 2.5 mH
D) zero
Explanation
(a) Capacitance of wire
$C = 0.014 \times 10^{- 6} \times 200$
$= 2.8 \times 10^{- 6}F = 2.84\mu F$
For impedance of the circuit to be minimum
$X_{L} = X_{C}$
$\Rightarrow \ \ \ \ \ \ \ \ 2\pi vL = \frac{1}{2\pi vC}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L = \frac{1}{4\pi^{2}v^{2}C}$
$= \frac{1}{4(3.14)^{2} \times \left( 5 \times 10^{3} \right)^{2} \times 2.8 \times 10^{- 6}}$
$= 0.35 \times 10^{- 3}H = 0.35\ mH$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.