Question

A telephone wire of length 200 km has a capacitance of 0.014 μF 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

Answer: (A)

0.35 mH

Answer 2

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 $X_{L} = X_{C}$ ⇒

$2\pi\nu L = \frac{1}{2\pi\nu C}$

⇒ $L = \frac{1}{4\pi^{2}\nu^{2}C} = \frac{1}{4(3.14)^{2} \times (5 \times 10^{3})^{2} \times 2.8 \times 10^{- 6}} = 0.35 \times 10^{- 3}H = 0.35mH$