A telephone wire of length 200 km has a capacitance of 0.014 μ F km - 1 . If it carries an AC frequency of 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?

Question

A telephone wire of length 200 km has a capacitance of 0.014 μ F km - 1 . If it carries an AC frequency of 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?, 0.35 mH, 3.5 mH, 2.5 mH, zero

MARKS App · Accepted Answer

Capacitance of wire C = 0.014 × 10 - 6 × 200 = 2.8 × 10 - 6 F = 2.8 μF For impedance of the circuit to be minimum X L = X C ⇒ 2 π f L = 1 2 π f C L = 1 4 π 2 f 2 C = 1 4 3.14 2 × 5 × 10 3 2 × 2.8 × 10 - 6 = 0.35 × 10 - 3 H = 0.35 m H

Search any question & find its solution

Looking for more such questions to practice?