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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?

Options

  1. A0.35 mH
  2. B3.5   mH
  3. C2.5   mH
  4. Dzero

Correct answer

A. 0.35 mH

Step-by-step solution

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

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