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A telephone wire of length 200 km has a capacitance of 0.014 bcF 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?

1. 0.35 mH
2. 35 mH
3. 3.5 mH
4. Zero

Correct answer: 0.35 mH

Solution

Minimum impedance in a series AC circuit occurs at resonance, when inductive reactance equals capacitive reactance. First find total capacitance: \(C=0.014\,\mu\text{F/km}\times 200\,\text{km}=2.8\,\mu\text{F}\). Then use \(L=1/(\omega^2 C)\) with \(\omega=2\pi f\), which gives \(L\approx 0.35\,\text{mH}\).

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