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

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0.35 mH

35 mH

3.5 mH

Zero

answer is A.

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Detailed Solution

Capacitance of wireC=0.014×10−6×200=2.8×10−6F=2.8μFFor impedance of the circuit to be minimum XL=XC⇒2πνL=12πνC ⇒L=14π2ν2C=14(3.14)2×(5×103)2×2.8×10−6 =0.35×10−3H=0.35 mH

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