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Q.

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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a

0.35 mH

b

35 mH

c

3.5 mH

d

Zero

answer is A.

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Detailed Solution

Capacitance of wire,C=0.014 × 10−6 × 200=2.8 × 10−6 F=2.8 μFFor impedance of the circuit to be minimum, XL=XC⇒ 2πvL=12πvC⇒L=14π2v2C=14(3.14)2×5×1032×2.8×10−6=0.35 × 10−3 H=0.35 mH
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