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  • A telephone wire of length 200 km has a capacitance of per km. If it carries an ac of frequency 5 k

A telephone wire of length 200 km has a capacitance of 0.014 \mu F 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

Option: 1

0.35 mH


Option: 2

35 mH


Option: 3

3.5 mH


Option: 4

Zero


Answers (1)

best_answer

Capacitance of wire

\mathrm{\mathrm{C}=0.014 \times 10^{-6} \times 200=2.8 \times 10^{-6} \mathrm{~F}=2.8 \mu \mathrm{F}}

For impedance of the circuit to be minimum \mathrm{X_L=X_C \Rightarrow 2 \pi v L=\frac{1}{2 \pi v C}}

\mathrm{\Rightarrow \mathrm{L}=\frac{1}{4 \pi^2 \mathrm{v}^2 \mathrm{C}}=\frac{1}{4(3.14)^2 \times\left(5 \times 10^3\right)^2 \times 2.8 \times 10^{-6}}=0.35 \times 10^{-3} \mathrm{H}=0.35 \mathrm{mH}}

Posted by

HARSH KANKARIA

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