Electro Magnetic Induction And Alternating Currents Question 141

Question: A telephone wire of length 200 km has a capacitance of 0.014 mF 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

Options:

A) 0.35 mH

B) 35 mH

C) 3.5 mH

D) Zero

Show Answer

Answer:

Correct Answer: A

Solution:

Capacitance of wire

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

For impedance of the circuit to be minimum $ X_{L}=X_{C} $

Þ $ 2\pi \nu L=\frac{1}{2\pi \nu C} $

Þ $ L=\frac{1}{4{{\pi }^{2}}{{\nu }^{2}}C}=\frac{1}{4{{(3.14)}^{2}}\times {{(5\times 10^{3})}^{2}}\times 2.8\times {{10}^{-6}}} $ $ =0.35\times {{10}^{-3}}H=0.35,mH $



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