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 $
