A telegraph line of length 100 km has a capacity of 0.01 $\mu$F/km and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduced in series is _____________ mH. (if $\pi$ = $\sqrt{10}$)
Answer (integer)
100
Solution
<p>Total capacitance = 0.01 $\times$ 100 = 1 $\mu$F</p>
<p>$\omega$ = 500 $\times$ 2$\pi$ = 1000$\pi$ rad/s</p>
<p>$\omega L = {1 \over {\omega C}}$</p>
<p>$$ \Rightarrow L = {1 \over {{\omega ^2}C}} = {1 \over {{{10}^6}{\pi ^2} \times {{10}^{ - 6}}}} = {1 \over {10}}H$$ = 100 mH</p>
About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
This question is part of PrepWiser's free JEE Main question bank. 115 more solved questions on Alternating Current are available — start with the harder ones if your accuracy is >70%.