A 100$\Omega$ resistance, a 0.1 $\mu$F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.
Solution
C = 0.1 $\mu$F = 10<sup>$-$7</sup> F<br><br>Resonant frequency = 60 Hz.<br><br>${\omega _0} = {1 \over {\sqrt {LC} }}$<br><br>$2\pi {f_0} = {1 \over {\sqrt {LC} }} \Rightarrow L = {1 \over {4{\pi ^2}f_0^2C}}$<br><br>by putting values $L \simeq 70.3$ Hz.
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%.