Alternating Current

JEE Main level, single correct, medium

106 of 190

Question 106 of 190

A 100Ω100\Omega resistance, a 0.1μF0.1\mu F capacitor and an inductor are connected in series across a 250V250V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60Hz60Hz.

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Answer: (D) 70.3H70.3H

C=0.1μF=10−7FC=0.1\mu F={10}^{-7}F
Resonant frequency =60Hz=60Hz
ωo=1LC{\omega }_{o}=\dfrac{1}{\sqrt{LC}}
2πf0=1LC⇒L=14π2f02C2\pi {f}_{0}=\dfrac{1}{\sqrt{LC}}\Rightarrow L=\dfrac{1}{4{\pi }^{2}{f}_{0}^{2}C}
by putting values L≃70.3HzL≃70.3Hz.

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