An LCR circuit contains resistance of 110$\Omega$ and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45$^\circ$. If on the other hand, only inductor is removed the current leads by 45$^\circ$ with the applied voltage. The rms current flowing in the circuit will be :
Solution
Since $\phi$ remain same, circuit is in resonance<br><br>$\therefore$ ${I_{rms}} = {{{v_{rms}}} \over z}$ = ${{{v_{rms}}} \over R}$<br><br>$= {{220} \over {110}}$<br><br>$\Rightarrow$ ${I_{rms}} = 2A$
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About this question
Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C
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