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Question: Answered & Verified by Expert
An $L-C-R$ series circuit with a resistance of $100 \Omega$ is connected to 200 V (AC source) and angular frequency $300 \mathrm{rad} / \mathrm{s}$. When only the capacitor is removed, then the current lags behind the voltage by $60^{\circ}$. When only the inductor is removed the current leads the voltage by $60^{\circ}$. The average power dissipated in original $L-C-R$ circuit is
PhysicsAlternating CurrentJIPMERJIPMER 2016
Options:
  • A 50 W
  • B 100 W
  • C 200 W
  • D 400 W
Solution:
1066 Upvotes Verified Answer
The correct answer is: 400 W
Given, $R=100 \Omega, V=200 \mathrm{~V}, \theta=60^{\circ}, \omega=300 \mathrm{rad} / \mathrm{s}$
$\therefore$ Phase angle, $\tan \theta=\frac{X_L}{R}=\frac{X_C}{R}$
$\Rightarrow \quad \tan 60^{\circ}=\frac{X_L}{R}=\frac{X_C}{R}$
$\therefore \quad X_L=X_C=\sqrt{3} R$
i.e.
$Z=\sqrt{R^2+(\sqrt{3} R-\sqrt{3} R)^2}=R$
So average power,
$P=\frac{V^2}{R}=\frac{200 \times 200}{100}=400 \mathrm{~W}$

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