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  • An L-C-R series circuit with a resistance of  is connected to an AC source of <img alt="200 \mathrm

An L-C-R series circuit with a resistance of 100 \Omega is connected to an AC source of 200 \mathrm{~V}(\mathrm{rms} ) and angular frequency 300 \mathrm{rad} \mathrm{s}^{-1}. When only the capacitor is removed, 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 LC-R circuit is:

Option: 1

50 W


Option: 2

100 W


Option: 3

200 W


Option: 4

400 W


Answers (1)

Phase angle,
\mathrm{\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}} }
\mathrm{=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}} }
\mathrm{\Rightarrow \tan 60^{\circ}=\frac{X_L}{R}=\frac{X_C}{R} \Rightarrow X_L=X_C=\sqrt{3} R }
i.e., \mathrm{\quad Z=\sqrt{R^2+(\sqrt{3} R-\sqrt{3} R)^2} \Rightarrow Z=R }
So, average power, \mathrm{\mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}=\frac{200 \times 200}{100}=400 \mathrm{~W} }

Posted by

Ramraj Saini

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