Question

In a series $L-C-R$ circuit, $R=200\Omega$ and the voltage and the frequency of the main supply is $220\text{V}$ and $50\text{Hz}$ respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by ${30}^{\circ }$. On taking out the inductor from the circuit, the current leads the voltage by ${30}^{\circ }$. The power dissipated in the $L-C-R$ circuit is

• A. $305\text{W}$
• B. $210\text{W}$
• C. Zero
• D. $242\text{W}$

Answer 1

The correct option is D $242\text{W}$

If capacitor is removed, current lags by ${30}^{\circ }$
$\tan {30}^{\circ }=\frac{X_L}{R}\Rightarrow X_L=\frac{R}{\sqrt{3}}$
If inductor is removed, current leads by ${30}^{\circ }$
$\tan {30}^{\circ }=\frac{X_C}{R}\Rightarrow X_C=\frac{R}{\sqrt{3}}$
i.e $X_L=X_C$
So, Phase difference of complete circuit is $\varphi =0^{\circ }$

Power dissipated in the circuit is
$P=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi$
$P=\frac{V_{rms}^2}{R}=242\text{W}$