Question

In a series $LCR$ 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 $LCR$ circuit is-

• A. $305\text{W}$
• B. $210\text{W}$
• C. $0\text{W}$
• D. $242\text{W}$

Answer 1

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



When capacitor is removed,

$\tan \varphi =\frac{X_L}{R}=\tan {30}^{\circ }$

$X_L=R\tan {30}^{\circ }=\frac{R}{\sqrt{3}}$

When the inductor is removed,

$\tan \varphi =\frac{X_C}{R}=\tan {30}^{\circ }\Rightarrow X_C=\frac{R}{\sqrt{3}}$

Here, $X_L=X_C\Rightarrow Z=R$

Hence, power dissipated in the circuit is,

$P=\frac{V^2}{R}=\frac{(220{)}^2}{200}=242\text{W}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.