Question

A series LCR circuit with $100\Omega$ resistance is connected to an AC source of $200\text{V}$ and angular frequency $300\text{rad/s}$. When only the capacitance is removed, the current lags behind the voltage by ${60}^{\circ }$. When only the inductance is removed, the current leads the voltage by ${60}^{\circ }$. The average power dissipated in the LCR circuit is -

• A. $100\text{W}$
• B. $200\text{W}$
• C. $300\text{W}$
• D. $400\text{W}$

Answer 1

The correct option is D $400\text{W}$

When capacitance is removed, then

$\tan \varphi =\frac{X_L}{R}$

$\Rightarrow \tan {60}^{\circ }=\frac{X_L}{R}$

$\Rightarrow X_L=\sqrt{3}R...\left(1\right)$

When inductance is removed, then

$\tan \varphi =\frac{X_C}{R}$

$\Rightarrow \tan {60}^{\circ }=\frac{X_C}{R}$

$\Rightarrow X_C=\sqrt{3}R...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$,

$X_L=X_C$

$\Rightarrow$ LCR circuit is in resonance.

So, $Z=R$

Therefore,

$i_{\text{rms}}=\frac{V_{\text{rms}}}{Z}=\frac{200}{100}=2\text{A}$

Further, power dissipated in the LCR circuit,

$P_{\text{avg}}=V_{\text{rms}}{i}_{\text{rms}}\cos \varphi$

$\Rightarrow P_{\text{avg}}=200\times 2\times \cos 0^{\circ }[\text{At resonance,}\varphi =0^{\circ }]$

$\Rightarrow P_{\text{avg}}=400\text{W}$

Hence, option $\left(D\right)$ is the correct answer.