Question

An L-C-R series circuit with a resistance of $100\Omega$ is connected to $200V$ (AC source) and angular frequency $300rad/s$. When only the capacitor is removed, then the current lags behind the voltage by ${60}^o$. When only the inductor is removed the current leads the voltage by ${60}^o$. The average power dissipated in original L-C-R circuit is :

• A. $50W$
• B. $100W$
• C. $200W$
• D. $400W$

Answer 1

The correct option is D $400W$

Given,

$R=100\Omega ,V=200V,\theta ={60}^o,\omega =300rad/s$

$\therefore$ Phase angle, $\tan \theta =\frac{X_L}{R}=\frac{X_C}{R}$

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

$\therefore X_L=X_C=\sqrt{3}R$

i.e. $Z=\sqrt{R^2+{(\sqrt{3}R-\sqrt{3}R)}^2}=R$

So, average power,
$P=\frac{V^2}{R}=\frac{200\times 200}{100}=400W$