Question

An $LCR$ circuit contains of $100\Omega$ resistance and an $AC$ supply of $200\text{V}$ at $300\text{rad/s}$ angular frequency. When only capacitance is taken out from the circuit, the current lags the voltage by ${60}^{\circ }$. If, on the other hand, only the inductor is taken out, the current leads the applied voltage by ${60}^{\circ }$. The current flowing in the original circuit is,

• A. $1\text{A}$
• B. $1.5\text{A}$
• C. $2\text{A}$
• D. $2.5\text{A}$

Answer 1

The correct option is C $2\text{A}$

$\text{Case 1}$ : When only capacitor is removed,

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


$\text{Case 2}$ : when only inductor is removed,

$\tan {60}^{\circ }=\frac{X_C}{R}=\frac{1}{\omega CR}$


From the above conditions, we get

$X_L=X_C$

$\therefore$ For the original circuit,

$Z=R=100\Omega [\because Z=\sqrt{R^2+(X_L-X_C{)}^2}]$

$\Rightarrow i=\frac{E_{rms}}{R}=\frac{200}{100}=2\text{A}$

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