Question

An $\text{LCR}$ series circuit with $100\Omega$ resistance is connected to an $A.C.$ Source of $200\text{V}$ and angular frequency $300{\text{rad s}}^{-1}$. When only the capacitor is removed, the current lags behind the voltage by ${60}^{\circ }$. When only inductor is removed, the current leads the voltage by ${60}^{\circ }$. If all elements are connected, the current in the circuit is

• A. $2.5\text{A}$
• B. $2\text{A}$
• C. $1.5\text{A}$
• D. $0.5\text{A}$

Answer 1

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

$\text{Case-I}:$ $L-R$ circuit (C-removed)

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

Substituting the data gives,

$\tan {60}^{\circ }=\frac{X_L}{R}\Rightarrow X_L=100\sqrt{3}......\left(1\right)$

$\text{Case-II}:$ $C-R$ circuit (L-removed)

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

Substituting the data gives,
$\tan {60}^{\circ }=\frac{X_C}{100}\Rightarrow X_C=100\sqrt{3}...\left(2\right)$

$\text{Case-III}:$ Since $X_L=X_C$, circuit is in resonance

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

Hence, option (b) is the correct answer.