An LCR series circuit with  $100\text{Ω}$   resistance is connected to an a.c. source of r.m.s. voltage  $200\text{ V}$  and angular frequency  $300\text{ rad}/\text{sec}$. 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 ${60}^{\circ }$ . Calculate the r. m. s. current (in ampere) in the LCR series circuit. 

When capacitance is removed, the circuit contains only inductance and resistance. Phase difference  $\theta$  between the current and voltage is then given by  $\text{tan}\theta =(\omega \text{L}/\text{R})$, or,  $\omega \text{L}=\text{Rtan}\theta =100\text{tan}{60}^{\circ }$
When the circuit contains only capacitance and resistance, the phase difference between the voltage and current is given by  $\text{tan}\varphi =\frac{1}{\text{RC}\omega }$
 $\therefore \frac{1}{\text{C}\omega }=\text{Rtan}\varphi =100\text{tan}{60}^{\circ }$
The impedance of the LCR circuit is given by
 
 $Z=\sqrt{R^2+{(\omega b-\frac{1}{\omega c})}^2}$    $=\sqrt{{\text{R}}^2+{(100\text{tan}{60}^{\circ }-100\text{tan}{60}^{\circ })}^2}=\text{R}=100\text{Ω}$
The r.m.s. current is given by  $\text{I}=\frac{\text{V}}{\text{R}}=\frac{200}{100}=2\text{ A}$