Question

An $\text{LCR}$ circuit contains resistance of $10\Omega$ and a supply of $200\text{V}$ at $300\text{rad}$ angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by ${60}^{\circ }.$ If, on the other hand, only inductor is taken out the current leads by ${60}^{\circ }$ with the applied voltage. The current flowing in the $\text{LCR}$ circuit (in $\text{A}$)is (integer only)

• A. 20.00
• B. 20
• C. 20.0

Answer 1

Answer: (A), (B), (C)

Given,
$R=10\Omega ;V_{rms}=200\text{V};\omega =300\text{rad}$

When, capacitor is removed,



$\tan \varphi =\frac{V_L}{V_R}=\frac{\omega L}{R}[\because \varphi ={60}^{\circ }]$

$\tan {60}^{\circ }=\frac{\omega L}{R}.......\left(1\right)$

When, the inductor is removed,



$\tan \varphi =\frac{V_C}{V_R}=\frac{1}{\omega CR}[\because \varphi ={60}^{\circ }]$

$\tan {60}^{\circ }=\frac{1}{\omega CR}.......\left(2\right)$

From (1) and (2),

$\therefore \omega L=\frac{1}{\omega C}\text{(case of resonance)}$

Now, $Z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}=R=10\Omega$

$\therefore I_{rms}=\frac{V_{rms}}{Z}=\frac{200\text{V}}{10\Omega }=20\text{A}$