Q. In an $LCR$ circuit $R = 100\ ohm$. When capacitance $C$ is removed, the current lags behind the voltage by $\pi/3$. When inductance $L$ is removed, the current leads the voltage by $\pi/3$. The impedance of the circuit is
A) 50 $ohm$
B) 100 $ohm$
C) 200 $ohm$
D) 400 $ohm$
Explanation
(b) When $C$ is removed circuit becomes $RL$ circuit hence
$\tan\frac{\pi}{3} = \frac{X_{L}}{R}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
When $L$ is removed circuit becomes $RC$ circuit hence
$\tan\frac{\pi}{3} = \frac{X_{C}}{R}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From equation (i) and (ii) we obtain $X_{L} = X_{C}$. This is the condition of resonance and in resonance $Z = R = 100\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.