Q. An $LCR$ series circuit with $R = 100\Omega$ is connected to a $200\ V,\ 50\ Hz$ a.c. source when only the capacitance is removed, the current leads the voltage by $60{^\circ}$. When only the inductance is removed, the current leads the voltage by $60{^\circ}$. The current in the circuit is
A) $2A$
B) $1A$
C) $\frac{\sqrt{3}}{2}A$
D) $\frac{2}{\sqrt{3}}A$
Explanation
(a) If the capacitance is removed, it is an $L - R$ circuit $\phi = 60{^\circ}$
$\tan\phi = \frac{X_{L}}{R} = \tan{60{^\circ}} = \sqrt{3}$
If inductance is removed, it is a capacitative circuit or $R - C$ circuit. $|\phi|$ is the same
$\therefore L\omega = \frac{1}{C\omega}$ This is a resonance circuit
$Z = R;I_{rms} = \frac{E_{rms}}{R},\ E_{rms} = 200\ V$
$\therefore I_{rms} = \frac{200V}{100\Omega} = 2A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.