Question

In an $L-C-R$ circuit $R=100\Omega$. When capacitance $C$ is removed, the current lags behind the voltage by $\frac{\pi }{3}$., when inductance $L$ is removed, the current leads the voltage by $\frac{\pi }{3}$., The impedance of the circuit is

• A. $50\Omega$
• B. $100\Omega$
• C. $200\Omega$
• D. $400\Omega$

Answer 1

Answer: (B)

When $C$ is removed, circuit becomes $R-L$ circuit.
Hence, $\tan \frac{\pi }{3}=\frac{X_L}{R}\dots$ (i)
When $L$ is removed, circuit becomes $R-C$ circuit.
Hence, $\tan \frac{\pi }{3}=\frac{X_C}{R}$..(ii)
From Eqs. (i) and (ii),
we obtain $X_L=X_C$;
this is the condition of resonance
$\therefore Z=R=100\Omega$