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} \ldots $ (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
$\therefore Z=R=100 \,\Omega$