In an LCR circuit R = 100 Omega. When capaci | vedclass

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(B) When $C$ is removed,the circuit becomes an $RL$ circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = \pi /3$,so $

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$100$

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(B) When $C$ is removed,the circuit becomes an $RL$ circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = \pi /3$,so $	an(\pi /3) = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{R} \implies X_L = R\sqrt{3}$.When $L$ is removed,the circuit becomes an $RC$ circuit. The phase angle $\phi$ is given by $	an \phi = \frac{X_C}{R}$.Given $\phi = \pi /3$,so $	an(\pi /3) = \frac{X_C}{R} \implies \sqrt{3} = \frac{X_C}{R} \implies X_C = R\sqrt{3}$.Since $X_L = X_C$,the circuit is in a state of resonance.In an $LCR$ circuit at resonance,the impedance $Z = R$.Given $R = 100 \ \Omega$,therefore $Z = 100 \ \Omega$.

List-$I$	List-$II$
$A$. Purely capacitive circuit	$I$. $I$ leads $V$ by $90^{\circ}$
$B$. Purely inductive circuit	$II$. $I$ and $V$ are in phase
$C$. $LCR$ series at resonance	$III$. $V$ leads $I$ by angle $\theta$
$D$. $LCR$ series circuit	$IV$. $V$ leads $I$ by $90^{\circ}$

In an $LCR$ circuit $R = 100 \ \Omega$. 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......$\Omega$.

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