In an electrical circuit,R, L, C and an AC vo | vedclass

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(A) When $L$ is removed,the circuit becomes an $RC$ circuit. The phase difference $\phi_1$ is given by $	an \phi_1 = \frac{X_C}{R}$.Given $\phi_1 = \

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(A) When $L$ is removed,the circuit becomes an $RC$ circuit. The phase difference $\phi_1$ is given by $	an \phi_1 = \frac{X_C}{R}$.Given $\phi_1 = \pi / 3$,so $	an(\pi / 3) = \frac{X_C}{R} \implies X_C = R \sqrt{3}$.When $C$ is removed,the circuit becomes an $RL$ circuit. The phase difference $\phi_2$ is given by $	an \phi_2 = \frac{X_L}{R}$.Given $\phi_2 = \pi / 3$,so $	an(\pi / 3) = \frac{X_L}{R} \implies X_L = R \sqrt{3}$.Since $X_L = X_C$,the circuit is in resonance.In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Substituting $X_L = X_C$,we get $Z = \sqrt{R^2 + 0} = R$.The power factor is $\cos \phi = \frac{R}{Z} = \frac{R}{R} = 1$.

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