In an LCR series circuit,when L is removed fr | vedclass

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(A) In an $LCR$ series circuit,the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.When $L$ is removed,the circuit becomes an $RC$ circuit. The phase di

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(A) In an $LCR$ series circuit,the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$.When $L$ is removed,the circuit becomes an $RC$ circuit. The phase difference $\phi$ is given by $	an \phi = \frac{X_C}{R}$. Given $\phi = \frac{\pi}{3}$,so $	an \frac{\pi}{3} = \sqrt{3} = \frac{X_C}{R}$,which implies $X_C = \sqrt{3}R$.When $C$ is removed,the circuit becomes an $RL$ circuit. The phase difference $\phi$ is given by $	an \phi = \frac{X_L}{R}$. Given $\phi = \frac{\pi}{3}$,so $	an \frac{\pi}{3} = \sqrt{3} = \frac{X_L}{R}$,which implies $X_L = \sqrt{3}R$.In the original $LCR$ circuit,$X_L = X_C$,so the circuit is at resonance.At resonance,the impedance $Z = R$.The power factor $\cos \phi = \frac{R}{Z} = \frac{R}{R} = 1$.

In an $LCR$ series circuit,when $L$ is removed from the circuit,the phase difference between voltage and current is $\frac{\pi}{3}$. If $C$ is removed from the circuit instead of $L$,the phase difference is again $\frac{\pi}{3}$. The power factor of the circuit is $(\tan 60^{\circ}=\sqrt{3})$.

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