A series LCR circuit with resistance R=500 O | vedclass

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(D) When the capacitance is removed,the circuit becomes an $LR$ circuit. The phase difference is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = 60

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$500 \ \Omega$

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(D) When the capacitance is removed,the circuit becomes an $LR$ circuit. The phase difference is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = 60^{\circ}$,we have $	an 60^{\circ} = \frac{X_L}{R} = \sqrt{3}$,so $X_L = R\sqrt{3}$.When the inductance is removed,the circuit becomes an $RC$ circuit. The phase difference is given by $	an \phi = \frac{X_C}{R}$.Given $\phi = 60^{\circ}$,we have $	an 60^{\circ} = \frac{X_C}{R} = \sqrt{3}$,so $X_C = R\sqrt{3}$.Since $X_L = X_C$,the circuit is in resonance.At resonance,the impedance $Z = R$.Given $R = 500 \ \Omega$,the impedance $Z = 500 \ \Omega$.

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