In an LCR circuit,R = 100 Omega. When the ca | vedclass

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(B) For the $RL$ circuit,the phase angle $\phi = \pi / 3$. The tangent of the phase angle is given by $	an \phi = \frac{X_L}{R}$.Thus,$	an(\pi / 3)

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(B) For the $RL$ circuit,the phase angle $\phi = \pi / 3$. The tangent of the phase angle is given by $	an \phi = \frac{X_L}{R}$.Thus,$	an(\pi / 3) = \frac{X_L}{R} \implies \sqrt{3} = \frac{X_L}{100} \implies X_L = 100\sqrt{3} \ \Omega$.For the $RC$ circuit,the phase angle $\phi = \pi / 3$. The tangent of the phase angle is given by $	an \phi = \frac{X_C}{R}$.Thus,$	an(\pi / 3) = \frac{X_C}{R} \implies \sqrt{3} = \frac{X_C}{100} \implies X_C = 100\sqrt{3} \ \Omega$.In the $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.Since $X_L = X_C$,the term $(X_L - X_C) = 0$.Therefore,$Z = \sqrt{R^2 + 0} = R = 100 \ \Omega$.

In an $LCR$ circuit,$R = 100 \ \Omega$. When the capacitor $C$ is removed,the current lags behind the voltage by a phase of $\pi / 3$. When the inductor $L$ is removed,the current leads the voltage by a phase of $\pi / 3$. What is the impedance of the circuit in $\Omega$?

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