An LCR series circuit with 100 Omega resistan | vedclass

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Question

(A) In an $LCR$ series circuit,the impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.When the capacitor is removed,the circuit becomes an $LR$ ci

Vedclass · Accepted Answer

$400 W$

Vedclass · Answer

(A) In an $LCR$ series circuit,the impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.When the capacitor is removed,the circuit becomes an $LR$ circuit. The phase angle $\phi$ 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 inductor is removed,the circuit becomes an $RC$ circuit. The phase angle $\phi$ 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 a state of resonance.At resonance,the impedance $Z = R = 100 \Omega$.The power dissipated in the $LCR$ circuit is $P = \frac{V^2}{R} = \frac{(200)^2}{100} = \frac{40000}{100} = 400 W$.

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