An LCR series circuit with a resistance of 10 | vedclass

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(D) In an $LCR$ series circuit, when the capacitor is removed, the circuit becomes an $LR$ circuit. The phase angle is given by $	an \phi = \frac{X_L

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) In an $LCR$ series circuit, when the capacitor is removed, the circuit becomes an $LR$ circuit. The phase angle is given by $	an \phi = \frac{X_L}{R}$. Given $\phi = 60^o$, we have $	an 60^o = \frac{X_L}{R} \Rightarrow X_L = R \sqrt{3}$.When the inductor is removed, the circuit becomes an $RC$ circuit. The phase angle is given by $	an \phi = \frac{X_C}{R}$. Given $\phi = 60^o$, we have $	an 60^o = \frac{X_C}{R} \Rightarrow X_C = R \sqrt{3}$.Since $X_L = X_C$, the circuit is in resonance when both components are present.At resonance, the impedance $Z = R = 100 \, \Omega$.The average power dissipated is $P = \frac{V^2}{R} = \frac{200^2}{100} = \frac{40000}{100} = 400 \, W$.

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