An LCR series circuit with a resistance of 10 | vedclass

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(D) Given: $R = 100\,\Omega$,$V_{rms} = 200\,V$.$1$. When the capacitor is removed,the circuit becomes an $LR$ series circuit. The phase angle $\phi_1

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(D) Given: $R = 100\,\Omega$,$V_{rms} = 200\,V$.$1$. When the capacitor is removed,the circuit becomes an $LR$ series circuit. The phase angle $\phi_1 = 60^\circ$ (current lags voltage).$	an \phi_1 = \frac{X_L}{R} \Rightarrow 	an 60^\circ = \frac{X_L}{100} \Rightarrow X_L = 100\sqrt{3}\,\Omega$.$2$. When the inductor is removed,the circuit becomes an $RC$ series circuit. The phase angle $\phi_2 = 60^\circ$ (current leads voltage).$	an \phi_2 = \frac{X_C}{R} \Rightarrow 	an 60^\circ = \frac{X_C}{100} \Rightarrow X_C = 100\sqrt{3}\,\Omega$.$3$. Since $X_L = X_C$,the circuit is in a state of resonance.At resonance,the impedance $Z = R = 100\,\Omega$.$4$. The average power dissipated in the circuit is given by:$P = \frac{V_{rms}^2}{R} = \frac{200^2}{100} = \frac{40000}{100} = 400\,W$.

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