An L-C-R series circuit with 100 , Omega resi | vedclass

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Question

(A) Given: Resistance $R = 100 \, \Omega$, Voltage $V = 100 \, V$.Case $1$: When the capacitor is removed, the circuit becomes an $L-R$ series circuit

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$1$

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(A) Given: Resistance $R = 100 \, \Omega$, Voltage $V = 100 \, V$.Case $1$: When the capacitor is removed, the circuit becomes an $L-R$ series circuit. The current lags behind the voltage by $\phi = 45^{\circ}$.$	an \phi = \frac{X_L}{R} \implies 	an 45^{\circ} = \frac{X_L}{100} \implies 1 = \frac{X_L}{100} \implies X_L = 100 \, \Omega$.Case $2$: When the inductor is removed, the circuit becomes a $C-R$ series circuit. The current leads the voltage by $\phi = 45^{\circ}$.$	an \phi = \frac{X_C}{R} \implies 	an 45^{\circ} = \frac{X_C}{100} \implies 1 = \frac{X_C}{100} \implies X_C = 100 \, \Omega$.Since $X_L = X_C$, the circuit is in resonance. In resonance, the impedance $Z = R$.$Z = 100 \, \Omega$.The current flowing in the circuit is $I = \frac{V}{Z} = \frac{100}{100} = 1 \, A$.

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