An LCR circuit contains resistance of 110 , O | vedclass

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(D) Given: $R = 110 \, \Omega$,$V = 220 \, V$,$\omega = 300 \, rad/s$.When capacitance is removed,the circuit is an $LR$ circuit. The phase angle is g

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$2$

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(D) Given: $R = 110 \, \Omega$,$V = 220 \, V$,$\omega = 300 \, rad/s$.When capacitance is removed,the circuit is an $LR$ circuit. The phase angle is given by $	an \phi = \frac{X_L}{R}$. Since current lags by $45^{\circ}$,$	an 45^{\circ} = \frac{X_L}{R} = 1$,so $X_L = R$.When the inductor is removed,the circuit is an $RC$ circuit. The phase angle is given by $	an \phi = \frac{X_C}{R}$. Since current leads by $45^{\circ}$,$	an 45^{\circ} = \frac{X_C}{R} = 1$,so $X_C = R$.Since $X_L = R$ and $X_C = R$,it implies $X_L = X_C$.In an $LCR$ circuit,when $X_L = X_C$,the circuit is in resonance.At resonance,the impedance $Z = R$.The rms current $I = \frac{V}{Z} = \frac{V}{R} = \frac{220}{110} = 2 \, A$.

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