The figure shows an LCR series AC circuit. If | vedclass

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(C) In an $LCR$ series circuit,the phase angle $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.When $L$ is removed,the circuit becomes an $RC$ ci

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(C) In an $LCR$ series circuit,the phase angle $\phi$ is given by $	an \phi = \frac{X_L - X_C}{R}$.When $L$ is removed,the circuit becomes an $RC$ circuit. The current leads the voltage by $45^o$,so $\phi = -45^o$. Thus,$	an(-45^o) = -\frac{X_C}{R}$,which implies $\frac{X_C}{R} = 1$,or $X_C = R$.When $C$ is removed,the circuit becomes an $RL$ circuit. The current lags the voltage by $45^o$,so $\phi = 45^o$. Thus,$	an(45^o) = \frac{X_L}{R}$,which implies $\frac{X_L}{R} = 1$,or $X_L = R$.Since $X_L = R$ and $X_C = R$,it follows that $X_L = X_C$. This means the circuit is at resonance.At resonance,the impedance of the circuit is $Z = R = 100 \, \Omega$.The current in the original circuit is $I = \frac{V}{Z} = \frac{100 \, V}{100 \, \Omega} = 1 \, A$.

The figure shows an $LCR$ series $AC$ circuit. If $L$ is removed from the circuit,the current leads the voltage by $45^o$,while if $C$ is removed,the current lags the voltage by $45^o$. The current passing in the original circuit is......$A$.

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