In an LRC series circuit at resonance,the cur | vedclass

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(A) Initially at resonance: $X_L = X_C \Rightarrow Z = R$.Therefore,the peak current is $I_0 = \frac{V_0}{R} = 10\sqrt{2} \ A$.When the current lags b

Vedclass · Accepted Answer

Frequency must be increased and current after the change is $10 \ A$.

Vedclass · Answer

(A) Initially at resonance: $X_L = X_C \Rightarrow Z = R$.Therefore,the peak current is $I_0 = \frac{V_0}{R} = 10\sqrt{2} \ A$.When the current lags by $45^o$,the circuit must be inductive,meaning $X_L > X_C$. This implies the frequency $\omega$ must be greater than the resonant frequency $\omega_0 = \frac{1}{\sqrt{LC}}$. Thus,the frequency must be increased.For a phase angle $\phi = 45^o$,we have $	an(45^o) = \frac{X_L - X_C}{R} = 1$,which means $X_L - X_C = R$.The new impedance is $Z' = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + R^2} = R\sqrt{2}$.The new current is $I' = \frac{V_0}{Z'} = \frac{V_0}{R\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10 \ A$.

In an $LRC$ series circuit at resonance,the current in the circuit is $10\sqrt{2} \ A$. If the frequency of the source is changed such that the current now lags by $45^o$ behind the applied voltage,which of the following is correct?

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