When resonance is produced in a series LCR circuit, then which of the following is not correct?

In a series LCR circuit, resonance occurs when the inductive reactance $$X_L$$ equals the capacitive reactance $$X_C$$. At this frequency, the circuit behaves purely resistively. Let's evaluate each option step by step.

Option A states that the current in the circuit is in phase with the applied voltage. At resonance, the phase angle $$\phi$$ is given by $$\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$$. Since $$X_L = X_C$$, $$\phi = 0$$, meaning the current and voltage are in phase. Thus, option A is correct.

Option B states that inductive and capacitive reactances are equal. This is the defining condition for resonance in a series LCR circuit, so $$X_L = X_C$$. Therefore, option B is correct.

Option C states that if resistance $$R$$ is reduced, the voltage across the capacitor will increase. At resonance, the impedance $$Z = R$$ (minimum), so the current amplitude $$I_0 = \frac{V_0}{R}$$, where $$V_0$$ is the peak applied voltage. The voltage across the capacitor is $$V_C = I \cdot X_C$$. Substituting, $$V_C = \left(\frac{V_0}{R}\right) \cdot X_C$$. Since $$X_C = \frac{1}{\omega C}$$ remains constant at the same resonant frequency, $$V_C$$ is inversely proportional to $$R$$. If $$R$$ decreases, $$V_C$$ increases. Thus, option C is correct.

Option D states that the impedance of the circuit is maximum. The impedance in a series LCR circuit is $$Z = \sqrt{R^2 + (X_L - X_C)^2}$$. At resonance, $$X_L = X_C$$, so $$Z = \sqrt{R^2 + 0} = R$$. This is the minimum possible impedance, not the maximum. Therefore, option D is not correct.

Hence, the correct answer is Option D.