Given below are two statements : Statement I : In an LCR series circuit, current is maximum at resonance. Statement II : Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source. In the light of the above statements, choose the correct from the options given below :

Both Statement I and Statement II are true

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Both Statement I and Statement II are false

Alternating Current PYQ — JEE Main 2024 · 6 Apr · Shift 1 · Q74 with Solution

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Correct answer: C

Analyze Statement I

In a series LCR circuit, the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$ where $X_L = \omega L, \qquad X_C = \frac{1}{\omega C}.$

The current is $I = \frac{V}{Z}.$

At resonance, $X_L = X_C,$ so the impedance becomes minimum: $Z = R.$

Therefore, the current becomes maximum: $I_{\max} = \frac{V}{R}.$

So, Statement I is true.
Analyze Statement II

For a purely resistive circuit connected to the same voltage source, current is $I_R = \frac{V}{R}.$

For a series LCR circuit, $I_{LCR} = \frac{V}{\sqrt{R^2 + (X_L - X_C)^2}}.$

Since $\sqrt{R^2 + (X_L - X_C)^2} \ge R,$ we get $I_{LCR} \le \frac{V}{R} = I_R.$

Equality occurs only at resonance.

Hence, the current in a purely resistive circuit can never be less than that in the corresponding series LCR circuit for the same source voltage.

So, Statement II is true.
Conclusion
- Statement I: True
- Statement II: True
Therefore, the correct option is: $\boxed{\text{C}}$