Given below are two statements: Statement I : When the frequency of an a.c source in a series LCR circuit increases, the current in the circuit first increases, attains a maximum value and then decreases. Statement II : In a series LCR circuit, the value of power factor at resonance is one. In the light of given statements, choose the most appropriate answer from the options given below.

Both Statement I and Statement II are False.

Statement I is incorrect but Statement II is true.

Statement I is correct but Statement II is false.

Alternating Current PYQ — JEE Main 2023 · 12 Apr · Shift 1 · Q59 with Solution

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Correct answer: C

In a series LCR circuit, the impedance is

$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$

So the current is

$I = \frac{V}{Z}$

As frequency increases:

At low frequency, capacitive reactance $X_C = \dfrac{1}{\omega C}$ is large, so $|X_L - X_C|$ is large and current is small.
As frequency approaches resonance, $X_L = X_C$ , so impedance becomes minimum: $Z = R$ hence current becomes maximum.
Beyond resonance, inductive reactance $X_L = \omega L$ becomes large, so impedance increases again and current decreases.

Therefore, current first increases, becomes maximum at resonance, and then decreases.

So, Statement I is true.

Power factor in an AC circuit is

$\cos\phi = \frac{R}{Z}$

At resonance in a series LCR circuit,

$X_L = X_C \implies Z = R$

Thus,

$\cos\phi = \frac{R}{R} = 1$

So, the power factor at resonance is one.

Therefore, Statement II is true.

Since both Statement I and Statement II are true, the correct option is:

$\boxed{\text{C}}$

Stored correct answer: C

My derived answer: C

They agree.