Given below are two statements: Statement I : Maximum power is dissipated in a circuit containing an inductor, a capacitor and a resistor connected in series with an AC source, when resonance occurs Statement II : Maximum power is dissipated in a circuit containing pure resistor due to zero phase difference between current and voltage. In the light of the above statements, choose the correct answer from the options given below:

Both Statement I and Statement II are true

Statement I is false but Statement II is true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Alternating Current PYQ — JEE Main 2023 · 10 Apr · Shift 1 · Q61 with Solution

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Correct answer: D

For a series $RLC$ circuit connected to an AC source, 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}

The average power dissipated in the circuit is

P = VI\cos\phi = I^2R

where

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

At resonance,

X_L = X_C

Z = R

which is the minimum impedance of the circuit. Therefore current becomes maximum:

I_{\max} = \frac{V}{R}

Hence,

P = I^2R

becomes maximum at resonance.

So, Statement I is true.

For a pure resistor in an AC circuit, current and voltage are in phase, so

\phi = 0

Therefore,

\cos\phi = 1

which is the maximum possible value of power factor.

Average power is

P = VI\cos\phi

Thus, for a given $V$ and $I$ , power is maximum when

\phi = 0

That is exactly the case for a pure resistor.

So, Statement II is true.

Therefore, the correct option is:

\boxed{\text{D}}