Correct Option: (C)

Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]

In a series LCR circuit, the power loss (average power dissipated) is given by the formula:

$$P = V_{rms} I_{rms} \cos \phi$$

Where:

• $V_{rms}$ is the rms voltage of the ac source.
• $I_{rms}$ is the rms current in the circuit.
• $\cos \phi$ is the power factor, where $\phi$ is the phase angle between voltage and current.

Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]

At resonance, the inductive reactance ($X_L = \omega L$) is equal to the capacitive reactance ($X_C = \frac{1}{\omega C}$), which implies:

$$X_L - X_C = 0$$

The impedance of the circuit $Z$ becomes purely resistive:

$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$$

Consequently, the phase angle $\phi = 0$, making the power factor $\cos \phi = \cos(0) = 1$. The current at resonance is:

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

Substituting these into the power formula:

$$P = V \cdot I \cdot (1) = V \cdot \left( \frac{V}{R} \right)$$

Step 3 (Final Concluded Answer): [1/3 Mark]

The power loss at resonance is:

$$P = \frac{V^2}{R}$$

Thus, the correct option is (C).

💡 EL10 Concept Insight:

At resonance, the circuit's "push-pull" forces (inductor and capacitor) perfectly cancel each other out, leaving only the resistor to fight the current. Since the resistor is the only component that actually consumes energy as heat, the circuit acts like a simple DC circuit where power is just voltage squared divided by resistance.