Correct Option: (D) \(\frac{i_0 v_0}{2} \cos \phi\)

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

The instantaneous power \(p\) in an AC circuit is the product of instantaneous voltage \(v\) and instantaneous current \(i\). The average power \(P_{avg}\) over a complete cycle \(T\) is given by:

$$P_{avg} = \frac{1}{T} \int_{0}^{T} v \cdot i \, dt$$

Given: \(v = v_0 \sin \omega t\) and \(i = i_0 \sin (\omega t + \phi)\).

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

Substitute the expressions into the power formula:

$$P_{avg} = \frac{1}{T} \int_{0}^{T} (v_0 \sin \omega t) \cdot (i_0 \sin (\omega t + \phi)) \, dt$$

Using the trigonometric identity \(\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]\):

$$P_{avg} = \frac{v_0 i_0}{2T} \int_{0}^{T} [\cos(-\phi) - \cos(2\omega t + \phi)] \, dt$$

Since the average value of a cosine function over a full cycle is zero (\(\int_{0}^{T} \cos(2\omega t + \phi) \, dt = 0\)) and \(\cos(-\phi) = \cos \phi\):

$$P_{avg} = \frac{v_0 i_0}{2T} \int_{0}^{T} \cos \phi \, dt = \frac{v_0 i_0}{2T} \cdot \cos \phi \cdot [t]_0^T$$

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

Simplifying the expression:

$$P_{avg} = \frac{v_0 i_0}{2T} \cdot \cos \phi \cdot T = \frac{v_0 i_0}{2} \cos \phi$$

Thus, the average power consumed is \(\frac{i_0 v_0}{2} \cos \phi\).

💡 EL10 Concept Insight: Think of power as the "effective" energy flow; because the voltage and current aren't perfectly in sync (due to the phase angle \(\phi\)), we multiply the peak values by 1/2 to get the average, and then multiply by \(\cos \phi\) to account for how much they actually "overlap" in time.