Write the given voltage and current

$V = 20\sin(200\pi t)$ $I = 10\sin\left(200\pi t + \frac{\pi}{3}\right)$

So,

• Voltage amplitude: $V_0 = 20\,\text{V}$
• Current amplitude: $I_0 = 10\,\text{A}$
• Phase difference: $\phi = \frac{\pi}{3}$

Since current leads voltage by $\frac{\pi}{3}$, the power factor is $\cos\phi = \cos\frac{\pi}{3} = \frac{1}{2}$

Use the average power formula in AC

Average power dissipated is $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi$

where $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{20}{\sqrt{2}}$ $I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}}$

Substitute the values

$P_{\text{avg}} = \left(\frac{20}{\sqrt{2}}\right)\left(\frac{10}{\sqrt{2}}\right)\left(\frac{1}{2}\right)$

$P_{\text{avg}} = \frac{20\times 10}{2}\times \frac{1}{2}$

$P_{\text{avg}} = 100 \times \frac{1}{2} = 50\,\text{W}$

Check options

• A: $21.6\,\text{W}$ ❌
• B: $200\,\text{W}$ ❌
• C: $173.2\,\text{W}$ ❌
• D: $50\,\text{W}$ ✅

Conclusion

The average power dissipated is $\boxed{50\,\text{W}}$