Power in AC and Power Factor Questions in English

(C) The given equations are $E = 50 \sin(500t)$ and $I = 600 \sin(500t + \frac{\pi}{3}) \text{ mA}$.
Peak voltage $E_0 = 50 \text{ V}$.
Peak current $I_0 = 600 \text{ mA} = 0.6 \text{ A}$.
Phase difference $\phi = \frac{\pi}{3} = 60^{\circ}$.
The average power dissipated in an $AC$ circuit is given by $P = E_{rms} I_{rms} \cos \phi$.
$E_{rms} = \frac{E_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.
$P = \frac{E_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \times \cos \phi = \frac{E_0 I_0}{2} \cos \phi$.
Substituting the values: $P = \frac{50 \times 0.6}{2} \times \cos 60^{\circ}$.
$P = \frac{30}{2} \times 0.5 = 15 \times 0.5 = 7.5 \text{ W}$.
Wait, checking the calculation: $50 \times 0.6 = 30$. $30 / 2 = 15$. $15 \times 0.5 = 7.5 \text{ W}$.
Re-evaluating the options provided, there might be a typo in the question's options. Given the standard format, if $I = 600 \text{ mA} = 0.6 \text{ A}$, the result is $7.5 \text{ W}$. If $I = 400 \text{ mA}$, it would be $5 \text{ W}$. Assuming the closest logical answer based on standard textbook problems of this type, $7.5 \text{ W}$ is the calculated value.

(D) The average power dissipated in an $AC$ circuit is given by the formula: $P_{avg} = V_{rms} I_{rms} \cos \phi$.
Here, $V_{rms} = \frac{V_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$.
The phase difference $\phi$ between the voltage and current is $\frac{\pi}{3}$ (or $60^{\circ}$).
Substituting these values into the formula:
$P_{avg} = \left(\frac{V_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos 60^{\circ}$.
$P_{avg} = \frac{V_0 I_0}{2} \times 0.5$.
$P_{avg} = \frac{V_0 I_0}{2} \times \frac{1}{2} = \frac{V_0 I_0}{4}$.