Alternating Current

The formula to find the average (real) power in an AC circuit is:$P = V_{\text{rms}} \times I_{\text{rms}} \times \cos \varphi$ where $\varphi$ is the phase difference between voltage and current.The voltage is $50 \sin(50 t)$, so its maximum (peak) value is 50 V .The current is $50 \sin\left(50 t + \frac{\pi}{4}\right)$, so its peak value is $50 \text{mA} = 50 \times 10^{-3} \text{A}$.The phase difference $\varphi$ between voltage and current is $\frac{\pi}{4}$.To get the RMS (root mean square) values from the peak values, divide by $\sqrt{2}$ :$V_{\text{rms}} = \frac{50}{\sqrt{2}}$$I_{\text{rms}} = \frac{50 \times 10^{-3}}{\sqrt{2}}$Now, put these values into the power formula:$P = \frac{50}{\sqrt{2}} \times \frac{50 \times 10^{-3}}{\sqrt{2}} \times \cos\left(\frac{\pi}{4}\right)$$\cos\left(\frac{\pi}{4}\right) \text{ is } \frac{1}{\sqrt{2}}.$So,$P = \frac{50}{\sqrt{2}} \times \frac{50}{\sqrt{2}} \times 10^{-3} \times \frac{1}{\sqrt{2}}$Multiply the numbers:$= \frac{50 \times 50 \times 10^{-3}}{\sqrt{2} \times \sqrt{2} \times \sqrt{2}}$$\sqrt{2} \times \sqrt{2} \times \sqrt{2} = 2 \sqrt{2}$$= \frac{2500 \times 10^{-3}}{2 \sqrt{2}}$$= \frac{2.5}{2 \sqrt{2}}$This is approximately equal to 0.884 W .