Discussion Forum

The equation of AC voltage  is  $E= 220 sin (\omega t+\frac{\pi}{6})$ and AC current $I=10\sin (\omega t+\frac{\pi}{6})$ . The average  power dissipated is 

Answer:

Explanation:

 We know that ,   $Z=\frac{E_{0}}{I_{0}}$

 E₀= 220 V and I₀=10 A so Z= 220/10=  22 ohm

 $\phi =\left[ \frac{\pi}{6}-(-\frac{\pi}{6})\right]=\frac{\pi}{3}$

  $P_{a}=\frac{E_{0}}{\sqrt{2}}\times\frac{I_{0}}{\sqrt{2}}\times\cos\phi$

  $=\frac{220}{\sqrt{2}}\times\frac{10}{\sqrt{2}}\times\cos\frac{\pi}{3}=550 W$