Question

The equation of A.C. voltage is $E=220 \sin (\omega t+\pi / 6)$ and the A.C. current is $I=10 \sin (\omega t-\pi / 6)$. The average power dissipated is :

• A. 150W
• B. 550W
• C. 250W
• D. 50W

Answer 1

Answer: (B)

We know that, $ Z = \frac{E_0}{I_0}$
Given, $E_0 = 220V $ and $I_0 = 10 A$
so $Z = \frac{220}{10} = 22 \, ohm$
$ \phi =\left[\frac{\pi}{6} - \left(-\frac{ \pi }{6}\right)\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 $