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 $I_0=10A$
so $Z=\frac{220}{10}=22ohm$
$\varphi =\left[\frac{\pi }{6}-\left(-\frac{\pi }{6}\right)\right]=\frac{\pi }{3}$
$p_a=\frac{E_0}{\sqrt{2}}\times \frac{I_0}{\sqrt{2}}\times \cos \varphi$
$=\frac{220}{\sqrt{2}}\times \frac{10}{\sqrt{2}}\times \cos \frac{\pi }{3}=550W$