Electro Magnetic Induction And Alternating Currents Question 165
Question: The equation of AC voltage is $ E=220,\sin ,(\omega t+\pi /6) $ and the A.C. current $ I=10,\sin ,(\omega t+\pi /6) $ . The average power dissipated is
Options:
A) 150 W
B) 550 W
C) 250 W
D) 50 W
Show Answer
Answer:
Correct Answer: B
Solution:
-
We know that, $ Z=\frac{E_{0}}{I_{0}} $
Given, $ E_{0}=220 $ and $ I_{0}=10 $
So $ Z=\frac{220}{10}=22,ohm $
$ \phi =[ \frac{\pi }{6}-( -\frac{\pi }{6} ) ]=\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} $
