Electro Magnetic Induction And Alternating Currents Question 90
Question: The peak value of an alternating e.m.f. E is given by $ E=E_{0}\cos \omega ,t $ is 10 volts and its frequency is 50 Hz. At time $ t=\frac{1}{600}sec $ , the instantaneous e.m.f. is [MP PMT 1990; MP PET 2004]
Options:
A) 10 V
B) $ 5\sqrt{3},V $
C) 5 V
D) 1 V
Show Answer
Answer:
Correct Answer: B
Solution:
$ E=E_{0}\cos \omega t=E_{0}\cos \frac{2\pi t}{T} $ $ =10\cos \frac{2\pi \times 50\times 1}{600}=10\ \cos \frac{\pi }{6} $ $ =5\sqrt{3}\ volt. $
