Question

The peak value of an alternating e.m.f. E is given by $ E={{E}_{0}}\,\cos \omega t $ is 10volt and its frequency is 50 Hz. At time $ t=\frac{1}{600} $ sec, the instantaneous e.m.f. is:

• A. 1 V
• B. 5 V
• C. 10 V
• D. $ 5\sqrt{3}V $

Answer 1

Answer: (D)

Instantaneous value of E is given by $ E={{E}_{0}}\cos \omega t $ Here: $ {{E}_{0}}=10\,\text{volt,} $ $ t=\frac{1}{600}\sec , $ $ f=50\,Hz $ $ \therefore $ $ E={{E}_{0}}\cos \frac{2\pi t}{T} $ or $ E=10\cos \frac{2\pi \times 50\times 1}{600} $ $ =10\,\cos \frac{5\pi }{3}\times \frac{1}{10} $ $ =10\,\cos \frac{\pi }{6} $ $ =10\cos {{30}^{o}} $ $ =10\times \frac{\sqrt{3}}{2} $ $ =5\sqrt{3}\,\text{volt}\text{.} $