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: $E_0=10\text{volt,}$ $t=\frac{1}{600}\sec ,$ $f=50Hz$ $\therefore$ $E=E_0\cos \frac{2\pi t}{T}$ or $E=10\cos \frac{2\pi \times 50\times 1}{600}$ $=10\cos \frac{5\pi }{3}\times \frac{1}{10}$ $=10\cos \frac{\pi }{6}$ $=10\cos {30}^o$ $=10\times \frac{\sqrt{3}}{2}$ $=5\sqrt{3}\text{volt}\text{.}$