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 $50Hz.$ At time $t = \frac{1}{600}sec,$ the instantaneous e.m.f. is

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

Answer 1

Answer: (B)

$5\sqrt{3}V$

Answer 2

By using $E = E_{0}\sin\omega t$ = 10 cos 2πν t = 10

cos 2π × 50 × $\frac{1}{600}$ ⇒ E = 10 cos $\frac{\pi}{6} = 5\sqrt{3}V$