Question

The peak value of an alternating e.m.f. E is given by E = E₀ cos ω t is 10 volts and its frequency is 50 Hz. At time $t=\frac{1}{600} sec$, the instantaneous e.m.f. is
(a) 10 V
(b) 5√3 V
(c) 5 V
(d) 1 V

Answer 1

answer : option (b) 5√3 V

explanation : The peak value of an alternating e.m.f. E is given by $E=E_0cos\omega t$

or, $E=E_0cos\frac{2\pi t}{T}$

or, $E=E_0cos2\pi t\eta$

here, E is the instantaneous emf of alternating current in ac circuit , $E_0$ is the peak value , T is time period and $\eta$ is frequency.

given, $E_0$ = 10volts
$\eta$ = 50 Hz
and t = 1/600 sec

now, instantaneous emf , E = 10cos(2π × 1/600 × 50)
= 10cos(100π/600)
= 10cos(π/6) = 10 × √3/2 = 5√3 V

hence, instantaneous emf = 5√3V

Answer 2

Answer:

Answer is (c) 5Volts

Explanation:

I hope the answer is correct ....