The peak value of an alternating emf ' e ' given by e = e₀ cos ω t is 10 volt and its frequency is 50 Hz. At time t = 1/600 s, the instantaneous e.m.f is (cos π/6 = sin π/3 = √(3)/2) (a) 10 V (b) 1/√(3) V (c) 5 V (d) 5√(3) V

Answer: (d). ω t = 2π × 50 × 1/600 = π/6 e = 10cos π/6 = 10 × √(3)/2 = 5√(3) V So the answer is (d) 5√(3) V.

The peak value of an alternating emf ' e ' given by e = e₀ cos ω t is… | MHT CET 2021 (24 Sep, Shift 1)

Q29MHT CET 2021 · 24 SEP · SHIFT 1Easy

AC Basics: Peak, RMS & Average Values

The peak value of an alternating emf ' $e$ ' given by $e = e_0 \cos \omega t$ is 10 volt and its frequency is $50\text{ Hz}$ . At time $t = \frac{1}{600}\text{ s}$ , the instantaneous e.m.f is $\left(\cos \frac{\pi}{6} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\right)$

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Answer: (d)

$\omega t = 2\pi \times 50 \times \frac{1}{600} = \frac{\pi}{6}$
$e = 10\cos\frac{\pi}{6} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ V
So the answer is (d) $5\sqrt{3}$ V.

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