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The peak value of an alternating e.m.f. is 10 V and its frequency is 50 Hz. At time \(t = \frac{1}{600}\) s, the instantaneous e.m.f. is
- 10 V
- \(5\sqrt{3}\) V
- 5 V
- 1 V
Correct answer: \(5\sqrt{3}\) V
Solution
The emf is \(E = E_0\cos(\omega t)\), where \(E_0 = 10\) V and \(\omega = 2\pi f = 100\pi\) rad/s. At \(t = 1/600\) s, \(\omega t = 100\pi/600 = \pi/6\), so \(E = 10\cos(\pi/6) = 10\cdot \sqrt{3}/2 = 5\sqrt{3}\) V.
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