Chapter: AC Circuits· Physics

An alternating voltage is given by \(e = 8\sin 628.4t\). Find(i) peak value of e.m.f.(ii) frequency of e.m.f.(iii) instantaneous value of e.m.f. at time t = 10 ms.

Q.26 An alternating voltage is given by \(e = 8\sin 628.4t\). Find
(i) peak value of e.m.f.
(ii) frequency of e.m.f.
(iii) instantaneous value of e.m.f. at time t = 10 ms.

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

1 answer

Best answer

Answer: (i) 8 V (ii) 100 Hz (iii) 0 V

Comparing with \(e = e_0\sin\omega t\):

(i) Peak value \(e_0 = 8\) V.

(ii) \(\omega = 2\pi f = 628.4\) rad/s, so \(f = \dfrac{628.4}{2\times3.142} = \dfrac{628.4}{6.284} = 100\) Hz.

(iii) At \(t = 10\) ms \(= 0.01\) s: \(\omega t = 628.4\times0.01 = 6.284\) rad \(= 2\pi\) rad.

\(e = 8\sin(2\pi) = 0\) V. (After one full period, \(T = 1/f = 0.01\) s, the e.m.f. is back to zero.)

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

An alternating voltage is given by \(e = 8\sin 628.4t\). Find(i) peak value of e.m.f.(ii) frequency of e.m.f.(iii) instantaneous value of e.m.f. at time t = 10 ms.

1 answer

Your answer