The alternating current in a circuit is given by I=50 sin314t. The peak value

The Alternating Current In A Circuit Is Given By I50 Physics Question

The alternating current in a circuit is given by I=50 sin314t. The peak value and frequency of the current are

Options

(a) I₀=25 A and f=100 Hz
(b) I₀=50 A and f=50 Hz
(c) I₀=50 A and f=100 Hz
(d) I₀=25 A and f=50 Hz

Correct Answer:

I₀=50 A and f=50 Hz

Explanation:

From standard equation, we have I = I₀ sin ωt —–(i)

Given, I = 50 sin 31 4t —-(ii)

Comparing equation (i) and (ii), we get I₀ = 50 A, ω = 2πf = 314

⇒ f = 314 / (2 × 3.14) = 50 Hz

Related Questions:

  1. An electric charge 10⁻³ μC is placed at the origin (0,0) of X-Y co-ordinate system.
  2. In the elastic collision of objects
  3. A metal wire of circular cross-section has a resistance R₁. The wire is now stretched
  4. There are four light-weight-rod samples A,B,C,D seperately suspended by threads
  5. Which of the following is the function of the step-up transformer?

Topics: Alternating Current (96)
Subject: Physics (2479)

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*