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JEE Main 202429 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2024 · 29 January · Evening Shift · Physics · Alternating Current

In an a.c. circuit, voltage and current are given by: V = 100 (...

In an a.c. circuit, voltage and current are given by: V=100sin⁡(100t)VandV = 100\sin ( 100t ) V andI = 100sin(100t+π3)mA\left(100t + \dfrac{\pi}{3}\right)mA respectively. The average power dissipated in one cycle is :

Options

  1. A5W5W
  2. B10W10W
  3. C2.5W2 . 5W
  4. D25W25W

Correct answer

C. 2.5W2 . 5W

Step-by-step solution

Standard expression of voltage is given by, V=V0sinV = {V}_{0}sin\left(\omegat\right)andforcurrentisgivenby,and for current is given by,I = {I}_{0}\sin \left(\omegat + ϕ\right).Clearly,. Clearly,ϕisthephasedifferencehere.Fromthegivenexpression,is the phase difference here. From the given expression,ϕ = \dfrac{\pi}{3}.</p><p>Averagepowerdissipatedinonecycleisgivenby,</p><p>.</p><p>Average power dissipated in one cycle is given by,</p><p>{P}_{avg} = {V}_{rms}{I}_{rms}\cos ( ϕ )</p><p></p><p>= \dfrac{100}{\sqrt{2}} \times \dfrac{100 \times {10}^{- 3}}{\sqrt{2}} \times \cos \left(\dfrac{\pi}{3}\right)</p><p></p><p>= \dfrac{{10}^{4}}{2} \times {10}^{- 3} \times \dfrac{1}{2}</p><p></p><p>= \dfrac{10}{4} = 2 . 5W

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