JEE Main 2019PhysicsAlternating CurrentMediumMCQ

JEE Main 2019 — Alternating Current Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

An alternating voltage V(t)=220sin100πt volt is applied to a purely resistive load of 50 Ω . The time taken for the current to rise from half of the peak value to the peak value is:

Choose an option

Show full solutionCorrect option: D
Correct answer
D 3.33 ms

Step-by-step explanation

i=VR=22050sin⁡100πt
i=imaxsin⁡100πt
For i=imax2: 
imax2=imaxsin⁡100πt
sin⁡100πt=12 ⇒100πt=π6
t=1600  s
For i=imax:
 imax=imaxsin⁡100πt
sin⁡100πt=1 ⇒100πt=π2
⇒t=1200 s

so time for  imax2 to imax is 
t(imax)-timax2=1200-1600=1300=3.33 ms

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About this question

This is a previous-year question from JEE Main 2019, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.