JEE Main 2019PhysicsAlternating CurrentMediumMCQ

JEE Main 2019 — Alternating Current Question with Solution

JEE Main 2019 (12 Jan Shift 2)

Question



In the above circuit, C= 3 2 μF,  R 2 =20 Ω,  L= 3 10 H and R 1 =10 Ω. Current in L-R1 path is I1 and in C-R2 path it is I 2 . The voltage of AC source is given by, V=200 2  sin( 100 t ) volts. The phase difference between I1 and I2 is:

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Show full solutionCorrect option: D
Correct answer
D150o

Step-by-step explanation



Inductive reactance XL=ωL=100310 Ω

=10 3 Ω

Capacitive reactance

Xc=1ωC=1100×32×10-6 Ω=200003 Ω

For i1, phase difference between i1 & voltagetanϕ1 =xLR=10310=3

⇒ ϕ1 = 60o,  current lagging

For i2 , phase difference

tanϕ2=XCR=200003×20=10003

⇒  ϕ1 ≈ 90o,  current leading

∴ difference in phase =90o+60o=150o 

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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.