JEE Main 2019 Physics, Alternating Current — question figure In the above circuit, C= √ 3 2 μ F, R_2 =20 Ω , L= √ 3 10 H and R_1 =10 Ω . Current in L…
JEE Main 2019 — Physics Electromagnetism
2019mcqmedium
In the above circuit, C=23μF,R2=20Ω,L=103H and R1=10Ω. Current in L-R1 path is I1 and in C−R2 path it is I2. The voltage of AC source is given by, V=200 2 sin( 100 t ) volts. The phase difference between I1 and I2 is:
Official previous-year question
Held on 12 Jan 2019 · Verified 6 Jul 2026.
Options
A
60o
B
0o
C
30o
D
150o
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Solution
Inductive reactance XL=ωL=100103Ω
=103Ω
Capacitive reactance
Xc=ωC1=100×23×10−61Ω=320000Ω
For i1, phase difference between i1 & voltagetanϕ1=RxL=10103=3
⇒ϕ1=60o, current lagging
For i2 , phase difference
tanϕ2=RXC=3×2020000=31000
⇒ϕ1≈90o, current leading
∴ difference in phase =90o+60o=150o
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