JEE Main 12 Jan 2019 Evening Question 8
Question: In the given circuit, $ C=\frac{\sqrt{3}}{2}\mu F,R _2=20\Omega , $ and $ R _1=10\Omega . $ Current in $ L-R _1 $ path is $ I _1 $ and in $ C-R _2 $ path it is $ I _2. $ The voltage of A.C source is given by, $ V=200\sqrt{2}\sin (100t) $ volts. The phase difference between $ I _1 $ and $ I _2 $ is
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 0
B) $ 30{}^\circ $
C) $ 90{}^\circ $
D) $ 60{}^\circ $
E) None of these
Show Answer
Answer:
Correct Answer: E
Solution:
-
For current $ I _1 $ $ \tan \phi =\frac{X _{L}}{R _1}=\frac{\omega L}{R _1}=\frac{100\times \frac{\sqrt{3}}{10}}{10}=\sqrt{3} $ $ \phi =60^{o}; $
V leads $ I _1. $
For current $ I _2, $ $ \tan \phi ‘=\frac{X _{C}}{R _2}=\frac{1}{\omega CR _2}=\frac{1}{100\times \frac{\sqrt{3}}{2}\times {10^{-6}}\times 20}=\frac{1000}{\sqrt{3}} $ $ \phi ‘\simeq 90^{o};V;ags,I _2. $The required phases difference between $ I _1 $ and $ I _2 $ is $ \phi +\phi ‘=60^{o}+90^{o}=150^{o} $
*None of the given options is correct
