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JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2019
- Hard
- MCQ
In the below circuit, \(C\, = \,\frac{{\sqrt 3 }}{2}\,\mu F,\,R = 20\,\,\Omega ,\,L = \frac{{\sqrt 3 }}{{10}}H\) and \({R_1} = 10\,\,\Omega .\) Current in \(L-R_1\) path is \(I_1\) and \(C-R_2\) path is \(I_2.\) The voltage of \(A.C\) source is given by, \(V\, = \,200\sqrt 2 \,\sin \,(100\,t)\,volts.\) The phase difference between \(I_1\) and \(I_2\) is

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) None of these
Step-by-step Solution
Detailed explanation
For the first branch:
\( \tan \phi_{1} =\frac{-X_{c}}{R_{2}}=\frac{\frac{1}{\left(100 \times \frac{\sqrt{3}}{2} \times 10^{-6}\right)}}{20} \approx-10^{-3}\)
\(\Rightarrow \quad \phi_{1} \approx-90^{\circ}\)
For the second branch:
\(\tan \phi_{2}=\frac{X_{L}}{R}=\sqrt{3}\)
\(\Rightarrow \quad \phi_{2} \approx 60^{\circ}\)
Phase difference between current in branch \(1\) and \(2=150^{\circ} .\)
No option is correct.
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