question_answer

In the given circuit diagram, a wire is joining points B and D. The current in this wire is:

[JEE MAIN Held on 09-01-2020 Morning]

           

A) 0.4 A

B) 4 A

C) 2 A

D) zero

Correct Answer: C

Solution :

[c]

\[I=\frac{E}{{{R}_{eq}}}=\frac{E}{2},\operatorname{Re}q=2\Omega \]

\[I=10A\]

\[{{I}_{1}}=\frac{4}{5}I,\,\,{{I}_{2}}=\frac{1}{5}I\]

\[{{I}_{1}}=8A,\,\,{{I}_{2}}=2A\]

\[I_{1}^{'}=\frac{3}{5}\times 10=6A\]

\[I_{2}^{'}=4A\]

\[\Rightarrow {{I}_{BC}}=86=2A\]