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] |
\[I=\frac{E}{{{R}_{eq}}}=\frac{E}{2},\operatorname{Re}q=2\Omega \] |
\[I=10A\] |
\[{{I}_{1}}=\frac{4}{5}I,\,\,{{I}_{2}}=\frac{1}{5}I\] |
\[{{I}_{1}}=8A,\,\,{{I}_{2}}=2A\] |
\[I_{1}^{'}=\frac{3}{5}\times 10=6A\] |
\[I_{2}^{'}=4A\] |
\[\Rightarrow {{I}_{BC}}=86=2A\] |
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