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

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

1. $0.4\mathrm{}\mathrm{A}$
2. $2\mathrm{}\mathrm{A}$
3. $4\mathrm{}\mathrm{A}$
4. zero

Solution

$R_{eff}=\frac{4}{5}+\frac{6}{5}=2\mathrm{\Omega }$
$i=\frac{20}{2}$
$=10A$
$I=\frac{4i}{5}-\frac{3i}{5}=+\frac{i}{5}=2A$

Asked in: JEE Main 2020 (09 Jan Shift 1)

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