Question bank/JEE Main/Physics

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

AzeroB2 AC0.4 AD4A

Show answer and solution▾

$i = \frac{20}{2} = 10A$
$I = \frac{4i}{5} - \frac{3i}{5} = + \frac{i}{5} = 2A$

Source: Grass-G / P-19 Physics paper 28 Nov Fourth.docx