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▾

$i = \frac{20}{2} = 10A$
$1 = \frac{41}{5} - \frac{3i}{5}$
$= + \frac{i}{5} = 2A$

Source: Grass-G / P-12 Physics paper 2 Nov.docx