In the given circuit assuming poiny A to at zero potential use kirchhoff's rules to determine the potential at point B

Answer

`2V`

Explanation

According to kichhoff's first in arm
`DC = 3 - 1 = 2A`
pot diff across `2 Omega = 2 xx 2 = 4 V`
As point A is at zero potential so potential point C is `V_(C) = 2 V`
Now `V_(D) - V_(C) = 4 or V_(D) - 2 = 4 or V_(D) = 6V`
Here`V_(D) - V_(B) = 4 orV_(B) = V_(D) - 4 = 6 - 4 = 2V`