In the circuit shown in the figure, if potential at point A is taken to be zero, the potential at point B is 

 

Finding potential difference between A and B
(following the path ACDB)
$\begin{array}{l}{\mathrm{V}}_{\mathrm{A}}+1+2\times 1-2={\mathrm{V}}_{\mathrm{B}} ,{\mathrm{V}}_{\mathrm{A}} = 0 \mathrm{V} \\ {\mathrm{V}}_{\mathrm{B}}=1\mathrm{V}\end{array}$
(Current through 2Ω resistance can be calculated by applying KCL at junction C)