If V${}_A$ – V${}_B$ = 6 V then V${}_C$ – V${}_D$ is

V${}_A$ – V${}_B$ = 6 V

So current in the branch will be

I${}_{AB}$ = 6/2 = 3 A

We can see, that the circuit is a one part circuit looking from terminal BD as shown below



For a one port network current entering one terminal, equals the current leaving the second terminal. Thus the outgoing current from A to B will be equal to the incoming current from D to C as shown.

i.e. I${}_{DC}$ = I${}_{AB}$ = 3 A



The total current in the resistor 1 Ω will be

I₁ = 2 + I${}_{DC}$

= 2 + 3 = 5 A

V${}_{CD}$ = 1 × (-I₁)

= -5 V