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In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point $$B$$ with respect to the point $$A$$ is:
Applying KCL at junction $$C$$: $$I_{AC} + I_{DC} = I_{CF}$$
$$1 + I_{DC} = 2 \implies I_{DC} = 1\text{ A}$$
Writing KVL from $$A$$ to $$B$$ via path $$A \to C \to D \to B$$:
$$V_A + 1 - I_{DC}(2) + 2 = V_B$$
$$V_A + 1 - 1(2) + 2 = V_B \implies V_B - V_A = 1\text{ V}$$
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