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Kirchhoff's Laws and Potential Divider
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In the circuit shown, if a conducting wire is connected between points A and B, the current in this wire will-
(Two branches are connected in parallel across the source V. The upper branch has $4\,\Omega$ and $4\,\Omega$ in series with A as their junction; the lower branch has $1\,\Omega$ and $3\,\Omega$ in series with B as their junction.)
[add image]Before the wire is connected, find the potentials of A and B. Let the left end of both branches be at potential $V$ and the right end at zero.
Current in the upper branch $= \dfrac{V}{4 + 4} = \dfrac{V}{8}$, so $V_A = V - \dfrac{V}{8} \times 4 = \dfrac{V}{2}$
Current in the lower branch $= \dfrac{V}{1 + 3} = \dfrac{V}{4}$, so $V_B = V - \dfrac{V}{4} \times 1 = \dfrac{3V}{4}$
$V_A - V_B = \left(V - \dfrac{V}{8} \times 4\right) - \left(V - \dfrac{V}{4} \times 1\right) = -\dfrac{V}{2} + \dfrac{V}{4} = -\dfrac{V}{4}$
$\Rightarrow V_B \gt V_A$, whatever the value of $V$.
Current flows from the higher potential to the lower potential, so in the connecting wire it will flow from B to A.