The equivalent resistance and potential difference between $A$ and $B$ for the circuit is respectively
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The equivalent resistance between $C$ and $D$ is
$\frac{1}{{R'}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{3} = \frac{2}{3}$ or $R' = \frac{3}{2} = 1.5\,\Omega $
Now the equivalent resistance between $A$ and $B$ as $R' = 1.5\,\Omega $ and $2.5\,\Omega $ are connected in series, so
Now by ohm's law, potential difference between $A$ and $B$ is given by ${V_A} - {V_B} = iR = 2 \times 4.0 = 8\,\,volt$
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