As the P.D. between 4 $\Omega$ and 3 $\Omega$ (in parallel), are the same,

4 × 1 amp = 3 × i₁ $\Rightarrow {i_1} = {4 \over 3}A$

Total resistance of 4 $\Omega$ and 3 $\Omega$ = 12/7 $\Omega$.

Current in MQP (upper one) = $1 + {4 \over 3} = {7 \over 3}A$

$\therefore$ $P.D. = {{12} \over 7} \times {7 \over 3} = 4V$

Current in MNP = ${4 \over {1.25}} = {{4 \times 4} \over 5} = {{16} \over 5}A$

$\therefore$ P.D. across $1\Omega = {{16} \over 5}A \times 1\Omega = {{16} \over 5}volt$

$\Rightarrow$ P.D. across 1$\Omega$ = 3.2 volt.