In fig .current through 3$\mathrm{\Omega }$ resistor is 0.8 amp

then potential drop through 4$\mathrm{\Omega }$ resistor is

Voltage across 3 $\mathrm{\Omega }$ resistance = 3 x 0.8= 2.4V
This voltage is the same across 6 $\mathrm{\Omega }$ resistance. Hence current through this resistance
$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{2\cdot 4}{6}=0\cdot 4\mathrm{amp}$
Total current in the circuit
= 0.8 + 0.4= 1.2amp
Voltage across 4 $\mathrm{\Omega }$ resistance
            = 4x1.2= 4.8 volt