A network of resistances is connected to a 16V battery with internal resistance of 1 $\mathrm{\Omega }$as shown in fig. 

The equivalent resistance of the network and the voltage drops ${\mathrm{V}}_{\mathrm{AB}},{\mathrm{V}}_{\mathrm{BC}}\text{ and }{\mathrm{V}}_{\mathrm{CD}}$ are 

$\mathrm{Resistance} \mathrm{between} \mathrm{AB} 4\parallel 4, \mathrm{between} \mathrm{BC} 1\mathrm{\Omega }, \mathrm{between} \mathrm{CD} 12\parallel 6$

$$\begin{gathered} \mathrm{So} \mathrm{total} \mathrm{resistance} \mathrm{of} \mathrm{the} \mathrm{network} \\ 2+1+4=7 \mathrm{\Omega } \end{gathered}$$

$\mathrm{total} \mathrm{resistance} \mathrm{of} \mathrm{the} \mathrm{circuit} =7+1=8 \mathrm{\Omega }$

$\mathrm{current} \mathrm{in} \mathrm{the} \mathrm{circuit} =\frac{16}{8}=2 \mathrm{A}$

$\mathrm{So} {\mathrm{V}}_{\mathrm{AB}, }{\mathrm{V}}_{\mathrm{BC}} \mathrm{and} {\mathrm{V}}_{\mathrm{CD}}\mathrm{will} \mathrm{be} 4\mathrm{V}, 2 \mathrm{V} \mathrm{and} 8 \mathrm{V} \mathrm{respectively}$