In the circuit shown in fig.  E, F, G and H are cells of e.m.f. 2, 1, 3 and 1 volt and their internal resistances

are 2, 7,3 and 1 ohm respectively. The potential difference between B and D is

shows the current distribution

Applying Kirchhoff's first law at point D, we have
$\mathrm{i}={\mathrm{i}}_1+{\mathrm{i}}_2$…………(1)
Applying Kirchhoff second law to mesh,DBA, we have
$2\mathrm{i}+1\mathrm{i}+2{\mathrm{i}}_1=2-1=1$
or $3\mathrm{i}+2{\mathrm{i}}_1=1$……..(2)
Applying Kirchhoff second law to mesh DCBD we get
$3{\mathrm{i}}_2+1{\mathrm{i}}_2-2{\mathrm{i}}_1=3-1$
or $4{\mathrm{i}}_2-2{\mathrm{i}}_1=2$…….(3)
Solving eqs. (1), (2) and (3), we get
${\mathrm{i}}_1=-\frac{1}{13}\mathrm{amp}.,{\mathrm{i}}_2=\frac{6}{13} \mathrm{amp}.\mathrm{and} \mathrm{i}=\frac{5}{13} \mathrm{amp}.$
Potential difference between B and D
$=2{\mathrm{i}}_1=2\left(\frac{1}{13}\right)=\frac{2}{13} \mathrm{volt}$