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

shows the current distributionApplying Kirchhoff's first law at point D, we havei=i1+i2…………(1)Applying Kirchhoff second law to mesh,DBA, we have2i+1i+2i1=2−1=1or 3i+2i1=1……..(2)Applying Kirchhoff second law to mesh DCBD we get3i2+1i2−2i1=3−1or 4i2−2i1=2…….(3)Solving eqs. (1), (2) and (3), we geti1=-113amp.,i2=613 amp.and i=513 amp. Potential difference between B and D=2i1=2113=213 volt