In the circuit shown in fig. ${\mathrm{E}}_1=3\mathrm{V},{\mathrm{E}}_2=2\mathrm{V},{\mathrm{E}}_3=1\mathrm{V},\mathrm{R}={\mathrm{r}}_1={\mathrm{r}}_2={\mathrm{r}}_3=1\mathrm{ohm}$ .Find the potential difference between the points A and B and the currents through each branch.

Effective potential difference across three batteries connected in parallel i.e  V_PQ= $\frac{{\displaystyle \frac{3}{1}}+{\displaystyle \frac{2}{1}}+{\displaystyle \frac{1}{1}}}{3}=2V$

Since PD across PQ = 2V and E₂ = 2 V , hence current through the battery = 0 , I₃ = 0 

V_PQ   =2 =  3 - I₁ 

I₁ = 1 A which flows through E₃ also. 

Hence I₁ = I₃  = 1A