Two batteries with e.m.f. 12   V and 13   V are connected in parallel across a load resistor of 10…

As shown in figure two cells are connected in parallel with the load resistance of $R=10 \mathrm{\Omega }$. Let current $i_1, i_2$ drawn from both cells as shown, from $K.C.L.$  $i = i_1+ i_2 ...\left(1\right)$

Apply $K.V.L.$ for loop $ACDF$, 

$$\begin{gathered} E_1-i R-i_1 r_1=0 \\ \Rightarrow 12-10i-i_1=0 ...\left(2\right) \end{gathered}$$

From equations $\left(1\right)$ and $\left(2\right)$

$12-11 i_1-10i_2=0 ...\left(3\right)$

Now, for loop $BCDE$, 

Apply $K.V.L.$

$$\begin{gathered} E_2-10i-i_2{r}_2=0 \\ \Rightarrow 13-10 i-2i_2=0 \\ \Rightarrow \left(\because i=i_1+i_2\right) \\ \therefore 13-10 i_1-12 i_2=0 ...\left(4\right) \end{gathered}$$

From equation $\left(3\right)$ and $\left(4\right)$, 

$i_1=\frac{7}{16} \mathrm{A}, i_2=\frac{23}{32} \mathrm{A}$

So, net current in external load $R=10 \mathrm{\Omega }$ is

$I=i_1+i_2=\frac{7}{16}+\frac{23}{32}=\frac{37}{32} \mathrm{A}$

Then potential difference for a given load, 

$V=IR=\frac{37}{32}\times 32=11·56 \mathrm{V}$

So, the potential difference lies between, $11·5 \mathrm{V}$ and $11·6 \mathrm{V}$.