Two batteries with e.m.f \(12\ V\) and \(13\ V\) are connected in parallel across a load resistor of \(10\,\Omega\) . The internal resistances of the two batteries are \(1\,\Omega\) and \(2\,\Omega\) respectively. The voltage across the load lies between

\(\frac{V-12}{1}+\frac{V-13}{2}+\frac{V-0}{10}=0\)

[Let potential at \(\mathrm{P}, \mathrm{Q}, \mathrm{U}=0\) and at \(\mathrm{R}=\mathrm{V}\)

\(\Rightarrow \frac{\mathrm{V}}{1}+\frac{\mathrm{V}}{2}+\frac{\mathrm{V}}{10}=\frac{12}{1}+\frac{13}{2}+\frac{0}{10}\)

\(\Rightarrow \frac{10+5+1}{10} \mathrm{V}=\frac{24+13}{2}\)

\(\Rightarrow \mathrm{V}\left(\frac{16}{10}\right)=\frac{37}{2}\)

\(\Rightarrow \mathrm{V}=\frac{37 \times 10}{16 \times 2}=\frac{370}{32}=11.56\, \mathrm{volt}\)