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
JEE MAIN 2018, Medium
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Using Kirchhoff's law at $P$ we get
$\frac{V-12}{1}+\frac{V-13}{2}+\frac{V-0}{10}=0$
[Let potential at $\mathrm{P}, \mathrm{Q}, \mathrm{U}=0$ and at $\mathrm{R}=\mathrm{V}$
$\Rightarrow \frac{\mathrm{V}}{1}+\frac{\mathrm{V}}{2}+\frac{\mathrm{V}}{10}=\frac{12}{1}+\frac{13}{2}+\frac{0}{10}$
$\Rightarrow \frac{10+5+1}{10} \mathrm{V}=\frac{24+13}{2}$
$\Rightarrow \mathrm{V}\left(\frac{16}{10}\right)=\frac{37}{2}$
$\Rightarrow \mathrm{V}=\frac{37 \times 10}{16 \times 2}=\frac{370}{32}=11.56\, \mathrm{volt}$
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