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

Q: 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

a 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 $\mathrm{R}=\mathrm{V}$ $\Rightarrow \frac{\mathrm{V}}{1}+\frac{\mathrm{V}}{2}+\frac{\mathrm{V}}{10}=\frac{12}{1}+\frac{13}{2}+\frac{0}{10}$ $\Rightarrow \frac{10+5+1}{10} \mathrm{V}=\frac{24+13}{2}$ $\Rightarrow \mathrm{V}\left(\frac{16}{10}\right)=\frac{37}{2}$ $\Rightarrow \mathrm{V}=\frac{37 \times 10}{16 \times 2}=\frac{370}{32}=11.56\, \mathrm{volt}$

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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

A$11.5$ $ V$ and $11.6$ $ V$
B$11.4$ $ V$ and $11.5$ $ V$
C$11.7$ $V$ and $11.8$ $ V$
D$\;$ $11.6$ $ V$ and $11.7$ $ V$

JEE MAIN 2018, Medium

STD 11 Science
JEE
STD 12 - 3. current electricity
Physics

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