question_answer

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 Online 08-04-2018]

A) \[\text{11}\text{.4 V and 11}\text{.5 V}\]

B) \[\text{11}\text{.7 V and 11}\text{.8 V}\]

C) \[\text{11}\text{.6 V and 11}\text{.7 V}\]

D) \[\text{11}\text{.5 V and 11}\text{.6 V}\]

Correct Answer: D

Solution :

[d]

\[{{E}_{eq}}=\frac{\frac{12}{1}+\frac{13}{2}}{\frac{1}{1}+\frac{1}{2}}=\frac{37}{3}\]

\[\frac{1}{{{r}_{eq}}}=\frac{1}{1}+\frac{1}{2}=\frac{3}{2}\Rightarrow {{r}_{eq}}=\frac{2}{3}\]

\[I=\frac{\frac{37}{3}}{\frac{2}{3}+10}=\frac{37}{32}\]

Voltage across the load\[=\left( \frac{37}{32} \right)(10)=\frac{370}{32}=11.56\,\,volt\]