JEE Main On 08 April 2018 Question 4
Question: 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]
Options:
A) $ 11\text{.4 V and 11}\text{.5 V} $
B) $ 11\text{.7 V and 11}\text{.8 V} $
C) $ 11\text{.6 V and 11}\text{.7 V} $
D) $ 11\text{.5 V and 11}\text{.6 V} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ 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 $ =( \frac{37}{32} )(10)=\frac{370}{32}=11.56volt $
