Q 12-03-293JEE MainJEE Main 2018 (8 Apr)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:
Answer: (C) $11.5$ V and $11.6$ V
Equivalent cell for cells in parallel:
$$E_{eq} = \frac{\frac{12}{1} + \frac{13}{2}}{\frac11 + \frac12} = \frac{18.5}{1.5} = 12.33\ \text{V},\qquad r_{eq} = \frac{1\times2}{1 + 2} = \frac23\ \Omega$$
$$V = E_{eq}\frac{10}{10 + \frac23} = 12.33\times\frac{30}{32} \approx 11.56\ \text{V}$$
This lies between $11.5$ V and $11.6$ V.
Solution by Sreeraj P, M.Sc Physics