Q 12-03-217JEE MainJEE Main 2021 (27 Aug, Shift 1)Easy
First, a set of $n$ equal resistors of $10\ \Omega$ each are connected in series to a battery of E.M.F. $20$ V and internal resistance $10\ \Omega$. A current $I$ is observed to flow. Then, the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased $20$ times, then the value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 20
Series: $I = \dfrac{20}{10n + 10}$. Parallel: $I' = \dfrac{20}{\frac{10}{n} + 10}$.
$$\frac{I'}{I} = \frac{10n + 10}{\frac{10}{n} + 10} = \frac{n + 1}{\frac{1 + n}{n}} = n = 20$$
Solution by Sreeraj P, M.Sc Physics