Q 12-03-013NEETNEET 2023Top questionEasy
$10$ resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is :
Answer: (B) $100$
Series: $R_s = 10R$, so $I_s = \dfrac{E}{10R}$.
Parallel: $R_p = \dfrac{R}{10}$, so $I_p = \dfrac{10E}{R}$.
$$n = \frac{I_p}{I_s} = 100$$
Solution by Sreeraj P, M.Sc Physics