First, a set of n equal resistors of 10 Ω each are connected in… | JEE Main 2021 (27 Aug, Shift 1)

Q29JEE Main 2021 · 27 AUG · SHIFT 1Difficult

Cells: EMF, Internal Resistance & Combinations

First, a set of n equal resistors of 10 $\Omega$ each are connected in series to a battery of emf 20V 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 ................

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Answer: 20

Series: $I = \dfrac{20}{10n + 10}$ ; parallel: $I' = \dfrac{20}{10/n + 10}$ .
$\dfrac{I'}{I} = \dfrac{10n + 10}{10/n + 10} = \dfrac{10(n + 1)}{10(n + 1)/n} = n$
$\dfrac{I'}{I} = 20 \Rightarrow n = 20$

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