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Q.

First, a set of the n equal resistors of 10 $Ω$ each are connected in series to a battery of emf 20V and internal resistance 10 $Ω$ . 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 is 20.

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Detailed Solution

In series current, $i_{1} = \frac{20}{10 + 10 n} = \frac{2}{1 + n}$
Current in parallel will be $i_{2} = \frac{20}{\frac{10}{n} + 10} = \frac{2 n}{1 + n}$
$\frac{i_{2}}{i_{1}} = 20 \Rightarrow \frac{(\frac{2 n}{1 + n})}{(\frac{2}{1 + n})} = 20 \Rightarrow n = 20$

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First, a set of the n equal resistors of 10Ω each are connected in series to a battery of emf 20V and internal resistance 10Ω . 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 ________.