First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then, the n resistor are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ‘n’?

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answer is 10.

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Detailed Solution

In series combination of resistors, current I is given by $I = \frac{E}{R + n R^{'}}$

Whereas in parallel combination current 10I is given by $\frac{E}{R + \frac{R}{n}} = 10 I$

Now, according to problem, $\frac{1 + n}{1 + \frac{1}{n}} \Rightarrow 10 = (\frac{1 + n}{n + 1}) n \Rightarrow n = 10$

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