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A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is

1. 10
2. 11
3. 9
4. 20

Correct answer: 11

Solution

In the series case, the total resistance is nR + R, and the current is I = E / (nR + R). In the parallel case, the equivalent resistance is R/n, and the current becomes 10I = E / (R/n + R). Solving these equations gives n = 11.

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