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Q.

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 ________.

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answer is 20.

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Detailed Solution

In series current,i1=2010+10n=21+n  
Current in parallel will be i2=2010n+10=2n1+n
i2i1=20⇒2n1+n21+n=20⇒n=20

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