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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 resistors 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» When n resistors are in series, \(I=\frac{E}{R+nR};\) When n resistors are in parallel, \(\frac{E}{R\,+\,\frac{R}{n}}=10I\) \(\frac{1+n}{1+\frac{1}{n}}=10\) \(\Rightarrow\,\,\frac{1+n}{n+1}n=10\) \(\Rightarrow\,n=10.\) |
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