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Question
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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Solution
The equivalent resistance of series combination is in series with the internal resistance R of battery and in parallel combination of resistors, the equivalent resistance of parallel combination is also in series with the internal resistance of battery.
In series combination of resistors, current I is given by `I = E/(R + nR)`
Whereas in parallel combination current 10I is given by `E/(R + R/n) = 10I`
Now, according to problem,
`(1 + n)/(1 + 1/n)` ⇒ `10 = ((1 + n)/(n + 1))n`
⇒ n = 10
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