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’?

NCERT Class XII
Physics - Exemplar Problems

Chapter 3. Current Electricity

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2015-06-11T16:35:15+05:30

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Let the EMF be E. Then

Current when the Resistors are connected in series = \frac{E}{(nR + R)}
Current when the Resistors are connected in parallel = \frac{E}{(\frac{R}{n}+R)}

According to the question :
 \frac{10*E}{(nR+R)}= \frac{E}{(\frac{R}{n}+R)}\\ \\n=10
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