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Question

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

A
20
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B
10
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C
9
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D
11
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Solution

The correct option is B 10
Electrical resistance in series = nR
I=E(n+1)R ...............(1)
And in parallel 10 I=ERn+R=nER(n+1) ......(2)
Equating (1) and (2)
we get n=10


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