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NEET 2018 · Cells: EMF, Internal Resistance & Combinations

Cells: EMF, Internal Resistance & Combinations question from NEET 2018

Q35NEET 2018Medium

Cells: EMF, Internal Resistance & Combinations

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 10I. The value of n is

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Answer: (a)

  1. Series: I=EnR+R=E(n+1)RI = \frac{E}{nR + R} = \frac{E}{(n+1)R}
  2. Parallel: 10I=ER/n+R=nE(n+1)R10I = \frac{E}{R/n + R} = \frac{nE}{(n+1)R}
  3. Dividing: 10=n10 = n. So the answer is (a) 10.
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