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NEET generalGeneralEasy

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

B

11

C

10

D

9

Step-by-Step Solution

I=nRI = nR ...(i)

E=10I+nRE = 10 I + nR ...(ii)

  1. Dividing (ii) by (i),

(n+1)R10=1+1nR\frac{(n + 1)R}{10} = 1 + \frac{1}{n} R

After solving the equation, n=10n = 10

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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