CUET · PHYSICS · PYQ PAPER 2025
A set of n equal resistors, each of resistance R 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. The current drawn from the battery is 20 I. The value of n is
- A 10
- B 11
- C 20
- D 21
Answer & Solution
Correct Answer
(C) 20
Step-by-step Solution
Detailed explanation
\(I = \frac{E}{nR + R} = \frac{E}{(n+1)R}\) \(I' = \frac{E}{\frac{R}{n} + R} = \frac{E}{\frac{R(1+n)}{n}} = \frac{nE}{(n+1)R}\) \(I' = 20I \Rightarrow \frac{nE}{(n+1)R} = 20 \times \frac{E}{(n+1)R}\) \(n = 20\)
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