Q. 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
Explanation
[Solution] $I = \frac{E}{nR + R}$ and $10I = \frac{E}{R/n + R}$. Dividing: $10 = \frac{n+1}{1/n + 1} = n$, so n = 10. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.