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
- 10
- 11
- 20
- 9
Answer
(1) 10
Solution
Series: I = EnR + R = E(n + 1)R
Parallel: equivalent resistance = Rn
⇒ 10I = E Rn + R = nE(n + 1)R
Dividing: 10II = n
∴ n = 10
Answer: (1)