Question:
First, a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ‘n’?
Solution:
When the resistors are in series combination, the current is given as I = E/R+nR
When the resistors are in parallel combination, current is given as 10I = E/(R+R/n)
Solving the two equations, we get
n = 10