First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R and current I is observed to flow. Then, the resistors are connected in parallel to the same battery.It is observed that the current is increased 10 times. What is n?

# First a set of n equal resistors of R

**prasanna**#2

In series combination of resistors, current I is given by

I = E/E + nR

Whereas, in parallel combination current 10 I is given by

E/R + R/n = 10 I

====> E/R+R/n = 10 (E/R+nR)

Now, according to problem, 1+n/1+1/n =10

===>10 = (1+n/n+1)n

===> n = 10