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?
In series combination of resistors, current I is given byI = E/E + nRWhereas, 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