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'?

Explanation

When all resistors are connected in series, the equivalent resistances of the network will be given by
`R_("eqvt") = n R +R`
`rArr R_("eqvt") = (n+1)R`
Current flowing through the circuit,
`I = (E )/(R_("eqvt")) = (E )/((n+1) R)`
When all resistors are connected in parallel, the equivalent resistance of the network will be given by: Current flowing through the circuit in this case is:
`I = (E )/(R_("eqvt")) = (nE)/(R(n+1)) = (nE)/(R (n +1))`
Given `I. 10I`
`rArr (E(n))/((n+1)R) = 10 (E )/((n+1) R)`
`rArr n=10`