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

Current drawn from a battery when n resistors are connected in series isI=EnR+R                  .......(i)Current drawn from same battery when n resistors are connected in parallel is10I=ERn+R                .........(ii) On dividing eqn. (ii) by (i), 10=(n+1)R1n+1RAfter solving the equation, n=10