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Solution


Q.

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             [2018]
1 10  
2 11  
3 20  
4 9  

Ans.

(1)

Current drawn from a battery when n resistors are connected in series is

         I=EnR+R                         ...(i)

Current drawn from same battery when n resistors are connected in parallel is

       10I=ERn+R                       ...(ii)

On dividing eqn. (ii) by (i), 10=(n+1)R(1n+1)R

After solving the equation, n=10