Questions

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 10*I*. The value of *n* is

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a

10

b

11

c

20

d

9

detailed solution

Correct option is A

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

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