1.

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 arc connected in parallel to the same battery. Then the current drqwn from battery becomes 10 I. The value of 'n' is

Answer»

9
10
20
11

Solution :10
First SITUATION

`THEREFORE I = (E)/(R + nR) "" `....(1)
Second situation

`therefore I. = (E)/((R)/(N) + R) = (NE)/(R + nR) "" `... (2)
But I. = 10 I
`therefore (nE)/(R + nR) = (10 E)/(nR + R)`
`thereforen =10`


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