Question

A set of n equal resistor 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 resistance are connected in parallel to the same battery.then the current drawn from battery becomes 10 i. the value bof n is

Answer 1

will be the same as it is connected in parallel

Answer 2

Explanation:

I = E / R+ nr

 =E / R +nR = E/ R(n+1)

Given when the resistors are parallel arranged , I' = 10I

I' = E / R+ (R/n)  = 10I

E / R+ (R/n)  = 10 x E/ R(n+1)

10[ R(1 + 1/n )] = R (n+1 )

TIP :- For exam if binomial is hard just substitute each options for fast answering

OR

10 + 10/n = n +1

n = 10

                       Hope you understand  

 This is Habel Sabu ....... Isn't it the Brainliest