Question

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

Answer 1

Answer:

hope this will help you.

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  

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