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

HERE IS YOURS SOLUTION;

◆ The Key concept: The equivalent resistance of series combination is in series with the internal resistance R of battery and in parallel combination of resistors, the equivalent resistance of parallel combination is also in series with the internal resistance of battery.

So,

【 Refer the Attachment】

◆ So the answer is 10 ◆


HOPE IT HELPS

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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