Physics, asked by nantheta7, 1 year ago

An electric current of 5 amperes is divided into three branches, along three wire of same material
with same cross-section but with their lengths in the proportion of 1:2:3 then the current in the
middle branch will be :

Answers

Answered by huzurali453

we know that resistance =resistivity ×length/area of cross section

here only lengths are variable

so resistance will be directly proportional to the length of the conductor

lets assume the resistances are k :2k:3k

we need to find current in the middle branch so find the combining resistance of k and 3k

yes it is k×3k/k+3k=3k/4

therefore current flowing in the middle branch is =5×(3k/4)/(11k/4)

=15/11ampere

at last i used the formula of current in two branch you my know the formula

I(1)=I×R(2)/R(1)+R(2)

nantheta7: thanks for solution. but my book says, right answer is 15/11 A :///

huzurali453: yes i have mistaken now you can see the answer i have edited

nantheta7: thank you, in the last step how did you get 11k/4k?

huzurali453: ok see the answer again

nantheta7: thank you so much

huzurali453: here i converted the three resistances into two resistances

nantheta7: oh ok

huzurali453: Wc

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