What is the current in each branch? So far I have, 6V-12i1+12V-6(i1+i2) -2i1=0 1

Question

What is the current in each branch?

So far I have,

6V-12i1+12V-6(i1+i2) -2i1=0

18V-12i1-6i1-6i2-2i1=0

-20i1-6i2=-18V

12V-6(i2+i1)-4i2+6V-8(i2+i3)=0

18V-6i2-6i1-4i2-8i2-8i3=0

-6i1-18i2-8i3=-18V

6V-8(i3+i2)-10i3=0

-8i3-8i2-10i3=-6

-8i2-18i3=-6V

I broke it down so you could possibly see where I went wrong because as it stands I don't believe I can eliminate anyting. Please explain where I went wrong and the reasoning behind it as well as solve the problem, thank you.

Answers: i1=0.664A, i2=0.786A, i3=1.450A, i4=0.770A, i5=0.016A, i6=0.664A

Explanation / Answer

Equations are 100% correct.

-20i1-6i2=-18

-6i1-18i2-8i3=-18

-8i2-18i3=-6

by solving these equations we get,

i1 = 0.66432 A

i2 = 0.78559 A

i3 = -0.015817

Here, i1,i2,i3 are currents in three branches,

Now , current in R1 => i1 = 0.66432 A

i2 = 0.78559 A

i3 = i1+i2 = 0.66432+0.78559 = 1.4499

i4 = i2 + i3 = 0.78559 -0.015817 = 0.76977 A

i5 = i3 = 0.015817 A

i6 = 0.664A

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