- #1

kstorm19

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## Homework Statement

The problem asks to solve for i1, i2, and i3 using the mesh method in the following circuit:

I will also include the link, because I noticed I am having trouble attaching the image to the post: https://imgur.com/a/bBvuLXM

I would also like to confirm the answer by solving using the node analysis.

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2. Homework Equations

2. Homework Equations

V=IR

## The Attempt at a Solution

First of all, I solved for all three currents using the mesh analysis.

i1 = 5 A

For mesh 2, this is the equation I came up with:

12i2 + 11(i2-i3) + 13 (i2-i1) = 0

And so, using the Node 3:

i3 = 5+ 1/3 Vx

But Vx = 13i3

So i3 = 5 + 13/3 i3

i3= -1.5 A

Going back to equation for mesh 2 and solving for i2:

i2= -1.35 A

So I have:

i1 = 5 A

i2= -1.35 A

i3= -1.5 A

I wanted to confirm these results using the node analysis.

For node 1

(V1-V2)/12 + V1/11 + (V1-V3)/13 = 0

For node 2

(V2-V1)/12 + V2/13 = 5

For node 3

(V3-V1)/13 + 1/3 Vx + 5 = 0

So in this last equation I got a little confused, since when I was setting it up, V3 has the highest potential. If Vx = V1-V3 by looking at the arrow, should I write it that way in the equation, or should I put instead V3-V1?

I figured that maybe it should be written using V3-V1, so this is how I wrote the equation for node 3:

(V3-V1)/13 + 1/3 (V3-V1) + 5 = 0

Solving these 3 equations, I got:

V1 = 6.036 V

V2 = -0.061 V

V3 = -18.23 V

Now, replacing for Vx, which should be -19.5 V since Vx = 13i3 and i3=-1.5 A, I don't get that answer.

I also tried (V3-V1)/13 + 1/3 (V1-V3) + 5 = 0 for the third node equation in case I was wrong and I got:

V1 = -6.64 V

V2 = -3.21 V

V3 = 12.58 V

Now, subtracting V1-V3 gets me -19.22 V (almost -19.5 V from the mesh analysis).

But looking to confirm that i2=-1.35 A, (V2-V1)/12 doesn´t get me anywhere near that value. Where did I go wrong?

Thank you so much guys for your time and help.

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