Engineering Node/Mesh Analysis of a Circuit

[rkum99]

Homework Statement

The following is a link to the problem diagram and my work: http://imgur.com/a/9tppH
The goal is to determine the power through the 20V voltage source.

Homework Equations

KCL and KVL: Currents entering a node sum to zero. Voltage drops around a loop sum to zero.

The Attempt at a Solution

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I tried to use mesh analysis as seen in the link above, but the answer I receive is incorrect. Its not in the images above, but the final power through the 20V voltage source would be P = IV or P = (1.66 A)(20V) = ~33.3 W, which is an incorrect answer.

I appreciate any help.

 

[gneill]

Your problem and work should be visible in-thread. Helpers should not have to go fishing for it off-site. Trying to point out issues in work that is in an image is difficult because individual lines can't be quoted. You may find that many helpers are not willing to make the effort; you will get much better responses if you take the time to make it easy for helpers to give you help.

 

[rkum99]

My bad, I hoped that having everything in one place would make it simpler. Thank you for letting me know. I will relist everything here:

Homework Statement

Diagram:

The goal is to determine the power through the 20V voltage source.

Homework Equations

KCL and KVL: Currents entering a node sum to zero. Voltage drops around a loop sum to zero.

The Attempt at a Solution

I attempted using mesh analysis.

Starting with Mesh 3: 4(i₃ - i₁) + 6(i₃) - 10 = 0
Mesh 2: 12(i₂- i₁) + 3(i₂) - 10 = 0
Mesh 1: 4(i₃ - i₁) +12(i₂ - i₁) - 20 = 0

Simplifying the above equations, I get:

10i₃ - 4i₁ = 10
15i₂ - 12i₁ = 10
-16i₁ + 12i₂ + 4i₃ = 20

Solving it out, I get:

i₁ = -1.66 A
i₂ = -0.66 A
i₃ = 0.333 A

Since I want to find the power through the 20V voltage source:

P = IV = (I₁) (20V) = ~33.333

This is an incorrect answer, so I suppose I went wrong somewhere. If I had to guess, I would assume I improperly accounted for the 10 V voltage source in my equations? I would appreciate any help.

 

[gneill]

The Attempt at a Solution

I attempted using mesh analysis.

Starting with Mesh 3: 4(i₃ - i₁) + 6(i₃) - 10 = 0
Mesh 2: 12(i₂- i₁) + 3(i₂) - 10 = 0
Mesh 1: 4(i₃ - i₁) +12(i₂ - i₁) - 20 = 0

The mesh 3 equation looks fine.

In your mesh 2 equation you are summing potential drops around the loop. Is the 10 V source a drop or a rise? Check the sign you've assigned to it.

In your mesh 1 loop you seem to have switched to summing potential rises. Again, check the sign you've assigned to the voltage source.

 

[rkum99]

So, in the mesh 2 equation, the 10V source is a drop because the current goes through it from + to -.

Why would the signs change in the mesh 1 equations, however? In the mesh 3 equation, I subtract the voltage source as the current goes from - to +. Wouldn't it be the same in the mesh 1 equation - the current goes through the 20V source from - to +.

Sorry if I'm looking over something obvious - I'm having a hard time keeping track of signs in these KCL/KVL problems.

 

[gneill]

In the mesh 1 equation you changed from summing drops to summing rises. This must change the signs of everything. Look carefully at how you wrote the equation:

$4(i_3 - i_1) +12(i_2 - i_1) - 20 = 0$

Isolate just the $i_1$ terms: $-4 i_1 -12 i_1$. Note how they are both negative? When you did your "KVL walk" around the loop in the direction of the loop's mesh current you wrote the potential changes for resistors as a result of that current as negative vaues. So you are writing potential drops as negative values. This means you should also write potential rises as positive values. That would include the potential rise due to the 20 V source since you pass through it from - to + while moving clockwise around the circuit in the direction of the mesh current.

You may find it helpful to get in the habit of always writing your equations in terms of either potential rises or potential drops, and not mixing them from equation to equation. Being consistent will avoid having to think about the signs of things every time; there will be one consistent rule to apply every time.