# Mesh-current method

1. Sep 16, 2007

### Luke1294

1. The problem statement, all variables and given/known data

Use the mesh-current method to find the power dissipated in the 20 ohm resistor

2. Relevant equations

$$p=vi^2$$

3. The attempt at a solution

Looking at the picture above, you can see how I set my current flows, in hopes of making the most currents flow together. If the picture isn't entirely clear, voltage supplied at the far left of the circuit is dependent upon the current flowing from left to right on the 3 ohm resistor. I'm assuming I have some major sign-error issues and that's what's keeping me from the correct answer, provided to be 259.2 W.

My equations-

Current 1-
$$135 + 2I_1 +20(I_1+I_2) + 3(I_1+I_2) = 0$$
$$25I_1 + 20I_2 + 3I_3 = -135$$

Current 2-

$$10i_0 + I_2 + 20(I_2+I_1) - 4(I_2-I_3) = 0$$
$$I_1+I_3 = i_0$$
$$10(I_1 + I_2) + 20(I_2+I_1) - 4(I_2-I_3) = 0$$
$$30I_1 + 17I_2 +14I_3 = 0$$

Current 3-

$$5I_3 + 4(I_3 - I_2) + 3(I_1+I_3) = 0$$
$$5I_1 - 4I_2 + 12I_3 = 0$$

Okay, so I throw everything into a matrix, solve, for I1, I2, and I3. I got some very funky answers, so I know SOMETHING is wrong based on that... I got

$$I_1 = \frac{35100}{373} I_2 = \frac{-42930}{373} I_3 = \frac{-23085}{373}$$

So to find the actual current going through the 20 ohm resistor, I would sum I1 and I2. I wind up with -7830/373...When I plug that value into my power equation, I get 8813.245, a far cry from the expected value.

Any pointers?

$$Last edited: Sep 16, 2007 2. Sep 16, 2007 ### CEL You should have: Current 1- [tex] 135 + 2I_1 +20(I_1+I_2) + 3(I_1+I_3) = 0$$
$$25I_1 + 20I_2 + 3I_3 = -135$$
I think those errors are typos.

But in the second equation:
Current 2-

$$10i_0 + I_2 + 20(I_2+I_1) + 4(I_2-I_3) = 0$$
$$I_1+I_3 = i_0$$
$$10(I_1 + I_3) + 20(I_2+I_1) + 4(I_2-I_3) = 0$$
$$30I_1 + 24I_2 +6I_3 = 0$$

3. Sep 16, 2007

### Luke1294

Yes, the first equation was a typo. Will correct.

Why would it be +4(I2-I3)? The current is flowing from the negative side of a resistor to the positive side...wouldn't that give it a - sign?

Even if that is incorrect, those values unfortunatly do not lead me where I need to be :(

4. Sep 16, 2007

### l46kok

When you do mesh-currents on a circuit, you assume the directions of the current. Meaning you can ignore the polarity of the resistors. Don't worry, the algebra will take care of itself and show if the direction of the current is inversed (You will see a negative sign in front of it, if that is the case)

You are writing a mesh current equation from I2 side, which turns out to be 4(I2-I3), because I3 is going in the opposite direction of I2.

5. Sep 16, 2007

### Luke1294

Okay, I wasn't aware of the polarity thing. I understood the I2-I3, but the positive 4 was what threw me off. At any rate, I'm still not arriving at the correct answer.

With a matrix that looks like this...

25 20 3 | -135
30 24 6 | 0
5 -4 12 | 0

I wind up with
I1 = -87.75
I2 = 92.8125
I3 = 67.5.

But to find the branch current, you have to sum I1 and I2 together, which would give me a total current of 5.0625 running through that branch. Using my power expression, p=vi^2, I'm getting 512.57 W....according to the back of the book, I'm looking for 259.2 W. Do you see any other errors in there?

Thank you guys for your help!

6. Sep 16, 2007

### CEL

The third equation should be:
$$3 I_1 - 4 I_2 + 12 I_3 = 0$$
and not
$$5 I_1 - 4 I_2 + 12 I_3 = 0$$

7. Sep 16, 2007

### Luke1294

I don't know why I wrote p=vi^2..it's p=i^2r, and that is the formula I have been using all along. Even with the noted corrections, I'm still not quite where I should be...hmpf.

8. Sep 16, 2007

### Luke1294

Well, I found my error, with some help from the allaboutcircuits fourm. There are two main things- first, I have a typo in my first equation. The final term should be I3, not I2 again. The main problem, however, lies in the 2nd equation-

I went from
$$10i_0 + I_2 + 20(I_2+I_1) - 4(I_2-I_3) = 0$$
to
$$10(I_1 + I_2) + 20(I_2+I_1) - 4(I_2-I_3) = 0$$

I totally dropped the second term from the first equation- the I2 standing alone. With that, everything checks out, I get 18/5 for the branch current, and the correct value for power dissipated. Thank you all for your help, and I am sorry that my carelessness caused so many problems.

9. Sep 17, 2007

### CEL

I don´t know how introducing two errors in your equation did you get the right answer. According to Kirchoff's voltage law (KVL) the sum of all voltages across the loop must be zero.
The voltage across the 4 ohm resistor is $$+4(I_2 - I_3)$$ and you cannot drop the voltage across the 1 ohm resistor.
$$30 I_1 + 25 I_2 + 6 I_3 = 0$$
You should not trust unconditionally the answer given by the book and try to adjust your equations to get the "right" result.

10. Sep 17, 2007

### Luke1294

No, you misunderstand. Thats is the final equatation i arrive at. When working earlier, I had dropped the one ohm resistor term. When I realized and corrected this, I found the equation you had written, and get the correct answer. I should have been more clear.

11. Sep 17, 2007

### CEL

I got the book's answer using the correct equations. For
$$A= \left[ \begin{array}{ccc} 25 & 20 & 3 \\ 30 & 25 & 6\\ 3 & -4 & 12 \end{array} \right]$$

$$b = \left[\begin{array}{c}-135\\0\\0 \end{array} \right]$$

$$I = A^{-1} b$$
I got
$$I_1 = -64.8000$$
$$I_2 = 68.4000$$
$$I_3 = 39.0000$$

$$I_L = I_1 + I_2 = 3.6$$
$$p = 20 I_L^2 = 259.2$$