# Mesh-Current

Discussion in 'Homework Help' started by lukepfister, Sep 16, 2007.

1. ### lukepfister Thread Starter New Member

Sep 16, 2007
7
0
Find the power consumed by the 20 ohm resistor using the mesh-current technique.

So, here's what I got

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

Current 2-
$10i_0 + I_2 + 20(I_2+I_1) + 4(I_2-I_3) = 0$
$I_1+I_3 = i_0$
$10(I_1 + I_3) + 20(I_2+I_1) + 4(I_2-I_3) = 0$
$30I_1 + 24I_2 +6I_3 = 0$

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

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 errors in there?

Thank you guys for your help!

The final, correct equations are-

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

Mesh 2

$10i_0 + I_2 + 20(I_2+I_1) + 4(I_2-I_3) = 0$
$I_1+I_3 = i_0$
$10(I_1 + I_3) + I_2 +20(I_2+I_1) + 4(I_2-I_3) = 0$
$30I_1 + 25I_2 +6I_3 = 0$

Mesh 3

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

The final matrix would look like...

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

2. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
I'm unclear on the nature of the dependent voltage source. It appears to be defined as 10*i0. What I am unclear on is where is i0 defined?

I withdraw the question about i0. I just noticed it. The text is a bit light so I missed it the first scan. I'm good now.

hgmjr

3. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
I think there is an error in the final expression for current 3. Check the coefficient of I1.

hgmjr

4. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
Double check this expression for power against the one in your textbook.

hgmjr

5. ### lukepfister Thread Starter New Member

Sep 16, 2007
7
0
I have no idea why I wrote i^2v, I have been using the expression p=i^2R to find power. It was a typo.

As for the third equation, I found an error and now have it as
$3 I_1 - 4 I_2 + 12 I_3 = 0$ But am still getting incorrect results.

6. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
Is there any chance that there is an error in the diagram?

hgmjr

7. ### lukepfister Thread Starter New Member

Sep 16, 2007
7
0
No, the diagram is correct. The resistor polarity's were assigned by me, as were the mesh current/directions. Other than that, it is straight out of the book.

Can you solve and see what you get for the power at the resistor? Maybe that is where I am making mistakes.

First, I solve for the mesh currents I1 and I2 using a matrix. I1 has a negative value. I then sum I1 and I2 together to get the actual value of the current flowing through the resistor, and then plug it into p=I^2R.

8. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
I believe your expression for current 1 has a typo and sign error. In your expression for current 2 you accidentally dropped a term and you have a sign error as well.

Even after making the corrections I elude to above I don't get 259.2 Watts as the answer.

hgmjr

9. ### lukepfister Thread Starter New Member

Sep 16, 2007
7
0
I see the typo...no clue on the sign error/dropped terms. Can you give a hint?

10. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
As for the dropped term in current 2, look closely and you will see that there is a term in the first equation that got dropped in the third equation in the series due to oversight no doubt.

hgmjr

11. ### lukepfister Thread Starter New Member

Sep 16, 2007
7
0
I believe you are talking about the I2 term that mysteriously vanished? Good eye! With that, I get the correct answer! Thank you very, very much. It is amazing how one simple mistake like that has cost me several hours of frustration.

Once again, MANY THANKS!

12. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
If you have not already done so, I recommend you correct your equations in your original post for the benefit of others that might be interested in this thread.

hgmjr