Mesh Current: Cannot see my mistake in my approach?

[teh_dude]

Mesh Current: Cannot see my mistake in my approach?

Homework Statement

Find currents I1, I2, and I3 using Mesh Current Analysis in this circuit:

Homework Equations

KVL

The Attempt at a Solution

Initial equations:
Mesh 1: -120 + 20k(I1) + 40k(I1 - I2) = 0
Mesh 2: 40k(I2 - I1) - 4ix + 10k(I2 - I3) = 0
Mesh 3: 10k(I3 - I2) + 25k(I3) + 100 = 0

As seen in the circuit, ix = 40k(I1 - I2), so the equation for Mesh 2 can be rewritten as:

New Mesh 2: 40k(I2 - I1) - 160k(I1 - I2) + 10k(I2 - I3) = 0

After rearranging and consolidating, I come up with this system ready to put in an augmented matrix:

60k(I1) - 40k(I2) + 0(I3) = 120
-200k(I1) + 210k(I2) - 10k(I3) = 0
0(I1) - 10k(I2) + 35k(I3) = -100

The answer must be submitted online (yay for unlimited attempts!), and in this instance, must be in mA.

I get the following using the "rref()" function in my TI-84:

I1 = 5.3548387 mA

I2 = 5.0322581 mA

I3 = -1.4193548 mA

I get the following different answers by hand:

I1 = .003168316832 mA

I2 = 2.3762376 mA

I3 = 21.7821782 mA

All of these are incorrect according to the website, but I don't see an error in my equations (although my attempt by hand probably has errors).

FYI, this is an extra credit problem that I am doing solely for extra practice before the test. Generally I'm pretty proficient at this, but this one just isn't making any sense to me--either that or maybe I'm right and the site's wrong? lol...

 

[gneill]

4ix is a current source, not a voltage source.

 

[teh_dude]

...and therefore I2 = -4ix, and my set of equations must reflect that accordingly, right? ah HA! Thanks!

EDIT: ...aaaaand I2 therefore is a forced mesh that needs no equation of its own! derpa herp derp...lol. Whoops. Again, thanks.

 

[gneill]

By George, I think he's got it

 

[teh_dude]

Nope. Perhaps I'm taking the right approach now, but I'm still getting it wrong. I think before I post my new attempts, I'm going to go eat supper, go to my night class, and come back with a (hopefully) fresher mind that isn't low on blood sugar.

 

[teh_dude]

Okay, well, still no dice, and I still don't see what I'm doing wrong, especially with taking what should be the correct approach with the dependent source. The original problem is above, so I'm going to type out my latest attempt. There were two versions of it, because an approximation was easier, and should be close enough, but I'll post the exact numbers here:

Mesh 1: -120 + 20000I1 + 40000(I1 + 4ix) = 0
Mesh 2: I2 = -4ix
Mesh 3: 10000(I3 + 4ix) + 25000I3 + 100 = 0

ix = 40000(I1 + 4ix)



And now, the work, in detail:

ix = 40000I1 + 160000ix
-40000I1 = 159999ix
ix = (-40000/159999)I1

-120 + 20000I1 + 40000I1 + 160000ix = 0
-120 + 60000I1 + 160000(-40000/159999)I1 = 0
60000I1 - 40000.25I1 = 120
19999.75I1 = 120
I1 = 0.006000075 A = 6.000075 mA

10000I3 + 40000ix + 25000I3 + 100 = 0
35000I3 + 40000(-40000/159999)I1 + 100 = 0
35000I3 - 10000.0625I1 = -100
35000I3 - 60.00112501 = -100
35000I3 = -39.99887499
I3 = -0.001142825 A = -1.142825 mA

-4ix = -4(-40000/159999)I1 = -4(-40000/159999)(.006000075) = I2 = .0060001125 A = 6.0001125 mA

These answers for I1, I2, and I3 are all wrong according to the website. What am I doing wrong? Is there just some glaring algebraic fallacy that I'm blind to?

 

[gneill]

Your equation ix = 40000(I1 + 4ix) is not correct (you're equating a current to a voltage). Drop the 40000 Ohm resistance from it and it'll be fine.

 

[teh_dude]

I now see. I was trying to say that I = I*R, which is, for all intents and purposes, bull. lol...*sigh* I'll post back with the result either way. Thanks again..

 

[teh_dude]

*ahem*...it worked this time. Thank you for being the invaluable second mind and second set of eyes.

After I realized what you were saying, that I was trying to break Ohm's law, I realized I was doing the same thing in another problem. OOPS. So glad these are just extra credit/practice. lol. Thank you very much.

 

[gneill]

Glad it all worked out. Cheers.

 

[teh_dude]

GAAAAHHH! Now I've got another one--that other one that I mentioned--except I'm not sure that I have the right approach for part of it. It's another one listed as a mesh current problem.

Circuit:

Problem: Find iy, io, and vx.

General approach:

Clockwise supermesh around the top two meshes because of the 3A current source in between them.

The bottom mesh, well, there's a dependent current source there, so I figured the current in that bottom mesh was equal to 4iy, and I'm for sure not treating current sources like voltage sources this time lol, but this approach isn't working for me. I believe it probably has something to do with the fact that I'm not including the 120 V source in my equations. Thing is, I don't know how. Here's what I've tried that failed:

Supermesh equation: -230 + 3*I1 + 2*I2 + 2*vx + 3(I1 - 4iy) = 0

Others:

3A = I2 - I1 ---> I2 = 3 + I1

iy = I1

io = -I2 = -3 - I1

vx = 3(I1 - 4iy) = 3*I1 - 12iy = 3*I1 - 12*I1

I got some answers, but they're wrong.

I understand that I cannot directly determine the current at an independent voltage source, nor the voltage at a current source. So how does the 120 V source apply??

I was going to write KVL for the bottom mesh, but I don't see how it's possible.

 

[gneill]

Check the polarity indicated for Vx and then the equation you wrote for it.

 

[teh_dude]

Yeeeeeep that did it. lol. The overwhelming majority of my tutoring problems lately have been someone else going "fix this <fill in overly simple tiny mistake that derails the whole damn problem> and you're good". Thanks again. So I don't have to include the 120V source in the equations, and I'm not sure why, but maybe if I attacked it with superposition then I'd see the proof. Buuuuut I'm too tired right now for that. Sleepy time. Thanks again so much, kind sir.