# Mesh analysis problem with dependent sources

#### inkyvoyd

Joined Dec 6, 2011
25
Hello,
I've been trying to learn about other analysis methods, but have also come back to mesh analysis to make sure I have it down. I have recently encounted a problem with 4 sources, 2 dependent, and 2 independent. The problem can be found here (second half of the video;I know it's node voltage analysis but I'm trying to solve it with mesh). Alternatively, I have taken a screenshot for convenience. I have formulated the following system of equations with meshes ia-if (starting from top left and going right like reading a letter)
ia+1(ia-id)+i(ia-ib)=0
1(ib-ie)+1(ib-ia)-6)=0
6+2(ie-if)+1(ic-if)=0
-12+1(id-ia)+2ia=0
2ia+1(ie-ib)+1(ie-if)=0
1(if-ie)+1(if-ic)+if=0
I have used:
I_x=ie-if
V_x=1*ia
My solutions are: ia=24/7,ib=12/7,ic=-390/7,id=60/7,ie=-6,if=-144/7 (solved with mathematica)
These don't agree with those found in the video.

May someone please check over my answers or tell me what I did incorrectly? I have used kiloOhms as units, but any place value changing still seems to have differing answers.

Thank you very much!

#### WBahn

Joined Mar 31, 2012
25,090
What direction are your mesh currents? CW or CCW?

Have you tried verifying your answers by seeing if they are consistent with the problem? For instance, use your currents to calculate the node voltages along different paths and see if they agree.

#### inkyvoyd

Joined Dec 6, 2011
25
What direction are your mesh currents? CW or CCW?

Have you tried verifying your answers by seeing if they are consistent with the problem? For instance, use your currents to calculate the node voltages along different paths and see if they agree.
my mesh currents are CW. I can tell that my answers are wrong because the dependent current source is inconsistent - I'll check over my work again, but I'm having a lot of trouble catching what I'm doing wrong.
Thanks.

Wow - and now the answers mathematica are giving are different - but still wrong. I'll look over this tomorrow I guess. I probably fail harded somewhere.

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#### The Electrician

Joined Oct 9, 2007
2,751
Hello,
I've been trying to learn about other analysis methods, but have also come back to mesh analysis to make sure I have it down. I have recently encounted a problem with 4 sources, 2 dependent, and 2 independent. The problem can be found here (second half of the video;I know it's node voltage analysis but I'm trying to solve it with mesh). Alternatively, I have taken a screenshot for convenience. I have formulated the following system of equations with meshes ia-if (starting from top left and going right like reading a letter)
ia+1(ia-id)+i(ia-ib)=0
1(ib-ie)+1(ib-ia)-6)=0
6+2(ie-if)+1(ic-if)=0
-12+1(id-ia)+2ia=0
2ia+1(ie-ib)+1(ie-if)=0
1(if-ie)+1(if-ic)+if=0
I have used:
I_x=ie-if
V_x=1*ia
My solutions are: ia=24/7,ib=12/7,ic=-390/7,id=60/7,ie=-6,if=-144/7 (solved with mathematica)
These don't agree with those found in the video.

May someone please check over my answers or tell me what I did incorrectly? I have used kiloOhms as units, but any place value changing still seems to have differing answers.

Thank you very much!
You need to make these changes:

ia+1(ia-id)+1(ia-ib)=0
1(ib-ie)+1(ib-ia)-6)=0
2(ie-if)-ic=0
-12+1(id-ia)-2ia=0
2ia+1(ie-ib)+1(ie-if)=0
1(if-ie)+1(if-ic)+if=0

I mentioned to you in another thread that if you have a mesh where a current source is in a controlling branch, which you have here in mesh 3, you just use an equation for that mesh which sets the mesh current equal to that current source.

And if you're going to use k ohms then your answers will be in milliamps.

• inkyvoyd

#### WBahn

Joined Mar 31, 2012
25,090
ia+1(ia-id)+i(ia-ib)=0
You units don't work out.

ia has units of current.
1(ia-id) has units of voltage (assuming your 1 is actually 1kΩ).
i(ia-ib) has units of current-squared. And what current does 'i' refer to?

Since you mesh currents are CW, then your equation implies that you are summing up voltage drops around the loop. Fine, but be sure that you are consistent.

ALWAYS track your units throughout your work -- it is perhaps the single most powerful error detection and correction tool you have. If the units are wrong, you KNOW the answer is wrong -- no need to go any further until the units work out.

This equation should be:

1kΩ(ia) + 1kΩ(ia-id) + 1kΩi(ia-ib) = 0

1(ib-ie)+1(ib-ia)-6)=0
What is the final closing parenthesis match up with? Engineering is a very exacting and precise undertaking and it tends to be very unforgiving of sloppiness. Now, we all make mistakes, but we always need to be as diligent as we can to avoid them, detect them, and correct them.

Again, track your units. This should be:

1kΩ(ib-ie) + 1kΩ(ib-ia) - 6V = 0

6+2(ie-if)+1(ic-if)=0
And here is where failing to track your units bites you!

6V + 2(ie-if) + 1kΩ(ic-if) = 0

Now examine the units:
6V volts
2(ie-if) current
1kΩ(ic-if) volts

You can't add voltage and current! But because you don't track your units, you put yourself in a position where you couldn't detect this error. In the real world, people can and do die because engineers can't be bothered to track their units in their work. Personally, I think it should be considered criminal negligence. After all, if a doctor failed to do the most basic and simple things to verify the correctness of their diagnosis and someone dies as a result, it would be a major malpractice suit, if not worse. Yet doctors generally kill people one at a time while engineers do it in job lots.

Now, let's assume for the moment that you had tracked your units and you had detected this error. What would it have told you? Simple, the problem is that you need the voltage across the dependent current source, and you have no way to determine what it is. But wait, the point of the mesh equation is to determine the mesh currents. But you KNOW what this particular mesh current is, it is 2Ix, which you can express in terms of ie and if. So this equation should have simply been:

ic = 2(ie-if)

-12+1(id-ia)+2ia=0
Again your units don't work out. This should be:

-12V + 1kΩ(id-ia) - 2Vx = 0
Vx = -1kΩ(ia)
-12V + 1kΩ(id-ia) - 2[-1kΩ(ia)] = 0
-12V + 1kΩ(id-ia) + 2kΩ(ia) = 0

See how that '2' (from the 2Ix) does not have units but the '2' in the final equation does? Don't throw that information away.

2ia+1(ie-ib)+1(ie-if)=0
Same issues. This should be

-2Vx + 1kΩ(ie-ib) + 1kΩ(ie-if) = 0
-2[-1kΩ(ia)] + 1kΩ(ie-ib) + 1kΩ(ie-if) = 0
2kΩ(ia) + 1kΩ(ie-ib) + 1kΩ(ie-if) = 0

1(if-ie)+1(if-ic)+if=0
This one just needs the units included:

1kΩ(if-ie) + 1kΩ(if-ic) + 1kΩ(if) = 0

I_x=ie-if
V_x=1*ia
The first is correct. But the second is off by a minus sign. If ia is positive, going CW, then Vx is negative.

Ix = (ie-if)
Vx = -1kΩ(ia)

My solutions are: ia=24/7,ib=12/7,ic=-390/7,id=60/7,ie=-6,if=-144/7 (solved with mathematica)
These don't agree with those found in the video.
Don't let Mathematica do all your thinking for you. This is an easy enough set of equations to solve; do it by hand.

• inkyvoyd

#### The Electrician

Joined Oct 9, 2007
2,751
Again your units don't work out. This should be:

-12V + 1kΩ(id-ia) - 2Vx = 0
Vx = -1kΩ(ia)
-12V + 1kΩ(id-ia) - 2[-1kΩ(ia)] = 0
-12V + 1kΩ(id-ia) + 2kΩ(ia) = 0
There's a sign error here; it should be:

-12V + 1kΩ(id-ia) + 2Vx = 0
Vx = -1kΩ(ia)
-12V + 1kΩ(id-ia) + 2[-1kΩ(ia)] = 0
-12V + 1kΩ(id-ia) - 2kΩ(ia) = 0

• inkyvoyd

#### WBahn

Joined Mar 31, 2012
25,090
Yep, you are correct. Thanks for catching it.

#### inkyvoyd

Joined Dec 6, 2011
25
Thank you for taking the time to go through this problem and my mistakes step by step. I greatly appreciate the feedback - and in the future I won't omit units >.<
You need to make these changes:

ia+1(ia-id)+1(ia-ib)=0
1(ib-ie)+1(ib-ia)-6)=0
2(ie-if)-ic=0
-12+1(id-ia)-2ia=0
2ia+1(ie-ib)+1(ie-if)=0
1(if-ie)+1(if-ic)+if=0

I mentioned to you in another thread that if you have a mesh where a current source is in a controlling branch, which you have here in mesh 3, you just use an equation for that mesh which sets the mesh current equal to that current source.

And if you're going to use k ohms then your answers will be in milliamps.
Thanks - I must've glossed over that worrying about the other equations.

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