# Does mesh analysis give no solution?

#### MrAl

Joined Jun 17, 2014
9,785
ok, here is the steps to a solution:

Define mesh equations:

View attachment 275003
Simplify:

View attachment 275004

Add 2 times last equation to top equation to eliminate I2 and then solve for I3:

View attachment 275005
Solving for I2:

View attachment 275006
The voltage across R1 is: R1*I2=10*4/5=8V
The voltage across R2 is: R2*(I2-I3)=(4/5-3/5)*10=2V
The voltage across R3 is: R3*I3=20*3/5=12V

No issues becasue:
The voltage around loop 2 is:
-10+8+2=0
The voltage around loop 3 is:
-10+-2+12=0

Verified.

But I2 is 800mA and that is going through R1 and must be coming from V1.
And I3 is 600mA and is going through R3
The current through R2 is 800mA-600mA which is 200mA and thus a 2 volt drop is correct.

The loop currents are what I expect. Not sure what you are trying to say?

The voltage across R3 is not to surprising being when you look at the outer loop the the two voltage sources are in series allowing for up to 20V in the circuit.

View attachment 275007

Yep 800mA is the current through R1 and that happens to be the I2 current that I calculated so it is the I2 current which is 800mA?

So WBahn is correct: The calculated I2 current is 800mA and that is the current going through R1.

Also because both currents have positive values, the assumed direction of current flows (clockwise) is correct.
Well i understand how to perform the mechanics behind mesh analysis, but notice that 800ma does not 'flow' through R2. If 800ma "really" flows through R2 then we could not have 600ma flowing through R3 because the sum of currents entering a node is the same as the sum leaving the node. 800ma through R1 and 800ma Through R2 and 600ma through R3 means we have a deficit of 600ma. 800-800-600=-600, so something there cant be real Resistor R2 is not a calculating machine, it does not know how to add and subtract. If you look inside you do not see 800ma going one way and 600ma going the other way, you just see 200ma going one way. What really happens is at the top central node the 800ma splits into two currents, one 600ma and one 200ma. At the bottom node the 600ma combines with the 200ma and produces 800ma. So there is no addition or subtraction going on in R2.

But as i said before, that's not the only thing we do in circuit analysis that is not 'real', even with other circuits, so this starts to look like a common thing.

Someone mentioned the square root of minus 1: i=sqrt(-1) as being not real but the mesh current is real. But we also use sqrt(-1) in lots of analysis and come out with real values. How is that possible? Because it only has to be 'imaginary' for part of the calculation, and then if it still appears at the end there becomes a different interpretation of that 'quantity'.

We use things that we might call imaginary in lots of calculations so this should not be that big of a surprise.

If you would still like to call it 'real' that's your decision. In the end we still recognize mesh analysis to be a useful tool just like many other types of analysis we do.

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• dcbingaman

#### WBahn

Joined Mar 31, 2012
28,186
Ok why do you have to call it a 'mesh' current then. Why cant you just call it a 'current'.
Really? First, the claim is that writing a control signal in terms of mesh currents renders the circuit invalid. Then the value of a branch current that only borders one mesh can't be the same as that mesh current. Now we have an issue with calling it a mesh current? What's next?

Feel free to rename it if you want. How about "analysis method #42 current". Or perhaps call it a Kirchhoff current since it is defined in such a way as to guarantee that KCL is satisfied at every node. The name 'mesh' is merely descriptive of the visual image created when they are drawn on a circuit. But it is desirable to have a distinct name so that just by using that name, we know the characteristics that they have. All we have to say is, "Define a set of mesh currents." What would you prefer? "Define a set of currents that are cyclic and nonoverlapping and in the same direction and such that at least one such current passes through every branch." Isn't it just so much easier to use a name that tells you what analysis technique is being used and then get on with it?

Perhaps we shouldn't use the term "Thevenin voltage" and instead just call it "voltage"? Same reason -- the name tells us the approach that is being used.

My argument is that if you show a current path with an arrow then that current has to be present in all parts of that path. It cant be present in R2 because if it was we would have 800ma though R1 splitting into 800ma and 600ma which is 1400ma. So saying that I2 'flows' through R2 can not be right. The sum of the currents entering a node has to be the same as the sum leaving the nodes. If 800ma is going through R1 and R2 and 600ma through R3 then the sum would be 800ma-800ma+600ma at the central node.
Huh? I guess this is what's next. Now, apparently, mesh currents don't satisfy KCL at node junctions. If I2 is 800 mA, then you have 800 mA entering the node and 800 mA leaving the node. If I3 is 600 mA, then you have 600 mA entering the node and 600 mA leaving the node. How does that not satisfy the requirement that the sum of the currents entering a node has to be be the same as the sum of the currents leaving it? A mesh current, by the very nature of being cyclic, ensures that any junction it enters it MUST leave, hence guaranteeing that KCL is satisfied.

So my calling it a 'mesh' current we pretend it's real, that's all. The math works because of the principle of superposition.
How many times have I said that mesh currents are a mathematical equivalent that only apply to linear circuits because they rely on the validity of superposition?

But in superposition, once we combine quantities we can no longer distinguish what the two quantities were.
So? How are mesh current ANY different?

Why aren't you arguing that we can't use a Thevenin voltage source in a circuit because there's no physical voltage source in the circuit with that value? Or why, when doing an analysis, can we replace two resistors in parallel with a resistor that is smaller than either and carrying more current than either when there is no physical resistor in the circuit carrying that current? Why are these techniques that rely on mathematical equivalencies okay but equivalencies based on mesh currents aren't?

#### MrAl

Joined Jun 17, 2014
9,785
Really? First, the claim is that writing a control signal in terms of mesh currents renders the circuit invalid. Then the value of a branch current that only borders one mesh can't be the same as that mesh current. Now we have an issue with calling it a mesh current? What's next?

Feel free to rename it if you want. How about "analysis method #42 current". Or perhaps call it a Kirchhoff current since it is defined in such a way as to guarantee that KCL is satisfied at every node. The name 'mesh' is merely descriptive of the visual image created when they are drawn on a circuit. But it is desirable to have a distinct name so that just by using that name, we know the characteristics that they have. All we have to say is, "Define a set of mesh currents." What would you prefer? "Define a set of currents that are cyclic and nonoverlapping and in the same direction and such that at least one such current passes through every branch." Isn't it just so much easier to use a name that tells you what analysis technique is being used and then get on with it?

Perhaps we shouldn't use the term "Thevenin voltage" and instead just call it "voltage"? Same reason -- the name tells us the approach that is being used.

Huh? I guess this is what's next. Now, apparently, mesh currents don't satisfy KCL at node junctions. If I2 is 800 mA, then you have 800 mA entering the node and 800 mA leaving the node. If I3 is 600 mA, then you have 600 mA entering the node and 600 mA leaving the node. How does that not satisfy the requirement that the sum of the currents entering a node has to be be the same as the sum of the currents leaving it? A mesh current, by the very nature of being cyclic, ensures that any junction it enters it MUST leave, hence guaranteeing that KCL is satisfied.

How many times have I said that mesh currents are a mathematical equivalent that only apply to linear circuits because they rely on the validity of superposition?

So? How are mesh current ANY different?

Why aren't you arguing that we can't use a Thevenin voltage source in a circuit because there's no physical voltage source in the circuit with that value? Or why, when doing an analysis, can we replace two resistors in parallel with a resistor that is smaller than either and carrying more current than either when there is no physical resistor in the circuit carrying that current? Why are these techniques that rely on mathematical equivalencies okay but equivalencies based on mesh currents aren't?
Hello,

For some reason you keep talking about something that was discussed and verified. In the diagram with the three resistors i quoted the fact that I2-I3 can be real. I even used Jony's approach with mesh currents to solve for the voltage at the top of R1 in that second circuit he posted. Not sure why you brought this up again.

I said that if we have a REAL current of 800ma shown by the I2 current arrow in R1, then it would also go through R2, and that would mean that there would be a deficit in currents of 600ma because all of that 800ma would go DOWN through R2, none left for I3.
It doesnt matter if they add or subtract in R2 because R2 can not add nor subtract.
If we used pipes and marbles we'd see the same thing if we designed the pipes to act like resistances. The marbles going through #1 pipe would not all go though #2 pipe.

So i think you agree that we dont see 800ma through R2. However, you said you can measure "I2" by measuring the current through R1, and just because you measure 800ma doesnt mean you measured I2. The reason for that is because I2 (800ma) does not flow through R2 and you agreed on the quantity. Since 800ma does not flow through R2, we can not draw a current arrow outside of R2 that shows it is 800ma. When we draw the arrow there, we have to specify 200ma. And yes 200ma is the algebraic sum, but the current arrow for I2 does just outside of BOTH R1 and R2, yet only R1 has 800ma. So the mesh current is not real. You agreed that the mesh current is not real, but you still said you measured it. So which is it, is it real or not real.

So that's the real crux of this argument. If you say you measured I2 then the current arrow for I2 must be incorrect because when an arrow goes through one resistor and something else, that arrow indicates the current in that branch. It's not incorrect though even the arrow is just a simplification mechanism. In fact, we specify that the currents are "mesh" currents.

Besides all that, i would rather just agree that it doesnt make any difference than go on with this because you will never see my point of view.

#### dcbingaman

Joined Jun 30, 2021
855
Well i understand how to perform the mechanics behind mesh analysis, but notice that 800ma does not 'flow' through R2. If 800ma "really" flows through R2 then we could not have 600ma flowing through R3 because the sum of currents entering a node is the same as the sum leaving the node. 800ma through R1 and 800ma Through R2 and 600ma through R3 means we have a deficit of 600ma. 800-800-600=-600, so something there cant be real Resistor R2 is not a calculating machine, it does not know how to add and subtract. If you look inside you do not see 800ma going one way and 600ma going the other way, you just see 200ma going one way. What really happens is at the top central node the 800ma splits into two currents, one 600ma and one 200ma. At the bottom node the 600ma combines with the 200ma and produces 800ma. So there is no addition or subtraction going on in R2.

But as i said before, that's not the only thing we do in circuit analysis that is not 'real', even with other circuits, so this starts to look like a common thing.

Someone mentioned the square root of minus 1: i=sqrt(-1) as being not real but the mesh current is real. But we also use sqrt(-1) in lots of analysis and come out with real values. How is that possible? Because it only has to be 'imaginary' for part of the calculation, and then if it still appears at the end there becomes a different interpretation of that 'quantity'.

We use things that we might call imaginary in lots of calculations so this should not be that big of a surprise.

If you would still like to call it 'real' that's your decision. In the end we still recognize mesh analysis to be a useful tool just like many other types of analysis we do.
Thank you for the detailed response. I agree with everything you had to say. I provided the mesh analysis math just to make sure we were on the same page. I agree that R2 cannot perform the math and find the difference between I2 and I3. The idea that we can take the difference or look at it as the difference between two currents it just for convenience. I have found the more ways I have to look at a situation, the more understanding I take away from it.

#### MrAl

Joined Jun 17, 2014
9,785
Thank you for the detailed response. I agree with everything you had to say. I provided the mesh analysis math just to make sure we were on the same page. I agree that R2 cannot perform the math and find the difference between I2 and I3. The idea that we can take the difference or look at it as the difference between two currents it just for convenience. I have found the more ways I have to look at a situation, the more understanding I take away from it.
Hi,

Oh yes that's good too because we can check our results using more than one method and catch mistakes. Both ways must agree.

• dcbingaman

#### The Electrician

Joined Oct 9, 2007
2,916
Hi,
I have this "simple" circuit that contains a constant current source, resistor, and a current-controlled voltage source.
View attachment 274586

I decided to use a mesh analysis.

View attachment 274594

And my mesh equations

Ia = Iin (1)

(Ix - Ia)*R1 - V + Ix*R2 = 0 (2)

(Ix - Iin)*R1 - p * (Ix - Iin) + Ix*R2 = 0

Ix = (Iin (p - R1))/(p - R1 - R2)

And the values are Iin = 4mA; R1 = 1kΩ; R2 = 500Ω; p = 1kΩ
And the problem begins when I set R2 to. And try to find Ix current. And I don't get any solution at all. Why is that?
It seems that "right" answer is Ix = Iin = 4mA (R2 = 0Ω). Maybe I'm misinterpreting this.
What do you think about it?
Transforming the dependent voltage source and R2 to a dependent current source, we have this circuit: Notice that the control law for the dependent current source is (p*I1)/R2. The R2 in the denominator means that if we set R2 to zero, we may run into difficulties.

This circuit can be solved in the usual way with 3 mesh currents, i1, i2 and i3; Ia is the same as i1 and Ix is the same as i2. Keeping only R2 as a symbolic variable we get this solution for the 3 mesh currents, followed by 2 numeric solutions with R2 first set to 500 ohms, and then with R2 set to zero: For the case where R2=0, the equations admit of a solution where Ia is 4 mA and Ix is also 4 mA. This means that I1 is zero, and the control law for the dependent current source reduces to 0/0, an indeterminate value. But if we choose that value to be 2 mA, then Iy is 2 mA and Iz is 2 mA; the circuit equations are satisfied.

Taking this back to the original circuit, the solution is Ia= 4 mA and Ix= 4 mA.

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• Jony130

#### WBahn

Joined Mar 31, 2012
28,186
For the case where R2=0, the equations admit of a solution where Ia is 4 mA and Ix is also 4 mA. This means that I1 is zero, and the control law for the dependent current source reduces to 0/0, an indeterminate value. But if we choose that value to be 2 mA, then Iy is 2 mA and Iz is 2 mA; the circuit equations are satisfied.

Taking this back to the original circuit, the solution is Ia= 4 mA and Ix= 4 mA.
While the solutions equations "admit of a solution" where Ia and Ix are both 4 mA, that is not "the" solution, only "a" solution. You chose to make a particular indeterminate parameter a specific value. But why must it be that value?

Go back to the original circuit and, with R2 = 0, choose to set I1 to be 0 mA. That means that the voltage output of the dependent source is 0 V and that the current source is pushing all 4 mA through it. So, in the second diagram, Iz Ix = 4 mA and Ix = 0.

Or set I1 = -10 mA. That means that Ia = 4 mA and Ix = -6 mA. With p = 1 kΩ that means that the output of the CCVS is -10 V, which is exactly what is needed to drive -10 mA through R1.

So ANY value of Ix is a solution.

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#### The Electrician

Joined Oct 9, 2007
2,916
While the solutions "admit of a solution" where Ia and Ix are both 4 mA, that is not "the" solution, only "a" solution. You chose to make a particular indeterminate parameter a specific value. But why must it be that value?
The solutions don't admit of a solution; the equations admit of a solution. I didn't say it is "the" solution; I said it was "a" solution. Please don't misquote me.

The circuit I gave in post #66 can be solved using the 3 meshes I show and I especially like the fact that the control law for the dependent current source has R2 in the denominator. Thus it's easy to see why trying to solve the circuit will probably lead to unusual results if R2 is zero.

The way to do it is to solve with R2 as a symbolic variable. Then if the limit of the solution is taken as R2→0 we obtain numerical values for i1, i2 and i3. I explained that I made the indeterminate control law a particular value because then the mesh equations are satisfied with i1=4 mA, i2=4 mA and i3=2 mA, and those are the only currents that do satisfy the mesh equations.

Any set of currents that satisfy the mesh equations of my circuit in post #66 will give values for Ix that will also work in the circuit of post #1, but there is only one value of Ix that satisfies the post #66 mesh equations. Not every solution to the circuit in post #1 is also a solution to my circuit in post #66.

The circuit of post #66 showed us right away why we were likely to get indeterminate results, and also led us immediately to a value of Ix that works in the circuit of post #1, and also works in a circuit (post #66) which is equivalent to it.

Perhaps the fact that there is only one value of Ix that satisfies the 3 mesh equations of post #66 means that that value should be preferred as the solution to post #1?  Go back to the original circuit and, with R2 = 0, choose to set I1 to be 0 mA. That means that the voltage output of the dependent source is 0 V and that the current source is pushing all 4 mA through it. So, in the second diagram, Iz = 4 mA and Ix = 0.
There is no variable "Iz" in the second diagram. The variable Iz has only been used in this thread by me in post #66.

Or set I1 = -10 mA. That means that Ia = 4 mA and Ix = -6 mA. With p = 1 kΩ that means that the output of the CCVS is -10 V, which is exactly what is needed to drive -10 mA through R1.

So ANY value of Ix is a solution.
Any value of Ix may be a solution to the circuit of post #1, but not every value of Ix is a solution (satisfies all 3 mesh equations) to the circuit of post #66, and it is that circuit that I concerned myself with, knowing that a found value for Ix would also satisfy the post #1 circuit.

#### MrAl

Joined Jun 17, 2014
9,785
Transforming the dependent voltage source and R2 to a dependent current source, we have this circuit:

View attachment 275188

Notice that the control law for the dependent current source is (p*I1)/R2. The R2 in the denominator means that if we set R2 to zero, we may run into difficulties.

This circuit can be solved in the usual way with 3 mesh currents, i1, i2 and i3; Ia is the same as i1 and Ix is the same as i2. Keeping only R2 as a symbolic variable we get this solution for the 3 mesh currents, followed by 2 numeric solutions with R2 first set to 500 ohms, and then with R2 set to zero:

View attachment 275187

For the case where R2=0, the equations admit of a solution where Ia is 4 mA and Ix is also 4 mA. This means that I1 is zero, and the control law for the dependent current source reduces to 0/0, an indeterminate value. But if we choose that value to be 2 mA, then Iy is 2 mA and Iz is 2 mA; the circuit equations are satisfied.

Taking this back to the original circuit, the solution is Ia= 4 mA and Ix= 4 mA.
So what we end up with is a shorted out resistor (R1) with a current source (Iin) All the current goes through the short and none through the resistor.

• dcbingaman

#### The Electrician

Joined Oct 9, 2007
2,916
So what we end up with is a shorted out resistor (R1) with a current source (Iin) All the current goes through the short and none through the resistor.
For the case of R2 = 0, the same thing happens in the circuit of post #66, and in the circuit of post #1 when Ix = 4 mA. The 4 mA from Iin all goes through Ix, bypassing R1, and thence through the shorted R2.

#### WBahn

Joined Mar 31, 2012
28,186
The solutions don't admit of a solution; the equations admit of a solution. I didn't say it is "the" solution; I said it was "a" solution. Please don't misquote me.
Sorry. Let me quote your exact words:

Taking this back to the original circuit, the solution is Ia= 4 mA and Ix= 4 mA.