# Does mesh analysis give no solution?

#### Jony130

Joined Feb 17, 2009
5,435
Hi,
I have this "simple" circuit that contains a constant current source, resistor, and a current-controlled voltage source.

I decided to use a mesh analysis.

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?

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#### MrAl

Joined Jun 17, 2014
9,633
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 - r * (Ix - Iin) + Ix*R2 = 0

Ix = (Iin (r - R1))/(r - 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?
H Jony,

Im glad you drew it all out so well it's hard to follow some questions because the drawings are so bad sometimes and unclear, and i havent used mesh in a long time now.

What it looks like could be the problem is that sometimes you can not just 'set' a resistor value to zero when you want a highly theoretical result from an unusual circuit like that. Instead you can try "allowing" the resistor value to tend to zero and take the limit. This would be written something like this:
I=Limit of Function(R) as R goes to zero.
although there are usually symbols for all that.
Anyway, see if that helps.
BTW this would be an approach if you wanted to know what the circuit did with a short and started with a resistor because it's hard to start with a perfect short in some places sometimes.

These highly theoretical question sometimes lead to highly theoretical results too. You may end up with a discontinuity at R=0.

Oh i see the limit approach may not help here.
I did it a different way and found your solution to be right, that as the currents balance out V goes to zero, so all of the input current would flow though V. That means no net current through R1. This came about by considering that V is controlled by the net current through R1, not really by separate currents adding or subtracting.
Ill try one more thing too next.

I tried another way and it also led to the voltage across R1 coming out to zero. That can only happen if the two currents are equal and opposite in R1.

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#### WBahn

Joined Mar 31, 2012
27,886
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?
Thanks for presenting the problem and your work very clearly.

You say that you don't get any answer at all for Ix, but yet you then show that you get the answer Ix = In = 4 mA. That's an answer. The question is whether that answer makes sense, and that is the question you should always ask.

So let's see if we can determine whether it makes sense. The wonderful thing about most engineering problems is that the correctness of the answer can be determine from the problem itself.

If Ix = 4 mA, what is I1?

For that value of I1, what is the voltage at the output of the current source?

For that value of I1, what is the voltage output of the CCVS?

What is the voltage across R2?

Is this voltage across R2, combined with the voltage output of the CCVS, consistent with the voltage previously determined at the output of the current source.

#### Jony130

Joined Feb 17, 2009
5,435
You say that you don't get any answer at all for Ix, but yet you then show that you get the answer Ix = In = 4 mA. That's an answer.
(Ix - Ia)*R1 - V = (Ix - Ia)*R1 - (Ix - Iin)*P =
= (Ix - 4mA)*1kΩ - (Ix - 4mA)*1kΩ =
= (Ix - 4)*1 - (Ix - 4)*1 = Ix - 4 - Ix + 4.

And from this, I wrongly concluded that Ix = 0.

#### WBahn

Joined Mar 31, 2012
27,886
(Ix - Ia)*R1 - V = (Ix - Ia)*R1 - (Ix - Iin)*P =
= (Ix - 4mA)*1kΩ - (Ix - 4mA)*1kΩ =
= (Ix - 4)*1 - (Ix - 4)*1 = Ix - 4 - Ix + 4.

And from this, I wrongly concluded that Ix = 0.
The actual conclusion, of course, is just that Ix = Ix.

This implies one of two things. Either not all of the equations were used to achieve this result, or possibly that Ix can be anything.

Since both equations were used, let's explore the latter case.

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

Setting R2=0 yields

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

The "obvious" answer is to cancel out the (p - R1) leaving

Ix = Iin

and the conclusion that Ix = 4 mA is the answer (and the only answer).

While this is A valid answer, it is not the ONLY valid answer.

The need to take extra care whenever the denominator might become zero was a focus of a recent thread.

Well, if R2 = 0, that means that the denominator becomes (1 kΩ - 1 kΩ) = 0 and what we end up with is

Ix = Iin * (0/0)

And greater care needs to be given.

Let's first see if it's true that Ix can be ANY current whatever. So let's set Ix = Io and see if this works.

That means that I1 = (Io - Iin) and that V = (1 kΩ)(Io - Iin).

So what is the voltage at the top-left node (I'll call it Vin)?

Going from 0V through R1 it is

Vin = - I1 * 1 kΩ = -(Io - Iin)*(1 kΩ)

Going from 0V through the CCVS it is

Vin = -V = -(1 kΩ)(Io - Iin)

So, yes, if R2 = 0, the value of Ix can be anything.

Now, does this make sense?

Remember that an ideal current source will produce any voltage necessary to make the programmed current flow through it, while an ideal voltage source will source or sink any amount of current in order maintain the programmed voltage across it.

By taking R2 to 0, the current source and the voltage source are placed in parallel. That decouples them in the sense that the current source will put out whatever voltage is needed to match the voltage source and the voltage source will output whatever current is needed to make the current source happy.

So now it comes down to the voltage source and R1.

I1 is the current flowing in R1 and it is equal to

I1 = V/R1

But the output of the voltage source is controlled by the current in R1 such that

V = p*I1
V = p*V/R1
V = V*(p/R1)

If p = R1, then we have

V = V

Note that there's no problem dividing p/R1 since both p and R1 are finite.

The conclusion is that, if p = R1 (and R2 = 0, don't forget that constraint), that the current source produces an indeterminate voltage because ANY voltage it produces will result in exactly the correct amount of current in R1 to sustain that output voltage.

Now, this actually reveals another interesting case. What if p does NOT equal R1?

Then we have

V = V*(p/R1)

and the factor in parens does NOT go to 1. In that case, the only way that the equation can be satisfied is if

V = 0

Does this make sense?

If V = 0, the I1 = 0 and this results in V = 0. Everything is good. So what about Iin? Remember that we have an ideal voltage source, so that 4 mA will merely flow through the that source and back to the current source and everyone is happy.

Bottom line:

If R2=0, then V=0 and Ix=Iin UNLESS p = R1, in which case both V and Ix are indeterminate.

If R2 is NOT zero, then R2 serves to couple the current source to the voltage source such that they each affect the entire circuit.

Finally, note that you have messed up the units in your final line:

(Ix - Ia)*R1 - V = (Ix - Ia)*R1 - (Ix - Iin)*P =
= (Ix - 4mA)*1kΩ - (Ix - 4mA)*1kΩ =
= (Ix - 4)*1 - (Ix - 4)*1 = Ix - 4 - Ix + 4.
You can divide the next-to-last line by 1 kΩ to get

= (Ix - 4mA) - (Ix - 4mA)

But can't throw away the mA units, because then you are claiming that you are subtracting 4 from a couple of currents? 4 what? Amperes? Milliamperes? Quatloos?[/QUOTE][/QUOTE]

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#### WBahn

Joined Mar 31, 2012
27,886
WBahn although on the simulation Jn = 0mA and not 4mA

https://tinyurl.com/2mhs52fl
Simulators are only good within certain bounds and, in general, they can't deal with indeterminate solutions properly. It comes down to how they determine convergence and that is highly dependent on how they determine and deal with initial conditions. If they can converge on a stable solution, that's what they will yield as a result. But if you force the initial conditions to be something else, they will converge and be happy yielding a different result.

In that falstad simulator can you set initial conditions on the current in the resistor? If so, set it to, say 100 mA and see what happens. It's still an indeterminate case, so I don't know how the simulator will respond.

#### Xenon02

Joined Feb 24, 2021
355
I also tried this one as a solution :

But it was pretty wierd because I had something like: I*(-R1+r) = 0, R1=r = 1k
I*(0) = 0
So I = 0/0 ?

To be honest this is pretty tricky.

It showed me 2 different solutions so :

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#### WBahn

Joined Mar 31, 2012
27,886
I also tried this one as a solution :

But it was pretty wierd because I had something like: I*(-R1+r) = 0, R1=r = 1k
I*(0) = 0
So I = 0/0 ?
So look at your next to last step:

I*(0) = 0

Do you see how ANY value of I is a solution to this equation?

Also consider what you would have gotten if r was NOT equal to R1:

I*(something-not-zero) = 0

which would only be satisfiable if I is exactly zero.

It showed me 2 different solutions so :
Some simulators use random number generators as part of their initialization process. It looks like maybe the falstad simulator might. There's pros and cons to doing so.

But it underscores the need to not just take whatever a simulator says as being absolute truth. They are powerful and useful tools, but they are not capable of asking if their answer makes sense. We must still do that.

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#### MrAl

Joined Jun 17, 2014
9,633
Here is a related question see what you guys can come up with.
'i' is the current through the current controlled voltage source,
What is the solution for V as R tends to zero?

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#### WBahn

Joined Mar 31, 2012
27,886
V can't equal i. That has no more meaning that saying that someone's height is equal to their weight.

What is the transimpedance of this CCVS? Do you mean that V = (1 Ω)*i ?

If R is non-zero, then if the transimpedance of the CCVS is not exactly equal to R, the only solution is i = 0 which means V = 0. If the transimpedance is equal to R, then any value of i is a solution.

If R is zero, the V MUST be zero since V is also the voltage across R. This then requires that i = 0 if the transimpedance is non-zero.

For R non-zero but tending toward zero, you have the same situation. If R is not equal to the transimpedance, the V = 0.

The voltage is indeterminate ONLY for the case of R being equal to the a non-zero transimpedance. For all other values of R, including R=0, the voltage and the current are zero.

#### Xenon02

Joined Feb 24, 2021
355
Hmmm of course if R isn't equal r. The i is equal 0 , but it's hard to prove that if R is equal r then the same Solution which is i equal 0 is correct using basic equations like me.
He would do it like in the Picture which is i*0=0.
i=0/0 which doesn't make sense.

Is it possible to get this solition with basic equations like KVE or KCE

#### MrAl

Joined Jun 17, 2014
9,633
V can't equal i. That has no more meaning that saying that someone's height is equal to their weight.

What is the transimpedance of this CCVS? Do you mean that V = (1 Ω)*i ?

If R is non-zero, then if the transimpedance of the CCVS is not exactly equal to R, the only solution is i = 0 which means V = 0. If the transimpedance is equal to R, then any value of i is a solution.

If R is zero, the V MUST be zero since V is also the voltage across R. This then requires that i = 0 if the transimpedance is non-zero.

For R non-zero but tending toward zero, you have the same situation. If R is not equal to the transimpedance, the V = 0.

The voltage is indeterminate ONLY for the case of R being equal to the a non-zero transimpedance. For all other values of R, including R=0, the voltage and the current are zero.
V=i means that V=1*i.

Well see this circuit was meant to bring out a fact about circuits like this that is either obvious or not obvious at all. It's meant to usher in the idea of realism.

Any engineer who looks at this circuit would know right away that the voltage and current MUST be zero at all times. The impedance does not change anything. So there is no way that just any value of i can be a solution.
It's just that sometimes i think we get a little too deep into a simple problem. We cant have any current flow without something active to start that current.

I guess we could impose some initial power conditions to show how things might change, but in the absence of a power source of some kind nothing is going to happen.

There was another question on another site that asked about power sharing between isolated circuits that were connected only with a single ground wire. The level of theory we have to go into changes with the problem sometimes.

I think this is more important when dealing with real life circuits but it's still interesting to think about.
For isolated circuits we may end up with a similar result. The original circuit brings this to mind because we might look at the current source as being isolated from the voltage source even though Ia is within the voltage source.
In pure math if a solution can be anything, then it is usually taken to mean no solution. Maybe that's going too far though.

#### MrAl

Joined Jun 17, 2014
9,633
Hello again,

Here is another view of the circuit that might help. Iin is split into two parts, one part flows through R1 and the other through V. See if that helps.

#### WBahn

Joined Mar 31, 2012
27,886
V=i means that V=1*i.
This is still meaningless. What does it mean for a voltage to be equal to 100 mA?

If I told you that there are three people in the room and that, for each person, w=1*h, meaning their weight is equal to their height. The first person is 6 ft, the second person is 72 inches, and the third person is 183 cm. How much does each person weigh?

Well see this circuit was meant to bring out a fact about circuits like this that is either obvious or not obvious at all. It's meant to usher in the idea of realism.

Any engineer who looks at this circuit would know right away that the voltage and current MUST be zero at all times. The impedance does not change anything. So there is no way that just any value of i can be a solution.
It's just that sometimes i think we get a little too deep into a simple problem. We cant have any current flow without something active to start that current.
So, according to your reasoning, if there is an inductor that is shorted with a wire across it that has been sitting undisturbed for a month, any engineer who looks at it would know right away that the voltage and current MUST be zero at all times. Therefore changing the impedance by cutting the wire does not change anything and there is no way that just any value of i can be a solution.

I will then show you where you can go to most major hospitals and walk up to such an inductor and if that engineer takes your approach and cuts that wire that they will die a rather gruesome death because they have just open-circuited an MRI machines superconducting magnet that has been sitting there, short-circuited, with an indeterminate amount of current flowing in it (within limits, such as the critical current density limit at a given temperature, magnetic field, and stress). If we know the specifics of the magnet, we can determine the current by measuring the magnetic field strength and mapping that back to a magnet current.

In pure math if a solution can be anything, then it is usually taken to mean no solution. Maybe that's going too far though.
That's going too far. As I've stated before, when you have a solution such as that, you need to look further. If you are trying to model a real system, it may well mean that your model is insufficient and you need to come up with a better model. But your model may be just fine. If it is a theoretical exercise, then just going, "well, any engineer knows that it must be zero" is wrong. The task was to answer the problem as given, not as you would like it to be or to impose arbitrary conditions as you believe must be the case. Indeterminate is indeterminate and if the answer is indeterminate, then that is the answer that should be given. To do otherwise is to give a wrong answer that can lead to unintended consequences. If the answer is stated as indeterminate, then the reviewer has the necessary heads up that something more needs to be considered.

#### MrAl

Joined Jun 17, 2014
9,633
This is still meaningless. What does it mean for a voltage to be equal to 100 mA?

If I told you that there are three people in the room and that, for each person, w=1*h, meaning their weight is equal to their height. The first person is 6 ft, the second person is 72 inches, and the third person is 183 cm. How much does each person weigh?

So, according to your reasoning, if there is an inductor that is shorted with a wire across it that has been sitting undisturbed for a month, any engineer who looks at it would know right away that the voltage and current MUST be zero at all times. Therefore changing the impedance by cutting the wire does not change anything and there is no way that just any value of i can be a solution.

I will then show you where you can go to most major hospitals and walk up to such an inductor and if that engineer takes your approach and cuts that wire that they will die a rather gruesome death because they have just open-circuited an MRI machines superconducting magnet that has been sitting there, short-circuited, with an indeterminate amount of current flowing in it (within limits, such as the critical current density limit at a given temperature, magnetic field, and stress). If we know the specifics of the magnet, we can determine the current by measuring the magnetic field strength and mapping that back to a magnet current.

That's going too far. As I've stated before, when you have a solution such as that, you need to look further. If you are trying to model a real system, it may well mean that your model is insufficient and you need to come up with a better model. But your model may be just fine. If it is a theoretical exercise, then just going, "well, any engineer knows that it must be zero" is wrong. The task was to answer the problem as given, not as you would like it to be or to impose arbitrary conditions as you believe must be the case. Indeterminate is indeterminate and if the answer is indeterminate, then that is the answer that should be given. To do otherwise is to give a wrong answer that can lead to unintended consequences. If the answer is stated as indeterminate, then the reviewer has the necessary heads up that something more needs to be considered.
For the solution set we can say "no unique solution" but sometimes that means we dont have a solution we can use.

But V=1*i why does that not make sense? You are the one always making sure everyone uses dimensional analysis, why did you stop here? Maybe you have a good reason i dont know.
But here:
V=1*i
on the left we have 'volts' so on the right we must also have 'volts.
Since there is a current 'i' and we need to keep the units the same on the right side, that means that '1' must be in Ohms. So what's the problem. Are you saying i have to actually write out the units for something this simple?
V(volts)=1(Ohms)*i(amps)
That seems unnecessary but feel free to do it not like it's wrong or anything.
We can also declare '1' to be a gain and if it's a current controlled voltage source doesnt that mean we MUST mutliply a current by a constant to get the output voltage? Is that wrong? How would you do it?

Oh a shorted inductor will maintain it's current indefinitely as you know, as long as we are talking pure theory (or maybe a superconductor). But that current had to come from somewhere originally. So the answer is no but would be nice to see where it came from. So an engineer i would think would say NO as long as he could notice the current. Unfortunately, once he measures the current the current could decrease.
The real thing i think is that the dependent source is not an inductor and it doesnt have an initial current unless we assign one, and then i think we end up with a current that keeps increasing to infinity. It looks like positive feedback.

I got a little interested in the current that goes though R1 and that goes through V from the original source of Iin so i created that new drawing. Not sure if it helps but maybe you can figure out what those two currents are, IinA and IinB. That may make the problem simpler. In any case, i think it would be interesting to see how the current splits (or whatever) and i have to admit i am feeling a bit lazy right now

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#### MrAl

Joined Jun 17, 2014
9,633
Hello again,

Ok i have another solution to offer. This is in the form of three equations:

Vx=(2*p*I1*R1)/(R1+p) (voltage at top of R1, referenced to ground 0v)
Ix=(I1*R1-p*I1)/(R1+p) (this flows through the voltage source left to right in the original schematic)
iR1=(2*p*I1)/(R1+p) (current down through R1)
(I1 is Iin in the original schematic, and R2=0 as requested)

If you want to try to find something wrong with these that's great. They seem to work.
I think the only unusual part of this is that the voltage source affects the current negatively yet it senses the input current negatively too.

I didnt try a huge number of values so maybe something will turn up. It's obvious though that if p=-R1 then a zero in the denominator will cause infinite values, so i am thinking maybe stick to positive values of p for now, although some others may work. This does put to rest the idea that the voltage at the top of R1 is zero unless maybe certain conditions are met.

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#### WBahn

Joined Mar 31, 2012
27,886
For the solution set we can say "no unique solution" but sometimes that means we dont have a solution we can use.

But V=1*i why does that not make sense? You are the one always making sure everyone uses dimensional analysis, why did you stop here? Maybe you have a good reason i dont know.
I didn't stop here. This equation is not dimensionally consistent. You are saying that a voltage is equal to a pure number multiplied by a current. This is the exact same thing as saying that some objects weight is equal to some pure number multiplied by its height. It's a meaningless statement.

But here:
V=1*i
on the left we have 'volts' so on the right we must also have 'volts.
Since there is a current 'i' and we need to keep the units the same on the right side, that means that '1' must be in Ohms. So what's the problem.
The problem is that you are CHANGING the equation based on what you WANT it to be.

Again, this is NO different then me giving you the equation

w = 1*h

and then telling you that I am 6 ft tall and you then using that equation to determine how much I weigh. You won't do so, no matter how many times I ask, because you know that this is a meaningless equation. The bottom line is that you can't guess what the units on that 1 are supposed to be, and so you dodge the question but you THINK you know what the units on the 1 in the other equation somehow "must" be and so you are okay with it.

Are you saying i have to actually write out the units for something this simple?
V(volts)=1(Ohms)*i(amps)
That seems unnecessary but feel free to do it not like it's wrong or anything.
It IS wrong. V and i are both symbolic quantities and thus carry their own units. V has units of energy per units charge and i has units of charge per unit time. But 1 is not a symbolic parameter, it is a specific quantity and quantities have both magnitude and units. You are claiming that the units on this particular quantity are dimensionless, when they aren't. They need to have dimensions of energy-time per charge. Ohm's is just one of many possible units that have that dimension.

Consider the distance-rate-time formula:

d = r*t

Each of those is a symbolic parameter and they each carry their own dimensions. The 'd' has the dimension of length, the 'r' has dimensions of distance per unit time, and the 't' has units of time.

Now let's say that I constrain 'r' to be a particular speed whose magnitude is 60. You are claiming that that should be written as

d = 60*t

Okay, so if t is then 2 minutes, how many feet has the object moved?

Are you going to claim that the units on that 60 must be feet per minute and then claim that the object moved 120 feet?

But what if, instead, I wrote that equation correctly as

d = (60 mph)*t

and then ask you the same exact question: If t is then 2 minutes, how many feet has the object moved?

Do you see that the '60' needs to have units? If they are left out, the equation is dimensionally inconsistent because the notion that distance is somehow equation to time multiplied by a pure number is meaningless.

Do you also see that not only is it not necessary, but it is wrong to say something like

d(feet) = (60 mph)*t(minutes)

What if 't' is in weeks, or seconds, or days? Why can't 'd' be in inches, or meters, light-seconds?

The following is perfectly correct:

10,560 ft = 60 mph * 2 minutes

That is every bit as correct as equation such as

72 inches = 6 feet

We can also declare '1' to be a gain and if it's a current controlled voltage source doesnt that mean we MUST mutliply a current by a constant to get the output voltage? Is that wrong? How would you do it?
The gain of a current-controlled voltage source is a transimpedance. For some amount of current measured at the control point, the source produces a certain voltage out someplace else (hence the 'trans' in the name). The units therefore have dimensions of voltage per unit current. They might be volts/amp or mV/kA or kV/mA. It would be very natural to express the transimpedance as something like 1 V/A, but 1 V/A is also known as 1 Ω.

In a very real sense, a resistor is a current-controlled voltage source. Pass a current through it and it produces an output voltage. A 10 kΩ resistor produces 10 V/mA, or 10,000 V/A, or 10 mV/uA. A resistor, however, is not a 'trans'-impedance device, but rather just an impedance device.

#### WBahn

Joined Mar 31, 2012
27,886
Hello again,

Ok i have another solution to offer. This is in the form of three equations:

Vx=(2*p*I1*R1)/(R1+p) (voltage at top of R1, referenced to ground 0v)
Ix=(I1*R1-p*I1)/(R1+p) (this flows through the voltage source left to right in the original schematic)
iR1=(2*p*I1)/(R1+p) (current down through R1)
(I1 is Iin in the original schematic, and R2=0 as requested)

If you want to try to find something wrong with these that's great. They seem to work.
I think the only unusual part of this is that the voltage source affects the current negatively yet it senses the input current negatively too.

I didnt try a huge number of values so maybe something will turn up. It's obvious though that if p=-R1 then a zero in the denominator will cause infinite values, so i am thinking maybe stick to positive values of p for now, although some others may work. This does put to rest the idea that the voltage at the top of R1 is zero unless maybe certain conditions are met.
Could you please supply a schematic in which all of your parameters are clearly defined? You seem to be changing the definitions of some (such as I1, which is defined in the original schematic as the current flowing upward through R1) and using others that aren't defined in the original schematic at all (such as Ix). I'm not going to guess whether or not your Ix refers to the same thing as the Ix the TS introduced in their second diagram, since you are already feeling free to change the supplied definitions of things.

#### MrAl

Joined Jun 17, 2014
9,633
I didn't stop here. This equation is not dimensionally consistent. You are saying that a voltage is equal to a pure number multiplied by a current. This is the exact same thing as saying that some objects weight is equal to some pure number multiplied by its height. It's a meaningless statement.

The problem is that you are CHANGING the equation based on what you WANT it to be.

Again, this is NO different then me giving you the equation

w = 1*h

and then telling you that I am 6 ft tall and you then using that equation to determine how much I weigh. You won't do so, no matter how many times I ask, because you know that this is a meaningless equation. The bottom line is that you can't guess what the units on that 1 are supposed to be, and so you dodge the question but you THINK you know what the units on the 1 in the other equation somehow "must" be and so you are okay with it.

It IS wrong. V and i are both symbolic quantities and thus carry their own units. V has units of energy per units charge and i has units of charge per unit time. But 1 is not a symbolic parameter, it is a specific quantity and quantities have both magnitude and units. You are claiming that the units on this particular quantity are dimensionless, when they aren't. They need to have dimensions of energy-time per charge. Ohm's is just one of many possible units that have that dimension.

Consider the distance-rate-time formula:

d = r*t

Each of those is a symbolic parameter and they each carry their own dimensions. The 'd' has the dimension of length, the 'r' has dimensions of distance per unit time, and the 't' has units of time.

Now let's say that I constrain 'r' to be a particular speed whose magnitude is 60. You are claiming that that should be written as

d = 60*t

Okay, so if t is then 2 minutes, how many feet has the object moved?

Are you going to claim that the units on that 60 must be feet per minute and then claim that the object moved 120 feet?

But what if, instead, I wrote that equation correctly as

d = (60 mph)*t

and then ask you the same exact question: If t is then 2 minutes, how many feet has the object moved?

Do you see that the '60' needs to have units? If they are left out, the equation is dimensionally inconsistent because the notion that distance is somehow equation to time multiplied by a pure number is meaningless.

Do you also see that not only is it not necessary, but it is wrong to say something like

d(feet) = (60 mph)*t(minutes)

What if 't' is in weeks, or seconds, or days? Why can't 'd' be in inches, or meters, light-seconds?

The following is perfectly correct:

10,560 ft = 60 mph * 2 minutes

That is every bit as correct as equation such as

72 inches = 6 feet

The gain of a current-controlled voltage source is a transimpedance. For some amount of current measured at the control point, the source produces a certain voltage out someplace else (hence the 'trans' in the name). The units therefore have dimensions of voltage per unit current. They might be volts/amp or mV/kA or kV/mA. It would be very natural to express the transimpedance as something like 1 V/A, but 1 V/A is also known as 1 Ω.

In a very real sense, a resistor is a current-controlled voltage source. Pass a current through it and it produces an output voltage. A 10 kΩ resistor produces 10 V/mA, or 10,000 V/A, or 10 mV/uA. A resistor, however, is not a 'trans'-impedance device, but rather just an impedance device.

Hello again,

I like your definition of volts/unitcurrent very much, but you next need to show how you would write this.
I wrote:
V=1*i
because we had a current controlled voltage source. The '1' is often referred to as a "gain" and that's accepted in many contexts. But if you dont like that it's fine, you may want to mention that to the people that make circuit simulators, but aside from that, show me how YOU would write it.
I think you are making too much out of this, but if you have more information that's always good.

Again, i wrote:
V=1*i
now show me how you would write it and maybe i could change my writing then.