# Does mesh analysis give no solution?

#### MrAl

Joined Jun 17, 2014
9,785
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.
Hello again,

The circuit is the same except for one important 'change' that i assumed everyone would assume but now that you mention it it looks like nobody else did, and i'll mention this in a minute.

Isnt Ix in the original? That's the current though the voltage source pointing left to right.
I1 is the input current, originally Iin.

I can redraw the schematic but it's the very same except for the variables, except for one major thing i am seeing now that may have been mistaken in the original question.
That is, the definition of the voltage source. It was:
V=(Ix-Ia)*p

This was changed to:
V=(Ix-I1)*p
or as per original drawing:
V=(Ix-Iin)*p

The reason for this is because the definition:
V=(Ix-Ia)*p

immediately sets the circuit to be an impossible circuit that needs no analysis of any currents or voltages because the current "Ia" is a mesh current and therefore not a real current. That means that current "Ia" can never be measured and therefore the voltage source definition:
V=(Ix-Ia)*p

is impossible and therefore the question is ill posed. That would immediately disqualify this circuit as anything that could be created in real life or even in theory because theory rejects a controlled device based on a fictitious current. It has to be a real current. This might be more apparent in my previous drawing where i tried to draw out the way the assumed currents would (possibly) flow.

Thus i was assuming that Ia=Iin or in my equations Ia=I1. That's about the only way we would have anything to analyze in the first place.

But i am still interested in seeing how you would write out that expression:
V=1*i

and how we could explain this to circuit simulator software engineers. If you have a really good reason for this, it could be very useful and beneficial.

#### Jony130

Joined Feb 17, 2009
5,440
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)
What was the initial equation you have used to get these results?

#### WBahn

Joined Mar 31, 2012
28,186
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.
I have written it many times in this thread.

The purely symbolic relation is

V = k*i

where k is the transimpedance of the CCVS. If k has a concrete value of 1 V/A, which is the same as 1 Ω, then simply replace the k with its concrete value.

V = (1 V/A)*i

or

V = (1 Ω)*i

The parens are optional, but I think they make it more readable.

Since we have had airliners full of people run out of fuel in midflight and have slammed hundred-million dollar space probes into planets because people couldn't bothered to track their units, I don't think I'm making too much out of this. I personally witnessed someone die about three feet away from me because they didn't feel any need to track their units. So, no, I don't think I'm making too much out of this.

• dcbingaman

#### WBahn

Joined Mar 31, 2012
28,186
Hello again,

The circuit is the same except for one important 'change' that i assumed everyone would assume but now that you mention it it looks like nobody else did, and i'll mention this in a minute.

Isnt Ix in the original? That's the current though the voltage source pointing left to right.
I1 is the input current, originally Iin.
This is the original problem schematic: Notice that there is NO Ix in there. Ix is a mesh current defined in the second image where the analysis is being set up.

The reason for this is because the definition:
V=(Ix-Ia)*p

immediately sets the circuit to be an impossible circuit that needs no analysis of any currents or voltages because the current "Ia" is a mesh current and therefore not a real current. That means that current "Ia" can never be measured and therefore the voltage source definition:
V=(Ix-Ia)*p

is impossible and therefore the question is ill posed.
The question is perfectly well-posed. The circuit definition is as given above. The second schematic is NOT the problem definition. It is setting up the circuit analysis using a particular technique in which the physical currents are represented by superimposed mesh currents.

and how we could explain this to circuit simulator software engineers. If you have a really good reason for this, it could be very useful and beneficial.
Circuit simulators, like most programming languages, do not provide for the proper treatment of units since they are strictly number crunchers. Most of them will allow you to tack on units to many of the parameters because SPICE allows it, but these are completely ignored. When I make a schematic, if the software allows proper units to at least be annotated, I include them in the value. If not, then I put a piece of text there so that it is properly documented.

There have been a number of programming languages that have offered units support since the 1960's, but I don't think any of them have ever gained much traction. VHDL provides some utility for this, but it's pretty clunky.

#### MrAl

Joined Jun 17, 2014
9,785
What was the initial equation you have used to get these results?

Hello again,

I'll post more about that, but first i want to point out that when you did the analysis using mesh, you can not use the mesh current Ia within the definition of the voltage source which was V=p*(Ix-Ia) because the mesh current Ia is not a real current. That could be why you had trouble getting an expression.
For a current to be within a control element, it has to be measurable, and since Ia can not be measured it can not be used in that way.
If you want to set it up differently, i was thinking you really meant V=p*(Ix-Iin) which is perfectly valid. That's actually what i used for those three equations, and they are not very hard to develop.
I'll show the derivation if you like but it will be using Iin as one of the control variables.

I see why you replaced the definition of V but right off that wont or at least should not work.
So maybe mesh analysis does not work with this circuit as you suspected. Mesh analysis is purported to fail with non planar networks, so maybe placing Ia within the controlled source violates that requirement too.

If you would like to see more about why this is so, try making a schematic with V sensing Ix and Ia. You can get it to sense Ix but since Ia is not measurable, there is no wire you can put the current shunt in to measure it.
It's interesting that we may be able to measure I1 (first circuit) so we can try that if you like, but that would have to entail using a different type of analysis i think, unless maybe we can bend the rules a little and just put the voltage at the top of R1 divided by R1: vR1/R1 to specify the sense current.

There are some rules that we dont hear about too often but they kick in now and then.

Here is what i get for the original circuit, the first one, for the voltage at the top of R1:
vR1=(Iin*R1*R2)/(R2+R1-p)
but that comes from nodal analysis not mesh and using the original I1 as in the first drawing as the current sense for V.
The resistors are all the same as in the original schematic.
Now as R2 goes to zero it looks like vR1 goes to zero but that should be investigated a little more.
When R2 is small but not zero, it has a normal effect that any resistor would have.

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#### MrAl

Joined Jun 17, 2014
9,785
This is the original problem schematic:

View attachment 274832

Notice that there is NO Ix in there. Ix is a mesh current defined in the second image where the analysis is being set up.

The question is perfectly well-posed. The circuit definition is as given above. The second schematic is NOT the problem definition. It is setting up the circuit analysis using a particular technique in which the physical currents are represented by superimposed mesh currents.

Circuit simulators, like most programming languages, do not provide for the proper treatment of units since they are strictly number crunchers. Most of them will allow you to tack on units to many of the parameters because SPICE allows it, but these are completely ignored. When I make a schematic, if the software allows proper units to at least be annotated, I include them in the value. If not, then I put a piece of text there so that it is properly documented.

There have been a number of programming languages that have offered units support since the 1960's, but I don't think any of them have ever gained much traction. VHDL provides some utility for this, but it's pretty clunky.
Ok then replace V=1*i with V(volts)=1(V/A)*i(Amps).

Yes it was nutty about that one spacecraft.

You had shown the original circuit and it's different than the one with the mesh currents, so that second circuit is the one i did but i used Iin in place of Ia because i thought that was what he meant.
But since the actual original circuit is also shown, i can work with that one next.
It appears that the original circuit is as you say and that's good to point out, but that doesnt have mesh currents so it does not help answer his original question about mesh analysis.

Since that was the question, the answer should have been related to that circuit. With that in mind, there can not be a solution because the seconds circuit question is ill posed, and again that is because it uses a mesh current in a controlled source.

The actual original circuit should be cake now.

#### WBahn

Joined Mar 31, 2012
28,186
Ok then replace V=1*i with V(volts)=1(V/A)*i(Amps).
This is still not correct. V and i are symbolic variables and, as such, they carry their own units. It should be written as

V = (1 V/A) * i

You are still in this rut of thinking that units are something that are tacked on when convenient for notational purposes. They are not -- they are a fundamental part of the value. Whatever value 'V' end up having, it has dimension of energy per unit charge. That may or may not be volts. It might be calories/electron. The same with 'i'. It will have dimensions of charge per unit time, which may or may not be amperes.

You had shown the original circuit and it's different than the one with the mesh currents, so that second circuit is the one i did but i used Iin in place of Ia because i thought that was what he meant.
But since the actual original circuit is also shown, i can work with that one next.
It appears that the original circuit is as you say and that's good to point out, but that doesnt have mesh currents so it does not help answer his original question about mesh analysis.

Since that was the question, the answer should have been related to that circuit. With that in mind, there can not be a solution because the seconds circuit question is ill posed, and again that is because it uses a mesh current in a controlled source.
There is NO problem with using mesh currents in analyzing this circuit. If there is, then you can't use mesh currents for any circuit analysis. After all, how can you say that the physical, measurable voltage across a resistor is equal to the resistance multiplied by the difference of two mesh currents, neither of which may exist or be measurable?

#### MrAl

Joined Jun 17, 2014
9,785
This is still not correct. V and i are symbolic variables and, as such, they carry their own units. It should be written as

V = (1 V/A) * i

You are still in this rut of thinking that units are something that are tacked on when convenient for notational purposes. They are not -- they are a fundamental part of the value. Whatever value 'V' end up having, it has dimension of energy per unit charge. That may or may not be volts. It might be calories/electron. The same with 'i'. It will have dimensions of charge per unit time, which may or may not be amperes.

There is NO problem with using mesh currents in analyzing this circuit. If there is, then you can't use mesh currents for any circuit analysis. After all, how can you say that the physical, measurable voltage across a resistor is equal to the resistance multiplied by the difference of two mesh currents, neither of which may exist or be measurable?
If you cant say that there are units of volts on the left side of that little equation then i dont know what else to tell you.

It's not the mesh current that is a problem. It is the mesh current in the controlled voltage source. It's not a real current.
For example, if you try to draw this circuit so that the current Ia a mesh current can be measured, you cant do it directly. There is no place to put the current shunt, which is the sense element for the current Ia so that the voltage source can be controlled at least in part by Ia.
So try to draw the circuit where something senses that current Ia and post it right here.
I believe the only way to do it is to transform it into a REAL current.
But hey if you can sense Ia directly then show it i would be happy to see it.
There may be a way around it using a trick that transforms it into a real quantity (not necessarily a current), but i dont want to say it works yet until i test it.

I hate to say it and i know i am not perfect dont get me wrong, but you really need to do some reading my friend Again, dont get the idea that i am perfect just because i point something out.
The old old saying, "Pobody's nerfect".
This wouldnt be complete though unless i add that i have liked the posts you have posted in this thread.

#### MrAl

Joined Jun 17, 2014
9,785
Hello again,

Ok i think i have a proof of why Ix-Ia does not work besides the fact that Ia is fictitious.

We can't measure Ia directly because there is no place to put the shunt. That's a red flag.
But besides that, to get a calculation of Ia alone, we can measure Ix and -I1.
Ix is the second circuit current on the right, and -I1 is the current down though R1.
Now since -I1 runs down R1 and Ix runs up R1, that means Ix is subtracted from -I1 whereas Ia is not subtracted.
So we end up with:
Ia=-I1+Ix
That is simply adding Ix back to -I1 so that what remains has to be Ia.
Now that we have that, when we write the expression for V in the second circuit (without the p for now), we get:
V=Ix-Ia=Ix-(-I1+Ix), which equals I1.

So the only way we can get a real Ia is in an indirect way by measuring two currents and summing them.
Of course the final result there is V=I1*p which was the original spec.
This means Ia can not be used in the expression for the voltage source, directly, we have to calculate it and use the calculation rather than the symbolic Ia.

Will mesh work like this? Try it and see if you can get a reasonable result. Not sure if it will help but at least we know we cant put Ia in the voltage source expression directly.

#### WBahn

Joined Mar 31, 2012
28,186
If you cant say that there are units of volts on the left side of that little equation then i dont know what else to tell you.
It doesn't matter whether the units on the left hand side have to be volts are not. Whatever the units are, they are carried by the symbolic parameter 'V'.

But, for the record, they do NOT have to be volts. They can be ANY units that are commensurable with energy per unit charge, including something like calories per electron.

V = k*i

is a mathematical relationship relating three quantities. This is NO different than the familiar

d = s * t

that relates distance, speed, and time for an object traveling at constant velocity.

The variable 'd' can have ANY units that are commensurable with length. the variable 's' can have ANY units that are commensurable with length per unit time, and 't' can have ANY units that are commensurable with time.

Whatever units they happen to have, they are carried by the symbolic variables.

But, let's use your reasoning and go back to the question you keep dodging.

The relationship between a particular object's weight, w, and height, h, is

w = 1 * h

If the height of the object is 72 inches, what is its weight?

You insist that that 1 has no units, so clearly you don't need them to tell us all what the weight of that object is. So what is it?

Do you not see that that relation is fundamentally NO different than

V = 1 * i

It's not the mesh current that is a problem. It is the mesh current in the controlled voltage source. It's not a real current.
The CCVS is NOT being controlled by a mesh current, it is being controlled by (Ix - Ia), which IS a physically measurable quantity, namely the current I1 in the original diagram. The CCVS does not EVER need to sense Ix or Ia individually, it ALWAYS is controlled by (Ix-Ia), which IS the sensible current flowing upward in R1. The use of mesh currents is a contrivance that is equivalent mathematically (in linear circuits) via superposition.

For example, if you try to draw this circuit so that the current Ia a mesh current can be measured, you cant do it directly. There is no place to put the current shunt, which is the sense element for the current Ia so that the voltage source can be controlled at least in part by Ia.
So try to draw the circuit where something senses that current Ia and post it right here.
I believe the only way to do it is to transform it into a REAL current.
But hey if you can sense Ia directly then show it i would be happy to see it.
There may be a way around it using a trick that transforms it into a real quantity (not necessarily a current), but i dont want to say it works yet until i test it.
Again, it doesn't matter whether you can separately sense Ix and Ia, because the CCVS never has to do that. But, in this circuit, if you really, really, really felt the need, then put a shunt in series with the current source to sense Ia and put a shunt in series with R2 and then use a diff amp to take the difference. But why would you do that. Just measure the current in R1, is IS the quantity (Ix-Ia) that is controlling the source.

#### MrAl

Joined Jun 17, 2014
9,785
It doesn't matter whether the units on the left hand side have to be volts are not. Whatever the units are, they are carried by the symbolic parameter 'V'.

But, for the record, they do NOT have to be volts. They can be ANY units that are commensurable with energy per unit charge, including something like calories per electron.

V = k*i

is a mathematical relationship relating three quantities. This is NO different than the familiar

d = s * t

that relates distance, speed, and time for an object traveling at constant velocity.

The variable 'd' can have ANY units that are commensurable with length. the variable 's' can have ANY units that are commensurable with length per unit time, and 't' can have ANY units that are commensurable with time.

Whatever units they happen to have, they are carried by the symbolic variables.

But, let's use your reasoning and go back to the question you keep dodging.

The relationship between a particular object's weight, w, and height, h, is

w = 1 * h

If the height of the object is 72 inches, what is its weight?

You insist that that 1 has no units, so clearly you don't need them to tell us all what the weight of that object is. So what is it?

Do you not see that that relation is fundamentally NO different than

V = 1 * i

The CCVS is NOT being controlled by a mesh current, it is being controlled by (Ix - Ia), which IS a physically measurable quantity, namely the current I1 in the original diagram. The CCVS does not EVER need to sense Ix or Ia individually, it ALWAYS is controlled by (Ix-Ia), which IS the sensible current flowing upward in R1. The use of mesh currents is a contrivance that is equivalent mathematically (in linear circuits) via superposition.

Again, it doesn't matter whether you can separately sense Ix and Ia, because the CCVS never has to do that. But, in this circuit, if you really, really, really felt the need, then put a shunt in series with the current source to sense Ia and put a shunt in series with R2 and then use a diff amp to take the difference. But why would you do that. Just measure the current in R1, is IS the quantity (Ix-Ia) that is controlling the source.

Hello again,

I think for the subject of 'units' we are talking different contexts that's all it is. In some contexts we would have to be more explicit but if someone doesnt know that you can take that '1' to be similar to a gain then they are reading out of context. In this thread maybe you make a good point but i dont really want to debate that particular issue anymore because i know you are very strict about your units and apparently the context means nothing to you except for the purpose of 'correcting' them.
But see here's the thing ... if you knew how to 'correct' the units oringinally used then you must have been able to interpret the units BEFORE the correction, otherwise you could never suggest a correction because you wouldnt know what the "real" units were. In other words, you saw V=1*i and say that's wrong, but you were able to specify the units. How could you specify units if you didnt know what they were by seeing V=1*i . The only answer is that you could read them by the context, which is the circuit as well as the expression, not just the expression alone. When you isolate one thing by itself out of a massive amount of material that means you may be taking something out of context.
I wont say you are wrong though because i have seen papers written where the context is not clear. In those cases it may be much harder to interpret what units are being used. Some units are more important too such s distance, but i think volts are volts around the world, i could be wrong though.
But i dont think that is the main issue anyway. The main issue is the second drawing where a mesh current is being used as PART of a controlled source. We can talk about units anytime. So let me digress now by going back to the issue of a mesh current, real or not, measurable or not.

For your last paragraph, you would need to sense Ia if you put Ia into the expression for the voltage source.
If you put I1 into the expression then you can sense I1.

It's a known fact that you can not measure Ia, and because Ix is also a mesh current, you can not measure that either.

If you disagree with Ix being non measurable, then simply show me where you put the current shunt to measure that. if you cant do that, show me where you can put two current shunts to measure Ix, and if you cant do that, show me where you could put three current shunts to measure Ix, etc., etc.
If you rather stick to 'measuring' Ia, then do the same for that, use any number of current shunts you wish.
If you do not wish to draw the diagram with the shunt(s) then just tell me what line(s) you would like to see current shunts in to measure either Ia or Ix, but you'll have to show the polarity of the shunt(s). For example, if you put a shunt in series with Iin (perfectly reasonable btw) and you have the positive end of the shunt connected to the current source Iin, then the output voltage from the shunt will be positive at the end that connects to the top of the current source Iin.

"Again, it doesn't matter whether you can separately sense Ix and Ia, because the CCVS never has to do that. But, in this circuit, if you really, really, really felt the need, then put a shunt in series with the current source to sense Ia and put a shunt in series with R2 and then use a diff amp to take the difference. But why would you do that. Just measure the current in R1, is IS the quantity (Ix-Ia) that is controlling the source."

you seem to be implying that we put one current shunt in series with Iin such that the positive output would be taken at the top of Iin, and so anything that Iin is the shunt measures, then a shunt in series with R2 which i am not sure what polarity you would use, but lets say you put the positive end at the top of R1, then subtract.
If that's true (and i think i have the polarities right), then what you are measuring is, again, NOT Ia you would be measuring either I1 or -I1 (original, first diagram) because Iin minus iR2 equals I1 (or -I1 depending on polarities).
What that means is if you substituted this into the expression for V where Ia goes you get:
V=(Ix-I1)*p or you get V=(Ix+I1)*p
which would not be the right control value. The right value is V=I1*p with the polarity shown in the first diagram or V=-I1*p if you use the opposite polarity to measure I1.

So what this leads to is the statement that, you can never draw a schematic that shows a way to measure Ia (or Ix).
Are there some circuits that are contrary to that? There could be some symmetry that in some circuit it works, but i dont think so, and certainly not for this circuit, but you can try maybe you can come up with something. Personally i think it would be cool if you did.

Last but not least, the solution for vR1 for the original circuit using I1 as the control value is:
vR1=(Iin*R1*R2)/(R2+R1-p)

and those are all original values.
I guess we might want to specify that R1+R2 can not equal p unless maybe we allow for an infinite voltage at the top of R1. I dont think a matrix method would work then though.

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#### Jony130

Joined Feb 17, 2009
5,440
I'll post more about that, but first i want to point out that when you did the analysis using mesh, you can not use the mesh current Ia within the definition of the voltage source which was V=p*(Ix-Ia) because the mesh current Ia is not a real current. That could be why you had trouble getting an expression.
For a current to be within a control element, it has to be measurable, and since Ia can not be measured it can not be used in that way.
Ok but if I used a classic way Iin + I1 - Is = 0 (1)

V- VR2 - VR1 = 0 (2)

p*I1 - Is*R2 - I1*R1 = 0

The solution is exactly the same as my previous solution based on the mesh analysis.

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

I1 = (Iin R2)/(p - R1 - R2)

#### MrAl

Joined Jun 17, 2014
9,785
Ok but if I used a classic way

View attachment 274906

Iin + I1 - Is = 0 (1)

V- VR2 - VR1 = 0 (2)

p*I1 - Is*R2 - I1*R1 = 0

The solution is exactly the same as my previous solution based on the mesh analysis.

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

I1 = (Iin R2)/(p - R1 - R2)
so what do you get for the voltage at the top of R1?
I thought you said you got no solution with R2=0?
How did Is get into that? Oh ok you are using that no problem.
But in your first post you wrote:
Ia = Iin (1)
and that's not true. In this new solution you recognize the Iin also flows through V.

I am not saying that the original circuit can not be solved. I got:
vR1=(Iin*R1*R2)/(R2+R1-p)

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#### Jony130

Joined Feb 17, 2009
5,440
I thought you said you got no solution with R2=0?
I wrongly interpreted the solution.

(Ix - Ia)*R1 - V = (Ix - Ia)*R1 - (Ix - Iin)*P =
= (Ix - 4mA)*1kΩ - (Ix - 4mA)*1kΩ =
= (Ix - 4mA)*1 - (Ix - 4m)*1 = Ix - 4 - Ix + 4.

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

so what do you get for the voltage at the top of R1?
V = (Iin R2)/(R1 + R2 - r)

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#### MrAl

Joined Jun 17, 2014
9,785
I wrongly interpreted the solution.

(Ix - Ia)*R1 - V = (Ix - Ia)*R1 - (Ix - Iin)*P =
= (Ix - 4mA)*1kΩ - (Ix - 4mA)*1kΩ =
= (Ix - 4mA)*1 - (Ix - 4m)*1 = Ix - 4 - Ix + 4.

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

V = (Iin R2)/(R1 + R2 - r)
Hello again Jony,

Thanks for clarification.
Is that your new solution V=Iin*R2/(R1+R2-p)?
That's not correct but i believe if you try again you will get it right because i miss quoted one fact. It is true that a mesh current is not real as my previous drawing shows, but the combination of two mesh currents as in this circuit which is Ia-Ix or Ix-Ia is real. This means as WBaln suggested, the difference should be valid in the controlled source.

I am pretty sure if you repeat this calculation you will get the right solution because you started out doing it the right way, and when you specified Ia as Ia=Iin you were not insinuating that Ia was real just that the assignment for Ia was usable as a mesh current. This may be why it gets a little confusing.
But in reality the reason we can state that Iin can be used for the mesh current Ia is because we consider the current source to be a current source for current but an open circuit for voltage. I guess that's not too important for this circuit, but in a circuit with a current source in one of the central branches we would have to look at it in that way.

So anyway, i did it your way as in the second circuit and i got the right result for vR1 which is the voltage at the top of R1. I am sure you will get the correct result if you just look at this again.

This was a pretty interesting circuit ha ha. It brought out some interesting things.
BTW where did you get this circuit?

#### Jony130

Joined Feb 17, 2009
5,440
Ok, but how you would calculate (proof) that when R2 is 0Ω we have Iin = Is.
What method would you use?
BTW where did you get this circuit?
From Xenon02 circuit theory course.

#### MrAl

Joined Jun 17, 2014
9,785
Ok, but how you would calculate (proof) that when R2 is 0Ω we have Iin = Is.
What method would you use?

From Xenon02 circuit theory course.
Hello again,

Oh ok, do you know who the authors were, and is that the book that tells you that Iin=Is when R2=0?

Well the formula i found was:
vR1=(Iin*R1*R2)/(R2+R1-p)
and vR1 is at the top of R1.

When R2 is zero in the denominator, we still have R1-p which for a moment we consider non zero, but when R2 is also zero in the numerator, we have 0/(R1-p) which is equal to zero.
When R1=p then we end up with vR1=Iin*R1, and when p=R1 we end up with vR1=Iin*p, but since they are then both the same it doesnt matter which we use.

Is that what you were looking for or did you have something else in mind?

Side note:
I made the mistake of saying that Ix-Ia could not be used for anything but since that is real it can be used, in theory.
However, there's no way to draw a circuit that has both Ia and Ix separate in the voltage source.
We often accept things that are not real in circuit analysis though maybe this should not be that much of a surprise. We just have to (usually) make sure they come out to real things at some point.