Solving Thevenin's Theorm with Superposition

Thread Starter

project_science

Joined Sep 14, 2018
21
Hi,

I've worked through the attached problem as best I can. The goal of the problem is to find the Thevenin's equivalent of the network and graph its i-v relationship.

I have 3 questions, and they are clearly marked on my work as Q1, Q2, and Q3, but the first two are very similar questions.

Q1: Is it correct to take out R2? I assumed no current would flow across it without a driving voltage. (My other idea would be that there would actually be some current, and it would be i3, which branches from i2", such that i2" = i3 + i4, where i3 splits to the right at the bottom of R1, and i4 returns to the left at the bottom of R1.)

Q2: Is it correct to take out R2 here? I assumed no current would go through it as Isc" would take all the current from current source, I.

Q3: Does this i-v relationship graph make sense?
 

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WBahn

Joined Mar 31, 2012
24,692
Hi,

I've worked through the attached problem as best I can. The goal of the problem is to find the Thevenin's equivalent of the network and graph its i-v relationship.

I have 3 questions, and they are clearly marked on my work as Q1, Q2, and Q3, but the first two are very similar questions.

Q1: Is it correct to take out R2? I assumed no current would flow across it without a driving voltage. (My other idea would be that there would actually be some current, and it would be i3, which branches from i2", such that i2" = i3 + i4, where i3 splits to the right at the bottom of R1, and i4 returns to the left at the bottom of R1.)

Q2: Is it correct to take out R2 here? I assumed no current would go through it as Isc" would take all the current from current source, I.

Q3: Does this i-v relationship graph make sense?
You have some problems even before you get to your first question, so let's start there.

Midway down the first page, you make two assertions.

i2' = i + V/R2
i2' = i + I

Do you see how these two equations, if true, REQUIRE that I = V/R2 ?

If so, then do you see that this REQUIRES that R2 = V/I ?

But I is the current output of an independent current source, which can be whatever we set it to be. Similarly, V is the output of an independent voltage source and it is whatever we set it to be. But since R2 is just a resistor that we can choose arbitrarily, it makes no sense to claim that it will always somehow end up being the ratio of V/I.

This is one thing you want to try to develop the habit of -- constantly ask yourself throughout your work if it makes sense. If not, then it almost always means you've done something wrong, which is the case here.

Now, looking at your drawing I see that you are using the symbol I multiple times to mean different things. This virtually always ends poorly. In any given problem, only use a symbol (variable) to mean one thing and one thing only. Also, when you define a current, be sure to define the polarity (direction). You did this with your second I but not with your Voc -- if Voc turns out to be positive, which terminal is more positive than the other? Now, looking at your drawing again, I see that you did define the polarity, but not where you defined Voc (you did that in the second drawing but defined the polarity in the third). Get in the habit of defining the polarity at the same place you define the symbol -- not only will it cut down on miscommunications, but you are far less likely to mess up and using it with one polarity in one part of your work and the other polarity in another part.

BTW -- You are doing a MUCH better job at presenting your work and including everything that's needed than most. It's just a matter of some fine tuning that you will find makes a huge improvement in your results (and grades).

So let'c call that second I something else -- how about Iv (the current out of the voltage source)?

Great. Glad you agree. :D

Your actual KCL equation

i2' = i + I

is therefore fine. But it is NOT the case that

I = V/R2.

You are making a very common mistake in applying Ohm's Law in that you are throwing the nearest I, V, and R at the equation. Remember that Ohm's Law relates the value of a resistance to the current through THAT resistor (which is I in this case) to the voltage across THAT resistor (which is NOT V in this case).

Now, getting to your questions.

Q1 -- No, you cannot just assume that no current flows in R2. Whatever voltage appears across R2 also appears across R1, so if there's no voltage to drive a current in R2, there isn't a voltage to drive a current in R1. If that's the case, then were does the current from the current source go. A common mistake is to assume that the voltage across a current source is somehow always zero -- but this is as silly as assuming that the current out of a voltage source is somehow always zero. Both are power supplies and both produce both voltage and current. The only difference is that an ideal voltage source will deliver whatever current is required in order to maintain a constant output voltage while an ideal current source will produce whatever voltage is required in order to maintain a constant output current.

Q2 -- Yes. The line of reasoning here is that the short circuit prevents the development of any voltage across R1 (or R2, since we've already established that in the circuit with the voltage source turned off that they are in parallel) and without any voltage across it, there can't be any current through it.

Q3 -- Close. The slope of the line can't be Rth because that graph has a negative slope. Remember that the slope of a line is rise over run. Your rise is (-Voc) for a run of (Iss). Also, you now appear to be using the symbol I to mean something else again -- be careful about that as it WILL cause you all kinds of grief.
 

RBR1317

Joined Nov 13, 2010
490
Another way to check your answer comes from the fact that here the Thevenin resistance is by inspection equal to the parallel combination of R1 & R2.
See https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/ "• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points."
 

WBahn

Joined Mar 31, 2012
24,692
Another way to check your answer comes from the fact that here the Thevenin resistance is by inspection equal to the parallel combination of R1 & R2.
See https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/ "• (2) Find the Thevenin resistance by removing all power sources in the original circuit (voltage sources shorted and current sources open) and calculating total resistance between the open connection points."
A much better way to think about this process is not "removing" all power sources, but rather turning all power sources off or, equivalently, setting their outputs to zero. The problem with the "removing" rule is that it leads many people to simply remove the voltage sources (since that's what they think they were supposed to do) leaving them as an open circuit. They make far fewer mistakes if they instead always think of setting the voltage on a voltage source to 0 V.
 

Thread Starter

project_science

Joined Sep 14, 2018
21
You have some problems even before you get to your first question, so let's start there.

Midway down the first page, you make two assertions.

i2' = i + V/R2
i2' = i + I

Do you see how these two equations, if true, REQUIRE that I = V/R2 ?

If so, then do you see that this REQUIRES that R2 = V/I ?

But I is the current output of an independent current source, which can be whatever we set it to be. Similarly, V is the output of an independent voltage source and it is whatever we set it to be. But since R2 is just a resistor that we can choose arbitrarily, it makes no sense to claim that it will always somehow end up being the ratio of V/I.

This is one thing you want to try to develop the habit of -- constantly ask yourself throughout your work if it makes sense. If not, then it almost always means you've done something wrong, which is the case here.

Now, looking at your drawing I see that you are using the symbol I multiple times to mean different things. This virtually always ends poorly. In any given problem, only use a symbol (variable) to mean one thing and one thing only. Also, when you define a current, be sure to define the polarity (direction). You did this with your second I but not with your Voc -- if Voc turns out to be positive, which terminal is more positive than the other? Now, looking at your drawing again, I see that you did define the polarity, but not where you defined Voc (you did that in the second drawing but defined the polarity in the third). Get in the habit of defining the polarity at the same place you define the symbol -- not only will it cut down on miscommunications, but you are far less likely to mess up and using it with one polarity in one part of your work and the other polarity in another part.

BTW -- You are doing a MUCH better job at presenting your work and including everything that's needed than most. It's just a matter of some fine tuning that you will find makes a huge improvement in your results (and grades).

So let'c call that second I something else -- how about Iv (the current out of the voltage source)?

Great. Glad you agree. :D

Your actual KCL equation

i2' = i + I

is therefore fine. But it is NOT the case that

I = V/R2.

You are making a very common mistake in applying Ohm's Law in that you are throwing the nearest I, V, and R at the equation. Remember that Ohm's Law relates the value of a resistance to the current through THAT resistor (which is I in this case) to the voltage across THAT resistor (which is NOT V in this case).

Now, getting to your questions.

Q1 -- No, you cannot just assume that no current flows in R2. Whatever voltage appears across R2 also appears across R1, so if there's no voltage to drive a current in R2, there isn't a voltage to drive a current in R1. If that's the case, then were does the current from the current source go. A common mistake is to assume that the voltage across a current source is somehow always zero -- but this is as silly as assuming that the current out of a voltage source is somehow always zero. Both are power supplies and both produce both voltage and current. The only difference is that an ideal voltage source will deliver whatever current is required in order to maintain a constant output voltage while an ideal current source will produce whatever voltage is required in order to maintain a constant output current.

Q2 -- Yes. The line of reasoning here is that the short circuit prevents the development of any voltage across R1 (or R2, since we've already established that in the circuit with the voltage source turned off that they are in parallel) and without any voltage across it, there can't be any current through it.

Q3 -- Close. The slope of the line can't be Rth because that graph has a negative slope. Remember that the slope of a line is rise over run. Your rise is (-Voc) for a run of (Iss). Also, you now appear to be using the symbol I to mean something else again -- be careful about that as it WILL cause you all kinds of grief.
Thanks for all your responses! I’ve redone the homework, and I think I’ve made some good improvements. I have two remaining questions which accompany my new homework attachment:

Q4. The algebra can obviously get messy, but does my solution towards Rth look right now? (page 3)
Q5. I am not sure how to draw and i-v relationship. When v goes up, i must, too (I think…). But you said it’s not equal to Rth
 

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RBR1317

Joined Nov 13, 2010
490
For Q4 the Thevenin resistance is shown as a function of (R2, i, I, V). If one were to find Rth by substituting circuit values for R2, i, I & V, what would be substituted for "i"? Looked for that but could not find it.

Also, we know that the Thevenin resistance can be found by zeroing the sources. So why are non-zero sources in the equation for Rth?
 

Thread Starter

project_science

Joined Sep 14, 2018
21
For Q4 the Thevenin resistance is shown as a function of (R2, i, I, V). If one were to find Rth by substituting circuit values for R2, i, I & V, what would be substituted for "i"? Looked for that but could not find it.

Also, we know that the Thevenin resistance can be found by zeroing the sources. So why are non-zero sources in the equation for Rth?
In the original circuit, "i" was shown to be coming from left-to-right in the circuit (it's how it was drawn in the text). Since "i" is unknown, and coming from the ports, I felt it had to be left as an unknown.

As for your second comment, I don't currently know how to answer that other than non-zero sources contribute to the equivalent resistance.
 

WBahn

Joined Mar 31, 2012
24,692
In the original circuit, "i" was shown to be coming from left-to-right in the circuit (it's how it was drawn in the text). Since "i" is unknown, and coming from the ports, I felt it had to be left as an unknown.

As for your second comment, I don't currently know how to answer that other than non-zero sources contribute to the equivalent resistance.
When you are finding the open-circuit voltage, you KNOW that i = 0 because it is open circuit (and also that v = Voc where v is the terminal voltage). When finding the short-circuit current, you know that i = -Isc (and also that v = 0).
 

RBR1317

Joined Nov 13, 2010
490
The attached diagram illustrates why the current "i" need not be considered when finding the Thevenin equivalent of the circuit. The current "i" is initially defined for finding the i-v characteristic of the circuit once the Thevenin equivalent has been derived. Obtaining the Thevenin equivalent prior to finding the i-v characteristic will simplify the math; however, as the TS has demonstrated, incorporating "i" into the Thevenin equivalent does not lead to simplification.
 

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Thread Starter

project_science

Joined Sep 14, 2018
21
Ok, I think I am getting this. I redid my previous homework (3rd time!) as I see that i = 0 when the circuit is open. It makes sense, but the initial problem through me off just by labeling it!

Anyhow, my main question is in 6B (p. 3). I have a voltage in my Rth eq., which I found as a question in 6A (p. 2).

According to member RBR1317, I can't have non-zero sources in my equation for Rth, which confuses me, since his work has "V" in the eq. for Isc, which is the same as I have.

Am I missing something from your explanation, or is this now correct?
 

Attachments

WBahn

Joined Mar 31, 2012
24,692
Ok, I think I am getting this. I redid my previous homework (3rd time!) as I see that i = 0 when the circuit is open. It makes sense, but the initial problem through me off just by labeling it!

Anyhow, my main question is in 6B (p. 3). I have a voltage in my Rth eq., which I found as a question in 6A (p. 2).

According to member RBR1317, I can't have non-zero sources in my equation for Rth, which confuses me, since his work has "V" in the eq. for Isc, which is the same as I have.

Am I missing something from your explanation, or is this now correct?
The Voc and Isc expressions will almost always have a source in them -- if they didn't there would be no voltage or current. But the RATIO of the two should not -- the sources should cancel out.
 

WBahn

Joined Mar 31, 2012
24,692
Ok, I think I am getting this. I redid my previous homework (3rd time!) as I see that i = 0 when the circuit is open. It makes sense, but the initial problem through me off just by labeling it!

Anyhow, my main question is in 6B (p. 3). I have a voltage in my Rth eq., which I found as a question in 6A (p. 2).

According to member RBR1317, I can't have non-zero sources in my equation for Rth, which confuses me, since his work has "V" in the eq. for Isc, which is the same as I have.

Am I missing something from your explanation, or is this now correct?
You are still using the same symbol to mean different things and I warned you that it would bite you sooner or later. It just bit you. When working with the voltage source you use 'I' for the current out of the voltage source. But then when you work with the current source you use 'I' for the current out of the current source. Then when you combine the results you treat them as the same 'I' -- they aren't!

Take care to always use any given symbol to mean one and only one thing throughout the entire problem. Call the current out of the current source Is and the current out of the voltage source Iv, for instance.

Similarly, you use i2" to mean both the current in R1 and the current in Rp -- these are not the same currents. In this case, since you are interested in the voltage across R1 and Rp and they happen to be the same, it doesn't bite you. But it easily could have.

You need to solve for both Iv and Is in terms of the circuit parameters (the source and resistance values).
 

Thread Starter

project_science

Joined Sep 14, 2018
21
Ok, I redid my HW a 4th time, and I believe I've fixed two issues. I see what you mean about:

1. voltage current, Iv
2. current from a current source, Ic

Labeling these individually, I see Rth (p.3) resulted with voltages and currents cancelling out from the ratio of Voc / Isc. This resulted where Rth = R1. Does that make sense that in this problem, R2 was eliminated in Thevenin's equivalent circuit?

If this is correct, then I just need to figure out Q5 (from p.3)
 

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WBahn

Joined Mar 31, 2012
24,692
Ok, I redid my HW a 4th time, and I believe I've fixed two issues. I see what you mean about:

1. voltage current, Iv
2. current from a current source, Ic

Labeling these individually, I see Rth (p.3) resulted with voltages and currents cancelling out from the ratio of Voc / Isc. This resulted where Rth = R1. Does that make sense that in this problem, R2 was eliminated in Thevenin's equivalent circuit?

If this is correct, then I just need to figure out Q5 (from p.3)
You're getting closer, but not quite there yet.

The fact that you even asked whether it made sense that R2 was eliminated is a very good sign -- and something that so few people do. Now you need to take the next step and see if you can answer that question for yourself.

Remember that another way (probably the first way that you learned) to find Rth is to zero the independent sources and then just find the equivalent resistance looking into the ports. If you do that for this circuit, what do you get?
 

WBahn

Joined Mar 31, 2012
24,692
Ok, I redid my HW a 4th time, and I believe I've fixed two issues. I see what you mean about:

1. voltage current, Iv
2. current from a current source, Ic

Labeling these individually, I see Rth (p.3) resulted with voltages and currents cancelling out from the ratio of Voc / Isc. This resulted where Rth = R1. Does that make sense that in this problem, R2 was eliminated in Thevenin's equivalent circuit?

If this is correct, then I just need to figure out Q5 (from p.3)
At the very bottom of page 2 you make the substitution that Iv = V/R2. Where does that come from?

You stopped your initial analysis before it was done. You need to find i2' in terms of the circuit parameters, namely I, V, R1, and R2. Iv is not a circuit parameter, it's just a variable that's been defined to aid in the analysis -- meaning that it can't survive the analysis.

You are very close.
 

Thread Starter

project_science

Joined Sep 14, 2018
21
At the very bottom of page 2 you make the substitution that Iv = V/R2. Where does that come from?

You stopped your initial analysis before it was done. You need to find i2' in terms of the circuit parameters, namely I, V, R1, and R2. Iv is not a circuit parameter, it's just a variable that's been defined to aid in the analysis -- meaning that it can't survive the analysis.

You are very close.
Ok, basically, I asked if it made sense that R2 is eliminated in Rth, and in my 5th try for my HW, it appears it doesn’t make sense to eliminate R2 (my Voc needed to be fixed), based on the feedback you’ve given me. To start, I’ve answered your two questions:

1. From the initial problem, if I just zeroed all the independent sources (current and voltage sources), R1 || R2, and thus, that is the equivalent resistance as viewed from the ports, where Rth = Rp = (R1*R2) / (R1 + R2)

2. Iv = (V / R2) comes from p.3 (in current HW) in the middle, left-side of the page, where the current source is zeroed so that voltage acts alone. Thus, Isc’ = Iv = (V / R2). (This has been updated in current HW attached.)

After doing the analysis, it appears that setting both independent sources to zero on the initial circuit, and solving for the equivalent resistance, Rth is the same as when I worked through 3 pages of a formal analysis. Correct?

If it’s correct, then Q5 is still to be completed.
 

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WBahn

Joined Mar 31, 2012
24,692
2. Iv = (V / R2) comes from p.3 (in current HW) in the middle, left-side of the page, where the current source is zeroed so that voltage acts alone. Thus, Isc’ = Iv = (V / R2). (This has been updated in current HW attached.)
If Iv is V/R2, that means that the voltage ACROSS R2 is V. If so, that means that the voltage across R1 is zero in order to satisfy KVL. But if the voltage across R1 is zero, then the current through R1 is zero. So were does the current Iv go once it leaves R2?

Remember, Ohm's Law requires that you use the voltage ACROSS the resistance in question -- not just some voltage value that happens to be conveniently close by.
 

Thread Starter

project_science

Joined Sep 14, 2018
21
I figured that if the circuit was shorted, denoted by Isc', then R1 is effectively removed from the circuit since current takes the path of least resistance. In that case, the circuit is reduced such that the only dependent element is R2. That's why I set Isc' = Iv = V / R2

(I made a small mistake at the bottom of p. 3 where I wrote Isc = V / R2. It should have been Isc' = V / R2.)
 

WBahn

Joined Mar 31, 2012
24,692
I figured that if the circuit was shorted, denoted by Isc', then R1 is effectively removed from the circuit since current takes the path of least resistance. In that case, the circuit is reduced such that the only dependent element is R2. That's why I set Isc' = Iv = V / R2

(I made a small mistake at the bottom of p. 3 where I wrote Isc = V / R2. It should have been Isc' = V / R2.)
Perhaps I misread your work -- I thought you did this when calculating the open circuit voltage.
 
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