Thevenin & Norton Circuit with Current source?

[Marcin H]

Homework Statement

Homework Equations

V=IR
Thevenin/Norton

The Attempt at a Solution

I am trying to find the Voc, but this circuit isn't making sense to me. So to start we can ignore the 9ohm resistor and that branch for now when Finding Voc so you are basically left with the square circuit on the left and I labeled my ground in the bottom right of that square. Now to find the Voc I have to find the voltage across the 3ix current source, but I am not sure how. How can I apply KVL or KCL to the loop if I don't know the resistance or voltage of the current source? And if you do KCL at the Top middle node then you get ix = 3ix which would give you 1=3 . Should I find I norton first and then I can find Rt which would just be 11ohms and then do V= InRt?

 

[gneill]

As you've noticed, if ab is open then only the one loop is involved, hence only one current.

For what possible value of $i_x$ can $i_x = 3 i_x$? What potential at the central node will guarantee this value of current?

 

[Marcin H]

As you've noticed, if ab is open then only the one loop is involved, hence only one current.

For what possible value of #i_x$can$i_x = 3 i_x##? What potential at the central node will guarantee this value of current?

Your text is a bit glitchy for me, but are you asking what value of ix will give you ix = 3ix? 0 is the only value, but I don't really get how that works because if you take that equation and divide by ix you get 1=3 which gives you nothing. and how does a current of 0 in this circuit make sense? You have a voltage source and a current source.

Also, if it is 0, I am not sure how to find the potential that will make that true. Do I just use KCL at the node?

 

[gneill]

Your text is a bit glitchy for me, but are you asking what value of ix will give you ix = 3ix? 0 is the only value, but I don't really get how that works because if you take that equation and divide by ix you get 1=3 which gives you nothing. and how does a current of 0 in this circuit make sense? You have a voltage source and a current source.

Also, if it is 0, I am not sure how to find the potential that will make that true. Do I just use KCL at the node?

I did fix the text, undoubtedly shortly after you opened the post!

Yes, the only value of $i_x$ that can satisfy the requirement is zero. What must be the potential difference across the 2 Ω resistor in that case? So what is the potential at the center node?

 

[Marcin H]

I did fix the text, undoubtedly shortly after you opened the post!

Yes, the only value of $i_x$ that can satisfy the requirement is zero. What must be the potential difference across the 2 Ω resistor in that case? So what is the potential at the center node?

It would also have to be 0

EDIT* So if ix is 0 then does that mean the 3ix current source is also at 0A meaning that the voltage across there has to be 0 so that would mean Voc is also 0. So the thevenin voltage is 0?

 

[cnh1995]

It would also have to be 0

EDIT* So if ix is 0 then does that mean the 3ix current source is also at 0A meaning that the voltage across there has to be 0 so that would mean Voc is also 0. So the thevenin voltage is 0?

No.
Use KVL when Ix=0.

 

[Marcin H]

Ok so how could I show that exactly using some equations? Can I right Voc =V1 = 3(ix)R = 0. V1 is what I labeled my middle node.

Also, I am not sure how I could find my norton equivalent now. Since I have Vthevenin (0V) and Rthevenin (11ohms) can I just say Inorton = Vth/Rth = 0/11 = 0? This seems way to easy. I can get some node equations when I put a wire from a to b to find Isc but it looks like In would be 0 if you use KCL at the node. ix=3ix+In and if ix =0 then IN=0. I have never had a problem like this and it just doesn't seem right.

 

[cnh1995]

Ok so how could I show that exactly using some equations?

Show what? Vth=0? It is not zero..I edited my post a couple of minutes after posting.

 

[cnh1995]

I have never had a problem like this and it just doesn't seem right.

Dependent sources can be a bit confusing in some cases. You got Ix=0 right. But since Ix=0, what is the voltage across the 2 ohm resistor? Apply KVL to the circuit and see.

 

[Marcin H]

Show what? Vth=0? It is not zero..I edited my post a couple of minutes after posting.

Oh. I didn't see your edit. If I use KVL (starting in the bottom left corner) I would get 0 = -8V +2ix + 3ix. But that doesn't make sense. how can I use KVL when there is a current source with an unkown resistance. I don't know how to find the voltage across the current source. That equation also does't make sense because if you plug in 0 for ix you will get -8V = 0 which is wrong.

EDIT*

Dependent sources can be a bit confusing in some cases. You got Ix=0 right. But since Ix=0, what is the voltage across the 2 ohm resistor? Apply KVL to the circuit and see.

If ix is 0 then V=IR = (0)*(2ohms) = 0

 

[cnh1995]

If ix is 0 then V=IR = (0)*(2ohms) = 0

Right.
Current through the dependent source is 0. In fact, no current is flowing I am any branch.
What does this tell you about the current source?

 

[Marcin H]

Right.
Current through the dependent source is 0. In fact, no current is flowing in any branch.
What does this tell you about the current source?

Is the current source off or something? I am not too sure what that tells me about the current source. Would the coltage across it be 8V then? Is it acting just like its shorted? I can just remove that component?

 

[gneill]

EDIT* So if ix is 0 then does that mean the 3ix current source is also at 0A meaning that the voltage across there has to be 0 so that would mean Voc is also 0. So the thevenin voltage is 0?

Your conclusion that if the current source is 0 Amps that the potential across it must be 0 Volts is not correct. A current source produces whatever potential difference is required in order to achieve and maintain its designated current. That can be any voltage, positive, negative, or zero.

In this case its goal is achieved if the current $i_x$ is zero, making $3i_x$ zero also. In order for that to happen the current through the 2 Ω resistor must be zero. So what what can you say about the potentials (with respect to your chosen reference point) at both ends of that resistor?

 

[Marcin H]

Your conclusion that if the current source is 0 Amps that the potential across it must be 0 Volts is not correct. A current source produces whatever potential difference is required in order to achieve and maintain its designated current. That can be any voltage, positive, negative, or zero.

In this case its goal is achieved if the current $i_x$ is zero, making $3i_x$ zero also. In order for that to happen the current through the 2 Ω resistor must be zero. So what what can you say about the potentials (with respect to your chosen reference point) at both ends of that resistor?

Would it just be 8V then?

 

[gneill]

Would it just be 8V then?

Well, confirm your hypothesis. Would KVL be satisfied for the loop?

 

[Marcin H]

-8V+8V=0 so I guess it will. Hmm. So Vth is 8V.

SO if Vth is 8V I can just use Rth which is 11ohms and use that to find Inorton right? In = (8/11) A

 

[gneill]

-8V+8V=0 so I guess it will. Hmm. So Vth is 8V.

SO if Vth is 8V I can just use Rth which is 11ohms and use that to find Inorton right? In = (8/11) A

How do you conclude that Rth is 11 Ω? Remember, you can't suppress a controlled source to find the Thevenin resistance: the actions of controlled sources will affect what the load "sees" as the impedance of the network, since they muck with the V-I characteristics that appear at the output.

Finding the open circuit voltage and the short circuit current is the way to finding the Thevenin or Norton equivalent. The Thevenin (or Norton) resistance is the ratio of V_oc/I_sc.

 

[Marcin H]

How do you conclude that Rth is 11 Ω? Remember, you can't suppress a controlled source to find the Thevenin resistance: the actions of controlled sources will affect what the load "sees" as the impedance of the network, since they muck with the V-I characteristics that appear at the output.

Finding the open circuit voltage and the short circuit current is the way to finding the Thevenin or Norton equivalent. The Thevenin (or Norton) resistance is the ratio of V_oc/I_sc.

I thought we could find Rt by shorting all the sources and just looking at the resistors. I remember that rule from somewhere, but I'm not sure it's right. But if we said that the current throught the 2 ohm and the current source is 0 then Wouldn't In also be 0? or would the voltage source be providing a current to IN now?

 

[gneill]

I thought we could find Rt by shorting all the sources and just looking at the resistors. I remember that rule from somewhere, but I'm not sure it's right. But if we said that the current throught the 2 ohm and the current source is 0 then Wouldn't In also be 0? or would the voltage source be providing a current to IN now?

You can suppress fixed sources (shorting voltage sources, opening current sources), but NOT controlled sources. Controlled sources are active components and change their voltage and/or currents depending upon the conditions in the circuit (or externally applied conditions). In some ways they appear to act like variable resistors.

So your path forward right now is to determine the short circuit current.

 

[Marcin H]

You can suppress fixed sources (shorting voltage sources, opening current sources), but NOT controlled sources. Controlled sources are active components and change their voltage and/or currents depending upon the conditions in the circuit (or externally applied conditions). In some ways they appear to act like variable resistors.

So your path forward right now is to determine the short circuit current.

Would it not be 0? Can we use the KCL equation ix = 3ix + Isc and use the fact that ix=0 to show that In=0? Or is that not true once you put a wire from A to B. Here is what I came up with after shorting a-b. Is this correct?

 

[gneill]

No, Isc won't be zero. Once you short the output you have a new circuit, so your previous value of 8 V for V1 won't apply anymore. Analyse the new circuit from scratch. The node equation approach is convenient, but don't assume a value for V1. Instead, solve for it then obtain the short circuit current as the current through the 9 Ω resistor branch.

Hint: Identify the variable $i_x$ with the auxiliary equation $i_x = \frac{8 - V1}{2}$ in order to eliminate $i_x$ from the node equation.

 

[Marcin H]

No, Isc won't be zero. Once you short the output you have a new circuit, so your previous value of 8 V for V1 won't apply anymore. Analyse the new circuit from scratch. The node equation approach is convenient, but don't assume a value for V1. Instead, solve for it then obtain the short circuit current as the current through the 9 Ω resistor branch.

Hint: Identify the variable $i_x$ with the auxiliary equation $i_x = \frac{8 - V1}{2}$ in order to eliminate $i_x$ from the node equation.

Is my way from the picture wrong? Are you saying I should take the equation for ix ( the one you just wrote) and plug it into 3ix and then solve for V1? How will V1 help me find Isc?

 

[gneill]

Is my way from the picture wrong? Are you saying I should take the equation for ix ( the one you just wrote) and plug it into 3ix and then solve for V1? How will V1 help me find Isc?

When you write the node equation you have two unknowns: $V1$ and $i_x$. The auxiliary equation allows you to eliminate $i_x$ and solve for $V1$.

$I_{sc}$ is the current through the 9 Ω resistor. What's the potential across it?

 

[gneill]

Alternatively you could use mesh analysis to solve for the current in the second loop. But you'll need to use a supermesh, and another auxiliary equation tying the mesh currents together with the controlled source.

 

[Marcin H]

When you write the node equation you have two unknowns: $V1$ and $i_x$. The auxiliary equation allows you to eliminate $i_x$ and solve for $V1$.

$I_{sc}$ is the current through the 9 Ω resistor. What's the potential across it?

8V is the potential across so would the In just be I=V/R = 8V/9ohm = 8/9A?

 

[gneill]

8V is the potential across so would the In just be I=V/R = 8V/9ohm = 8/9A?

Are you sure that it's 9 V 8 V ? Remember, V1 in this circuit is not guaranteed to be the same as in the previous one where the output was open. Show your solution for V1.

[Edit: Fixed the voltage value to match the quote]

 

[Marcin H]

Are you sure that it's 9 V? Remember, V1 in this circuit is not guaranteed to be the same as in the previous one where the output was open. Show your solution for V1.

Oh, right. It won't be 9V. But it should be equal to V1 (the voltage across the 3ix current source) but I am not sure how to find that now after shorting ab. The way i did it in the picture of my solution is the only way I can think of it. I didn't learn mesh/super mesh yet so I have to do it using KCL and node voltage I think. Or loop analysis. I don't know how to find the potential across the 3ix source once the circuit is shorted.

 

[gneill]

Write a node equation for the junction of the 2 and 9 Ohm resistors. The auxiliary equation remains $i_x = \frac{8 - V_1}{2}$. Solve for $V_1$. The current through the short ab is the same as the current through the 9 Ω resistor, and you'll have the potential across it.

 

[Marcin H]

Write a node equation for the junction of the 2 and 9 Ohm resistors. The auxiliary equation remains $i_x = \frac{8 - V_1}{2}$. Solve for $V_1$. The current through the short ab is the same as the current through the 9 Ω resistor, and you'll have the potential across it.

View attachment 105734

Ok so my equation would be (V1-8)/2 = 3ix + (V1-0)/9

and plugging the ix in I get a value for V1 of 144/17

Then I=V/R = (16/17)A

 

[gneill]

Ok so my equation would be (V1-8)/2 = 3ix + (V1-0)/9

and plugging the ix in I get a value for V1 of 144/17

Then I=V/R = (16/17)A

Check your node equation. You've got terms for currents flowing out of the node on both sides of the equation. If you're going to do it that way, one side or the other should describe a currents flowing into the node.

The best way to write a node equation consistently is to write a sum that equals zero. Choose either all currents flowing into the node or all currents flowing out. Then write all the terms on one side of the equation and set it equal to zero. So in this case if we assume that all currents are flowing out of the node:
$$\frac{V_1 -8}{2} + \frac{V_1}{9} + 3 \left(\frac{8 - V_1}{2} \right) = 0$$

 

[Marcin H]

Check your node equation. You've got terms for currents flowing out of the node on both sides of the equation. If you're going to do it that way, one side or the other should describe a currents flowing into the node.

The best way to write a node equation consistently is to write a sum that equals zero. Choose either all currents flowing into the node or all currents flowing out. Then write all the terms on one side of the equation and set it equal to zero. So in this case if we assume that all currents are flowing out of the node:
$$\frac{V_1 -8}{2} + \frac{V_1}{9} + 3 \left(\frac{8 - V_1}{2} \right) = 0$$

Did I draw my arrows at V1 incorrectly? I just look at current in = current out and then write the equation. I'm not too sure why your equation is like that, but using it I get V1= -9V and that would give me IN= -1A

 

[gneill]

Did I draw my arrows at V1 incorrectly? I just look at current in = current out and then write the equation. I'm not too sure why your equation is like that, but using it I get V1= -9V and that would give me IN= -1A

In your equation (post #29) your terms all describe currents leaving the node. That means, for example, if you moved the term on the left side over to the right it would be negated and describe a current entering the node. Then your equation would say that some current entering the node plus all the currents leaving is zero. That's not KCL.

You seem to have made a sign error in solving the equation from post #30. Can you show your work?

 

[Marcin H]

In your equation (post #29) your terms all describe currents leaving the node. That means, for example, if you moved the term on the left side over to the right it would be negated and describe a current entering the node. Then your equation would say that some current entering the node plus all the currents leaving is zero. That's not KCL.

You seem to have made a sign error in solving the equation from post #30. Can you show your work?

I just double checked it. I got + 1A now.

 

[gneill]

Good! So now you have both the open circuit voltage and the short circuit current. What then is the Thevenin resistance?

 

[Marcin H]

Good! So now you have both the open circuit voltage and the short circuit current. What then is the Thevenin resistance?

RT = Vt/Isc = 8V/1A = 8ohms