sn00py23, if you'll change your settings so you can receive PMs, I'll message you.

 

(V3 - Vo)/10 + (V1 - Vo)/32

 

sn00py23, if you'll change your settings so you can receive PMs, I'll message you.

Sorry, was looking around and didn't know what I was changing.

 

In the first step, I get 12v at V2 (the noninverting input to the opamp), which is what you had earlier. V2=V3, so you have 12v at V3 as well. V0 should be 24v. Of course, none of the other voltages really matter, Vth is the 12v you are getting at V2. What values have you gotten for Rth in the second part?

Did you use Nodal to determine that 12V. I was at a friend's house the first time I did this and I believe I used what we were first working on but didn't think that could be right, so I left my notes there and won't be able to get them until I see them again Monday.

I'm used to using KVLs to determine what my open source voltage is and only used Node voltage to determine what all other unknowns are, so I'd like to be able to check if that 12V is true and would need V1 to do that. I'm not 'even' going the path of the 32 Ohm resistor because it's feeding back, if I read this correctly.

What I got at my friend's house was 1.something K Ohms for Rth. Is that close to what you got?

 

Okay, putting in your 3 equations and the last one of V2=V3, but changing the V2=V3 to one equation of (V3-Vo)/10 + (Vo)/10, I came up with Node voltages of 9V, 9V, 7.5V and 15. When I took V2-V3 = 0 I got 11V for three Nodes and then 22V at Vo.

So, I'm thinking I'm close if 12V is right, but still not sure how you're getting it.

 

Sorry, long night. I just got up a bit ago.

Did you use Nodal to determine that 12V?

Yes.

I'm used to using KVLs to determine what my open source voltage is...

I personally like nodal analysis over mesh in most cases.

...so I'd like to be able to check if that 12V is true and would need V1 to do that...

Once you have all of your equations set up you can readily solve for V1 and all of the other voltages using nodal analysis. However, recall The Electrician's hint that, with an ideal opamp, it is assumed that no current flows in or out of the inputs. Then ask yourself how much current is flowing through the eight ohm resistor, apply ohm's law, and you will know the relationship between V1 and V2. It is a bit of a trick though. If you set up all of the nodal equations it will work out just as well.

What I got at my friend's house was 1.something K Ohms for Rth. Is that close to what you got?

No. I worked the problem out my way and AAC's way, came up with the same answer both times, and it is not in the kΩ range. Remember that you have to use nodal analysis on two different circuits. Its a lot of work, but doesn't require any guessing about where current is flowing and how the circuit can be made simpler. Frequently, if you just plunge ahead with what you know will be a lot of work, instead of trying to find an easier solution, you end up saving time and effort in the end.

 

is that if there is no current flowing through the op amp that there is no current flowing through the 8 Ohm resistor? But, there will need to be one you put the 4 Ohm resistor back in, right?

I put his and my last equation in three times and go a different answer each time. I used my calculator and for some reason I don't get 12V for V2.

If I use (1 + Rg/Rf) and get 2 for Vout how did everyone else get 12V I get Vin as 1V

I realize I may be jumping the gun at some points because this is pretty new to me, but I'm looking for an approach overall. Setting aside everything for the moment and looking at what I did so far, I completed Nodal format eqs,. and I don't get 12V. So, either I did something wrong or my last equation is incorrect.

Too, if V2=V3 and I am using nodal to find the drops across the two 10 Ohm resistors upon plugging that in it should work out the same, no? But, mine isn't.

Right now, I'm not even interested in using Thevenin until I can get my voltage drops correct and I appear to be way off base.

I think I can see why V1 has to be the same as V2 so there is no current expected over the 8 Ohm resistor, right? Even so that means if I do a KVL from points A to B (where the 4 Ohm resistor was) I should be able to get Vth, but I don't. I even went so far as to convert the voltage source to a current source to perhaps assist me and still I don't get 12V.

Thanks for jumping in and trying to help.

 

No. I worked the problem out my way and AAC's way, came up with the same answer both times, and it is not in the kΩ range. Remember that you have to use nodal analysis on two different circuits. Its a lot of work, but doesn't require any guessing about where current is flowing and how the circuit can be made simpler. Frequently, if you just plunge ahead with what you know will be a lot of work, instead of trying to find an easier solution, you end up saving time and effort in the end.

Whoa, wait a minute, use Nodal on TWO different circuits??? I'm trying it with just one. TWO???? I don't remember the Electrician saying anything about TWO circuits. He gave me Node equations for three of the four nodes in this circuit. I went from confused to lost.

 

Whoa, wait a minute, use Nodal on TWO different circuits??? I'm trying it with just one. TWO???? I don't remember the Electrician saying anything about TWO circuits. He gave me Node equations for three of the four nodes in this circuit. I went from confused to lost.

Yes, you find the Thevenin voltage first with the 4Ω resistor removed and nothing else changed. Then you have to come up with a second set of equations with the voltage source removed and 1 amp injected into the circuit at V2 to find the Thevenin resistance.

 

Sorry, was looking around and didn't know what I was changing.

Go to your control panel.

Under "Settings and options", choose "Edit options".

Inside the big window labeled "Messaging and Notification", find a subsection labeled "Private Messaging".

Check the box that says "Enable Private Messaging".

 

is that if there is no current flowing through the op amp that there is no current flowing through the 8 Ohm resistor?

Correct. So applying Ohm's law, 0amps * 8Ω equals zero volts difference from one side of the resistor to the other. So V1=V2.

But, there will need to be one you put the 4 Ohm resistor back in, right?

Not into the ideal opamp, only through the four ohm resistor.

I put his and my last equation in three times and go a different answer each time. I used my calculator and for some reason I don't get 12V for V2.

It is easy to make mathematical errors with these simultaneous equations. Here is how I got the first part:

All of which results in:

If I use (1 + Rg/Rf) and get 2 for Vout how did everyone else get 12V I get Vin as 1V

That is the gain, not Vout. Multiply that by V2 and you get Vout (V0).

 

Yes, you find the Thevenin voltage first with the 4Ω resistor removed and nothing else changed. Then you have to come up with a second set of equations with the voltage source removed and 1 amp injected into the circuit at V2 to find the Thevenin resistance.

I did the first Nodal set of equations with the 4 Ohm resistor removed and still can't come up with 12V.

You state that there is no current in the op amp which give me none at the 8 Ohm resistor, but now I'm supposed to put 1A into the op amp which doesn't allow current flow just so I can make the equations work and you think "I'M" trying to look for an easy way to do this???

I'm just looking for something that makes sense to me and gives me an answer I can prove. I still haven't figured out how to get Rth and have come up with all kinds of things for that which, again, make no sense. I haven't heard from the Electrician so I think he gave up on me.

 

I did the first Nodal set of equations with the 4 Ohm resistor removed and still can't come up with 12V.

Look at my previous post.

You state that there is no current in the op amp which give me none at the 8 Ohm resistor, but now I'm supposed to put 1A into the op amp which doesn't allow current flow...

You are putting 1amp into V2, it does not have to go into the opamp...like this:

 

Okay, putting in your 3 equations and the last one of V2=V3, but changing the V2=V3 to one equation of (V3-Vo)/10 + (Vo)/10, I came up with Node voltages of 9V, 9V, 7.5V and 15. When I took V2-V3 = 0 I got 11V for three Nodes and then 22V at Vo.

So, I'm thinking I'm close if 12V is right, but still not sure how you're getting it.

You are so close!!!

You have: (V3-Vo)/10 + (Vo)/10, but this isn't an equation yet. You have to sum those two currents to zero. And the second term is wrong. The 10 ohm resistor that goes to ground is going from V3 to ground, not from Vo to ground.

To use the nodal method, you derive simple expressions for the current in each resistor connected to a node, and sum those currents to zero.

The current in the resistor connected between V3 and Vo is (V3-Vo)/10. That's because the voltage across that resistor is (V3-Vo) and the value of the resistance is 10Ω. Ohm's law tells you that the current in a resistor is just the voltage across it divided by the value of the resistance; in other words, the current is (V3-Vo)/10.

The voltage across the resistor connected from V3 to ground is (V3 - 0), and the value of the resistor is 10Ω, so the current in that resistor is (V3-0)/10, or V3/10.

Summing the two currents in the two 10Ω resistors to zero, we get the third equation: (V3-Vo)/10 + V3/10 = 0. There is no current into the opamp inputs , so these two are the only currents into the V3 node.

You have to remember to sum the currents to zero to get an equation.

Don't change any of the equations I already gave you!

I asked you to work on finding the third equation, and you have essentially got it, but you put it in place of the last equation instead of the third, (up till now, unknown) one!

I gave you:

Eq 1: (V1-6)/16 + (V1-V2)/8 + (V1-Vo)/32 = 0
Eq 2: (V2 - V1)/8 = 0
Eq 3: ???????
Eq 4: V2 - V3 = 0

If we add what you got (with corrections), we'll have:

Eq 1: (V1-6)/16 + (V1-V2)/8 + (V1-Vo)/32 = 0
Eq 2: (V2 - V1)/8 = 0
Eq 3: (V3-Vo)/10 + V3/10 = 0
Eq 4: V2 - V3 = 0

Solve these 4 equations and you will get the 4 node voltages. V2 will be the Vth you're looking for.

Give your results, and if you get the correct Vth, we'll move on the getting Rth.

 

I thought we were working with four Nodes and that zero you have in there has me for a loop.

The way we learned how to do nodal was to put the variables into our calculator and hit solve. I have almost identical to yours, but using yours with that zero in there gives me an error overall.

I took everything to zero and solved using Nodal format, what am I doing wrong?

 

So, I didn't have to solve for the Node voltage at Vo???

 

Hallelujah! If I hadn't of already fallen from a few churches, I'd go back today! I finally got it, but how come I wasn't solving for Vo which is what I was trying to do?

I need to get something to eat and allow my brain to cool before I begin with the Rth.

Thanks bunches!

 

I thought we were working with four Nodes and that zero you have in there has me for a loop.

The way we learned how to do nodal was to put the variables into our calculator and hit solve. I have almost identical to yours, but using yours with that zero in there gives me an error overall.

I took everything to zero and solved using Nodal format, what am I doing wrong?

I don't know what "zero" you're referring to. Do you mean the zero in (V3-0)/10 in this sentence?

"The voltage across the resistor connected from V3 to ground is (V3 - 0), and the value of the resistor is 10Ω, so the current in that resistor is (V3-0)/10, or V3/10."

It goes away in the end in the four equations:

Eq 1: (V1-6)/16 + (V1-V2)/8 + (V1-Vo)/32 = 0
Eq 2: (V2 - V1)/8 = 0
Eq 3: (V3-Vo)/10 + V3/10 = 0
Eq 4: V2 - V3 = 0

You'll notice that each equation is set equal to zero.

You may have to rearrange them to use them in the matrix solver on you calculator, if that's what you're using:

Eq 1: (1/8 + 1/16 + 1/32)*V1 + (-1/8)*V2 + (0)*V3 + (-1/32)*Vo = 6/16
Eq 2: (-1/8)*V1 + (1/8)*V2 + (0)*V3 + (0)*Vo = 0
Eq 3: (0)*V1 + (0)*V2 + (1/10 + 1/10)*V3 + (-1/10)*Vo = 0
Eq 4: (0)*V1 + (1)*V2 + (-1)*V3 + (0)*Vo = 0

All the coefficients in parenthesis form the elements of a matrix. I left the terms like (0)*V3 in there to serve a place holders.

What calculator are you using?

 

Hallelujah! If I hadn't of already fallen from a few churches, I'd go back today! I finally got it, but how come I wasn't solving for Vo which is what I was trying to do?

I need to get something to eat and allow my brain to cool before I begin with the Rth.

Thanks bunches!

Why would you want to solve for Vo? The problem didn't ask for that; it asked for the current i in the 4 ohm resistor, and you don't need to know Vo to get that. The solution to the 4 equations does give you Vo, as well as V1, V2 and V3, but only V2 is really needed to get Vth; the others are incidental to your real problem!

Instead of worrying individually about where the currents are going, or what voltage as one point affects some other voltage, you should just concentrate on getting the node equations right. Then everything will be solved for.

It's really easy at this point. Two tiny changes in the 4 equations and you have Rth.

What you have to do is set the battery voltage to zero, and inject a 1 amp current at V2.

It would help avoid confusion if you would keep the four equations in a list like this and see if you can show the two necessary changes (show the changes in red for clarity):

Eq 1: (V1-6)/16 + (V1-V2)/8 + (V1-Vo)/32 = 0
Eq 2: (V2 - V1)/8 = 0
Eq 3: (V3-Vo)/10 + V3/10 = 0
Eq 4: V2 - V3 = 0

 

The reason I ask about Vo is because I'd really like to know if what I have for it is correct and only because the gentleman earlier gave me the five equations and I was trying to work them together as he did yours and his.

But, I can't do his, but I can do yours - now. I guess I'm not sure why or how I can solve for a portion of the circuit (and yes, I know I'm only interested in one spot and that may be incidental to you, but I'm trying to understand this.)

Does the current source have to be 1A, can it be 1mA? Also, does it have to be drawn in the direction you show or does it make a difference if the current is pointing away from V2?

I'll work on the balance in a moment.

So, what you're saying is for me to use Nodal, again, to find Rth. But I thought to find Rth, all sources had to be removed?

One other question, since I'm looking for the current at the 4 Ohm resistor, would it have been good to use Norton?