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

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

When I tried to change colors a lot of stuff came up in brackets and then everything changed, so sorry if it doesn't work. And, 'this' gives me Rth??

Those are the only two changes and I think I answered one of my own questions about Vo. If I find Vin, I'll have Vo.

Is that right?

 

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

When I tried to change colors a lot of stuff came up in brackets and then everything changed, so sorry if it doesn't work. And, 'this' gives me Rth??

Those are the only two changes and I think I answered one of my own questions about Vo. If I find Vin, I'll have Vo.

Is that right?

Absolutely right What do you get when you solve the equations?

Your original problem statement wanted the current in the 4 ohm resistor using the Thevenin equivalent to find it. Why do you want Vo?

 

I hesitate to add my comments here since Electrician has done such a great & patient job of leading sn00py23 through the task - with good contribution by Heavydoody. I'm fearful of adding any additional confusion!

With respect to finding Rth it might be worth considering the notion that

Rth=Voc/Isc

If the +ve op-amp terminal is shorted to ground Isc will flow. This would be an alternative to driving the op-amp there with a 1A source.

The "beauty" of this approach is that if the +ve op-amp terminal is grounded then the op-amp output will also be at ground potential.

In that case you have the 6V source driving a simple resistive network comprising the 16Ω in series with a parallel combination of 8Ω and 32Ω with Isc flowing in the 8Ω. It's a simple matter of finding this Isc and subsequently Rth since Voc is already known [12V].

Electrician - you are surely in line for sainthood!

 

With respect to finding Rth it might be worth considering the notion that

Rth=Voc/Isc

If the +ve op-amp terminal is shorted to ground Isc will flow.

I mentioned this early on, but the idea didn't seem to get much notice. It looks like this (with nodal equations for proof):

 

I got 3.2, 11.2, 11.2 and 22.4, but I'm not sure if these are voltages for the Nodes "again" or that I'm trying to find something else. Or, am I taking the 12V from the first Nodal problem and now having solve for another voltage at V2 I need to take the difference and interject it into the equiv. circuit.

This is pretty confusing and thank you so much for helping (both of you). And, I appreciate your patience. Okay, I'm still a bit confused. I understand how you got Vth and I understand a bit more about an op amp (just a bit) and how it is being used in this circuit.

I was always taught to do any resistance checks you had to remove all the sources. Okay, I get it that the current source is from the meter, but I never utilized that premise to solve a circuit. I just worked back to that point taking into consideration all the loads, so that op amp still has me a bit confused, and although I'm looking for resistance I am solving for voltage again and I'm supposed to get Rth how?

 

I 'noticed' your dwg heavyduty but thought this was for solving Vth. I was confused with your post because you were solving for five Node voltages and Electrician was solving for four. Sorry for not giving credit where credit is due.

 

I got 1.924, .7328, .7328 and 1.46 for the answers to these which don't match up to what I got using the other equations that I got from Electrician.

 

I 'noticed' your dwg heavyduty but thought this was for solving Vth. I was confused with your post because you were solving for five Node voltages and Electrician was solving for four. Sorry for not giving credit where credit is due.

This is a different drawing from what I posted previously...in response to t_n_k's comment. As to the fifth equation, it will get you the current out of the opamp...which you didn't need. Whenever I set up a problem I label all unknown nodes and find their equations. Sometimes you wind up with more than you need.

There are two methods for finding the Rth. My last post was the second as pointed out by t_n_k (and myself previously). Everything you need to solve for Rth is there. Use equations 1 and 4 to solve for V0 (the result may surprise you). Insert that into equation 3 and you have V1. V1 minus zero divided by the 8 ohm resistor gets you the short circuit current. Divide your Vth by this current and you have Rth.

Sorry about the brevity here. I am working on a ton of statics homework right now, and I loath statics.

 

Are my solutions correct? They don't look it.

 

Look closely at equation 4 (2*V3=V0). Then notice that in equation 1 V3=0. 2*0=0.

 

...substituting 0 for V0 into equation 3 yields V1=12/7V...that minus 0 (the voltage drop across the 8Ω resistor) divided by 8 gets 3/14A for Isc...your original Vth (12V) divided by Isc gets Rth...

 

I'm using a TI-85

 

I saw eq. 4 and didn't understand what the extra zero by itself was for. Okay, and since I'm using a TI-85, all I key in are the inverted Ohmic values into the Simult function and wallah, I get an answer. I don't break it down to 1/5 or 1/8 or 2* anything and that is what's confusing me even more having to do that instead of just plugging in what is attached to the node. So, I'm looking carefully, but I'm not seeing your point.

 

Did I wear you guys out? I'm assuming my calculations weren't correct?

 

I saw eq. 4 and didn't understand what the extra zero by itself was for. Okay, and since I'm using a TI-85, all I key in are the inverted Ohmic values into the Simult function and wallah, I get an answer. I don't break it down to 1/5 or 1/8 or 2* anything and that is what's confusing me even more having to do that instead of just plugging in what is attached to the node. So, I'm looking carefully, but I'm not seeing your point.

When you respond to an earlier post, and not the one immediately preceding your response, it's difficult to tell what you're referring to.

It would help if you respond by pressing the "Quote" button, which is what I did here.

I think you may be referring to post #34. In that case, equation 4 is:

Eq 4: V2 - V3 = 0

There isn't any extra zero in there. The zero that's there isn't "extra"; it's just the zero that is needed for the right side of the equation.

You just need to solve these four equations:

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

If you do this, the voltage V2 will be numerically equal to Rth. Remember what I said in an earlier post? If you pass 1 amp through a resistor, the voltage across the resistor will be equal to the value of the resistor. It's just Ohm's law: R = E/I; if I is 1, the R = E, the resistance is equal in numeric value to the voltage.

So, if we pass 1 amp through the Thevenin resistance (with the 6 volt battery set to zero), Rth, the voltage across the Thevenin resistance will be equal to the value of the Thevenin resistance. The Thevenin resistance is the resistance seen at node V2, so if we inject 1 amp into V2 and solve the network for the node voltages, the voltage at V2 will be equal to Rth.

We will also get the voltages at the other nodes, V1, V3 and Vo, but they are incidental; we don't really need them.

After we have Vth and Rth, we have the Thevenin equivalent of a voltage source of Vth in series with a resistor of Rth. We then connect that combination to another 4Ω resistor and calculate the current in the 4Ω resistor and we have the answer to your original problem.

I think you're having trouble solving the four equations above on your TI-85. I can't help you there; you'll have to figure that out yourself, but to give some help knowing when you've got it. here are the values for V1 and Vo you'll get when you have correctly solved the four equations:

V1 = 48
V2 = ???? This value will be Rth.
V3 = ????
Vo = 112

Did you ever change your profile options so that I could send you a PM?

 

This might help:

http://www.infj.ulst.ac.uk/NI-Maths/Guides/TexTI-85/Ch2/THBMatri.html

Go down to the section titled "Solving a system of linear equations".

 

When you respond to an earlier post, and not the one immediately preceding your response, it's difficult to tell what you're referring to.

It would help if you respond by pressing the "Quote" button, which is what I did here.

I think you may be referring to post #34. In that case, equation 4 is:

Eq 4: V2 - V3 = 0

There isn't any extra zero in there. The zero that's there isn't "extra"; it's just the zero that is needed for the right side of the equation.

You just need to solve these four equations:

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

If you do this, the voltage V2 will be numerically equal to Rth. Remember what I said in an earlier post? If you pass 1 amp through a resistor, the voltage across the resistor will be equal to the value of the resistor. It's just Ohm's law: R = E/I; if I is 1, the R = E, the resistance is equal in numeric value to the voltage.

So, if we pass 1 amp through the Thevenin resistance (with the 6 volt battery set to zero), Rth, the voltage across the Thevenin resistance will be equal to the value of the Thevenin resistance. The Thevenin resistance is the resistance seen at node V2, so if we inject 1 amp into V2 and solve the network for the node voltages, the voltage at V2 will be equal to Rth.

We will also get the voltages at the other nodes, V1, V3 and Vo, but they are incidental; we don't really need them.

After we have Vth and Rth, we have the Thevenin equivalent of a voltage source of Vth in series with a resistor of Rth. We then connect that combination to another 4Ω resistor and calculate the current in the 4Ω resistor and we have the answer to your original problem.

I think you're having trouble solving the four equations above on your TI-85. I can't help you there; you'll have to figure that out yourself, but to give some help knowing when you've got it. here are the values for V1 and Vo you'll get when you have correctly solved the four equations:

V1 = 48
V2 = ???? This value will be Rth.
V3 = ????
Vo = 112

Did you ever change your profile options so that I could send you a PM?

Sorry, the reason is that I'm answering the other gentleman on here as well as you and I'm getting you confused. I believe this will help me. Thanks bunches.

 

You just need to solve these four equations:

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

here are the values for V1 and Vo you'll get when you have correctly solved the four equations:

V1 = 48
V2 = ???? This value will be Rth.
V3 = ????
Vo = 112

Don't know why I didn't get it before, BUT I have it now. THANKS!

Okay, I have Vth = 12V and Rth = 56 Ohms.

If I put that in the equivalent circuit I get an Rtotal eq. of 60 Ohms which will give me a current value of 200mA or .2A.

Is that correct?

 

Absolutely correct.

Good job, it's good to see that perseverance paid off for you!!!

 

Absolutely correct.

Good job, it's good to see that perseverance paid off for you!!!

My perseverance may equate to tenacity while I'm thinking that your time equates to kudos from heaven!

Take care and thanks, again!