The Node Voltage Equation

The Electrician

Joined Oct 9, 2007
2,970
Why did you start a new thread to show your work? You already had a thread going about this problem: http://forum.allaboutcircuits.com/t...-where-is-the-polarity-vx-pointing-to.129474/

Furthermore, you need to participate in the process when you ask for help. You asked for help with a circuit essentially identical to this one a month ago, but did not answer questions: http://forum.allaboutcircuits.com/t...eal-current-sources-using-kvl-and-kcl.128112/

We would like to help, but you have to meet us halfway.
 

Thread Starter

Mohamed Hesham Mohammed

Joined Oct 8, 2016
34
Why did you start a new thread to show your work? You already had a thread going about this problem: http://forum.allaboutcircuits.com/t...-where-is-the-polarity-vx-pointing-to.129474/

Furthermore, you need to participate in the process when you ask for help. You asked for help with a circuit essentially identical to this one a month ago, but did not answer questions: http://forum.allaboutcircuits.com/t...eal-current-sources-using-kvl-and-kcl.128112/

We would like to help, but you have to meet us halfway.
I don't your problem but in every post i see your name poping up and criticizing me if you are willing to help then thank you very much. If you see any insufficient Data tell me and i will provide.
 

WBahn

Joined Mar 31, 2012
30,072
This Question is so difficult for me to Solve I just want to make sure that my sloution is in the right Path
My work is Attached

Thanks For Help
You images are to light for me to make out reliably (at least not without more effort than I am willing to put forth).

If you have a set of answers, then you should be able to quickly verify that they are correct by plugging them back into the circuit to see if they are consistent with the circuit.
 

The Electrician

Joined Oct 9, 2007
2,970
Looking at the very last line of the second image showing your work, I see:

Va = -1.57
Vb = -.65
Vc = -3.45

I can't make out the value for Vd. It looks like -4.5.90, or possibly -4.5.10, or maybe -45.90, or -45.10
When you show your work you should make sure that your handwriting is very clear. When it isn't, the people here will be annoyed by how much work they have to do to help you.

Let's see if those values satisfy a simple relationship among them.

The circuit shows IΔ = (Va-Vb)/4. Using your values, I get IΔ= -.23 and 5*IΔ = -1.15

Next, the circuit requires that Vb-Vc = 5*(IΔ).

Using your values, Vb-Vc = -.65 - (-3.45) = 2.8, but 5*IΔ = - 1.15

Your values for Va, Vb and Vc don't satisfy the circuit constraints.

This is the kind of testing you can do yourself. As WBahn said in the previous post "If you have a set of answers, then you should be able to quickly verify that they are correct by plugging them back into the circuit to see if they are consistent with the circuit."

You still haven't answered my question: "The month old thread subject asked for the power supplied by the ideal current sources. What is the problem asking for now? Is it the same thing, the power supplied by the current sources?"

Edit: I just noticed that the problem image itself specifies what you're to find: Vx, IΔ and VΔ

I'm sorry if you're offended by my criticism, but when you ask for help here you can expect that you will experience criticism; the criticism is intended to be "constructive" criticism to help you improve your problem solving skills, and sometimes to improve your problem presentation.
 
Last edited:

The Electrician

Joined Oct 9, 2007
2,970
The last equation in the first image of your work is unclear.

It looks like V?/1 + (Va-Vb)/4 + (Va-Ve)/5 = 0

I can't make out the symbol I've replaced with a question mark. What is that? Is it Vx?

At any rate, whatever it is, you are apparently trying to sum the currents into node A to zero. I suppose the V?/1 is intended to be the current in the 1 ohm resistor. But what about the current in the 3Vx voltage source? That source is also a current path attached to node A; you can't ignore it. This error is propagating throughout all the rest of your calculations.

You should get rid of the 1 ohm and the 8 ohm resistors anyway. They are in parallel with voltage sources and they have no effect on the voltages at various nodes elsewhere in the circuit. Let your circuit be like this:

BigCircuit.jpg
 

Thread Starter

Mohamed Hesham Mohammed

Joined Oct 8, 2016
34
Looking at the very last line of the second image showing your work, I see:

Va = -1.57
Vb = -.65
Vc = -3.45

I can't make out the value for Vd. It looks like -4.5.90, or possibly -4.5.10, or maybe -45.90, or -45.10
When you show your work you should make sure that your handwriting is very clear. When it isn't, the people here will be annoyed by how much work they have to do to help you.

Let's see if those values satisfy a simple relationship among them.

The circuit shows IΔ = (Va-Vb)/4. Using your values, I get IΔ= -.23 and 5*IΔ = -1.15

Next, the circuit requires that Vb-Vc = 5*(IΔ).

Using your values, Vb-Vc = -.65 - (-3.45) = 2.8, but 5*IΔ = - 1.15

Your values for Va, Vb and Vc don't satisfy the circuit constraints.

This is the kind of testing you can do yourself. As WBahn said in the previous post "If you have a set of answers, then you should be able to quickly verify that they are correct by plugging them back into the circuit to see if they are consistent with the circuit."

You still haven't answered my question: "The month old thread subject asked for the power supplied by the ideal current sources. What is the problem asking for now? Is it the same thing, the power supplied by the current sources?"

Edit: I just noticed that the problem image itself specifies what you're to find: Vx, IΔ and VΔ

I'm sorry if you're offended by my criticism, but when you ask for help here you can expect that you will experience criticism; the criticism is intended to be "constructive" criticism to help you improve your problem solving skills, and sometimes to improve your problem presentation.
Thank you very much for Your concern.Appreciated
I will try again with better handwriting and substitute back into the circuit.
 

Thread Starter

Mohamed Hesham Mohammed

Joined Oct 8, 2016
34
The last equation in the first image of your work is unclear.

It looks like V?/1 + (Va-Vb)/4 + (Va-Ve)/5 = 0

I can't make out the symbol I've replaced with a question mark. What is that? Is it Vx?

At any rate, whatever it is, you are apparently trying to sum the currents into node A to zero. I suppose the V?/1 is intended to be the current in the 1 ohm resistor. But what about the current in the 3Vx voltage source? That source is also a current path attached to node A; you can't ignore it. This error is propagating throughout all the rest of your calculations.

You should get rid of the 1 ohm and the 8 ohm resistors anyway. They are in parallel with voltage sources and they have no effect on the voltages at various nodes elsewhere in the circuit. Let your circuit be like this:

View attachment 115563
Shouldn't I treat the 3vx as a supernode between a and the Ground
So
Va-0=3vx
 
V?=Va

Should i consider a Voltage source between a reference node and referenced node a super node
Yes, but what is your purpose with this equation: V?/1 + (Va-Vb)/4 + (Va-Ve)/5 = 0

It appears that you are adding up all the currents into node A; was that your purpose? But what about the current delivered by the 3Vx source? You can't ignore that current.
 

Thread Starter

Mohamed Hesham Mohammed

Joined Oct 8, 2016
34
Yes, but what is your purpose with this equation: V?/1 + (Va-Vb)/4 + (Va-Ve)/5 = 0

It appears that you are adding up all the currents into node A; was that your purpose? But what about the current delivered by the 3Vx source? You can't ignore that current.
What should be written for it Considering an ideal wire and ideal dependent source
 
What should be written for it Considering an ideal wire and ideal dependent source
You cannot write an equation summing currents into node A. The equation Va/1 + (Va-Vb)/4 + (Va-Ve)/5 = 0 is not true because it leaves out the current contribution of the 3Vx source. The only equation you can write at node A is this: Va = 3*Vx. You then need to substitute for Vx. You don't have Vx well defined on your circuit. I would suggest that you add a designator for another of your non-essential nodes so that you have something to use for Vx.

Here is your circuit with the rest of the nodes (which are non-essential) labeled.

BigCircuit X.jpg

This allows you to write Va = 3*(Ve-Vf)

You will probably need some more equations as well.
 

Thread Starter

Mohamed Hesham Mohammed

Joined Oct 8, 2016
34
You cannot write an equation summing currents into node A. The equation Va/1 + (Va-Vb)/4 + (Va-Ve)/5 = 0 is not true because it leaves out the current contribution of the 3Vx source. The only equation you can write at node A is this: Va = 3*Vx. You then need to substitute for Vx. You don't have Vx well defined on your circuit. I would suggest that you add a designator for another of your non-essential nodes so that you have something to use for Vx.

Here is your circuit with the rest of the nodes (which are non-essential) labeled.

View attachment 115575

This allows you to write Va = 3*(Ve-Vf)

You will probably need some more equations as well.
thank you very much that will help i will try and solve it and update you
 
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