node analysis and dependent sources

calculate current i

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The Electrician

Joined Oct 9, 2007
2,971
They would all be variants on a theme. Some people work with currents flowing into the node, some with currents flowing out. Some put all the terms on one side and some put the terms with unknowns on one side and knowns on the other.
I had a feeling that this was the case. My feeling about such variations is that they are, to borrow a phrase from the lawyers, a distinction without a difference.
 

WBahn

Joined Mar 31, 2012
29,979
I had a feeling that this was the case. My feeling about such variations is that they are, to borrow a phrase from the lawyers, a distinction without a difference.
I agree. But then again look at all the people that need that "wheel" that gives them V = IR, I = V/R, and R = V/I as distinct formulas to choose from.
 

RBR1317

Joined Nov 13, 2010
713
...a distinction without a difference.
Of course it is merely an alternate form of the same equation. But the difference is that I can look at a circuit and write the node equations without having to think much about it and without error when using the method I described. But if I tried to write the node equations in the form shown in post #20, I would fail the way I failed so many years ago because I just don't feel the connection between the circuit and that form of the equation.
 

MrAl

Joined Jun 17, 2014
11,396
I had a feeling that this was the case. My feeling about such variations is that they are, to borrow a phrase from the lawyers, a distinction without a difference.
[side note: RBR1317, there is a topological connection]

Hello again,

I just quoted that one post but i could have quoted the last three or more. There is a difference. It is similar to how we measure distances when there are several items spaced regularly. For example look at a ruler, if we look at the point where we see the 1 inch mark and the point where we see the 2 inch mark, we can either state that we measured the points themselves or the distance between them. If we look at two points in time 1 sec and 2 seconds we can either say we have those two points or we could say we have 1 second elapsed.
We might also note that there is an inherent property that always exists here, and that is that 2>1 always. There's no other possibility, and this means we have a binary reduction logic here, where because 2 is greater than 1, then 1 can not ever be greater than 2. Seems obvious but recognizing that fact brings up different ways of looking at nodal.

But the nodal thing has not been explained well enough to see this yet.
For example, isnt it just a little peculiar that in the more common method we seem to need to look at TWO nodes at the same time in order to derive the equation? For example:
(Vz-Va)/Ra+(Vz-Vb)/Rb

With that, we are looking at two nodes, using ONLY voltages to derive the partial equation and keeping the current as implied. Why cant we just look at one voltage and one current? Since the current is implied from position, we can also look at the voltage and position.
Along this line of thought, we also can note that if one voltage is positive then the voltage on the other side of the component must be negative, and that is where the binary logic reduction comes into play.

In the second method, we only have to look at one node at a time, and one current direction, because the 'other' voltage is implied. If Vz is positive, then Va, Vb, or Vc must be negative, or we could swap signs, so if Vz is negative then the others (in that same drawing) must be positive.

Also, it does not matter that the equations come out the same. If we use a different method to get to the same equations it's still a different technique. But the method has to be clearly pointed out to see this too, as it is not JUST a juxtaposition of variables that makes it different, it's an entirely different way of looking at it.
This is something you CAN NOT see from looking over someone's shoulder.

To start, we look at ONE node at a time and only one node, determine the current direction according to the topological positioning, then write one tiny fragment of the equation. For example:
-Vz/Ra

Next the topological positioning rule, which must be uniform, will tell us that (for this example) Vz is on the right side of component Ra so Vz is negative. The next positioning rule says that because the current must flow INTO Vz (it's negative and another binary logic reduction comes into play) Vz stays negative, or using signs, Vz*(-1)*(+1)=-Vz.
Next we can look at Vz connected to Rc. Here, Vz is to the left of Rc so it is positive, but that means current is flowing OUT of Vz and into Rc, so Vz ends up negative again:
-(Vz/Rc)

where the parens show that the negative sign was included because of the position of Vz relative to the topological position of Rc, which makes the current flow negative.

The same rules apply for Vz and Rb, but because that component is vertical we have to apply top and bottom rules, but for Vz only the bottom rule is needed. So we look at Vz, then decide where it is positionally relative to the component. We end up with the starting equation:
-Vz/Ra-Vz/Rb-Vz/Rc

but then we do the same thing for nodes Va, Vb, and Vc, and take note of the direction of the current of the current source.

The begging question here is why do all this?
The answer is, in making the matrix for the set of equations it is more convenient programmatically, and it is easier to arrange the matrix because you dont have to make two entries at the same time; with only one node to look at there can be only one matrix entry.
It is of course another technique on it's own though, and with a little practice helps eliminate the need to look at two nodes at once in consideration of how to get the equations.

So it's not just a juxtaposition of variables, it's an entirely different approach to nodal. It's like the culmination of a host of implications that leads to simplification.
 
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The Electrician

Joined Oct 9, 2007
2,971
Doing a nodal analysis and solving 5 simultaneous equations I get...

Vy = -0.625V
Vb = 26.625
Vp = 11.50
Vr = 41.25
Vo = 28.375

Vx = -12.87 V
I = 13.75

nodal voltage subscripts refer to the color of the node per post #4 and Vg =0
MrAl mentioned in post #6 that Vx is between 2 and 3 volts. I agree.

I solved the problem with both the nodal and mesh methods and got the same result from both:

Vx = 2.59375 V
i = 3.4375 A
 

WBahn

Joined Mar 31, 2012
29,979
Doing a nodal analysis and solving 5 simultaneous equations I get...

Vy = -0.625V
Vb = 26.625
Vp = 11.50
Vr = 41.25
Vo = 28.375

Vx = -12.87 V
I = 13.75

nodal voltage subscripts refer to the color of the node per post #4 and Vg =0
How many times must it be explained to you? Do NOT just give answers to homework questions in this forum!

Hints and suggestions are great. Doing as Mr Al did and give some bounds on the answers is perfectly fine. But don't just give out answers.
 

DGElder

Joined Apr 3, 2016
351
Relax. The post is a week old and the OP has not been back since. He/she is probably sipping Pina Coladas on summer break by now. In any case it just gives the solution not how to do it so it would not be admissible as a homework assignment.
 

DGElder

Joined Apr 3, 2016
351
"MrAl mentioned in post #6 that Vx is between 2 and 3 volts. I agree.

I solved the problem with both the nodal and mesh methods and got the same result from both:

Vx = 2.59375 V
i = 3.4375 A "


Hmmm
 

WBahn

Joined Mar 31, 2012
29,979
Relax. The post is a week old and the OP has not been back since. He/she is probably sipping Pina Coladas on summer break by now. In any case it just gives the solution not how to do it so it would not be admissible as a homework assignment.
Sadly, you might be surprised by how often only the numerical answers are checked when grading an assignment. The work presented might bear absolutely no resemblance to the problem, but as long as the numbers in the box at the bottom are correct, they get full credit.
 

The Electrician

Joined Oct 9, 2007
2,971
Sadly, you might be surprised by how often only the numerical answers are checked when grading an assignment. The work presented might bear absolutely no resemblance to the problem, but as long as the numbers in the box at the bottom are correct, they get full credit.
Since (as I think will be borne out) DGElder gave wrong answers, I thought it permissible to correct them so that subsequent readers of this post won't be misled.

Or moderation could remove the offending posts.
 

WBahn

Joined Mar 31, 2012
29,979
Since (as I think will be borne out) DGElder gave wrong answers, I thought it permissible to correct them so that subsequent readers of this post won't be misled.

Or moderation could remove the offending posts.
Pointing out that they are wrong is definitely reasonable (though, admittedly, leaving the door open for the TS to snarf up and turn in incorrect answers does have a certain appeal to it). Giving the actual correct answers would be, in general, a bit more debatable, but I'm not quibbling about that at all here.

The matter of when it becomes reasonable to post solutions (either answers or full up solutions) to questions that have been left unresolved is a very open one. Some would argue that they should never be solved (except, with assistance, by the TS) since those same questions are very likely to be given to other students later, either in that same course or at other schools. Personally, there are so many solutions to so many problems out in the wild on the Internet that I think instructors have to assume that solutions to a large fraction of problems they assign can be found (or bought). So I'm not too concerned about that. So the real issue, for me, is to try to make sure that a TS that refuses to put forth any effort not be able to benefit from a posted solution. Many schools give two weeks to work a problem set (some more, but not too many as far as I know). While most of the people that want us to just work their problems for them are posting the night before it is due, there are quite a few that actually put thought and effort into their cheating and so post their problems as soon as they are assigned hoping to tease out as many solutions as they can. Those are the people I really don't want to see helped. So my recommendation would be that any homework thread less than three weeks old should be considered live in terms of not giving out answers or solutions directly.
 

WBahn

Joined Mar 31, 2012
29,979
Hi,

That's great, but i wonder what happened to the OP.
All indications would point to him going away when he discovered that he wasn't going to get his homework done for him unless he actually put in some effort of his own.
 
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