Calculate Current, Power, EMS..

[evol_w10lv]

I finish this task:
I1 = (fi1-fi3+E1)/R1 = (-8.2877-21.6043+18.8)/523 = 0.0212 A
I2 = (fi2-fi1)/R2 = (0.0185-(-8.2877))/295 = 0.0282 A
I3 = (fi3-fi2)/R3 = (21.6043-0.0185)/766 = 0.0282 A
I4 = (fi1-E2)/R4 = (-8.2877+19)/217 = 0.0494 A
I5 = (E3-fi3)/R5 = (38.1-21.6043)/334 = 0.0494 A
I6 = fi2/R6 = 0.0185/805 = 0.0000229 A
Is it OK?

And now I hope that someone of you will chek the first part with the main task when you won't be busy.

 

[The Electrician]

Despite that I believe, that solution in post #14 is wright. I tried to solve it with KCL method.

I1*(R5+R01+R1) + I6*R6 = E1
-I6*R6 + I4*R4 + I2*(R02+R2) = E2
-I2*(R2+R02) + I3*(R03+R3) = -E2-E3
A] -I2 + I4 - I3=0
B] -I4 - I6 + I1 =0
C] -I1 + I6 + I2=0

I didn't include I01, I02, I03, I5, because
I1 = I01 = I5
I02 = I2
I3 = I03

But I guess something is missing to solve that system. Can you look at it, what's there is wrong? For me it's easier to try to use this KLC method, that's why I don't want use nodal solution, but of course we can try if it's necessary and you shouldn't suggest, what's wrong in KLC equations.

I'm not sure why you're doing this. In post #6, Laplace pointed out that it's inefficient to give multiple designations to the same current. For example, in this diagram I3 is the same as I03, I2 is the same as I02, et cetera. There's no point is doing this.

 

[evol_w10lv]

I just have a requirement to use KLC method to solve the task. Or I can use KLC method in other task, but then I have to use nodal method for this task.

If you want help on a nodal solution. you should add to your diagram some additional details. Let the right end of your voltmeter (bottom right corner of the schematic) be your ground (reference node); show a ground symbol there. Then show (and number) 3 nodes--top of R4, bottom of R4 and top of R5. Get rid of the red current loops. Post that picture and then we'll discuss a nodal solution.

A nodal solution is going to be much less appropriate because all your sources are voltage sources instead of current sources.

I less understand nodal method, but we can try to find solution.

 

[The Electrician]

I finish this task:
I1 = (fi1-fi3+E1)/R1 = (-8.2877-21.6043+18.8)/523 = 0.0212 A
I2 = (fi2-fi1)/R2 = (0.0185-(-8.2877))/295 = 0.0282 A
I3 = (fi3-fi2)/R3 = (21.6043-0.0185)/766 = 0.0282 A
I4 = (fi1-E2)/R4 = (-8.2877+19)/217 = 0.0494 A
I5 = (E3-fi3)/R5 = (38.1-21.6043)/334 = 0.0494 A
I6 = fi2/R6 = 0.0185/805 = 0.0000229 A
Is it OK?

And now I hope that someone of you will chek the first part with the main task when you won't be busy.

Your results look ok, but if you carried more digits you would see that i2 and I3 are not identical, for instance. They are slightly different.

 

[The Electrician]

Here in America, there is no KLC method.

There are two Kirchoff laws--the voltage law, "Kirchoff's voltage law", KVL

and Kirchoff's current law, KCL.

KVL is used in the mesh or loop method.

KCL is used in the nodal method.

 

[evol_w10lv]

Yes, there was my mistake with KLC.
Please, can you write equations with KCL for the nodal method if you didn't like my equations with Kirchoff Rules method?

 

[The Electrician]

Label the voltages at the three nodes Va, Vb and Vc.(or perhaps Ua, Ub and Uc). You need 3 equations, one for each node.

You must write an equation for the current through each branch that connects to a given node. The current through a branch is equal to the voltage between nodes plus or minus any voltage sources in the branch, divided by the total resistance in the branch. Then add all the branch currents connected to that node and set them equal to zero.

Give it a try and show your work.

 

[evol_w10lv]

I'm not confident at all about my equations..

-Va*(G4+G03+G3+G2+G02) + Vc*(G2+G02) + Vb*G4 = -E3*(G03+G3) + E2*(G2+G02)
-Vc*(G5+G02+G2+G6) + Vb*G6 + Va*(G02+G2) = -E*(G02+G2)
-Vb*(G4+G6+G01+G2) + Va*G4 + Vc*G6 = - E1*(G01+G1)

 

[The Electrician]

The thing to do now is to solve the equations and compare the results with the results from the mesh method.

 

[Laplace]

Curious as to why you choose the ground reference as an isolated node between two resistors instead of a node that is a common point for the most circuit branches. If you made node 'C' the circuit ground, then how many equations would it be necessary to write? Is it ever necessary to get the node equation for the ground point? Would fewer equations make it easier to solve?

Also, at this point I'm not sure what the component values are supposed to be.

 

[evol_w10lv]

I didn't know, where I have to put circuit ground, that's why I put it, where The Electrician suggested.
Here are the component values:
E1 = 40V
E2 = 50V
E3 = 30V
R01 = 0.1 Ohm
R02 = 0.2 Ohm
R03 = 0.2 Ohm
R1 = 3.9 Ohm
R2 = 3.8 Ohm
R3 = 2.9 Ohm
R4 = 4 Ohm
R5 = 5 Ohm
R6 = 4 Ohm

Can you chek my equations with your program or maybe you can offer your equations using nodal method for this task, please!

 

[The Electrician]

Curious as to why you choose the ground reference as an isolated node between two resistors instead of a node that is a common point for the most circuit branches. If you made node 'C' the circuit ground, then how many equations would it be necessary to write? Is it ever necessary to get the node equation for the ground point? Would fewer equations make it easier to solve?

Also, at this point I'm not sure what the component values are supposed to be.

He did choose that ground point on my advice.

The schematic doesn't show the voltmeter which had one end connected to the lower right corner. I think at some point he plans to put the voltmeter back and determine what it would read. I figured he could then note that the voltmeter reading would be exactly one of his calculated node voltages.

Anyway, it won't hurt him to do a little more work and note that the reference can be any node; 3 equations instead of 2.

 

[The Electrician]

I didn't know, where I have to put circuit ground, that's why I put it, where The Electrician suggested.
Here are the component values:
E1 = 40V
E2 = 50V
E3 = 30V
R01 = 0.1 Ohm
R02 = 0.2 Ohm
R03 = 0.2 Ohm
R1 = 3.9 Ohm
R2 = 3.8 Ohm
R3 = 2.9 Ohm
R4 = 4 Ohm
R5 = 5 Ohm
R6 = 4 Ohm

Can you chek my equations with your program or maybe you can offer your equations using nodal method for this task, please!

What do you get when you solve the equations? You should carry more digits in your solution, maybe 6 digits.

How are you solving the equations? Are you using a calculator, or a mathematical program like Mathcad?

If you don't have those you could use this online calculator:

http://www.bluebit.gr/matrix-calculator/linear_equations.aspx

 

[Laplace]

I have attached the MathCAD page illustrating how I would write the node equations for nodes A & B, first for the general case with node C active and then for the specific case with node C as the ground reference (Vc=0). Only provided the solution for the two node voltages since that is what the node equations are for. Note that the equation for any node is a statement of KCL, that the sum of all currents flowing from a node must equal zero. Each term of the equation is structured as the node voltage minus the adjacent node voltage divided by the resistance to the adjacent node.

 

[The Electrician]

I have attached the MathCAD page illustrating how I would write the node equations for nodes A & B, first for the general case with node C active and then for the specific case with node C as the ground reference (Vc=0). Only provided the solution for the two node voltages since that is what the node equations are for. Note that the equation for any node is a statement of KCL, that the sum of all currents flowing from a node must equal zero. Each term of the equation is structured as the node voltage minus the adjacent node voltage divided by the resistance to the adjacent node.

In your second equation, and before you set Vc=0, the last term has R5 in the denominator. If the reference node is the lower right corner, and Vc is active, then R5 shouldn't be in that equation, should it?

That means that if you were starting from the correct Vc active equation, you would have to do more than just set Vc=0 to get the equation for the case where Vc becomes the reference node, I think.

 

[Laplace]

... If the reference node is the lower right corner, and Vc is active, then R5 shouldn't be in that equation, should it?

What if the reference node is not the lower right corner? Suppose there is no reference node, or just a phantom reference? That would not invalidate the equations as written, but setting Vc=0 makes it easier to get a solution.

 

[The Electrician]

What if the reference node is not the lower right corner? Suppose there is no reference node, or just a phantom reference? That would not invalidate the equations as written, but setting Vc=0 makes it easier to get a solution.

What is a phantom reference?

My comments start out talking about the case where the reference node IS the lower right hand corner. I believe your second equation, left side was intended to be for that case. Then the R5 in the denominator of the last term was in error.

When the R5 element is present in that denominator, then that equation is not correct for the case where the reference is the lower right hand corner.

Then, if R5 is absent as it should be for a correct equation for the case where the right-hand lower corner is the reference, simply setting Vc to zero will not produce the correct second equation for the case where Vc becomes the reference. You would have to insert the R5 element.

But, by having the R5 term present, in error, it sets up a scenario where then setting Vc=0 did produce the correct equation for the case where Vc becomes the reference node. But, this was not a correct procedure.

Setting a node variable to zero is not the way, generally, to derive nodal equations for the case where that node becomes the reference node. It is an incorrect procedure, generally, and you got a correct equation on the right hand side because you applied an incorrect procedure to an incorrect equation (second equation, left hand side) and compensating errors gave you a correct result for the case where the reference node is now the former Vc node.

 

[Laplace]

What is a phantom reference?

A phantom reference is a sop for anyone who absolutely believes that a reference node is necessary in order to create the node equations. However, I have demonstrated that no reference node is required because every term in the node equation has the voltage difference between that node and every other adjacent node. A reference node is only necessary to express an absolute voltage but not necessary to express a voltage difference. That is why the node equations could be written as though node 'C' was active, but that is not how I derived the final node equations -- for that I made node 'C' the ground reference and wrote the node 'A' & node 'B' equations directly. Just thought it was interesting that the same result could be obtained from the phantom equations by setting Vc=0.

I was not claiming that it is correct procedure to proceed with no reference node, merely demonstrating that node equations can be valid even with no ground point. (Actually solving such a set of equations could be problematic though.) "Incorrect" is such a harsh word. Nevertheless, if you could apply an "incorrect" procedure to an "incorrect" equation, followed with compensating errors and produce the correct result, I'd think you were pretty clever!

 

[The Electrician]

A reference node is only necessary to express an absolute voltage but not necessary to express a voltage difference.

What is an absolute voltage? AFAIK, there's no such thing. Voltages must be measured with respect to some reference. That's why a voltmeter has two probes---one to connect to the point where you want to measure the voltage and one to connect to the reference.

I was not claiming that it is correct procedure to proceed with no reference node, merely demonstrating that node equations can be valid even with no ground point.

When you said "...how I would write the node equations for nodes A & B, first for the general case with node c active", I was led to believe that you were letting node c be a real variable (in the way textbooks usually teach it), not a dummy variable, and that you were showing how you would set up equations to solve for it.

You put Vc in your equations but then when you set it equal to zero, that's the same thing as letting node c be the reference node in the first place, so you do in fact have a "ground point". So instead of letting node c be the reference, and not appearing in the equations, you go through this two step process of taking it into account as though it were a real variable, and then setting it to zero, making it vanish from the equations. What advantage is there in this roundabout procedure?

If node c was really active (as I understand what "active" would mean in this context), then you need 3 equations. To me, saying that you are treating a node as "active" means that it's a real variable for which you are solving, but that's not what you were doing. If that were what you had been doing, then the second left hand equation was "incorrect" for that purpose. I should have realized when you didn't provide 3 equations that you meant something different when you said you were treating node c as active, than what I thought it meant.

(Actually solving such a set of equations could be problematic though.)

They're not hard to solve. The solution of your two left hand equations is:

Va = Vc - 1.394402
Vb = Vc + 6.70229

Vc remains an unknown, a dummy variable. What you're really doing here is using node c as your reference. You can call it a phantom reference, but it's still a reference.

"Incorrect" is such a harsh word. Nevertheless, if you could apply an "incorrect" procedure to an "incorrect" equation, followed with compensating errors and produce the correct result, I'd think you were pretty clever!

Only if it ALWAYS works out that way.

Anyway, I'm not sure how an OP who is struggling with these problems is helped by showing him a procedure using a "phantom" reference. It's probably best to stick with the standard methods.

 

[evol_w10lv]

Anyway, I'm not sure how an OP who is struggling with these problems is helped by showing him a procedure using a "phantom" reference. It's probably best to stick with the standard methods.

Yes, I would like to see standard methods. For now it is too complicated to use inovative solutions.

I calculated equations:
-Va*(G4+G03+G3+G2+G02) + Vc*(G2+G02) + Vb*G4 = -E3*(G03+G3) + E2*(G2+G02)
-Vc*(G5+G02+G2+G6) + Vb*G6 + Va*(G02+G2) = -E2*(G02+G2)
-Vb*(G4+G6+G01+G2) + Va*G4 + Vc*G6 = - E1*(G01+G1)


Va = 0.195342,
Vb = 1.36132,
Vc = 18.4131

Then:
I5 = 3.68262
I6 = (1.36132-18.4131)/4 = -4.262945
I2 = ((-18.4131)+0.195342+50)/(0.2+3.8) = 7.9455605
I1 = ((-1.36132)+50)/(0.2+3.8) = 12.15967
I4 = (18.4131-1.36132)/4 = 4.262945
I3 = (30-0.195342)/3.1 = 9.614405

And results are different with previous method:

I3=I03=Ia = 10.127A
I5=I1=I01=Ic = 3.7A
I6 = -Ib-Ic = -(-0.024)-3.7 = -1.676A
I4 = -Ib = -(-2.024) = 2.024A
I02 = I2= -Ib+Ia = -(-2.024)+10.127 = 12.151A

Can you explain, whats wrong? And maybe you can offer your equations, there is no more so much time for me to discuss and try to write equations correctly again and again. I guess, it won't take long time for you to write them.