Discussion in 'Homework Help' started by EE_Bob, Jul 21, 2009.

1. ### EE_Bob Thread Starter Member

Jul 21, 2009
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0
This is a fun problem given from Sadiku's Fundamentals of Electric Circuits text. The chapter covering supernodes gives a good explanation of how to use and solve circuits that would govern the use of a supernode. However, it fails to introduce how to solve a circuit that involves a supernode using more than two nodes.

This is the circuit pictured below.
I have gone about this in more than one way. My latest attempt came very close at the books given answer however was still not correct.

Using nodal analysis determine the three voltages.
Any help is greatly appreciated.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi EE_Bob,

I get i=0.4945A

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Some clues by the way ...

Keep in mind

V2 = V1-10;
V1=2 x i;

Start by writing the node equation at node 3 - you should find V3=(32/3) x i

The trick is then writing two other equations which allow you to eliminate the current in the 10V source and hence find the solution for i using the relationship for V3 found above.

4. ### Ratch New Member

Mar 20, 2007
1,068
3
EE_Bob,

Fairly straight forward. Ix = current through the 10 volt source, Iy = current through the dependent voltage source. Five equations, five unknowns.

(1/2)*V1+(V1-V3)*(1/6)-Ix = 0
(1/4)*V2+Ix+Iy = 0
(1/3)*V3-Iy+(V3-V1)*(1/6) = 0
v1 = V2+10
v3 = V2+5*V1*(1/2)

Solving we get:

V1 =3.043478261
V2 = -6.956521739
V3 = 0.6521739135
Ix = 1.920289855
Iy = -0.1811594203

Ratch

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi Ratch,

I have ...

V1 = 0.989V
V2= -9.011V
V3 = 5.275V
Ix = -.2198A (current flows contrary to voltage source driving direction)
Iy = 2.4725A

Looks like a dilemma - unless there is more than one solution!

Could be worth checking using simulation.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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This is correct, but I think you were supposed to use the supernode method.

This problem yields to the loop method directly, (without having to use a superloop), which provides a check.

7. ### Ratch New Member

Mar 20, 2007
1,068
3
The Electrician,

The OP requested a nodal analysis, but I really did it using the supernode method in that I found the currents through each of the voltage sources directly. A classic supernode solution would entail finding each of the currents separately and back substituting. I think that takes longer and is more prone to mistakes. I would be interested to observe a quicker nodal solution.

t_n_k,

The determinate of the system of linear equations is nonzero. Therefore the solution set is unique and only one of us is correct. Can you find any error in my published equations?

Ratch

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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tnk, try solving by the loop method and compare to your results, and to Ratch's results.

9. ### EE_Bob Thread Starter Member

Jul 21, 2009
16
0
Ratch does have the correct values. I appreciate all the attention and help that has been given so far. I am more interested in solving this problem using a supernode method. The book illustrates that such a problem is a supernode worthy circuit and should be solvable using supernode techniques.

Here is the page of the book that explains this. The original post shows the circuit in question and is Figure 3.14 as the book notes. Im interested in the part that a supernode can be made of more than just two nodes. My professor had given us examples with such a circuit, although I am unable to find the problem or how it was done.

In some way it must be feasible to have both the dependant and independant voltage sources within a supernode involving three nodes (V1-V3). I am unsure as how to write a KVL equation for such a supernode. I am also confused as whether to pay any attention to the top resistor in the circuit that would play in parallel with this three node supernode.

10. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
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Since both sources are so connected as to constitute only one supernode, there need only be 1 equation for that supernode.

The first row of the matrix is the supernode equation, and the next two are the constraint equations for the two sources.

$\left[ \begin{array}{3}\frac{1}{2}&\frac{1}{4} &\frac{1}{3}\\1&-1&0\\\frac{-5}{2}&-1&1\end{array}\right]*\left[ \begin{array}{4}V1\\V2\\V3\end{array}\right]=\left[ \begin{array}{4}0\\ 10\\0\end{array}\right]$

11. ### EE_Bob Thread Starter Member

Jul 21, 2009
16
0
The Electrician, thank you very much for helping me out with this problem.
I see that the supernode equation (KCL) does not involve the 6 ohm resistor. I suppose this is because it is in parallel with supernode. Does it not matter because if you were to perform KCL from either V1 to V3 or V3 to V1 the signs would cancel the current through the 6 ohm resistor to the supernode?

So to recap the method of solving a supernode equation whether it be 2 or more nodes is to write a KCL equation of all currents flowing in/out of the supernode and then writing KVL equations for each source present?

Last edited: Jul 21, 2009
12. ### Ratch New Member

Mar 20, 2007
1,068
3
The Electrician,

Bravo! Learn something every day. I notice that the northmost resistor (6 ohms) does not affect V1,V2,V3 , but it does change the current existing in the voltage sources.

Ratch

Last edited: Jul 21, 2009
13. ### steveb Senior Member

Jul 3, 2008
2,433
469
Have you guys all gone bonkers? Or, have I?

Why did you discard t_n_k's answer and accept Ratch's when it's clear that t_n_k's solution satisfies the dependent source equations Iy=5i and Ratch's does not?

Ratch's solution has V1=3.04V which means that the current i equals 1.502 Amps. Now multiply this times 5 and you should have Iy=7.5 amps! This does not match Ratch's answer Iy=-0.18 A.

t_n_k's solution has V1=0.989 V which means that the current i equals 0.4945 Amps. Now multiply this times 5 and you get 2.4725 A which is consistent with t_n_k's answer for Iy.

Are there two ways of interpreting that schematic? It makes me wonder because you all seem to agree this is right.

EDIT: I see the confusion. The dependent source is a voltage source. Yeah, I see why t_n_k did that. Stating that a voltage is 5i is very confusing because the units are not correct. It should be V= (5 ohms) *i . Looking quickly you can think it's a current source.

Last edited: Jul 21, 2009
14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi All,

Yep - Steve picked up my error. For some reason I saw it as a current source. All clear to me now.

All power to the collective wisdom of the Forum! Cheers.

15. ### Ratch New Member

Mar 20, 2007
1,068
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To the Ineffable All,

Doesn't a current source have an arrow showing assumed direction, and doesn't a voltage source have signs showing assumed polarity?

Ratch

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Exactly - I'm just dumb.

17. ### steveb Senior Member

Jul 3, 2008
2,433
469
Me too, I did the same thing.

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
The way I do it is to write the node equations as though they were not part of a supernode, then add the rows of the matrix which constitute the supernode, the sum of the rows becoming one row. For example, given V1, V2, and V3, nodes of the network, ignoring the supernode, we would have:

$\left[ \begin{array}{3}\frac{1}{2}+\frac{1}{6}&0 &\frac{-1}{6}\\0&\frac{1}{4}&0\\\frac{-1}{6}&0&\frac{1}{3}+\frac{1}{6}\end{array}\right]*\left[ \begin{array}{4}V1\\V2\\V3\end{array}\right]$

Add the rows into one row and you have:

$\left[ \begin{array}{3}\frac{1}{2}&\frac{1}{4} &\frac{1}{3}\end{array}\right]*\left[ \begin{array}{4}V1\\V2\\V3\end{array}\right]$

Add the constraints for the two voltage sources (one dependent), and the stimulus column vector, and you have the system to solve:

$\left[ \begin{array}{3}\frac{1}{2}&\frac{1}{4} &\frac{1}{3}\\1&-1&0\\\frac{-5}{2}&-1&1\end{array}\right]*\left[ \begin{array}{4}V1\\V2\\V3\end{array}\right]=\left[ \begin{array}{4}0\\ 10\\0\end{array}\right]$

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
At least you realized where I had gone wrong - I claim the dunce's hat!