NVM Solving for variables?

Thread Starter

thatsmessedup

Joined Feb 15, 2018
59
Here is the matrix you derived and its solution showing all of your variables, not just Ix. I checked the numbers in this matrix several times and I believe I'm using the same values shown on your work sheet.

View attachment 149422

To 8 digits, you got Ix = +0.37930552 A. I'm showing more digits than you did for a reason that will be explained. Notice that the sign which the matrix arithmetic delivers is positive even though you showed a result with a minus sign. Did your calculator actually return a result with a minus sign?

Each row in the matrix corresponds to one of your equations, i.e., row 1 is Eq. 1, row 2 is Eq. 2, etc.

In row 1 (equation 1) you left out the -12/20. This should have been added to the far right column.

In row 5, column 4 you have 1/22; it should be -1/22

In row 7, column 1 you have 1/20; it should be -1/20

In row 7, column 8 you have 12/20; it should be -12/20

If these changes are made this is the result:

View attachment 149423

Now the result for Ix is -0.37869988. Notice that if the magnitude of your result and of this result are rounded to 3 places, the result is .379

Apparently the errors you made sufficiently compensated for one another that your final result magnitude, rounded to 3 digits is the same as the rounded magnitude with your errors corrected. This could lead one to overlook the errors and think you had done everything correctly. The sign difference is another matter.
Thanks. I had actually put -1/20 in the bottom left as I had gotten the correct answer but a positive result and found my error in equation 7. I had just forgot to update that error in the table. I had also forgot to update (In row 7, column 8 you have 12/20; it should be -12/20) mistake in the table, but had it in the calculator. As for all the other I was unaware. It is really amazing how you can mess up so bad and not even realize. This has opend my eyes and I will attempt to self verify more often. Thanks again.
 
Take heart! It's not often that this many errors will compensate each other so well that you can't tell the error in the final result without examining more than 3 decimal places.

A technique that I think well worth doing is to solve the circuit with another method. Even though the problem required you to use nodal analysis, you don't have to tell the instructor that you also solved it another way. If you know both nodal and mesh analysis, solve the circuit both ways. Not only does this give you a check on your results, but it gives you practice in network solving.

Here's the mesh analysis solution assuming that Vy = -Ve and numbering the meshes starting at the lower left and going clockwise:

BigDep4.png

The Ix value is exactly the same as the nodal result.
 
Now that we have all of the setup equations written, the next (and most frequently overlooked step) is to go back and verify their correctness one at a time and term by term. ALL of the EE stuff related to this problem is captured by these equations -- everything from this point on is pure math. If we have made a mistake in any of these equations, the math can't catch it since we are simply solving the math associated with a different problem.
You don't have an expression relating ix to is2. Without that no solution is possible.

At this point we might notice that R1 appears in none of our setup equations. Does this make sense? If not, then NOW is the time to resolve this issue because, if we have made a mistake by not including it, then everything we do from this point on is pure wasted effort.

In this case, it DOES make sense, because it is connected to a current source and so the voltage at node a will become whatever is needed to make 3 A flow through R1 regardless of what (finite) value R1 happens to have.
Just because a resistor is connected to a current source doesn't cause that resistor to not appear in the equations. It must be connected in series with the current source as by a fortuitous coincidence it is in this circuit.
 

MrAl

Joined Jun 17, 2014
11,389
I can't begin to understand what you guys are talking about. I've solved the problem and have gotten the same answer as my teacher (ix = -0.379[A]). If there is a mistake in one of my labelings or maths please refer specifically to them in my work and I will be the wiser for it. Otherwise thank you for all your help in the matter.
Hi,

That result of -0.379 amps is very interesting because that's not what i get.

I get ix as -0.51248942 with my custom software and i get -0.51248940 in a simulator.
I got that first result and since you said your result was the same as your 'teacher' then i checked in the simulator. I got the same results in both experiments after rounding to 8 significant digits.

Someone else want to verify or disprove this?

EDIT:
I left out a resistor i see now, R8, which is zero ohms in my simulation.
So that's another exercise for ya :)

For the current example i get:
-0.37869988 amps
to 8 significant digits and
-0.378699877504352043745925
to 24 significant digits.
 
Last edited:

WBahn

Joined Mar 31, 2012
29,978
You don't have an expression relating ix to is2. Without that no solution is possible.
Yep -- I started listing all of the values for all of the sources and then decided to only list the controlling equations for the dependnent sources and screwed up. You can see that my headings don't match the expressions. I'll correct that. Thanks for catching it.

Just because a resistor is connected to a current source doesn't cause that resistor to not appear in the equations. It must be connected in series with the current source as by a fortuitous coincidence it is in this circuit.
Since I was discussing how it is connected in this circuit, I guess I assumed that the fact that I was talking about a series connection was rather obvious. But I've changed the post to make that explicit.
 
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