# thevenin equivalent problem.. need help how to solve ?

Discussion in 'Homework Help' started by swaggerz95, Dec 28, 2014.

Dec 6, 2014
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2. ### WBahn Moderator

Mar 31, 2012
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This is Homework Help, not Homework Done For You. This is YOUR homework, YOU are expected to show YOUR best effort to do YOUR homework. That is what gives us the starting point to HELP you move from where you are to where you need to get.

In this case, recognize that you actually have two problems. In the first you want to find Iu and in the second you want to find Io. Treat them as two separate problems. In each case, use the component that carries the desired current as the load and find the Thevenin equivalent for everything else.

3. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
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hi. this is what im done... im very confused at this time ... im use voltage test... sorry 4 my bad english

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,922
601
That is Mesh-Current Method.

You said that you wanted to use Thevenin Equivalent Circuit. You have not done so. Is the assignment to use Thevenin?

5. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
0
that is my problem.. i im very confused.
i refer to fundamental electric and electronic book...
like this...

6. ### WBahn Moderator

Mar 31, 2012
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You found Iu when you have a 1V test source applied and the independent sources turned off. So what is the next step in finding Rth?

You want to break most problems into three distinct pieces. The first is to clearly layout the problem with a properly annotated schematic (which you have done nicely, thank you). The second and most important part is setting up the equations. This is where ALL of the EE stuff is. Everything after that (the third piece) is just math. So put all of the EE stuff in one place so that you can check it easily because any mistakes here usually can't be caught and corrected later on because "later on" is just math and if your setup is wrong then it is merely describing some other circuit instead of the one you are really working with and the math doesn't care.

When setting up your equations you always want to verify them after you have written them down to ensure that they are right before you proceed because if they are wrong, then all of the math you do after that is wasted time and effort because the answer is virtually guaranteed to be wrong.

In your case, your very first equation (the Loop 1 initial equation) is wrong. Everything after that is wasted time and effort.

After writing down

$
5I_u \, + \, 1 \; = \; 80I_1 \, - \, 80I_3
$

You should have asked whether the two voltages on the left-hand side add as you go around the loop. Do they?

Another part of the first part is to see if you can reorganize the layout and perhaps simplify it to make the next parts easier to do. In this case there is a LOT of simplification that can be done.

Finally, you should always, always, ALWAYS track your units. Nearly all authors and instructors (most of whom have never worked in the real world and seen someone die because they couldn't be bothered to track their units) are inexcusably sloppy about units. But the simple fact is that you WILL make stupid math mistakes -- we all do -- and that most of those mistakes will screw up the units, provided they are there to be screwed up. If the units don't work out, the answer is wrong. Period. Tracking units is very probably the single most effective error detection tool available to the engineer. NEVER just tack the units that you WANT the answer to have onto the end result. Doing so, in my opinion, is criminally negligent and, at times, courts have agreed when deciding and awarding wrongful death and injury cases.

So let's look at that equation again:

$
5I_u \, + \, 1 \; = \; 80I_1 \, - \, 80I_3
$

The first term on the left is a current. The second term is just a number. You can't add a current and a number. Both terms on the right are currents. But is that really what they are? Remember that Mesh Current Analysis is nothing more than a systematic application of KVL in which you are summing up voltages. That '80' is not a number, it is supposed to be a resistance and a resistance times a current yields a voltage. Similarly, the '1' on the left hand side is your test source voltage, so it is 1V.

Also, that '5' in the first term is not just a number. We have a current controlled voltage source which means that we take a current and multiply it by a gain to get a voltage, so that gain MUST have units of voltage per current. Most authors will leave this out which, again, is really pretty inexcusable. But they do and so we have to patch up their work by assuming (something that engineering should NOT be about!) what the author meant, namely that it is 5V/A.

So your equation should have been

$
$$5\frac{V}{A}$$I_u \, + \, 1V \; = \; $$80\Omega$$I_1 \, - \, $$80\Omega$$I_3
$

Last edited: Dec 28, 2014
7. ### WBahn Moderator

Mar 31, 2012
20,075
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He's using Mesh-Current in order to solve for the Thevenin resistance, which is perfectly acceptable.

8. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
0

thank u !!! im not sure the answer. then i was constract the circuit at PSPICE and got this answer at " iprobe" . is it correct ?

9. ### WBahn Moderator

Mar 31, 2012
20,075
5,666
Well, one of them is probably wrong.

I'm not familiar with that current meter part, but I assume that the value displayed is either the current coming into the top of the meter (the end with the scale arc) or the current coming into the bottom of the meter (I'm guessing the top). In either case, one of your meters is not inserted into the circuit to give the correct polarity of the desired current.

Other than that I think you have the schematic done correctly.

Remember that, at the end of the day, this is just a circuit problem in which you are trying to find two particular currents. The fact that what you submit for grading has to be based on finding a Thevenin equivalent in no way prevents you from forgetting about Thevenin and solving for the currents using any method you know of. So start with the circuit and analyze it and get the answers. Then you have something to compare the results that you get using Thevenin equivalents to.

10. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
0
im confused with this problem... can you explain to me?

11. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
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oppsss.. im wrong with figure number one. it must flow at the cross wire, not trough the resistor

12. ### WBahn Moderator

Mar 31, 2012
20,075
5,666
I'm not sure how the first figure (the one on the left) relates to this problem -- perhaps it doesn't. But your correction is correct in that the current will flow through the wire and not the resistor in parallel with it. But your arrows seem to imply that all of the current at the tail of the left-most arrow will eventually flow out of the head of the right-most arrow. That isn't the case, even if you adjust the path to move the current through the wire short instead of the resistor, because it can branch at the current source.

As for the second picture, if you redraw the circuit slightly you will probably see exactly what the implication of the circled wire is.

Do you see how nothing has changed as a result of the tiny edit? But now it is clear how the four resistors on the top can be combined into a single equivalent resistor.

13. ### WBahn Moderator

Mar 31, 2012
20,075
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Once you show your work all of the way to getting Rth, I'll post a way to get it that is downright trivial. It all has to do with simplifying the circuit before you start.

14. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
0
the Rth is 40 ohm ? that right ? 1v/0.025A = 40 ohm

15. ### WBahn Moderator

Mar 31, 2012
20,075
5,666
That doesn't count as "showing your work".

16. ### swaggerz95 Thread Starter New Member

Dec 6, 2014
17
0
im done!!! thank u

17. ### WBahn Moderator

Mar 31, 2012
20,075
5,666
Attached is my solution for the first part. It's taken a few days for me to get a good enough internet connection to upload the files successfully.

Some things worth noting:

1) Be willing to redraw the circuit so that it is laid out in a more conventional fashion for what you are doing. Also, look for simplifications that you can make to reduce the number of equations you end up working with. And never forget that an adhoc approach can often be a more direct and simpler approach that blindly applying more formal methods.

2) Label your nodes and terminals on the original problem and then carry those labels to all of your subsequent drawings. That will help ensure that your circuits remain equivalent. I caught one error I had made because of this.

3) Remember that the "ground" (or "common reference") terminal is arbitrary. If it makes it easier for you to define a different node as your reference for a particular circuit, then do so.

4) Track your units. I made a mistake that the units caught immediately.

Note that, in this case, checking the answer for the first problem gives you the answer for the second problem, namely what is Io. So checking those results is trivial.

File size:
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• ###### Problem1_2.pdf
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130.3 KB
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Last edited: Jan 4, 2015
18. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,368
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If only tracking units would tip you off that you said VB = -60V + 6.18A*20Ω = 63.6V, when it's actually VA that equals 63.6V.

19. ### WBahn Moderator

Mar 31, 2012
20,075
5,666
Yep, that's a documentation typo that didn't get caught. I knew I was calculating Va and in my mind I was thinking Vb - IR and I just wrote Vb on the left of the equals sign since I was planning to use the numbers immediately. Notice that in the actual work I almost always start with a symbolic statement of the expression and plug the numbers in separately. I tend to relax the degree of formalism in the checks. Good catch. That's one of the reasons why I annotate the diagram with the voltages and currents and then use the diagram as my source of information for calculations.

Thanks for looking things over and, if that's the only criticism, then I'm thinking I did okay.

Nothing will guarantee that you will catch every mistake. Tracking units will catch most errors, but there are plenty of common errors that won't affect the units. Asking if the answer makes sense will catch most errors, but some errors are within the realm of reasonableness and other times it is nearly impossible to estimate what a reasonable answer should be (though in practice this is often easier than in homework type problems). Annotating daughter diagrams with node labels and such will prevent a lot of errors and make it possible to catch a lot of others. Even checking your answer at the end is not foolproof. Finally, there is always the possibility that you simply misunderstood the problem (and that happens in the real world, too). But having a suite of techniques that are aimed at error detection and correction is an invaluable tool that, sadly, most students are never adequately made aware of (if they are made aware of any of them at all).