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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.

 

hi. this is what im done... im very confused at this time ... im use voltage test... sorry 4 my bad english

 

hi. this is what im done... im very confused at this time ... im use voltage test... sorry 4 my bad english

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?

 

that is my problem.. i im very confused.
i refer to fundamental electric and electronic book...
like this...

 

hi. this is what im done... im very confused at this time ... im use voltage test... sorry 4 my bad english

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?

Don't worry about your bad English -- we'll muddle through. I assure you that your English is much better than my [insert your native language here].

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
\)

 

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?

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

 

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?

Don't worry about your bad English -- we'll muddle through. I assure you that your English is much better than my [insert your native language here].

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 second 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 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
\)

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

. is it correct ?

 

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.

 

oohh. i see.. thanks for your advice.. and one more thing...
im confused with this problem... can you explain to me?

 

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

 

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.

 

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.

 

the Rth is 40 ohm ? that right ? 1v/0.025A = 40 ohm

 

the Rth is 40 ohm ? that right ? 1v/0.025A = 40 ohm

That doesn't count as "showing your work".

 

im done!!! thank u

 

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.

5) Ask if the answers make sense as you go. You can do things like ask about limiting cases and whether your expressions limit properly. On numerical calculations estimate what the answer should be, and preferably upper and lower bounds, before you crank the numbers. We all make mistakes in calculations, whether it's by hand or using a calculator. The first time I calculate Iu I got an answer of 9.13 A. I immediately knew it was wrong because I had mentally said that the answer is roughly 180V/30Ω which is 6A and my answer should be a bit more than that. I had also done a quick upper bound check by using 200V/25A to know that the answer had to be less than 8A. Had I not done that, then I probably would not have noticed that I had screwed up the math when cranking the numbers into my calculator (though I would have caught it eventually because I routinely check my answers by plugging them back into the problem at the end).

6) Check your work. One of the nice things about engineering is that, most of the time even in the real world, you can verify that your answers are correct. At the bottom of my second page I took the answer I got and used it to walk around the circuit finding all of the voltages and currents. My main check was to figure out how much current was flowing into the current source, treating it as an unknown current, and checking if it actually worked out to be 20A. Using the answer to work out the other voltages and currents usually doesn't take long at all. In this case, the first time I did it it didn't work out because when I set up my final circuit I failed to note that Iu is going counter clockwise and so I got an answer of +6.18A. The reason this happened was because I got sloppy and did not originally label the 'a' and 'b' terminals on that diagram. Had it not been for doing the check, I would not have been any the wiser. There is seldom any excuse for turning in a solution that you don't KNOW is correct. If nothing else, imagine how much your grades will improve if you simply catch all of your little mistakes and turn in correct solutions.

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.

 

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.

 

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.

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).

 

The unit checking is a really good point. I know this is pretty late in the game, but Mathcad is a useful program for unit checking if you have it. It recognizes many common and not so common Engineering units and will keep track of the units for you as you go through a calculation. Just a helpful tip.