# Thevenin's theorem.What is wrong in this project?

Discussion in 'Homework Help' started by rac1, Apr 9, 2012.

1. ### rac1 Thread Starter New Member

Apr 8, 2012
5
0
Hi
i solved this question, for p=10, R4=0.5961 ohm
i solved this circuit that you can see in the picture.
But when i find a value, and put it to the circuit after i find VR4 and IR4.
P=VR4*IR4 should be true with in the question but it doesnt.
Am i wrong at finding Rthevenin solution?
could you help me?
thanks.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I guess you are trying to find the the Thevenin equivalent "looking into" the circuit at the R4 terminals with R4 removed.

Your derivation of the Thevenin voltage is almost correct to the extent that Va would be 16V and the voltage across R5 due to the presence of the controlled source would be 40V. The equivalent voltage magnitude would be the difference rather than the sum of these two values - i.e. 24V rather than 56V. The higher potential would be at the controlled source (right hand) side of R4.

The presence of the controlled source is problematic in finding a solution for R4 values to give the desired power levels or 10W, 20W or 40W. If R4 is anything other than an open circuit, then the controlled source value and hence the Thevenin equivalent changes from what you would have eventually found. I suspect the solution may even require an iterative approach for each case. At the moment I can't see an easy method of solution.

If you have access to a circuit simulation program this might provide some insights.

I did attempt this and found that for the 10W case one could have two values for R4 to achieve this - namely 0.6275 ohms or 46.204 ohms.

Last edited: Apr 9, 2012
3. ### mlog Member

Feb 11, 2012
276
36
It looks to me like your voltage should be -24 instead of +56.

4. ### WBahn Moderator

Mar 31, 2012
23,858
7,382
I agree with mlog.

Look at your expression for the open circuit voltage in diagram #2. You have:

Voc = 5Va/2 + Va

Now look at the polarities carefully and you will see that you first DROP 5Va/2 before picking up Va.

Make that adjustment and you should be fine.

Word to the wise: track your units throughout your work - don't just tack them on at the end. Failure to do so WILL result in mistakes not getting caught and also in mistakes being made BECAUSE you didn't track units. On the other hand, carefully tracking units throughout all steps of your work will let you catch the overwhelming majority of mistakes you will invariably and inevitably make. I'll forgive any engineer for being human and making human mistakes, but I'll fire any engineer that doesn't feel it is worth their time to utilize such a simple means of catching most of those mistakes at the point at which they are made. Right now it's just points on a homework problem, but you are learning how to work with real things in the real world that affect real people. Always remember, an incompetent or careless physican costs lives one at a time, while engineers kill people in job lots.

Getting back to the problem.

I assume you have been through how to pick a load resistance to get maximum power from a real source (i.e., source with an output resistance). If so, then you know that when the load resistance is equal to the source resistance, you will deliver the most power to the load that you can (let's only deal with real resistances, since I'm guessing you haven't gotten to impedances yet). If you choose a load resistance either above or below that, then you deliver less power to the load. In the extreme, you deliver zero power to an infinite resistance because there is no current through it and zero power to a short circuit because there is no voltage across it.

Since we have already told you that your Thevenin voltage is -24V and you've calculated the effective resistance to be 13.077 ohms, you can easily determine that the maximum power you can deliver to R4 will be (assuming these numbers are correct)

$
P_{max}=\frac{(-24V)^2}{4*13.08 \Omega} = 11.0W
$

Another way of getting this is to consider that, under max power transfer conditions, Vload = Vth/2 = 12V and Rload = 13.08ohm. You get the same number.

Consider this in light of what the question is asking for. Either we have made a mistake or some parts of the question is asking for something that isn't possible. The latter can't be ruled out because the point might be for you to recognize that there isn't a solution.

It should then not be surprising that for desired powers below the maximum possible load power that there will be two load resistances, one greater and one smaller than the 13.08 ohm equivalant resistance, that will work.

A good exercise (and perhaps even worth some brownie points on the homework) is to develop the quadratic equation giving the load resistance as a function of desired power into the load and then solve it. If you express it in terms of P/Pmax, then it is actually a pretty tame expression.

Taking it even a step further, you might give some thought as to which of the two possible load resistances is preferred (given no other knowledge about what factors might be at play) and why?

Last edited: Apr 10, 2012
5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Hopefully the OP will take note of my comments with respect to the role of the dependent current source.

If one "solves" the problem using the deduced Thevenin equivalent looking into the circuit with R4 removed then the 10W power level in R4 would presumably yield two values for R4 of 24.452Ω or 6.994Ω.

It's of interest to then replug those values back into the original circuit to determine if the power values are indeed 10W for each case. It turns out not to be the case for the reasons I outlined in my earlier post. With the higher value of 24.452Ω the power in R4 would be 15.82W and with the lower value of 6.994Ω the power value would be 26.29W.

One would then conclude that whilst the revised Thevenin equivalent derivation is correct for the case of R4 being an open circuit it does not suffice for the required calculations.

Last edited: Apr 10, 2012
6. ### WBahn Moderator

Mar 31, 2012
23,858
7,382
It's been so long (over twenty years) since I've done this kind of analysis that I clearly was not properly taking into account the dependent current source, either. Something didn't seem quite right as I was looking over the OP's work, but I couldn't put my finger on it clearly enough to have the alarms go off that should have.

I wrote a little Excel simulator (only uses 7 cells) so that I can use the Goal Seek function to solve it and I verified an open circuit voltage of -14V but got a short circuit current of 4.457A yielding, nominally, an equivalent resistance of 5.384ohms. Since it is a linear circuit, it should have a Thevenin equivalent and this should be a valid way to find it.

Using these values, I verified t_n_k's values of 46.204 ohms (I got 46.139 ohms) and 0.6275 ohms (I got 0.6276 ohms) as yielding 10W in R4.

I believe that one way to get at the equivalent resistance more directly is to turn off the sources but then apply a test voltage, Vx, to the output and see what the resulting current, Ix, is with the equivalent resistance being Vx/Ix. But that is probably more work than finding Vth and Inorton.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
The exact solution to find the true Thevenin equivalent can proceed as follows.

As previously discussed the Thevenin voltage is -24V.

The equation describing the conditions with R4 a short is

$\frac{26-V_a}{5}+\frac{V_a}{4}=\frac{V_a}{8||10}=\frac{V_a}{4.4444}$

from which the value of Va with R4 shorted is found as

Va=29.714V

The short circuit current magnitude is then found as

$I_{sc}=\frac{V_a}{4}-\frac{V_a}{10}=7.4286-2.9714=4.4571A$

This gives Rth=24/4.4571=5.3846Ω

These values are the same as those found by WBahn.

8. ### rac1 Thread Starter New Member

Apr 8, 2012
5
0
firstly i want to thank you very much.
I want to show my sol'n

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9. ### WBahn Moderator

Mar 31, 2012
23,858
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Good job!

Here are a couple of comments on your work that are meant purely as constructive feedback.

First, of course, I will always rant about units. I really do believe you will be amazed at how your scores will improve if you get in the habit of tracking units throughout your work because you will stop missing points here and there, big and small, for mistakes that would otherwise go uncaught. If I were grading this submission, at this stage in the semester, you would have gotten significant points deducted due to units. Of course, by this point in the semester all of my students have long since realized how serious I am and have gotten real good at tracking their units. I seldom have to deduct any units-related points after the first three homework assignments are graded, which is good because my red ink transfer pumps are usually ready to cry, "Uncle!" by that time.

The other big thing I noticed is that you are a bit sloppy in defining polarities. Whenever you draw a voltage (such as Voc) on a diagram, always indicate which side is positive. While using a single-headed arrow (as opposed to the double-headed arrow you used) is usually interpreted as having the positive end at the arrow head, it is generally preferred to make it explicit by putting at least a '+' sign on one side (and, in which case, you can still then use the double-headed arrow). I generally use double-headed arrows and both a '+' and a '-' sign. When indicating a current, draw the single-sided arrow in the direction of positive-current flow.

10. ### rac1 Thread Starter New Member

Apr 8, 2012
5
0
I like writing the units of the values, because i know that when i miss something or decide to something is wrong, units are very useful things to find my errors easily.Also Step by step solution is very useful to see where i am wrong.
for example
-(3Vx-2Vb)+3Vb=0
everybody looks and says this; "hmm it is easy i can do it without writing step by step."
but %10 of the people makes this;
-3Vx+Vb=0 why? because we are nervous in the exams and we want to solve a question quickly.After this part, every thing is wrong and we get "0" zero points or 5 points etc.
I am aware of this.
Im sorry for my english.I hope, you understand

11. ### WBahn Moderator

Mar 31, 2012
23,858
7,382
Being careful and step by step is definitely valuable. The example you give is the kind of error that tracking units won't help. For those, the second part of my old professor's teachings can frequently, but not always, catch. He always said, "The only thing that separates you and me from the folks getting the Nobel Prize are that those people always, always, always do two things: they always track their units and they always ask themselves if the answer makes sense." While clearly more than a bit of hyperbole, I have found that advice to be perhaps the most valuable and useful professional advice I have ever received.

In looking at your work, I must say that while you like writing the units, you in fact don't track them at all. For them to be useful, you have to track them completely, thoroughly, and without exception. The units are part of the number. Imagine someone leaving off the units digit (so they use 120 instead of 123) in half of their calculations and then tack them on at the end. You would expect a very high likelihood of their making a mistake. You would tell them that they have to have that '3' in EVERY place that that value is used. The same with units. They won't catch errors unless the units are forced to go through the same algebraic steps as the numbers.

Most textbook authors make this hard because they are sloppy themselves. For instance, in this problem the controlled current source was given as Va/4. Well, that has units of voltage, not current. It should have been specified as Va/4ohms. Then the units work out. A block that takes an input voltage and produces an output current is known as a transconductance amplifier and the gain has units of conductance, namely siemens or inverse resistance (1/ohms).

If I get a chance, I'll work and post my solution to the problem as though I were a student (or one of the ways I would expect my students to work it in order to get full credit).