I think I made a crucial mistake at the electronics test, please take a look

AI Thread Summary
The discussion revolves around a participant's struggle with an electronics test question involving voltage sources and circuit analysis. They express regret over their approach, mistakenly combining voltages instead of applying Kirchhoff's laws correctly. Participants suggest using Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) to set up equations for solving the circuit, emphasizing that multiple equations are necessary for accurate analysis. The conversation highlights the importance of practice with similar problems and understanding circuit principles, as well as the potential for learning from mistakes. Ultimately, the participant seeks to improve their skills in electronics despite feeling overwhelmed by the test.
  • #51
Femme_physics said:
I did fix the mixed current :)

Edit: That should be enough equations to solve, right?

Yep! :smile:
 
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  • #52
I got I2 = 8571 [A]

I can post my full solution of 3 eq 3 unknown but I have a feeling it's overkill?

Anyway, all I have to do is find the voltage drop after R1 and if I set Va = 0 then 9 - voltage drop at R2 will give me the answer! Right?
 
  • #53
Femme_physics said:
Are these the correct 3 eq 3 unknowns?

NO!
Your way of drawing currents at a node is wrong.
You drew I1 downward, flowing into the junction from above, but there is nothing above the junction. Keeping the directions, all currents flow downward through the resistors, therefore I1+I2+I3=0
Apply stevenb's method!
You must draw the arrow of current just beside the resistor it flows through!
An other hint to check if your result is physically possibe: divide the range of the highest emf with the lowest resistance, it is about 20 A here. 8000 A is nonsense! ehild
 
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  • #54
ehild said:
NO!

@ehild, I think you're confusing Fp here.

Her equations are right with the choices for the currents she made.

The resulting answer for the current is wrong though, but I'd have to see the calculation to find the mistake.
 
  • #55
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  • #56
I like Serena said:
@ehild, I think you're confusing Fp here.

Her equations are right with the choices for the currents she made.

The resulting answer for the current is wrong though, but I'd have to see the calculation to find the mistake.

Yes, my apologies, those arrows are OK for the bottom junction. But the figure for the currents is really confusing.


ehild
 
  • #57
ehild said:
Yes, my apologies, those arrows are OK for the bottom junction. But the figure for the currents is really confusing.


ehild

Forgot a decimal point! Forgot a decimal point!:blushing:
 
  • #58
Sorry, I misunderstood your drawing of the currents. It is all right, but I still suggest to draw the arrows in the circuit, beside the resistors and not somewhere else. See picture that I drew especially for you. :smile:
And I got different result for I2. Check the calculations. When you plug in I1=I2+I3 into eq3, the voltage is wrong.
ehild
 

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  • #59
I think I can just solve 3 eq 3 unknown using a calculator. Let's see, it shows

I1 = -12.846 [A] (why is it a minus?!?)
I2 = -12.308 [A]
I3 = -0.53846 [A]

Was that your result, ehild?
 
  • #60
No, do not complicate things with that calculator.
You substituted I1=I2+I3 into the equation 9-I1-0.5I2+10=0, didn't you? Do it again, please.
To solve an equation with a calculator, you need the constants on the RHS of the equation.

ehild
 
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  • #61
I redid it. Does I2 = 6.4615 [A] ?
 
  • #62
Femme_physics said:
I redid it. Does I2 = 6.4615 [A] ?

I get I2 = 5.714 A
(edit: made a mistake here though)

Btw, originally I though R1 was 0.8 ohms, with which you get "nice" round answers.
But I see now that you have R1 to be 0.5 ohms (are you sure?).
 
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  • #64
Femme_physics said:
I made up the figures so yes :)

So what is it this time?

The correct answer is I2 = 12.3 A

You made a mistake with the signs when you removed the parentheses. ;)
 
  • #65
If you keep I2 in fraction form, I2=160/13, you get a nice simple number for the voltage.

ehild
 
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  • #66
I like Serena said:
The correct answer is I2 = 12.3 A

You made a mistake with the signs when you removed the parentheses. ;)
If you keep I2 in fraction form, I2=160/13, you get a nice round number for the voltage.

ehild

Wait a second now, that means my calculator was correct!^^ Except the answer is in plus. You said my answers weren't correct. You mean the sign wasn't correct but the figures were, yes?
 
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  • #67
Femme_physics said:
Wait a second now, that means my calculator was correct!^^ Except the answer is in plus. You said my answers weren't correct. You mean the sign wasn't correct but the figures were, yes?

Yes. :)
 
  • #68
Yes, you calculator was correct, only you input wrong data. The sign was negative because the constants should be on the other side as the unknowns.

ehild
 
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  • #69
Yes, you calculator was correct, only you input wrong data. The sign was negative because the constants should be on the other side as the unknowns.

Aha! How curious :) Thanks. So here I'll go solving the actual problem:

, http://img32.imageshack.us/img32/1391/198dq.jpg

And...I have a feeling I'm off.
 
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  • #70
Femme_physics said:
Aha! How curious :) Thanks. So here I'll go solving the actual problem:

And...I have a feeling I'm off.

Almost. ;)

But your final Vab is not right.

How did you get Va and Vb?


Actually, what you should have done is calculate the voltage drop across the branch from a to b.
 
  • #72
Femme_physics said:
Start with Va. That's how I got Va ->

Sorry, but you can't do that.
These voltages do not add up like that.


Can you calculate the voltage drop across 1 branch from a to b?
 
  • #73
Femme_physics said:
Start with Va. That's how I got Va ->

http://img96.imageshack.us/img96/5859/bekeif.jpg

Voltages need a reference point. You are free to say that Va is 16 V, but not by using the method you used to combine voltages which has no basis at all. This Va of 16 V would be relative to some other reference point, which you don't know about yet. The value of Va relative to arbitrary points is completely irrelevant to the problem. What matters is the voltage Va relative to Vb, which we call Vab. Vab is the only quantity that matters here. So, it's best to not even introduce Va and Vb in this problem.

Remember what I posted earlier?

I1=(Vab-10V)/R1
I2=(Vab+9V)/R2
I3=(Vab-6V)/R3

and I1+I2+I3=0

Nowhere do you see Va or Vb, but you do see Vab.

You are really overcomplicating the process here. A simple substitution using the above equations would allow you to calculate Vab very simply. Then once Vab is known, the three current formulas above would give you I1, I2 and I3 instantly.

Another hint is that the numbers work out very simply here and Vab, I1, I2 and I3 are all nice integer numbers. Don't even post an answer like Vab=14.5 V or I1=3.45556. It will be wrong. :smile: (EDIT ! Scratch that, I read R1=0.8 Ohms instead of R1=0.5 Ohms)
 
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  • #75
Just take ONE path from b to a and sum the voltage rises/drops along the way.
 
  • #76
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  • #77
Femme_physics said:
Okay okay!

I got it right? o:)

Yes, that's correct if Va is your reference point and you're looking for Vba, the potential at b with respect to a.

For some reason I thought that you wanted Vab, the potential at a with respect to b. In other words, I thought you were to take b as the point of reference for potentials in the circuit. But if it's a, then fine.
 
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  • #78
Victory! But check the sign, how is Vab defined.

ehild
 
  • #79
w00t! You said "victory" that means I don't have to check or do anything more! :-p:approve:

Thank you all sooooooooooooooooooooooooooo much! o:) I wouldn't be able to get this far without the great helpers of this great forum!

*does victory dance!*EDIT: I'll read the Norton Equivalents stuff you posted later... I need a well needed break after this victory lap ;)
 
  • #80
Here's some more light reading... demonstration of three different methods to solve the problem.
 

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  • #81
Here is a fourth way. (by the way, sorry, but before I wrongly read R1 as 0.8 ohms instead of 0.5 ohms. the math works out much better with 0.8 ohms)

I don't present this 4th method with the intent of teaching it, but instead offer it as a museum piece. Before the days of computers, calculators and other such technology, and back in the stone age when we used slide rules and math tables, the signal flow graph (SFG) method was a very common tool used by circuit designers.

Very complex circuits could be solved by making a pictorial flowgraph and applying Mason's gain formula. The method automatically spit out the formulas for anything of interest in a linear system and also put the formula in a form that allowed rapid approximations, when the formulas got too unweildy. Another advantage is that feedback loops (which are critical to understand in complex systems) are readily seen in the flowgraph as internal loops.

For this relatively simple circuit, the solution for Vab is found without any intermediate steps. Simply draw a properly labeled schematic. Then draw the flowgraph by inspection of the schematic, - step by step starting at the output Vab and working back to the sources V1, V2 and V3. Then Mason's gain formula (known by heart) gives the answer directly, without algebra.
 

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  • #82
Oh, my, god, gneill! This is an incredible piece of work! I can't believe you did it. Thank you so much-- this will be cherished for sure! I wish I could give you more than just my mere thanks, but I assure you this will be used wisely. I will tomorrow go on on the business of reading it all. The posts, the attachments, and everything, to understand the other methods of solving the problem. I will definitely reply more.

EDIT: And steven, same thing said to you! I'll read everything tomorrow. You two are incredible! Everyone, really. ILS, ehild, everyone :)!

Incredibly cheered me up. I'll be doing at the end of this month term B of the test, hopefully I'll be acing it! :)
 
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  • #83
Glad to help, FP. I hope it proves useful.
 
  • #84
stevenb said:
Before the days of computers, calculators and other such technology, and back in the stone age when we used slide rules and math tables, the signal flow graph (SFG) method was a very common tool used by circuit designers.

Hi Stevenb,
I really grew up in the stone age using slide rules and math tables, but I did not hear about signal flow graphs - it looks interesting- but I do not understand how to draw such one and how the method works, can you suggest some material about it?


ehild
 
  • #85
ehild said:
Hi Stevenb,
I really grew up in the stone age using slide rules and math tables, but I did not hear about signal flow graphs - it looks interesting- but I do not understand how to draw such one and how the method works, can you suggest some material about it?


ehild

I'm actually post-stone-age myself, but not by much of a margin. I started studying engineering around 1980, so I was taught by some stone-age mentors, which I feel very fortunate about, by the way.

I was lucky to learn electronics from one of the old analog design masters, just before he retired. He was in his eighties at the time, and wearing a hearing-aid bigger than today's cell phones. I'm pretty sure he chiseled out a stone wheel or two in his time. I still have the notes from his class on analog circuit design, which I took as a senior in college. I was further blessed in grad school to be assigned twice as a TA for the lab associated with his lectures. There is no better way to learn than to try and teach others, and doing it under the piercing gaze of such a master just added to the pressure. So this was quite an education.

Anyway, Prof. H. was a firm believer that the SFG approach is the best for analog circuit design, even after the wider availabiltiy of computers. He felt it provides insight to the inner workings of feedback. Somehow he eventually convinced me of this and I've used this approach throughout my own career. The only thing I've added to the process is checking all formulas derived from Mason's gain rule, using symbolic processors like Mathematica, Maple and Maxima. Hence, I get the confidence of accuracy from the new ways, while keeping the insight of the old ways.

You are right. The SFG approach is one that most people have not heard about, for some reason. Perhaps it's because there is a pretty steep learning curve to get proficient enough to use it to it's full advantages. Hence, it's a hard sell to get people to appreciate it. There are so many ways to solve linear systems that many people think it's just another way, but few give design insight like this technique.

The original paper on this is by S. J. Mason, from which we get Mason's Gain Rule.

S. J. Mason, "Feedback Theory - Some Properties of Signal Flowgraphs", Proc. of the Inst. of Radeo Eng., 41, pp. 1144-1156; September 1953

His follow up paper is available on-line here:

http://dspace.mit.edu/bitstream/handle/1721.1/4778/RLE-TR-303-15342712.pdf

And some on-line documents are here. The second one looks really good.

http://www.ece.tufts.edu/~srout01/ee12-2008/pdfs/lecture14-ogata.pdf

http://www.ives.edu.mx/bibliodigital/Ingenierias/Libros%20Ingenieria/Automatica/the%20engineering%20handbook/Section%2016/Ch96.pdf

If you are interested, you are welcome to PM me your email, and I can send you a PDF of some pages from my electronics notebook (20 MByte file !). I wish I could post it here, but it's just too big.
 
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  • #86
Thank you! I'll send a PM.

ehild
 
  • #87
Sorry it took me a while to reply, the second term of the electronics test is only at the 12/7/11 and I got other tests before that, but I do wish to say a couple of things.

1) With respect to flowchart diagram, how useful is it? Should I really bother studying what appears to be a new complicated method to solve things that doesn't appear to be in our course material?

2) With respect to matrices in the solution. We haven't studied matrices in our math-preparation course, and I doubt we will. I'd hate to start extra math just to pick up on an "alternatve path" to solve this stuff. I am interetsted in understanding Norton Equivalents. Are matrices compulsory?

3)

Trying to understand what you did gneill at the attachement.

Plus, I don't see how you could've turned this:

http://img87.imageshack.us/img87/8149/this1b.jpg

To that!:

http://img232.imageshack.us/img232/3737/this2d.jpg Is it really the same circuit? Because I see different loops I can due in the latter circuit that I can't do in the original. I don't really see where the voltage sources are in here.
 
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  • #88
Femme_physics said:
1) With respect to flowchart diagram, how useful is it? Should I really bother studying what appears to be a new complicated method to solve things that doesn't appear to be in our course material?

I think the flowchart explanation was not intended for you, but for me. :smile:

Actually this thread contains tons of information that falls outside of the scope of what you need to learn for your finals test. :wink:
Femme_physics said:
2) With respect to matrices in the solution. We haven't studied matrices in our math-preparation course, and I doubt we will. I'd hate to start extra math just to pick up on an "alternatve path" to solve this stuff. I am interetsted in understanding Norton Equivalents. Are matrices compulsory?

No, you don't need matrices. It's an aside showing a short hand notation. You can ignore it.
Femme_physics said:
3) Is it really the same circuit? Because I see different loops I can due in the latter circuit that I can't do in the original. I don't really see where the voltage sources are in here.

It's a "Norton equivalent" circuit, but I prefer not to try to explain it.
IMHO it's really better if you learn the other 2 methods first.
 
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  • #89
Where can I find a good source of such problems/questions where I apply KLV and KLC?
 
  • #90
Femme_physics said:
Where can I find a good source of such problems/questions where I apply KLV and KLC?

Google?
For instance google "practice KVL and KCL".

Oh, and you could search in PF I think. There must be tons of similar problems here! :D


Here's a few google hits that you may like:

http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Basic/Basic4Ki.html

http://www.google.com/url?sa=t&sour...dHwx4GC-A&sig2=qig3JELuBhX3j7NIzS8j3g&cad=rja
 
  • #91
Fair enough, thanks :) I'll drop "Norton and friends" and go google some stuff to exercise the basics! Although, I'll do mechanics this month, next month electronics.
 
  • #92
Femme_physics said:
Fair enough, thanks :) I'll drop "Norton and friends" and go google some stuff to exercise the basics! Although, I'll do mechanics this month, next month electronics.

Oh, and I found another good one where you can practice KVL and KCL.
I just know you'll like it. :smile:
Here it is:
https://www.physicsforums.com/showpost.php?p=3262139&postcount=3
 
  • #93
Femme_physics said:
2) With respect to matrices in the solution. We haven't studied matrices in our math-preparation course, and I doubt we will. I'd hate to start extra math just to pick up on an "alternatve path" to solve this stuff. I am interetsted in understanding Norton Equivalents. Are matrices compulsory?

No, matrices are not compulsory. They provide a handy way to organize the calculations, and permit you to bring to bear all the techniques you may have learned in Linear Algebra, but no, they are not compulsory, merely a convenience.

3)

Trying to understand what you did gneill at the attachement.

Plus, I don't see how you could've turned this:

http://img87.imageshack.us/img87/8149/this1b.jpg

To that!:

http://img232.imageshack.us/img232/3737/this2d.jpg


Is it really the same circuit? Because I see different loops I can due in the latter circuit that I can't do in the original. I don't really see where the voltage sources are in here.

It's an equivalent circuit (or model, if you prefer). All the voltage sources in the original circuit have been transformed into current sources, so you won't find any voltage sources in the "new" circuit.

The individual voltage sources and their series resistors get transformed as in the figure in the attachment below, where the Norton current equivalent for one such voltage-resistance is shown. These two source models will behave in exactly the same fashion in any circuit in which they are inserted.

For the circuit images that you've provided, you should be able to identify the voltage/resistor and current/resistor pairs that correspond in the two diagrams.
 

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  • #94
I like Serena said:
@gneill: have you considered putting your explanation of Norton on the wikipedia page?
I think that would be a valuable addition. :)

@stevenb: have you considered putting the content of your guidebook of Kirchhoff's laws on the wikipedia page?
I think that would be a valuable addition. :)

The articles as they are now could use some improvement. o:)

Although Wiki does have many useful entries, we are working on our own https://www.physicsforums.com/library.php" of core
concepts on PF.. I don't see many EE topics covered yet. These two certainly would fit.
The Kirchoff notes that stevenb touched on, are typically taught in a 1st yr Linear
Circuits course, but I have not seen them summarized well on ref websites.
You may also compare entries on http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html" . For those who may not
have used hyperphysics before, they have excellent illustrative material.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/norton.html#c1" (be sure to scroll down their pages).
 
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  • #95
edit: made a mistake
 
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  • #96
There's another easy solution; superposition.

Solve the problem with two of the generators set to zero. Do this three times, once for each non-zero generator and add the voltages.
 
  • #97
OKay honestly there is much much easier way to do this. Use Mesh's method and you only have to do with 2 unknowns. If you are scared with simutaneous equations, use a matrix. ALso nice hair dude.

Also just for my own learning sake, do you just add the total voltage drop in the middle branch?

Say 9V + voltage drop across resistor R2?

It shuold be a negative voltage drop right?
 
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