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

In summary, the conversation revolved around a difficult circuit problem and the participants attempted to solve it using different methods. The problem involved finding the potential between two points, and there were discussions about combining voltage sources, using Kirchoff's current rule, and writing equations using KVL and KCL. Ultimately, it was suggested to use KCL at one node and KVL for the two loops, resulting in two equations with two unknowns. The conversation ended with one participant expressing regret for choosing such a difficult question, but also feeling motivated to improve their skills in electronics.
  • #1
Femme_physics
Gold Member
2,550
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This was my circuit, and they asked me to find the potential between a and bI had no clue how to combine the voltages, I stupidly added them...but the weird thing is that they appear to both contribute to the main voltage and counter each other!

I feel really bad now because I'm sure I made a mistake. *sighs*

I wish I had a similar circuit to practice on. :frown::frown:

Really feel down :frown:http://img405.imageshack.us/img405/4354/24804378.jpg

I've no clue...do I even need to combine them? What is it? What the hell is going on with these voltage source? Do I have multiple equations using KVL with them, or can I just use a combined voltage source?
 
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  • #2
Hi Fp! Glad to see you! :smile:

This is a pretty hard problem!
Were you going for a perfect score? :rolleyes:

This is typically a problem you solve with KVL and KCL.
(Can't combine the voltage sources, because they are neither in series nor parallel.)

This leads to 4 equations with 4 unknowns...
(Yeah, I know, such a beast! )
 
  • #3
My bad then, should've went with another question. Oh well. :(
 
  • #4
Yeah, you didn't get it. Sorry to say.

A simple way to do this is to use Kirchoff's current rule.

I1+I2+I3=0

Each current will depend only on Vab, the voltage source in that leg, and the resitance in that leg. This is because voltages in parallel are the same, so the current in each branch does not depend on the other branches (EDIT: other than through the constraint imposed by Kirchoff's current rule, that is).

Basically, you have 1 equation and 1 unknown if you do it this way. Do it out and you'll be surprised how low the voltage is.
 
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  • #5
Femme_physics said:
My bad then, should've went with another question. Oh well. :(

Any chance I can cheer you up? :shy:

For what it is worth, I don't really expect anyone in your class to have completed this problem...
 
  • #6
Basically, you have 1 equation and 1 unknown if you do it this way. Do it out and you'll be surprised how low the voltage is.

But I can't solve it with 1 eq 1 unknown right?

-------------

For what it is worth, I don't really expect anyone in your class to have completed this problem...
No one in my class had probably did it. I mistakenly chose it and now completely regret setting myself up and not going for an easier problem, but there was a weird clause in that other easier problem but I could answer for sure 4 of those 5 clauses. I just wanted to get a perfect score.

I'm an idiot. I'll have to mourn for a bit then. Thanks ILS, stevenb.
 
  • #7
TBH with you, since I have so little practice with doing KVL and building mutliple equations with multiple unknowns, good chance I'd flunk it too anyway if I did it this way. Argh!
 
  • #8
Femme_physics said:
But I can't solve it with 1 eq 1 unknown right?

Does that mean you want to know how to do it? :wink:
Femme_physics said:
I just wanted to get a perfect score

Isn't it enough to just have the highest score in class? :)
 
  • #9
Isn't it enough to just have the highest score in class? :)

Highest score doubtfully. We had 3 questions each worth 33.3 - flunking one means I got at most 66.66. If that's the highest score we're in bad shape. But I'm glad to be mistaken, get me some extra well-needed practice in electronics

Does that mean you want to know how to do it?

Well, if it's going to be long then maybe not today.. still grumpy after the test, but eventually :) with you everything is possible, that I know for sure
 
  • #10
Femme_physics said:
Highest score doubtfully. We had 3 questions each worth 33.3 - flunking one means I got at most 66.66. If that's the highest score we're in bad shape. But I'm glad to be mistaken, get me some extra well-needed practice in electronics

Well, did you write anything down anyway?
It would for instance be nice if you had written down something like KVL.
That would probably score you a few points...


Femme_physics said:
Well, if it's going to be long then maybe not today.. still grumpy after the test, but eventually :) with you everything is possible, that I know for sure

Well, I only wanted to cheer you up, making sure you know that you can do it! :smile:
I'm not really sure how long it would take.
If you are in a good mood and focused, it's short, otherwise it is a bit longer.
 
  • #11
Well the electric force is independent of path so no matter what path you take, the voltage between those two points will be the same (basically). Ok so there are three loops, 1 - the "outer" loop the "right" loop and the "left" loop. Draw currents in whatever direction you want through the wire segments. There are three wire segments so there are three currents. Then write KVL (three equations) and just plug the matrix into a calculator and then use the currents to solve for the voltage.
 
  • #12
I'll reply and look it at it tomorrow, getting late here. Thanks :)
 
  • #13
Stevenb, close but I think she may be more confused now :) The current law says simply that the current into a node is equal to the current leaving the node. There are only two nodes, A and B. Each branch does have its own current, label them I1, I2, I3. For each one you must guess the direction the current flows. It does not matter if they are wrong, but they must be consistent. Since the battery is inverted on the second node we can say that current is flowing opposite the other two. Now looking at the top node A, Current 1 and 3 are leaving while current 2 is entering. So the correct equation is I1 + I3 = I2.
 
  • #14
Three of the various ways to solve it (and all three can come in handy at times):

1. KCL at node (a) and KVL for the three branches; four equations in four unknowns (but you need only solve for Va).

2. KVL for the two loops; two equations in two unknowns (the loop currents). Solve for one of the loop currents and apply it to the appropriate branch that it flows through alone to calculate Va (the current causes a voltage drop across the branch resistor. Add that to the voltage supply in that branch).

3. Convert all the voltage supplies and their series resistances to their Norton current supply equivalents. They'll all be in parallel and will simply add (like FP desired to do with the voltage sources!). The Norton resistance is just the three resistors in parallel. You're left with a single current supply and one resistor. Voltage? Easy: Va = I*R.
 
  • #15
gneill said:
3. Convert all the voltage supplies and their series resistances to their Norton current supply equivalents. They'll all be in parallel and will simply add (like FP desired to do with the voltage sources!). The Norton resistance is just the three resistors in parallel. You're left with a single current supply and one resistor. Voltage? Easy: Va = I*R.

This one is the easiest and nicest solution, thank you, gneill!

ehild
 
  • #16
ehild said:
This one is the easiest and nicest solution, thank you, gneill!

ehild

That's why I saved it for last! Cheers.
 
  • #17
audiologies said:
Stevenb, close but I think she may be more confused now :) The current law says simply that the current into a node is equal to the current leaving the node. There are only two nodes, A and B. Each branch does have its own current, label them I1, I2, I3. For each one you must guess the direction the current flows. It does not matter if they are wrong, but they must be consistent. Since the battery is inverted on the second node we can say that current is flowing opposite the other two. Now looking at the top node A, Current 1 and 3 are leaving while current 2 is entering. So the correct equation is I1 + I3 = I2.

Well, yes. I was assuming that KCL is taught in this level class, and that a hint at the approach is enough. But, you are correct to specify it more clearly.

To clarify, I'm assuming all currents in the same direction, let's say toward the floor.

Femme_physics said:
But I can't solve it with 1 eq 1 unknown right?

Sure you can, if you set it up properly.

So I1+I2+I3=0 is one constraint. The other constraint is that Vab is the same across all branches, hence the following.

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

The rest is algebra. This is a nice simple way to solve. It uses the most basic principles you need to know (Ohms Law, Kirchoff's Voltage Law and Kirchoff's current Law). The main trick is seeing the quick way to solve, which is difficult when you are first learning the subject. As you said, having a similar practice problem ahead of time would have been helpful.
 
  • #18
@gneil: Nice summary of all the methods. :)

For myself I was (or am) gunning for method 1, and would be trying to present it as only 1 equation with 1 unknown (substituting the KVL branches in KCL), which is what stevenb suggested.

Btw, I didn't know about a Norton equivalent circuit yet. Just looked it up on wikipedia. Nice!
So I learn something new too. Thanks gneill! :smile:Edit: Ah, I just got overtaken by stevenb.
 
  • #19
Thanks for the replies. Finally awake and ready to tackle them :)

3. Convert all the voltage supplies and their series resistances to their Norton current supply equivalents. They'll all be in parallel and will simply add (like FP desired to do with the voltage sources!). The Norton resistance is just the three resistors in parallel. You're left with a single current supply and one resistor. Voltage? Easy: Va = I*R.
This one is the easiest and nicest solution, thank you, gneill!

Well, yes, thank you, but I never heard of this " Norton current supply equivalents"! It seems to be the key I was looking for but never had any practice in it. Do I just add voltage soruces in parallel the same way I do with resistors?
2. KVL for the two loops; two equations in two unknowns (the loop currents). Solve for one of the loop currents and apply it to the appropriate branch that it flows through alone to calculate Va (the current causes a voltage drop across the branch resistor. Add that to the voltage supply in that branch).

I never had to do KLF for a loop with one more voltage source. I'm not sure whether I set it in a minus or a plus.

I've made up these two circuits to see if I understand. Are my equations correct for each of them?

http://img89.imageshack.us/img89/9039/voltageeasy.jpg

http://img824.imageshack.us/img824/831/66065562.jpg
Well, did you write anything down anyway?
It would for instance be nice if you had written down something like KVL.
That would probably score you a few points...

No, I never thought he was going to give us a several unknowns problem so I was being stupidly lazy and combined them to a 25V source, again, stupidly. What can I say? I never had enough practice with those types of multiple voltage sources circuits.
If you are in a good mood and focused, it's short, otherwise it is a bit longer.

No doubt, I'm here to get the job done :approve: I was just tired and frustrated last night.

Sure you can, if you set it up properly.

So I1+I2+I3=0 is one constraint.

You call it a constraint, I call it an equation. How can one tell the difference?I'd also like to add another thing that I'm not sure is "legal". I looked at this circuit as though it was 3D. So I chose to swing its leg around and ended up with a circuit like that:

http://img848.imageshack.us/img848/3500/circuitredrawn.jpg

Is that legit?
 
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  • #20
Femme_physics said:
I never had to do KLF for a loop with one more voltage source. I'm not sure whether I set it in a minus or a plus.

I've made up these two circuits to see if I understand. Are my equations correct for each of them?

http://img824.imageshack.us/img824/831/66065562.jpg

Tbh, your equations are actually right, but I think you do misunderstand the signs and the directions. The fact that they can still be right is because you did not specify all your choices, leaving room for the choice of plus and minus signs.

You should mark the direction you choose for the current "I", by putting an arrowhead on the wire, labeled with "I". As yet you didn't do that.
And separately from that, you should indicate which direction you've chosen for your KVL-loop. You did that only in the second scan.

So I'm going to limit myself to the second scan, and I'm going to make my own choice for the direction of the current "I". I choose "I" to go up through the resistor R1.

The equation that you would get with these choices is:

ΣV=0: +V2 -V1 -IR1 = 0

That is, V2 goes "up" from the negative pole (the short one) to the positive pole.
V1 goes "down" from the positive pole to the negative one.
And since the current "I" flows in the direction of the chosen loop, the voltage must go "down" across the resistor with IR1.

As you can see I get exactly opposite signs as you have, which means the result will be the same, so your equation is not wrong, but I wonder if you understood properly what you were doing.

Femme_physics said:
No, I never thought he was going to give us a several unknowns problem so I was being stupidly lazy and combined them to a 25V source, again, stupidly. What can I say? I never had enough practice with those types of multiple voltage sources circuits.

No doubt, I'm here to get the job done :approve: I was just tired and frustrated last night.

Practice makes perfect! :smile:
Femme_physics said:
You call it a constraint, I call it an equation. How can one tell the difference?

No difference. ;)
Femme_physics said:
I'd also like to add another thing that I'm not sure is "legal". I looked at this circuit as though it was 3D. So I chose to swing its leg around and ended up with a circuit like that:

Is that legit?

Before I say anything about it, what did you think to achieve with it?
 
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  • #21
Tbh, your equations are actually right, but I think you do misunderstand the signs and the directions. The fact that they can still be right is because you did not specify all your choices, leaving room for the choice of plus and minus signs.

You should mark the direction you choose for the current I, but putting an arrowhead on the wire, labeled with "I". As yet you didn't do that.
And separately from that, you should indicate which direction you've chosen for your KVL-loop. You did that only in the second scan.

I actually did mark the loops for all of them I just didn't upload the right pic. Argh.

As you can see I get exactly opposite signs as you have, which means the result will be the same, so your equation is not wrong, but I wonder if you understood properly what you were doing.

I do understand now :) Thanks.

Before I say anything about it, what did you think to achieve with it?

What do you mean? To me it seems perfectly legit, because it really doesn't matter if you ask me at which side it corrects to the cross-section. So yea, I really do think it's legit and it's the same as the original circuit.

No difference. ;)

Then in my mind it's still considered multiple equations.

Practice makes perfect!

Wouldn't be here if I didn't adhere to that! :)
 
  • #22
Femme_physics said:
I actually did mark the loops for all of them I just didn't upload the right pic. Argh.

Then will you please also mark your chosen direction of the current "I"?
Femme_physics said:
What do you mean? To me it seems perfectly legit, because it really doesn't matter if you ask me at which side it corrects to the cross-section. So yea, I really do think it's legit and it's the same as the original circuit.

So you took a perfectly good battery, sliced it into 2 halves, and moved one half somewhere else, with the wire still trailing from it?
And then you expected everything to work the same?

Still, why would you want to do that?
In certain cases it might be legit...
Femme_physics said:
Wouldn't be here if I didn't adhere to that! :)

And I wouldn't be here if you didn't adhere to that! ;)
 
  • #23
But...but.. I already trashed the paper!...

Oh fine...

*takes the wrinkled thing out of the trash and takes a pic*

http://img402.imageshack.us/img402/1862/loopyloops.jpg

So you took a perfectly good battery, sliced it into 2 halves, and moved one half somewhere else, with the wire still trailing from it?
And then you expected everything to work the same?

Still, why would you want to do that?
In certain cases it might be legit...

Why not? What's the difference? It's as though I just lift it up in 3D world -->

http://img827.imageshack.us/img827/1231/circuitycircuits.jpg
 
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  • #24
Femme_physics said:
But...but.. I already trashed the paper!...

Oh fine...

*takes the wrinkled thing out of the trash and takes a pic*

:D

But where is the arrowhead labeled with "I" (3rd time - *sigh*)?

And what should the proper equation be?


Femme_physics said:
Why not? What's the difference? It's as though I just lift it up in 3D world -->

Oh now I get it!
Yes that is perfectly legit with a wire model! :smile:
 
  • #25
:D

But where is the arrowhead labeled with "I" (3rd time - *sigh*)?

I thought that was obvious, the current is flowing the same way I marked the loop.

And what should the proper equation be?

Proper equation? But you said that my "equations are actually right" earlier.
Oh now I get it!
Yes that is perfectly legit with a wire model!

I wasn't told anything about what "model" the problem is. I still don't get whether it's legit or not.
 
  • #26
Femme_physics said:
I thought that was obvious, the current is flowing the same way I marked the loop.

Ah, that is not obvious to me, and I'm still wondering if you understand the significance of it.

Especially in the problem that this thread is about, it's very important how you choose the direction of your currents.


Femme_physics said:
Proper equation? But you said that my "equations are actually right" earlier.

Yeah, I take that back.
Sorry, but the signs are wrong.


Femme_physics said:
I wasn't told anything about what "model" the problem is. I still don't get whether it's legit or not.

Hmm, now I'm not sure if you're pulling my leg or not. :confused:
Just to be on the safe side I'm going to explain.

Do you know what the long line and the short line of the voltage source represent?

The combination of the two lines represent a battery, and as such you cannot separate them.
The wiggly line represents a resistor, which is usually a small cilinder with 4 colored bands on it and with two wire endings.

What you have drawn in 3D is not an electronic circuit.
It's a paperclip model.

So no, it's not legit.
 
  • #27
Especially in the problem that this thread is about, it's very important how you choose the direction of your currents.

Yeah, I take that back.
Sorry, but the signs are wrong.
Oh, you mean it should be opposite to what I wrote here?


http://img51.imageshack.us/img51/6205/signswronghow.jpg


Such as:

http://img861.imageshack.us/img861/9264/tryout.jpg

?

Hmm, now I'm not sure if you're pulling my leg or not. :confused:
Just to be on the safe side I'm going to explain.

Do you know what the long line and the short line of the voltage source represent?

The combination of the two lines represent a battery, and as such you cannot separate them.
The wiggly line represents a resistor, which is usually a small cilinder with 4 colored bands on it and with two wire endings.

What you have drawn in 3D is not an electronic circuit.
It's a paperclip model.

So no, it's not legit.

The combination of the two lines represent a battery, and as such you cannot separate them.

Well I've seen circuits where the plus pole is separated then the minus pole. I thought that means that I can play around with them in 3D space...but no, eh?

http://img9.imageshack.us/img9/2613/thiscase.jpg

What you have drawn in 3D is not an electronic circuit.
It's a paperclip model.

ROFL @"paperclip model"

Well, I just thought it was possible because of that model. So what you're telling me is that I should never try to think of electronics stuff in 3D?
 
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  • #28
Femme_physics said:
Oh, you mean it should be opposite to what I wrote here?

Such as:

http://img861.imageshack.us/img861/9264/tryout.jpg

?

Yes! This is right!
Thank you! :smile:



Femme_physics said:
Well I've seen circuits where the plus pole is separated then the minus pole. I thought that means that I can play around with them in 3D space...but no, eh?

ROFL @"paperclip model"

Oh yes, you can play around with them in 3D and also in 2D.
There's actually a lot more that you can do. :)

What you cannot do, is split the symbol for a battery in 2 halves - that's not a battery any more. ;)



Femme_physics said:
http://img9.imageshack.us/img9/2613/thiscase.jpg

Well, I just thought it was possible because of that model. So what you're telling me is that I should never try to think of electronics stuff in 3D?

In this circuit there is no battery present.
Note that there is nowhere a symbol of a long thin line combined with a short fat line.

What you have is 2 loose wire ends, that would typically be wired to an external battery that is not drawn, and that is not part of the circuit.
 
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  • #29
Yes! This is right!
Thank you!

Nah-uh, thank *YOU*! o:)


Oh yes, you can play around with them in 3D and also in 2D.
There's actually a lot more that you can do. :)

What you cannot do, is split the symbol for a battery in 2 halves - that's not a battery any more. ;)

But why? I mean, you agree with me that a battery is meant to produce current, and that a current goes from plus pole to minus pole. Then if the current splits the same way in both circuits (example in drawing below) then why should it matter how I orient it? I didn't change anything when it comes to current flow, current split, voltage level, resistors, or anything. What exactly did I change by redrawing the circuit the way I did?

http://img36.imageshack.us/img36/5607/cir1.jpg

http://img834.imageshack.us/img834/4791/cir2.jpg


In this circuit there is no battery present.
Note that there is nowhere a symbol of a long thin line combined with a short fat line.

What you have is 2 loose wire ends, that would typically be wired to an external battery that is not drawn, and that is not part of the circuit.

I see. Thanks :smile:
 
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  • #30
Have you ever cut a real battery to two halves? Such ones as in the picture?:biggrin:

ehild
 

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  • #31
I see. So there really isn't a "space" between the long line and the shorter line?
 
  • #32
Femme_physics said:
I see. So there really isn't a "space" between the long line and the shorter line?

No, there isn't a space between the lines.

In reality this space is filled with murky chemical stuff. It's really dirty.
If you open up a battery, you'll have quite a mess on your table (I tried!)
And I couldn't get the battery to work anymore! :(
 
  • #33
So long story short, I can't do what I did and in reality things look different than they do in a sketch, and that doing what I did I will be ripping the battery to two halves and it wouldn't function anymore. Got it. Duly noted, checked, affirmed confirmed and assured :)

Now to the real business at hand, getting the answer. I want to choose the "combined voltage source" system. Do I just combine the voltages in parallel like I do with resistors?
 
  • #34
I take it you're choosing door number 3, the Norton equivalence circuit?

Well, I don't feel quite confident that I can explain that to you.
(It might be a reasonably simple method, but I don't find it very intuitive.)

Perhaps gneill can?
Or ehild?
 
  • #35
It depends how you cut it. Cutting across, so both electrodes remains, you still can measure potential difference. Attached is a very common cell (Leclanche).
And you can make a very nice battery from an apple and two big nails or screws made of different metals, but removing them from the apple, they will not give electricity to anything.

ehild
 

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