Finding potential between two points with opposing voltage sources

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I see. So there really isn't a "space" between the long line and the shorter line?
 
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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! :(
 
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?
 
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?
 
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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Femme_physics said:
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?

In this method you convert each series combination of voltage source and resistor into a parallel combination of current sources and resistors. Do you know how to make this type of Norton equivalent source?

I feel like you are trying to run through the woods as a shortcut before you have even learned how to walk on the path the long way.

Shortcuts and fast methods are great, but shouldn't be used to circumvent the learning of the fundamentals.

But, by all means learn this method because it is very useful to know it.

Once you have the parallel combination of resistors and current sources, then add the currents and parallel combine the resistors. You will be left with one current source and one resistor. Then just multiply resistance times current to get the voltage.
 
I like Serena said:
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?


The Norton equivalent is really simple in this special case. Imagine you have a black box, with two terminals and you can connect different loads across them and measure the terminal voltage and the current through the load. Till the device in the box is linear, the graph I(V) is a straight line. You can get the same I(V) characteristics from a single ideal current source and an internal resistance connected across it. The current source provides current equal to the short-circuit current Is measured with zero-resistance load. The internal resistance Rs is equal to the open-circuit voltage (Vo) divided by the short-circuit current (Is), or simply the resultant resistance inside the block between the terminals. It is analogous to the Thevenin equivalent but that replaces the black box with a voltage source of emf=Uo and a series internal resistance Rs=Uo/Is.

To FP, I suggest the two-loop KVL method. She knows it.

ehild
 

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Perhaps an elementary introduction to Norton current equivalents is in order. The basic premise is that a voltage source with a series resistance is in every way equivalent to a current source with a parallel resistance. They are interchangeable in a circuit being analyzed, and in a number of circumstances can make the analysis easier.

Have a look at the first attached figure. On the left hand side is a voltage source V with a series resistance R. On the right, it's equivalent in current-source form. The value of I that should be assigned to the current source is I = V/R. That is, it is equal to the current that would flow in the voltage source circuit if its output were short-circuited; it's the maximum current that you could ever draw from the voltage source V with that series resistance.

Note that the open circuit voltage at the open terminals of the voltage source circuit is V; when no current is being drawn from the circuit, the voltage at its output is equal to that of its source, V. Similarly, for the current source version, when no current is being drawn from its terminals all the current I from the source flows through the resistor R, developing output voltage V = I*R, which is the same as that of the voltage source circuit. If the terminals of the current source circuit are shorted together, then all of the current I flows through that short circuit. Recall that I was also the maximum current that the voltage source circuit could provide.

The second figure summarizes these properties.

How does this help us to analyze your circuit? Consider the third figure where there are two battery+resistance branches in parallel. Suppose we want to find the net voltage at the output terminals. If we convert the voltage+resistance branches to their Norton current equivalents, as in the second diagram in the figure, you can see that you then have current sources and resistances in parallel. These can be combined by adding up the parallel currents and calculating the net parallel resistance. The resulting simplified equivalent circuit gives you the output voltage very easily indeed.
 

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gneill, ehild, I'll get to your explanation of "Norton current equivalents" later, but you'd first recommended me to solve it more fundamentally using KVL, which I think I got figured out in terms of theory anyway -->

To FP, I suggest the two-loop KVL method. She knows it.
Shortcuts and fast methods are great, but shouldn't be used to circumvent the learning of the fundamentals.

Is this the fundamentals?

http://img36.imageshack.us/img36/1581/v111m.jpg


http://img193.imageshack.us/img193/5264/v222.jpg

And now for the grand daddy of all, solving it!

http://img580.imageshack.us/img580/8459/vlast.jpg

And I should get right about 20V = 0!

No, wait, what??

See that doesn't make any sense!

What am I doing wrong?
 
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You have mixed signs again. Write the currents directly beside the resistors. Show the "walking direction" in both loops in the same figure. And I do not see I3 in the equations.

ehild
 
I attached a little tutorial on using and sticking to sign/direction conventions when applying the KVL rule. Adhering to conventions perhaps the most critical thing to do with circuit problems. All the theory in the world does not help if you get even one sign wrong. Keep in mind that there are a few different accepted convention systems in use, and you are free to use whatever you have been taught in your class. But, whatever you use, stick to it and apply it very carefully.

Note that you can chose any current direction or loop direction (they don't even have to match each other), but once you choose it, don't change it. If your answer ends up being negative, then it simply means the current is going in the opposite direction to the arrow, which is no problem.
 

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Thank you stevenb! Great file. I'll print it and put it in my electronics notebook. Thank you also ehild (and gneill and ILS and everyone else who posted!). When I get back home I'll look at it more in depth (hopefully I won't postpone it for tomorrow).
 
Femme_physics said:
Thank you stevenb! Great file. I'll print it and put it in my electronics notebook. Thank you also ehild (and gneill and ILS and everyone else who posted!). When I get back home I'll look at it more in depth (hopefully I won't postpone it for tomorrow).

Hey Fp! :smile:

Did you already take a look at stevenb's guidebook?
 
@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:)
 
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. :)

Thanks for the boost :smile: I think that what I wrote above about Norton equivalents is a bit superficial for inclusion in Wikipedia. My intention was merely to draw attention to one of the more practical uses of the method; sometimes "dry theory" alone leaves one wondering of what possible interest a thing could be, and I hope that I may have given some indication of how Norton (and by extension, Thevenin) equivalents can be made to perform nifty tricks on occasion.
 
I like Serena said:
Hey Fp! :smile:

Did you already take a look at stevenb's guidebook?

I did yesterday but I didn't get to solving the problem again. It's more like like a guide "page" though! But an awesome one :)I'm looking back at my equations and I see where my error is. I accidentally did -10V in my first equation whereas it should be +10V, right?
@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. :)

I agree!

And when I get to read gneill's Norton explanation I'm sure I'll say the same thing as ILS.
 
Femme_physics said:
Are these the correct 3 eq 3 unknowns?

Looks right!
 
Good morning Fp! :smile:

Femme_physics said:
I'm looking back at my equations and I see where my error is. I accidentally did -10V in my first equation whereas it should be +10V, right?

I believe so.

Edit: No, that is not all. You also have wrong signs for voltage drops across resistors.
That is the consequence of not drawing and applying the current directions correctly and consistently.
Femme_physics said:
Are these the correct 3 eq 3 unknowns?

Almost. In your first equation you have mixed up a current, but that is all.
All the loops and signs are ok! :smile:
 
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I did fix the mixed current :)

Edit: That should be enough equations to solve, right?
 
Femme_physics said:
I did fix the mixed current :)

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

Yep! :smile:
 
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?
 
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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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.
 
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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
 
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:
 
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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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?
 
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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