How to treat internal resistance in calculations ?

[I like Serena]

Really? How does it work? One voltage source "charges" the other voltage source and in turn the other voltage source decides to change direction?

I don't see how the effect of another voltage source can change the direction of another voltage source.

I can't relate to your dynamo example, I don't know the first thing about cars!

Well, a voltage source doesn't "decide" to change direction.
It gets fed electrons more forcefully than it can spew them out.

Bad example with the car then.
How about a rechargeable battery? Like the one in a mobile phone?
You plug it in, in a wall socket, and the battery in the phone gets charged by the stronger wall power.
If current would only flow from the battery outward, in the time the mobile phone would be dead and the battery would have to be replaced.

Well, students in my class are struggling enough. There's no need to bombard them with new fancy methods the way I see it! But, an interesting idea, just less practical to start being too innovative while the rest of them still have question marks popping over their heads over Kirchhoff laws.

Wait then till your teacher starts teaching it.
You'll be ahead and people will be eager to learn from you! :)

I always like examples from mechanics :)

:)

Duly noted. I actually heard the term "superposition" said by our teacher, but he didn't bother to explain that. Teachers can throw a lot of fancy terms and then tell us "oh, nevermind, you don't know that. That would be too hard for you." *chuckles*

You'll be turning the tables on them. ;)
Be careful though, their self esteem might not take well to a student knowing better than a teacher. :)

But I thought I USE Kirchhoff laws with superposition, no? That way I get more equations I think (like isolating beams on a frame), than if I look at the circuit as a whole without isolating any component.

Usage of Kirchhoff's laws implies superposition. It's all rolled into 1 equation.

It's different however from looking at a whole system, as in mechanics.
As yet we're talking about isolated circuits that have no interaction with the real world.
It's like a mechanical system without smachims, but just standing loose.

That will come though, but later.
Then you'll be applying for instance an input signal to a circuit, and the circuit might produce another output signal (this is a kind of "filter").

 

[Femme_physics]

[edit] made a post just before yours :) [/edit]

Well, a voltage source doesn't "decide" to change direction.
It gets fed electrons more forcefully than it can spew them out.

Makes sense.

Bad example with the car then.
How about a rechargeable battery? Like the one in a mobile phone?
You plug it in, in a wall socket, and the battery in the phone gets charged by the stronger wall power.
If current would only flow from the battery outward, in the time the mobile phone would be dead and the battery would have to be replaced.

So the current in the wall MUST be stronger than the current in the battery, otherwise it would be the battery "charging" the wall, and that's just stupid!

Usage of Kirchhoff's laws implies superposition. It's all rolled into 1 equation.

So I can't use Kirchhoff's laws WITHOUT superposition?

It's different however from looking at a whole system, as in mechanics.

Hmm. In mechanics when we look at the whole system we look at "external forces."


Re:

Wait then till your teacher starts teaching it.
You'll be ahead and people will be eager to learn from you! :)

Maybe it's not in our material plan? How can I know for sure? Or, if it is, how can I know if it's for this semester? I seem to be struggling with basic concepts. I rather get them straight first!

You'll be turning the tables on them. ;)
Be careful though, their self esteem might not take well to a student knowing better than a teacher. :)

Let alone, a gurrrl student!
Meh, they already know I'm a studies freak, but I doubt I'll know better than my teacher! But, thanks for the super vote of confidence! :)

 

[I like Serena]

[edit] I'll slow down my responses a little to keep everything in sync then!

I think we're "superimposing" a second thread in this thread here! [/edit]

We have appeared to have decided, in equation#2, that the current will not flow through R₁. Based on what "math" can I know that? Or is it based on a principle?

No, current will flow through R₁, also in equation #2.

In equation #2 Kirchhoff's voltage law is applied to the loop in which R₁ is not present. This means the loop is treated as an isolated system with an unknown current leaking out to R₁, but which is not relevant to the equation.

Note that in equation #1 the voltage law is applied to the loop in which R₂ is not present.

 

[Femme_physics]

[edit] I'll slow down my responses a little to keep everything in sync then! [/edit]

As you please :)

No, current will flow through R₁, also in equation #2.

Aha! Then how come we didn't include it? Or, did we make a mistake?

In equation #2 Kirchhoff's voltage law is applied to the loop in which R₁ is not present. This means the loop is treated as an isolated system with an unknown current leaking out to R₁, but which is not relevant to the equation.[

Note that in equation #1 the voltage law is applied to the loop in which R₂ is not present.
/QUOTE]

That's right. So what we did in these equations is superposition with isolation?


Superposition is just using one voltage source. Isolation, well, that's isolating a loop. If I understand correctly.

 

[gneill]

Very well, then let me worry about the math. These are the equations to get the answer of I₁ when the circuit is closed.

http://img690.imageshack.us/img690/5826/solvingi.jpg

We have appeared to have decided, in equation#2, that the current will not flow through R₁. Based on what "math" can I know that? Or is it based on a principle?

Hmm. Perhaps it's me, but I seem to recall that R1 was 50Ω and R2 30Ω. But otherwise, your methodology looks fine.

What you've done is chosen two KVL loops that cover all the components in the circuit. Good. You can choose any set of loops that you like as long as all the components in the circuit are traversed by at least one of them. It's perfectly fine that the first loop does not pass through R2 and V2; the second loop does that.

 

[Femme_physics]

Thanks gneill :)

indeed R1 is 50 ohms and R2 is 30 ohms, but there's also internal resistance of 10 which I added in series (that's why you see 60 and 40 instead of 50 and 30). I thought that makes sense, no?

 

[gneill]

Thanks gneill :)

indeed R1 is 50 ohms and R2 is 30 ohms, but there's also internal resistance of 10 which I added in series (that's why you see 60 and 40 instead of 50 and 30). I thought that makes sense, no?

Ah! Yes, that makes sense.

 

[I like Serena]

I merged your 2 replies here to get back into sync. :)

So the current in the wall MUST be stronger than the current in the battery, otherwise it would be the battery "charging" the wall, and that's just stupid!

Actually, it's not the current, it's the voltage.
And yes, if the voltage of your battery is higher than the voltage of the wall power, you will charge the wall!
(This is actually done in practice as well!)

So I can't use Kirchhoff's laws WITHOUT superposition?

Hmm, that seems to be a little black and white.
Actually you can choose. :)

If you apply Kirchhoff's laws including all voltage sources, you're using the laws INCLUDING superposition.

If you apply Kirchhoff's laws with only 1 voltage source, you're using the laws WITHOUT superposition. That way you can do the superposition afterward.

Hmm. In mechanics when we look at the whole system we look at "external forces."

Yes in electronics we look at "external power supplies" that are not part of the circuit itself.

Maybe it's not in our material plan? How can I know for sure? Or, if it is, how can I know if it's for this semester? I seem to be struggling with basic concepts. I rather get them straight first!

Don't you have books or stuff? You can look it up! Or ask your teacher.

And I think you are doing a fine job grasping the basic concepts (you're just grasping a little more at the same time).
You'll see when you're in class again.

Let alone, a gurrrl student!
Meh, they already know I'm a studies freak, but I doubt I'll know better than my teacher! But, thanks for the super vote of confidence! :)

:)

Aha! Then how come we didn't include it? Or, did we make a mistake?

It's not part of the isolated loop. No mistake.

That's right. So what we did in these equations is superposition with isolation?

Huh? I don't understand.

Superposition is just using one voltage source. Isolation, well, that's isolating a loop. If I understand correctly.

Yes, these are 2 different things.

In the isolated loop everything is still superimposed.

 

[Femme_physics]

Ah! Yes, that makes sense.

:)
Appreciate it.

Huh? I don't understand.

It's not part of the isolated loop. No mistake.

If you apply Kirchhoff's laws including all voltage sources, you're using the laws INCLUDING superposition.

If you apply Kirchhoff's laws with only 1 voltage source, you're using the laws WITHOUT superposition. That way you can do the superposition afterward.

I figured there's a 3rd way to apply Kirchhoff's law. With only 1 voltage source and WITH superposition. Or is that not allowed?

Don't you have books or stuff? You can look it up! Or ask your teacher.

We don't. We only have what we copy from the board. Our material in electronics is very specific to our needs at the moment.

And I think you are doing a fine job grasping the basic concepts (you're just grasping a little more at the same time).
You'll see when you're in class again.

Thanks :)

 

[I like Serena]

I figured there's a 3rd way to apply Kirchhoff's law. With only 1 voltage source and WITH superposition. Or is that not allowed?

Not sure what you mean.
What do you have in mind?

We don't. We only have what we copy from the board. Our material in electronics is very specific to our needs at the moment.

That's an awkward way to learn!
Can't look ahead and forces you to do things exactly in the pace the teacher sets.

 

[Femme_physics]

Not sure what you mean.
What do you have in mind?

My bad. I forgot Kirchhoff law is applied only to a single closed loop. But why did we necessarily choose the right divergence to the 10V source? Would picking the left side divergence of the closed loop using the 10V source legit as well? (the loop marked in red)
http://img6.imageshack.us/img6/7064/isitokay.jpg

That's an awkward way to learn!
Can't look ahead and forces you to do things exactly in the pace the teacher sets.

Same way in materials science. We cope with what we get, and try to solve exercises often a bit our of our league. :-/

 

[I like Serena]

My bad. I forgot Kirchhoff law is applied only to a single closed loop. But why did we necessarily choose the right divergence to the 10V source? Would picking the left side divergence of the closed loop using the 10V source legit as well? (the loop marked in red)

Random choice.
But yes, you can also choose your red loop. :)

Same way in materials science. We cope with what we get, and try to solve exercises often a bit our of our league. :-/

Then it's good that you already know a bit more about electronics!
If you get an exercise now, it won't be out of your league, but it'll be easy peasy! :)

 

[Femme_physics]

Random choice.
But yes, you can also choose your red loop. :)

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NOW it's clear :) I feel so empowered to do math and electronics now! You wanted me to find I₂ though? Well, I have 3 equations with 3 unknowns, I just plug the result of I₁ to the equations to get what I need.

#1
-60 x 0.2 -20I_L+18 = 0
I_L = 0.3 [A]


Plugging that to the last equation:
I₁+I₂-I_L=0
0.2+I₂-0.3=0
I₂=0.1 [A]
There!

 

[I like Serena]

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+29xp points
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NOW it's clear :) I feel so empowered to do math and electronics now! You wanted me to find I₂ though? Well, I have 3 equations with 3 unknowns, I just plug the result of I₁ to the equations to get what I need.

Oh yeah, I'd completely forgotten about that! :)

#1
-60 x 0.2 -20I_L+18 = 0
I_L = 0.3 [A]


Plugging that to the last equation:
I₁+I₂-I_L=0
0.2+I₂-0.3=0
I₂=0.1 [A]
There!

Very good!

However, I'm afraid I miss remembered. I thought I₂ would come out negative, meaning that the 10 V battery would be charging.
Anyway, I think that point is kind of moot now.

 

[Femme_physics]

Anyway, I think that point is kind of moot now.

It'd probably come up in other problems :)

+29xp points
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:D This is by far the longest homework help thread I ever had! Much obliged, ILS, gneill. You rock^^

Never thought I'd pass the 70 replies!

 

[I like Serena]

This is by far the longest homework help thread I ever had! Much obliged, ILS, gneill. You rock^^

Never thought I'd pass the 70 replies!

Apparently you were given a problem that you wouldn't be able to solve with the knowledge they imparted on you.

And not only did you solve the problem anyway, but you learned an entire new field of science to handle all of these problems!
I'm sorry it took us over 70 posts to teach you this entire field. :P

I suspect you will be the only one in your class that will have solved this problem, unless another student has an unfair advantage.