How to treat internal resistance in calculations ?

[Femme_physics]

But it looks as if you're quite able to solve a set of equations!
And that is a good thing, especially since this one is more complex than what you did till now for mechanics!

Thanks! Looks like it.
And I didn't expect you to confirm the answer to the wrong equations I wrote, heh. But, I did work hard at solving it so I figured I might as well write the answer :P

To set up the equation according to the voltage law as you formulated it in this thread, you would need to set it up so that the sum of all voltage differences is zero.

That is 18 should be on the other side of the equal sign (=).
And the result should then be zero.

Fair enough and I'd do that from now on, but the equations as written above are still correct in terms of their numerical values, not in term of the law-convention correspondence, right?

But I did the switchero for them:

-I₁ x 10 - I₁ x 50 - I_L x 20 -18 = 0
-I₂ x 10 - I₂ x 30 - I_L x 20 -10 = 0

Hmm,...come to think about it, I'm not sure whether 18 or 10 should be in a minus or plus... it goes from plus to minus... so according to the convention of going from plus to minus...I think I was correct in putting a minus to it. On the other hand, there's no plus in the equation, so it doesn't make sense it equals 0! So, the voltages must be plus.

So if a current goes through plus to minus it's a minus
If voltage is considered from plus to minus, it's a plus

Interesting, that.

 

[I like Serena]

But I did the switchero for them:

-I₁ x 10 - I₁ x 50 - I_L x 20 -18 = 0
-I₂ x 10 - I₂ x 30 - I_L x 20 -10 = 0

Hmm,...come to think about it, I'm not sure whether 18 or 10 should be in a minus or plus... it goes from plus to minus... so according to the convention of going from plus to minus...I think I was correct in putting a minus to it. On the other hand, there's no plus in the equation, so it doesn't make sense it equals 0! So, the voltages must be plus.

So if a current goes through plus to minus it's a minus
If voltage is considered from plus to minus, it's a plus

Interesting, that.

Well, the 18 is not from a resistor but from a battery, which is a "voltage source".
When you finish the loop across the battery, the voltage goes up from the minus pole to the plus pole, so this should be a plus.
So the equations should be:

-I₁ x 10 - I₁ x 50 - I_L x 20 + 18 = 0
-I₂ x 10 - I₂ x 30 - I_L x 20 + 10 = 0[edit] Note that in these equations you have chosen I₁ and I₂ to go up, which is a logical direction being "away" from the plus pole of the respective batteries. [/edit]

 

[I like Serena]

Well, I expected another reply from you by now, so I could continue the thread of our conversation, but it seems you're away.

So I'll continue the thread by myself.You wrote:

I1 -I2 -IL = 0

Since we should have established by now that both I1 and I2 are currents that enter the crossroad at A, they should both have a plus sign.

So the equation should be:

I1 +I2 -IL = 0

 

[Femme_physics]

Since we should have established by now that both I1 and I2 are currents that enter the crossroad at A, they should both have a plus sign.

I beg your pardon! It seems to me that I1 enters A, and it leaves as I2 and IL! No?

Well, I expected another reply from you by now, so I could continue the thread of our conversation, but it seems you're away.

I'm here, I'm not as fast as you :)

 

[I like Serena]

I'm here, I'm not as fast as you :)

I beg your pardon.
My mistake , you're still here (but soon I wont' be).
I thought it might be some time before I saw you again, and I didn't want to leave the set of equations "hanging".

I beg your pardon! It seems to me that I1 enters A, and it leaves as I2 and IL! No?

It's a matter of choice.
If you choose to let I2 "leave" A, then you must switch its sign in the second equation, since it will be going "toward" the plus pole of the 10 V battery.

 

[Femme_physics]

If you choose to let I2 "leave" A, then you must switch its sign in the second equation, since it will be going "toward" the plus pole of the 10 V battery.

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh. I get it now :) It depends on which pole of the battery it's going to! And there are 2 batteries! What a complex problem, yet I learn a lot from this one problem (thanks to you!).

Okay, I'll get cracking at solving those 3 equations.

 

[I like Serena]

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh. I get it now :) It depends on which pole of the battery it's going to! And there are 2 batteries! What a complex problem, yet I learn a lot from this one problem (thanks to you!).

Okay, I'll get cracking at solving those 3 equations.

Good luck!

I'll be going now and you won't get my undivided attention for the rest of this day.

 

[Femme_physics]

Even your divided attention is worth it :) Thank you.

 

[Femme_physics]

I "think" I got it, do I get your ILS Seal of Approval?

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

 

[I like Serena]

I "think" I got it, do I get your ILS Seal of Approval?

Yep! You get my ILSe seal of approval!
Nice work.

 

[Femme_physics]

Yep! You get my ILSe seal of approval!
Nice work.

Huzzah! ^^

Thank you :)

 

[Femme_physics]

Something is still unclear to me here. I just had to get back at it.

It's a matter of choice.
If you choose to let I2 "leave" A, then you must switch its sign in the second equation, since it will be going "toward" the plus pole of the 10 V battery.

But the battery we use for the entire circuit is the combined battery of 8, and I2 is going towards its minus. I figured we're ignoring the other volt source now since we got the combined result of 8.I look here at crossroad A again and use the currents sum law.

http://img839.imageshack.us/img839/7600/90440798.jpg

http://img219.imageshack.us/img219/5952/crossroada.jpg

Uploaded with ImageShack.us

According to this drawing, both I₂ and I_L are LEAVING, and I₁ is ENTERING the crossroad. So, I₂ and I_L must be opposite sign to I₁1. I don't understand the logic of changing the sign of I₂ if it appears to contradict Kirchhoff's law for currents. What am I missing?

 

[I like Serena]

I like it that you didn't forget previous problems, just because you might have thought you were done with it!

Something is still unclear to me here. I just had to get back at it.

But the battery we use for the entire circuit is the combined battery of 8, and I2 is going towards its minus. I figured we're ignoring the other volt source now since we got the combined result of 8.

Nope. We did not combine the batteries.
We couldn't, because they're not in series (as opposed to the other circuit).
This is exactly the reason why you *have to* use Kirchhoff's laws!

I look here at crossroad A again and use the currents sum law.

According to this drawing, both I₂ and I_L are LEAVING, and I₁ is ENTERING the crossroad. So, I₂ and I_L must be opposite sign to I₁1. I don't understand the logic of changing the sign of I₂ if it appears to contradict Kirchhoff's law for currents. What am I missing?

Your drawing is off.
You treated the problem as if I₂ is going the other way (which I intended).

Could you finish this problem (or rather extend on it beyond what is asked) and calculate all the variables, and in particular I₂?

 

[Femme_physics]

I like it that you didn't forget previous problems, just because you might have thought you were done with it!

]

Never! I want to truly understand things to the core! :)

Your drawing is off.
You treated the problem as if I2 is going the other way (which I intended).

So in this case there's a current emerging from the voltage difference of 10V [I2], and a current emerging from the voltage difference of 18V [I1]?

Could you finish this problem (or rather extend on it beyond what is asked) and calculate all the variables, and in particular I₂?

Happily (or, "at your command"! :) ), I just want to grasp the basic concept of using kirchhoff current law when there are 2 voltage sources, first!

 

[I like Serena]

Never! I want to truly understand things to the core! :)

Good!

So in this case there's a current emerging from the voltage difference of 10V [I2], and a current emerging from the voltage difference of 18V [I1]?

Yes and no. But in the set of equations we set up, yes, that is the case.
We'll get to the fine points when you solve the equations for I₂...

Happily (or, "at your command"! :) ), I just want to grasp the basic concept of using kirchhoff current law when there are 2 voltage sources, first!

I can "command" you? W00t!

So what else do you want to know?

 

[Femme_physics]

Yes and no. But in the set of equations we set up, yes, that is the case.
We'll get to the fine points when you solve the equations for I2...

Fair enough!

I used Kirchhoff's voltage law and it worked (I think) to find I2, but I tried ohm's law for the same closed loop, and ran into a bump when the voltage drop was more than the voltage! It didn't make sense! So I stopped. Is it impossible to use ohm's law here?

http://img23.imageshack.us/img23/7667/i2i2i2i2.jpg

Uploaded with ImageShack.us

I can "command" you? W00t!

Only in your superhero's costume! ;)

 

[gneill]

I'm afraid that you can't combine the voltage sources in this fashion when the circuit under consideration is no longer a simple series connection; you now have two separate loops to consider, and each source will be contributing something to both loops, not just one.

 

[I like Serena]

Fair enough!

I used Kirchhoff's voltage law and it worked (I think) to find I2, but I tried ohm's law for the same closed loop, and ran into a bump when the voltage drop was more than the voltage! It didn't make sense! So I stopped. Is it impossible to use ohm's law here?

[edit] WAIT - I have a mistake in the signs, I'll correct it and rescan ![/edit]

Ohm's law still works. It's just not enough to solve this problem!

What kind of bump did you run into?
Perhaps I can help to illuminate the intuitive sense of it.

[EDIT] Oh, I just saw your drawing. You combined the batteries again, but really, you can't do that! It will make the equations come out wrong! [/EDIT]

Only in your superhero's costume! ;)

Will you sow me a costume then?
One with the letters "HH" on it and with undies over it?

 

[Femme_physics]

I'm afraid that you can't combine the voltage sources in this fashion when the circuit under consideration is no longer a simple series connection; you now have two separate loops to consider, and each source will be contributing something to both loops, not just one.

[EDIT] Oh, I just saw your drawing. You combined the batteries again, but really, you can't do that! It will make the equations come out wrong! [/EDIT]

Well, now there's a contradiction!

In my earlier problem I also had a circuit that had a parallel component!

So how come using the combined voltage with kirchhoff's voltage law in my earlier problem was okay:http://img859.imageshack.us/img859/9899/therei2.jpg

But using the combined voltage with kirchhoff's voltage law in this problem is not okay?

Ohm's law still works. It's just not enough to solve this problem!

What kind of bump did you run into?
Perhaps I can help to illuminate the intuitive sense of it.

I've written it on the paper. I got to the point where the "voltage drop" was more than the voltage. It's on my post before this one.

 

[gneill]

Earlier, you were considering the case when the switch was open. Then, the circuit contained a single loop and all the components were in series. Thus, for purposes of determining currents, you could simply combine the voltage supplies into a single "net" supply.

When the switch is closed, another loop is created. Now both supplies will have independent influences on what happens in it, and there will be an interaction between the loops that alters the overall currents and voltages. Stand by for simultaneous equations!

 

[I like Serena]

Well, now there's a contradiction!

In my earlier problem I also had a circuit that had a parallel component!

So how come using the combined voltage with kirchhoff's voltage law in my earlier problem was okay:

But using the combined voltage with kirchhoff's voltage law in this problem is not okay?

Perhaps it would have been better if you hadn't combined the voltage sources.
You'd include the separate voltage sources in the voltage sum.
And in this case you'd see that they do add up! ;)

I've written it on the paper. I got to the point where the "voltage drop" was more than the voltage. It's on my post before this one.

Yeah, the problem is that you're using a calculated I₁ that is based on an 18 V voltage source. With the 18 V the voltage drop does not exceed the voltage source. :)

 

[jhae2.718]

To throw in my $0.02, it's probably better to just use KVL and KCL* to solve the circuits without combining the voltage sources. Of course, you get some nasty systems of equations, but a little matrix algebra makes it easier.


*I actually prefer to use \int\mathbf{E}\cdot d\mathbf{r} = 0, as it makes for an easier transition to taking a time variant magnetic flux into account.

 

[Femme_physics]

Earlier, you were considering the case when the switch was open. Then, the circuit contained a single loop and all the components were in series. Thus, for purposes of determining currents, you could simply combine the voltage supplies into a single "net" supply

When the switch is closed, another loop is created. Now both supplies will have independent influences on what happens in it, and there will be an interaction between the loops that alters the overall currents and voltages. Stand by for simultaneous equations!

A) Love your enthusiasm lol! :DB) Hmm, so I can't take the result of the open switch and apply it to the closed switch (in this case, anyway)?

Perhaps it would have been better if you hadn't combined the voltage sources.
You'd include the separate voltage sources in the voltage sum.
And in this case you'd see that they do add up! ;)

So are you telling me that whenever I have 2 voltage sources and I want to use kirchhoff's law for voltages, I mustn't combine them?

To throw in my $0.02, it's probably better to just use KVL and KCL* to solve the circuits without combining the voltage sources. Of course, you get some nasty systems of equations, but a little matrix algebra makes it easier.

Woah. Hefty.
It's not in our course to solve things that way. I would be losing contact with my classmates if my solution papers would include high-level methods!

Yeah, the problem is that you're using a calculated I1 that is based on an 18 V voltage source. With the 18 V the voltage drop does not exceed the voltage source. :)

Hmm, okay, fine, but when I'm considering the 10V source, does its current split in the crossroad to both right and let, or just right? (and if it's just right, how am I supposed to know? I learned that a current flows from positive to negative. Now the current has 2 options of how to flow, by diverging left or by diverging right. [shown in drawing]

http://img849.imageshack.us/img849/8716/theloop.jpg

 

[gneill]

Unfortunately, until you write and solve the equations you can't be certain of how much current is going to be going where. It is possible that the 10V "source" is actually sinking current (current flowing into the + terminal of the battery, not out) rather than supplying it. In the single loop circuit, what was the direction of the current? Clockwise, right? So the current then was flowing out of the 18V supply and into the 10V supply.

For this "new" problem with two loops, there are several possible approaches depending upon the concepts you've already covered in class. You've mentioned KVL and KCL, the basic "workhorses" of analysis. A pure KVL approach using what are called "loop currents" will produce two equations in two unknowns to solve. A nodal analysis employing KCL and and a bit of KVL will result in 4 equations; one for each circuit branch and one for the top node where they meet (they're not particularly tricky to solve). Either way is fairly straightforward with this circuit.

 

[I like Serena]

B) Hmm, so I can't take the result of the open switch and apply it to the closed switch (in this case, anyway)?

No. :)

So are you telling me that whenever I have 2 voltage sources and I want to use kirchhoff's law for voltages, I mustn't combine them?

It's a matter of choice.
You can combine them, but only if they are in series with no branches in between.
It's a good (double or triple) check not to combine them straight away, but only if the algebra of Kirchhoff's laws shows that you can.

Woah. Hefty.
It's not in our course to solve things that way. I would be losing contact with my classmates if my solution papers would include high-level methods!

Yeah, well, you'll have to figure that one out on your own.

It might also give you the opportunity to pull your class along in new and unprecedented directions! :)
I expect you'd get the support from your teacher if you try. ;)

Hmm, okay, fine, but when I'm considering the 10V source, does its current split in the crossroad to both right and let, or just right? (and if it's just right, how am I supposed to know? I learned that a current flows from positive to negative. Now the current has 2 options of how to flow, by diverging left or by diverging right. [shown in drawing]

You don't know yet how the current will flow until you've solved the equations.
In mechanics, it's exactly the same. You won't know which way the forces will go until you've calculated them. A negative sign means it was actually the other direction.

Think of it this way.
If you look at an electron that can move freely in the wire, it feels the electric field from both batteries. The one battery pushes it in one direction, the other battery pushes it in the other direction. The way it will go depends on which force is strongest.

Actually, there's a concept called "superposition" that treats the circuit as if there's only one battery at a time, and calculate the effective currents.
Then the same with the other battery.
And finally adding everything up (which is why it is called superposition).

It's possible you'll get this method in class.
Either way, the method with Kirchhoff's law is easier, less work, and less prone to errors.

 

[Femme_physics]

You don't know yet how the current will flow until you've solved the equations.

But, I can know for sure it will flow from plus to minus and not the other way around. That's just a core principle of electronics with respect to voltage sources and how they flow. No?

Unfortunately, until you write and solve the equations you can't be certain of how much current is going to be going where. It is possible that the 10V "source" is actually sinking current (current flowing into the + terminal of the battery, not out) rather than supplying it.

I never heard about a sinking current! How is that possible? Again, from my conclusion above, flow must go from + to -. I was never told otherwise!

In the single loop circuit, what was the direction of the current? Clockwise, right? So the current then was flowing out of the 18V supply and into the 10V supply.

Is it necessarily clockwise? What about this case->

http://img15.imageshack.us/img15/2613/thiscase.jpg

Uploaded with ImageShack.us

Looks like it would go counterclockwise to me!

You can combine them, but only if they are in series with no branches in between.

I see, so in the other exercise they were combinable. Here they're not because crossroad B in the middle separates them

http://img69.imageshack.us/img69/9650/samecross.jpg

Uploaded with ImageShack.us

Correct?



(if I missed something, I'll repost later. Too much confusion over the stuff I've written above for now)

 

[I like Serena]

But, I can know for sure it will flow from plus to minus and not the other way around. That's just a core principle of electronics with respect to voltage sources and how they flow. No?

I never heard about a sinking current! How is that possible? Again, from my conclusion above, flow must go from + to -. I was never told otherwise!

No.

It can go the other way. This means the battery is being "charged".
Think of a car, where the battery is charged, when the engine is running, by a dynamo.
The dynamo is a second voltage source in the same circuit which gives off a higher voltage.

Note that in a circuit with only one voltage source, the current will always go from + to -, which I think is what you're referring to.

Is it necessarily clockwise? What about this case->

Looks like it would go counterclockwise to me!

You're quite right.
It is not necessarily clockwise! :)

I see, so in the other exercise they were combinable. Here they're not because crossroad B in the middle separates them

Correct?

Yep!

(if I missed something, I'll repost later. Too much confusion over the stuff I've written above for now)

I see little confusion in your questions.
Seems you already understand perfectly, but are as yet not ready to claim that.

 

[gneill]

For circuits with a single voltage supply the current direction is unambiguous -- current will flow out of the + terminal, wander through the circuit, and return to the - terminal.

When you have multiple sources they can "oppose" each other and force current to enter some sources while it leaves others. Note that when we force current into a battery in this "reverse" direction, the battery is said to be "sinking" the current -- that's how a real life battery gets recharged.

The direction of current flow around a given loop (clockwise or counterclockwise) depends upon the location and polarity of the voltage or current supply in the loop. You can always redraw the loop to change the direction that you perceive -- just flip the drawing over!

In your original drawing of the circuit when the switch was open and you had the single loop with the 18V and 10V batteries (and R1 and R2), it is clear that the 18V battery is going to "overwhelm" the 10V battery and force the current to flow clockwise in the circuit as drawn. Thus the 18V battery will be "sourcing" current, while the 10V battery will be "sinking" (taking in) current.

Don't worry overmuch about this. The math takes care of the details in the current direction. The important thing to remember is that ideal voltage supplies will source or sink ANY amount of current required to maintain their designated potential difference (voltage) across them.

 

[Femme_physics]

[edit] Gneill I didn't see your reply before I posted-- will reply to it later[/edit]

Note that in a circuit with only one voltage source, the current will always go from + to -, which I think is what you're referring to.

It can go the other way. This means the battery is being "charged".

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!

Yeah, well, you'll have to figure that one out on your own.

It might also give you the opportunity to pull your class along in new and unprecedented directions! :)
I expect you'd get the support from your teacher if you try. ;)

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.

You don't know yet how the current will flow until you've solved the equations.
In mechanics, it's exactly the same. You won't know which way the forces will go until you've calculated them. A negative sign means it was actually the other direction.

I always like examples from mechanics :)

Actually, there's a concept called "superposition" that treats the circuit as if there's only one battery at a time, and calculate the effective currents.
Then the same with the other battery.
And finally adding everything up (which is why it is called superposition).

It's possible you'll get this method in class.
Either way, the method with Kirchhoff's law is easier, less work, and less prone to errors.

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*

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.

 

[Femme_physics]

Don't worry overmuch about this. The math takes care of the details in the current direction. The important thing to remember is that ideal voltage supplies will source or sink ANY amount of current required to maintain their designated potential difference (voltage) across them.

Good! Glad that's clear.

When you have multiple sources they can "oppose" each other and force current to enter some sources while it leaves others. Note that when we force current into a battery in this "reverse" direction, the battery is said to be "sinking" the current -- that's how a real life battery gets recharged.

I'll take your word for it and won't worry about it as per your next post...->

Don't worry overmuch about this. The math takes care of the details in the current direction. The important thing to remember is that ideal voltage supplies will source or sink ANY amount of current required to maintain their designated potential difference (voltage) across them.

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 [/QUOTE]

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?