How to treat internal resistance in calculations ?

[Femme_physics]

Homework Statement

http://img217.imageshack.us/img217/654/circuit05.jpg Internal Resistance = 10 ohms (each)
RL = 20 ohms
R2 = 30 ohms
R1 = 50 ohmsA) When the switch is open, calculate I
B) When the switch is closed, calculate I

Homework Equations

The Attempt at a Solution

First I'll try to answer A.

Whenever I'm told a voltage source has an internal resistance, I get confused.
Where is the internal resistance located in my calculations? Exactly where the volt source is? That's my assumption here.

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

The current is 8 voltage, clockwise. (18 minus 10)
So now I just do

R_t = Ra₁ + Ra₂ + R₁ + R₂ = 100 ohms

I = 8/100
I = 0.08 [A]

Is that right?

 

[I like Serena]

I can't match your specifications of the resistances to the drawing.
Ra and R3 are not in the drawing.
RL is missing.
The internal resistances are not specified.

An internal resistance behaves as if it is inside the battery connected to the plus pole of the voltage source.
I think this is what you understood and calculated, so that is good!

You mention Ra1 and Ra2, so I'm assuming those are the internal resistances, and I'm also assuming they add up to 50 ohms.
In that case your calculation is correct!

 

[Femme_physics]

Oops! The print in my paper is really small, so I thought R1 is RL and R1 is R3. Let me correct myself

RL = 20 ohms
R2 = 30 ohms
R1 = 50 ohms

The internal resistances are not specified.

I did mention it equals 10 (Ra = 10 ohms). Sorry if I didn't mention that was the internal resistance. I thought "Ra" means internal resistance.

An internal resistance behaves as if it is inside the battery connected to the plus pole of the voltage source.
I think this is what you understood and calculated, so that is good!

Great! :)

So, can we say that if I have 3 resistors in a circuit (R1, R2, R3) and I'm told there's internal resistance, can I imagine that resistance just emerging from the +?

Such as
http://img534.imageshack.us/img534/1294/visualizehow.jpg

You mention Ra1 and Ra2, so I'm assuming those are the internal resistances, and I'm also assuming they add up to 50 ohms.
In that case your calculation is correct!

No no, R1 and R2 are resistors that are drawn in the circuit, you see them?

 

[gneill]

The internal resistance of a real battery is largely due to the resistivity of the substances involved in the chemistry making up the cells. Rather small contributions come from the internal "wiring" of the battery.

A suitable "equivalent circuit model" for a real battery consists of an ideal, resistanceless cell in series with a "lump sum" resistor representing the internal resistance. It doesn't matter what labels you put on the values for the cell voltage or resistance, so long as you understand what they represent. It also doesn't matter where you place the internal resistance, it can go at either pole of the cell. The important thing is that it be in series with the cell.

 

[I like Serena]

Oops! The print in my paper is really small, so I thought R1 is RL and R1 is R3. Let me correct myself

RL = 20 ohms
R2 = 30 ohms
R1 = 50 ohms

Ok, that's better!

I did mention it equals 10 (Ra = 10 ohms). Sorry if I didn't mention that was the internal resistance. I thought "Ra" means internal resistance.

Nope, I'm not aware of any convention that says that "Ra" means internal resistance.
If it had been named "Ri" I would have understood ("i" being an abbreviation of "internal").
Perhaps "a" would be a Hebrew abbreviation for "internal"?
Anyway, if you specify "Ra" as being the internal resistance, it becomes true. :)

Anyway, I'm assuming that both internal resistances are the same then, and that they are both 10 ohms.

So, can we say that if I have 3 resistors in a circuit (R1, R2, R3) and I'm told there's internal resistance, can I imagine that resistance just emerging from the +?

Yep!

No no, R1 and R2 are resistors that are drawn in the circuit, you see them?

Yes.
So the sum of R1, R2 and the 2 internal resistances is 100 ohm.
This means that your answer is still correct!

 

[Femme_physics]

Nope, I'm not aware of any convention that says that "Ra" means internal resistance.
If it had been named "Ri" I would have understood ("i" being an abbreviation of "internal").
Perhaps "a" would be a Hebrew abbreviation for "internal"?

No, and I would be highly objecting to the concept of using Heblish in our electronic notation! But thanks for clearing it up :)

Anyway, I'm assuming that both internal resistances are the same then, and that they are both 10 ohms.

Yes, sorry I've done a sloppy work of mentioning things, or the correct things! I did copy-paste this question from something I wrote earlier and didn't want to post because I had too many questions posted at that time (lol), so I must've forgot it still needs some proper editing, or maybe it's just a lame excuse and I was just being careless translating.

Yep!

Great!

Yes.
So the sum of R1, R2 and the 2 internal resistances is 100 ohm.
This means that your answer is still correct!

Terrific. I'll try and tackle question B soon :)

 

[I like Serena]

No, and I would be highly objecting to the concept of using Heblish in our electronic notation! But thanks for clearing it up :)

"Heblish", I had to google it and learned something new again!
Yes, let's not do Heblish in electronic notation. ;)

Yes, sorry I've done a sloppy work of mentioning things, or the correct things! I did copy-paste this question from something I wrote earlier and didn't want to post because I had too many questions posted at that time (lol), so I must've forgot it still needs some proper editing, or maybe it's just a lame excuse and I was just being careless translating.

Ah, so even you have trouble multitasking sometimes! :)

Terrific. I'll try and tackle question B soon :)

I'll be waiting!

 

[ashishsinghal]

Like any other resistance!

 

[Femme_physics]

I got my scanner now :) and I quit goofing around with my webcam (*coughs*) and am ready to solve B.

How did I fare? (attached)

Like any other resistance!

I see that now, but nice of you to participate :)

 

[I like Serena]

How did I fare? (attached)

Sorry, but you can't replace the 2 batteries like that. ;)
[edit] They'd have to be in series to do that, and they aren't. [/edit]

The way I'd do it, is to apply Kirchhoff's laws.

First make a round from the 18 V battery through R_L back to the 18 V battery (voltage law).
This will give you an equation with I₁ and I_L.

Then make a round from the 10 V battery through R_L back to the 10 V battery (voltage law).
This will give you a second equation with I₂ and I_L.

Then look at the crossing of the wires where current in must equal current out (current law).
This will give you a third equation with I₁, I₂ and I_L.

Solving 3 equations with 3 unknowns will give you the solution.

I got my scanner now :) and I quit goofing around with my webcam (*coughs*) and am ready to solve B.

[edit] With pictures like that you can keep goofing around a lot more as far as I'm concerned! [/edit]

 

[Femme_physics]

I'll work on it tomorrow. Too tired for one day to solve a 3 unknown 3 eq thing. Much appreciated :)

 

[Femme_physics]

First make a round from the 18 V battery through R_L back to the 18 V battery (voltage law).
This will give you an equation with I₁ and I_L.

Ain't that 3 unknowns in this equation?

So, I'm ignoring the parallel portion in the middle? In that case, it turns into a series circuit. Shouldn't there be only 1 I?

 

[I like Serena]

Ain't that 3 unknowns in this equation?

No, only 2 unknowns.

So, I'm ignoring the parallel portion in the middle? In that case, it turns into a series circuit. Shouldn't there be only 1 I?

Yes, we're ignoring the parallel portion in the middle for now, or rather, we're not looking at it presently. It is still there, so you can't turn it into a series circuit.
And there will still be a current going through the middle portion, we're just not looking at it right now.

The thing is, we look at a loop in the circuit, summing up all the voltage differences, which always must fit due to Kirchhoff's voltage law.

 

[Femme_physics]

With respect to number of unknowns, I'll leave it for later. Let's take it step by step...

Yes, we're ignoring the parallel portion in the middle for now, or rather, we're not looking at it presently. It is still there, so you can't turn it into a series circuit.
And there will still be a current going through the middle portion, we're just not looking at it right now.

The thing is, we look at a loop in the circuit, summing up all the voltage differences, which always must fit due to Kirchhoff's voltage law.

So,

R₁+R_L = 70 ohms
I = V/R
I = 18/70
I = 0.257 [A]

So that gives me the I of the series circuit we cut (this is like "method of section" in trusses heh, or rather method of joints)

Where are the "unknowns" again?

 

[I like Serena]

So,

R₁+R_L = 70 ohms
I = V/R
I = 18/70
I = 0.257 [A]

So that gives me the I of the series circuit we cut (this is like "method of section" in trusses heh, or rather method of joints)

Where are the "unknowns" again?

Sorry no.
R₁ and R_L are not in series, so you can't add them.

What you have is I₁ going through R_a1 and R₁, and I_L going through R_L.
The corresponding voltage differences must add up to 18 V.

The unknowns are I₁ and I_L.

And yes, actually it's a lot like mechanics, were you separate the system in subsystems, and each subsystem must be in equilibrium, yielding a couple of equations.
So here it's like the sum of moments (read: voltage differences) that must be zero.
A moment is a force times a distance.
A voltage difference is a current times a resistance.

 

[Femme_physics]

Sorry no.
R₁ and R_L are not in series, so you can't add them.

But you said we're ignoring the middle section for now, that makes them series, I figured.

 

[I like Serena]

But you said we're ignoring the middle section for now, that makes them series, I figured.

Look at it this way.
The crossing of wires is like a joint.
And current is like a force.

I we leave out the middle section, we still have a current dripping sideways out of the wire crossing.
However, for the "moment sum" this current is not relevant.

[edit] I do hope the analogies don't make it harder to understand. [/edit]

 

[I like Serena]

Perhaps I should show you how the first equation is set up.

The current I₁ goes through R_a1.
The corresponding voltage difference is I1 x 10

The current I₁ goes through R₁.
The corresponding voltage difference is I₁ x 50

The current I_L goes through R_L.
The corresponding voltage difference is I_L x 20

The battery gives a voltage difference of 18 V.

So the equation is:
18 = I₁ x 10 + I₁ x 50 + I_L x 20

 

[Femme_physics]

That makes sense, since we don't know what I_L is once I1 splits. Come to think about it, we don't know what I1 is! Okay, so we covered one equation.

By the way, is the full equation for the voltage:

I₁ x 10 + I₁ x 50 + I₂ x 30 +I₂ x 10 + I_L x 20 = Vtotal?

And yes, actually it's a lot like mechanics, were you separate the system in subsystems, and each subsystem must be in equilibrium, yielding a couple of equations.
So here it's like the sum of moments (read: voltage differences) that must be zero.
A moment is a force times a distance.
A voltage difference is a current times a resistance.

Now you're talking MY language! :D

Then make a round from the 10 V battery through RL back to the 10 V battery (voltage law).
This will give you a second equation with I2 and IL.

Vtotal (for that section) = I₂ x 10 + I₂ x 30 + I_L x 20Yes?

 

[I like Serena]

That makes sense, since we don't know what I_L is once I1 splits. Come to think about it, we don't know what I1 is! Okay, so we covered one equation.

Yep!

By the way, is the full equation for the voltage:

I₁ x 10 + I₁ x 50 + I₂ x 30 +I₂ x 10 + I_L x 20 = Vtotal?

Nope. There's no such thing as Vtotal for the entire circuit (since there are 2 batteries). ;)

Now you're talking MY language! :D

I thought it might help to make you enjoy it more and to bring it home. :D

Vtotal (for that section) = I₂ x 10 + I₂ x 30 + I_L x 20


Yes?

Yep!

But what is "Vtotal (for that section)"?

 

[Femme_physics]

But what is "Vtotal (for that section)"?

Hmm, come to think about it, it should be equal 10V! I believe.

(I'll get back to try and get the last equation and solve this darn thing soon -- they're flying airplanes over the coast as it's Israel's independence day ^^)

Thank you ever so much :)

 

[I like Serena]

Hmm, come to think about it, it should be equal 10V! I believe.

That's a difficult one... Let me think about it awhile... I think...
Yep!

(I'll get back to try and get the last equation and solve this darn thing soon -- they're flying airplanes over the coast as it's Israel's independence day ^^)

Thank you ever so much :)

Have fun watching (and listening) to the planes and see you later! :)

 

[Femme_physics]

It was kinda boring, and all my pics were blurry. I wished I stayed and practiced electronics with you^^

Well, I'm unable to get the 3rd equation since I can't grasp what you mean by

Then look at the crossing of the wires where current in must equal current out (current law).
This will give you a third equation with I₁, I₂ and I_L.

Which section of the circuit are we cutting this time? I'm also not sure what you mean by "crossing of the wires"...

 

[I like Serena]

It was kinda boring, and all my pics were blurry. I wished I stayed and practiced electronics with you^^

Yeah, I checked a couple of times if you were on, and was kind of disappointed that you weren't. ;)

Well, I'm unable to get the 3rd equation since I can't grasp what you mean by

Which section of the circuit are we cutting this time? I'm also not sure what you mean by "crossing of the wires"...

Ah, apparently my non-native English skills turn out to be a problem.
Perhaps "branching of the wires"?
This is not about resistors, this is about the point where the wires branch, and where I₁ and I₂ come in and I_L goes out.
Can you tell me again what Kirchhoff's current law states?

 

[Femme_physics]

Can you tell me again what Kirchhoff's current law states?

Are you testing me? *has a nervous breakdown and faints*

...

*wakes up with an IV needle*... Well, it states that the sum of all currents going in and out of a crossroad must equal 0, and that the algebraic sum of the voltage differences in any loop must equal zero.

This is not about resistors, this is about the point where the wires branch, and where I1 and I2 come in and IL goes out.

Hmm. That seems horrifically easy.

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

So sum of all I on "crossroad A" = I₁ -I₂ -I_L = 0
I think I got my 3 equations. Unless you stop me halfway pointing out an embarrassing error, I'm solving it!

 

[I like Serena]

Are you testing me? *has a nervous breakdown and faints*

*slap* *slap* *slap* (I find that mildly enjoyable ;)
Are you ok? :concerned:

*wakes up with an IV needle*... Well, it states that the sum of all currents going in and out of a crossroad must equal 0,

Ah good, I was getting worried.

Yes! That is the "current law".

and that the algebraic sum of the voltage differences in any loop must equal zero.

Uhh, no that is not the current law.

Oh, you meant to give both laws! Then it's ok!

Hmm. That seems horrifically easy.

So sum of all I on "crossroad A" = I₁ -I₂ -I_L = 0
I think I got my 3 equations. Unless you stop me halfway pointing out an embarrassing error, I'm solving it!

Only a matter of signs.
When you did the second equation you effectively chose a direction for the corresponding current. You need to maintain that direction (that is, sign) here.

Go cyber girl, go!

 

[Femme_physics]

*slap* *slap* *slap* (I find that mildly enjoyable ;)

The slapping of me? You're doing it hard enough! Slap me harder!

Only a matter of signs.
When you did the second equation you effectively chose a direction for the corresponding current. You need to maintain that direction (that is, sign) here.

Hmm. Wait a second. If a current flows from positive from negative and enters a resistor, is that a plus or a minus?

And, when we're doing kirchhoff law for currents, what goes in is a plus, and what goes out is a minus, yes? I figured that's the convention anyway (like in mechanics, clockwise and counterclockwise torque being minus and plus respectively)

 

[I like Serena]

The slapping of me? You're doing it hard enough! Slap me harder!

*Looking around*
Oh hello, were you all watching?

Hmm. Wait a second. If a current flows from positive to negative and enters a resistor, is that a plus or a minus?

What would be plus or minus?

I assume you mean the product of current and resistance, which is the voltage difference across the resistor.
Since the voltage goes down in the direction of the current, this would be a minus.

And, when we're doing kirchhoff law for currents, what goes in is a plus, and what goes out is a minus, yes? I figured that's the convention anyway (like in mechanics, clockwise and counterclockwise torque being minus and plus respectively)

That's a good choice. :)

 

[Femme_physics]

Btw, I solved for the 3 equations earlier with the original signs as I wrote them through the thread and got a minus:


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

This gave me
I_L = -0.3158

I assume you mean the product of current and resistance, which is the voltage difference across the resistor.
Since the voltage goes down in the direction of the current, this would be a minus.

Ah, excellent. And I love the fact you don't just answer, but you give a logical explanation. :)

So my 3 equations really ought to be


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


Right?

 

[I like Serena]

Btw, I solved for the 3 equations earlier with the original signs as I wrote them through the thread and got a minus:I₁ x 10 + I₁ x 50 + I_L x 20 = 18
I₂ x 10 + I₂ x 30 + I_L x 20 = 10
I1 -I2 -IL = 0

This gave me
I_L = -0.3158

I didn't do this set of equations myself, so I dunno.

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!

Ah, excellent. And I love the fact you don't just answer, but you give a logical explanation. :)

So my 3 equations really ought to be-I₁ x 10 - I₁ x 50 - I_L x 20 = 18
-I₂ x 10 - I₂ x 30 - I_L x 20 = 10
I1 -I2 -IL = 0Right?

Let's start with the first equation again.
I didn't expect this question and actually I set up the equation the same way you formulated the voltage law a number of threads ago.

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.

 

[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!