Parallel voltage sources circuit

In summary: But without an actual earth, you are free to choose any point to be the zero point. You can choose A and you can choose B (but only 1 of the two! :wink:). I still don't understand. :(
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  • #2
Your equations look fine, although you've provided more equations than are strictly necessary to solve the circuit.

EDIT: Actually, check the sign of the I1 term in your second equation. In your first equation you assumed that I1 was flowing from A to B, so you must maintain that convention throughout. \EDIT

What exactly are you trying to solve for in this circuit, and are you expected to use any particular method of solution?
 
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  • #3
I see, so the second equation really should be:

Sigma = 0; 20 - I2x2 - 14 +I1x4 = 0

Yes?
 
  • #4
Yes.
 
  • #8
Good! :smile:

I'll post another problem in your thread, since the second problem there probably won't hold you up for long! :biggrin:
 
  • #9
Hehe. I was actually hoping you'd let me find the voltage difference between two points! Can you come up with two points for me?
 
  • #10
Sure! VAB?

And can you also give me PAB?
 
  • #12
Sorry, but I can't follow your reasoning here.

Anyway, you can't calculate the voltage at a point without defining a reference voltage first (usually earth).
But then, you usually have no need to know the voltage at a point.

You're only interested in the voltage difference between 2 points.
That is what you use and get from KVL.
And also what you use and get from Ohm's law (the voltage "across" a resistor based on current).
 
  • #13
What I did is calculate the voltage drop from each voltage loop UNTIL point A, then I added all the remiain voltages together (since I've discounted all the voltage drops).

So when calclating voltage difference I have to set a zero point otherwise I can't solve the thing? And just what exactly is "earth"? Is it A? Is it B?
 
  • #14
Femme_physics said:
What I did is calculate the voltage drop from each voltage loop UNTIL point A, then I added all the remiain voltages together (since I've discounted all the voltage drops).

I still don't understand. :(


Femme_physics said:
So when calclating voltage difference I have to set a zero point otherwise I can't solve the thing?

:uhh: Noooo! You don't set a zero point at all. You don't need it to calculate a voltage difference.
(You only need it to calculate the voltage at a point, and you don't need that.)

Could you perhaps look at the resistor between A and B, look up the current "through" it, and apply Ohm's law to find the voltage "across" it? :shy:


Femme_physics said:
And just what exactly is "earth"? Is it A? Is it B?

Usually a circuit in reality is grounded.
That is, some point of the circuit is literally attached to earth.
We have a nice symbol for it (actually 2 symbols that mean the same thing).

But without an actual earth, you are free to choose any point to be the zero point.
You can choose A and you can choose B (but only 1 of the two! :wink:).
 
  • #15
I still don't understand. :(

If we look at each loop individually, the same loops that I composed my equations with to solve for the current, we can see that that there "voltage drops throughout the loops, yes?

For instance, if we take the first loop of the 36V, the voltage drop of that 5 ohm resistor will be Io x 5 = 31.58

When it reaches point A the 36 voltage LOST an amont of 31.58 voltage. Therefor, the remainder is 4.42.

Same with loop 2, until I reach point A, and same with loop 3, until I reach point A. Then I add the remaining voltages together.

Noooo! You don't set a zero point at all. You don't need it to calculate a voltage difference.
(You only need it to calculate the voltage at a point, and you don't need that.)

Could you perhaps look at the resistor between A and B, look up the current "through" it, and apply Ohm's law to find the voltage "across" it?

Before that let me ask you a question, please.

I present to you case (1) and case (2)

http://img692.imageshack.us/img692/6089/case13.jpg

In case 1, the same voltage that drops through R1 also drops through R2. In case 2, it's not the case, right? Or, otherwise, not "necessarily" the case, because there are other voltage sources complicating things up, aye?
Usually a circuit in reality is grounded.
That is, some point of the circuit is literally attached to earth.

You mean as long as the circuit is not on a airplane/spaceship or anything that flies?
 
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  • #16
Femme_physics said:
Before that let me ask you a question, please.

I present to you case (1) and case (2)

In case 1, the same voltage that drops through R1 also drops through R2. In case 2, it's not the case, right? Or, otherwise, not "necessarily" the case, because there are other voltage sources complicating things up, aye?

Aye! :smile:Now let me see if I can follow your reasoning.
Femme_physics said:
If we look at each loop individually, the same loops that I composed my equations with to solve for the current, we can see that that there "voltage drops throughout the loops, yes?
Yes. :)

Femme_physics said:
For instance, if we take the first loop of the 36V, the voltage drop of that 5 ohm resistor will be Io x 5 = 31.58
Wait! Stop!
Yes, so the voltage difference "across" the 5 ohm resistor is 31.58 V.

Femme_physics said:
When it reaches point A the 36 voltage LOST an amont of 31.58 voltage. Therefor, the remainder is 4.42.
Wait! Stop!
There! You have your voltage difference between points A and B! :smile:

Femme_physics said:
Same with loop 2, until I reach point A, and same with loop 3, until I reach point A. Then I add the remaining voltages together.
Now why would you want to do that?
Did you perhaps look at another solution where they did something like that?
Actually, I remember Gneill did something like that, but he also defined his currents differently.
His method was quite nice, eliminating an equation and an unknown before he even started.
But you have to choose, you shouldn't mix them.
Femme_physics said:
You mean as long as the circuit is not on a airplane/spaceship or anything that flies?

*grins*
Did you never see the long copper wire connecting a plane to the ground? :confused:

Actually on a plane, the circuits would be connected to the frame of the plane.
That is big enough to function as an earth.
 
  • #17
Wait! Stop!
Yes, so the voltage difference "across" the 5 ohm resistor is 31.58 V.

Is it not also the "voltage drop"?

Wait! Stop!
There! You have your voltage difference between points A and B!
What?!? How?!? I didn't even get to point B!

Now why would you want to do that?
Did you perhaps look at another solution where they did something like that?

No, I don't have any solutions, just trying to reason out things on my own...apparently not working out so well :tongue:

Oh, so by "earth" you don't really mean "the ground of planet earth"? It's also the "air of planet earth"?
 
  • #18
Femme_physics said:
Is it not also the "voltage drop"?

Yes. It's the same thing.
"voltage drop" does suggest a bit that the voltage is going down.


Femme_physics said:
What?!? How?!? I didn't even get to point B!

Whaaaat? Didn't you?
I think you did! :smile:


Femme_physics said:
No, I don't have any solutions, just trying to reason out things on my own...apparently not working out so well :tongue:

Keep reasoning! At some point it will say *click*, and then you'll breathe hotly *Aaaaaaaaaaaah*. :blushing:


Femme_physics said:
Oh, so by "earth" you don't really mean "the ground of planet earth"? It's also the "air of planet earth"?

"Earth" or "ground" is a sufficiently large metal (conducting) body.
Not air, but a plane will have metal stuff attached, that can be used.

It means electrons can flow more or less freely in or out as needed.
 
  • #19
Whaaaat? Didn't you?
I think you did!
Did I? Does that mean point B is my zero point, or my "ground point", if you will?

Keep reasoning! At some point it will say *click*, and then you'll breathe hotly *Aaaaaaaaaaaah*.

hotly eh? :wink:

I will keep reasoning, then :approve:

"Earth" or "ground" is a sufficiently large metal (conducting) body.

Really? That's how "earth" is defined? I didn't know that...I wish we were told that.

It means electrons can flow more or less freely in or out as needed.

So a circuit in space can't have a "grounding point"?
 
  • #20
Femme_physics said:
Did I? Does that mean point B is my zero point, or my "ground point", if you will?

Hmm, B is still not your zero point.
But since you have a voltage "difference", it will look a bit as if B is zero I guess.
That is, you also have the voltage that A would have if B were the zero point.




Femme_physics said:
Really? That's how "earth" is defined? I didn't know that...I wish we were told that.

There! :smile:
It should be somewhere in your textbook if you have one.


Femme_physics said:
So a circuit in space can't have a "grounding point"?

Space... *looking dreamily up*

Well, we'd need to build a spaceship then, wouldn't we?
Let's put some conducting metal in!
If only to ground our artificially intelligent computer with the hot voice! :approve:
Would you be willing to lend your voice to make it happen? :blushing:
 
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  • #21
I should be somewhere in your textbook if you have one.

Yes, I agree, "YOU" really should be somewhere in my textbook, popping up as a cardboard image, saying all sort of smart things. :wink:

Space... *looking dreamily up*

Well, we'd need to build a spaceship then, wouldn't we?
Let's put some conducting metal in!
If only to ground our artificially intelligent computer with the hot voice!
Would you be willing to lend your voice to make it happen?

:D Happily, but I still don't get it...so the spaceship would be our earth?

Hmm, B is still not your zero point.
But since you have a voltage "difference", it will look a bit as if B is zero I guess.
That is, you also have the voltage that A would have if B were the zero point.

Hmm, I think I begin to understand. So it's basically 4.42V is the "remainder" voltage after the drop of Io, and that's when it hit point A. Okay, can you give me 2 different points?
 
  • #22
Femme_physics said:
Yes, I agree, "YOU" really should be somewhere in my textbook, popping up as a cardboard image, saying all sort of smart things. :wink:

Aww. I made a typo and corrected it.
But it is funny! :smile:


Femme_physics said:
:D Happily, but I still don't get it...so the spaceship would be our earth?

Yes! We would have a small planet to ourselves! :tongue2:


Femme_physics said:
Hmm, I think I begin to understand. So it's basically 4.42V is the "remainder" voltage after the drop of Io, and that's when it hit point A. Okay, can you give me 2 different points?

The point at left-top and B?

And do you also know the power dissipated between A and B?
 
  • #23
Yes! We would have a small planet to ourselves!

:biggrin:

But if it was just a circuit flowing in space, would it have a ground point at all? Can a circuit in space function?

The point at left-top and B?

Could it possibly equal 4.42+20 = 24.42 [V] ?

And do you also know the power dissipated between A and B?

Excellent question! But let me take it step by step if you will^^ Thanks
 
  • #24
Femme_physics said:
:biggrin:

But if it was just a circuit flowing in space, would it have a ground point at all? Can a circuit in space function?

No, it wouldn't really have a ground point.
It will still function, because a ground is not required.
Let's just say it will function less well.


Femme_physics said:
Could it possibly equal 4.42+20 = 24.42 [V] ?

Uhh... yes, it could! :smile:

Ok, one more to make sure!

The point left-bottom and A?
 
  • #25
No, it wouldn't really have a ground point.
It will still function, because a ground is not required.
Let's just say it will function less well.

So the "ground point" is essentially the point touching the surface from what I understand. But what if all the points in the circuits are touching the ground?

Ok, one more to make sure!

The point left-bottom and A?
Well let's see, if I set Va = 0 then the potentials difference you asked equals 20 - 5.211 x 2 = 9.578 [V]

Yes?
 
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  • #26
Femme_physics said:
So the "ground point" is essentially the point touching the surface from what I understand. But what if all the points in the circuits are touching the ground?

Remember that with "ground" we mean a metal conductor.

If all the points in the circuit touch this ground, we have shortcircuited it, and it won't do anything anymore.
A fuse will probably blow somewhere, since a fuse is usually included to protect against shortcircuiting.



Femme_physics said:
Well let's see, if I set Va = 0 then the potentials difference you asked equals 20 - 5.211 x 2 = 9.578 [V]

Yes?

Yes! I mainly wanted to check that you were aware of the sign, and that you had to subtract instead of add.
And you were! :smile:
 
  • #27
Yes! I mainly wanted to check that you were aware of the sign, and that you had to subtract instead of add.
And you were!

w00t! This is kinda fun :biggrin:

So, I guess I should find the power dissipated now. But, I don't get what it really means. Do you want me to find the power? I have a formula for power, not for "power dissipation". Which, is really funny, because THAT IS EXACTLY WHY I skipped a really easy question in my electronic test and tried tackling this kind of circuit which I didn't have enough practice on, and exactly why I got 80 instead of 95 :frown:

Remember that with "ground" we mean a metal conductor.

If all the points in the circuit touch this ground, we have shortcircuited it, and it won't do anything anymore.
A fuse will probably blow somewhere, since a fuse is usually included to protect against shortcircuiting.

So let me try and summerize.

If we put a circuit in space, it has no grounding point.
If we connect all the points in the circuit to the ground point, it will be shortcircuited and it won't do anything anymore
In reality we only ground a single point, and that's the actual point that's touching whatever metal is connected to the circuit.

Is this all true?
 
  • #28
Femme_physics said:
w00t! This is kinda fun :biggrin:

So, I guess I should find the power dissipated now. But, I don't get what it really means. Do you want me to find the power? I have a formula for power, not for "power dissipation". Which, is really funny, because THAT IS EXACTLY WHY I skipped a really easy question in my electronic test and tried tackling this kind of circuit which I didn't have enough practice on, and exactly why I got 80 instead of 95 :frown:

The formula for power is the formula for the "power dissipation".



Femme_physics said:
So let me try and summerize.

If we put a circuit in space, it has no grounding point.
If we connect all the points in the circuit to the ground point, it will be shortcircuited and it won't do anything anymore
In reality we only ground a single point, and that's the actual point that's touching whatever metal is connected to the circuit.

Is this all true?

Yes. :smile:
 
  • #29
The formula for power is the formula for the "power dissipation".

Is "power dissipation" the power equivalent to "voltage drop"?

Yes

Great! :smile:
 
  • #30
Could you perhaps look at the resistor between A and B, look up the current "through" it, and apply Ohm's law to find the voltage "across" it?
Hi PF! :smile:
I see you are working on some challenging linear circuits.
You've done a marvelous job working through to find VAB[ with one method.
Back in post #14, ILS made a suggestion (see above). This may be a good
time to go back solve for VAB using that idea and see if you get the same result. :rolleyes:
(there is a reason I recommend this as you will find later).
 
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  • #31
Femme_physics said:
Is "power dissipation" the power equivalent to "voltage drop"?

Yes.
 
  • #33
Let me see...
The voltage across the 5 ohm resistor is 31.58 V.
So the power dissipated in the 5 ohm resistor is... exactly what you wrote! :wink:

But... the 5 ohm resistor is not between A and B... :confused:

So no, you did not get it.

But the unit you deduced for the power looks very powerful indeed! :smile:
 
  • #34
I like Serena said:
Let me see...
The voltage across the 5 ohm resistor is 31.58 V.
So the power dissipated in the 5 ohm resistor is... exactly what you wrote! :wink:

But... the 5 ohm resistor is not between A and B... :confused:

So no, you did not get it.

But the unit you deduced for the power looks very powerful indeed! :smile:

ILS, perhaps you should have specified the particular component (or components, or path) for which you wanted the power dissipation. Technically, ALL of the components are between nodes A and B, as there is are paths that will take you from A to B through all of them.
 
  • #35
But the unit you deduced for the power looks very powerful indeed!
:biggrin:

But... the 5 ohm resistor is not between A and B...

Ah, I see the errors of my ways :smile:

I know what to do! *rubs hang together in an evil fashion*

Let me find Ptotal first. Then I shall have the answer, and all the smurfs will be mine!

((PS. Rtotal ends up being 6.33333 ohms ))
http://img824.imageshack.us/img824/8286/hastobeit.jpg
 
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