# Parallel voltage sources circuit

## Answers and Replies

gneill
Mentor
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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Gold Member
I see, so the second equation really should be:

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

Yes?

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Neat! I like how you carefully worked out the system of equations!
You're getting good at this! And what are I2 and I?

For reference, here's the PF thread:
https://www.physicsforums.com/showthread.php?t=395017

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Good! I'll post another problem in your thread, since the second problem there probably won't hold you up for long! Gold Member
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?

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Sure! VAB?

And can you also give me PAB?

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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).

Gold Member
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?

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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. :(

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:

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! ).

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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 [Broken]

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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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! Now let me see if I can follow your reasoning.
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. :)

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.

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! 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.

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? Actually on a plane, the circuits would be connected to the frame of the plane.
That is big enough to function as an earth.

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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"?

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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.

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

Whaaaat? Didn't you?
I think you did! 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*. 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.

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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? I will keep reasoning, then "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"?

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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.

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

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

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! Would you be willing to lend your voice to make it happen? Last edited:
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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. 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?

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Yes, I agree, "YOU" really should be somewhere in my textbook, popping up as a cardboard image, saying all sort of smart things. Aww. I made a typo and corrected it.
But it is funny! :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:

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?

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Yes! We would have a small planet to ourselves! 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

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Homework Helper 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.

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

Uhh... yes, it could! Ok, one more to make sure!

The point left-bottom and A?

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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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