Parallel voltage sources circuit

AI Thread Summary
The discussion revolves around solving a circuit with parallel voltage sources and determining the currents through resistors R1, R2, and R3. Participants confirm the accuracy of equations used to analyze the circuit, emphasizing the importance of maintaining consistent current direction conventions. The conversation highlights the necessity of defining a reference point for voltage calculations, clarifying that voltage differences can be assessed without needing a specific zero point. Additionally, the concept of grounding in circuits is explored, noting that grounding can vary based on the circuit's environment, such as in space. Overall, the dialogue fosters a deeper understanding of circuit analysis and voltage relationships.
  • #51
Femme_physics said:
More like a typo! I've accidentally linked the wrong file

Yep! You found the smurf! :)

Femme_physics said:
Interesting. So you get different answers if you pick different direction for I? I didn't know that!

If you pick the direction of a force in mechanics in the other direction, don't you get a different answer too?
 
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  • #52
Yep! You found the smurf! :)

No, no! I already had the smurf, I just mispointed on him! Like, it's totally not fair that you think I had to correct myself I already fixed that typo! grrr! I wrote the answer at the second line instead of doing the equal sign and stuff...ah nevermind you wouldn't ever believe me anyway! *storms off*

*comes back*

Oh right there's an exercise to solve. But if I didn't have...if I didn't...I'd so storm off right now!:mad:


As if!

If you pick the direction of a force in mechanics in the other direction, don't you get a different answer too?

You get the same value, just with a minus or plus
 
  • #53
Hmm... would it be reasonable to ask an easier parallel voltage sources problem before I try this one? I really want more practice on basic circuits with parallel voltages first...unless you think it would be redundant to me now? Whatever you tell me :smile:
 
  • #54
I've really decided it's too complex since we're not going to have something THAT difficult on the test!
 
  • #55
I just love it when you're mad cyber grrr! :!)
*Sees you storming off*
Wait! Come back! :shy:

And do you recall Sauron who mispointed on a small one?
It cost him everything!


And yes, the difference is just a minus or a plus.
Since you usually get dramatically different results if you make a mistake with one, it's an important part of the answer.
 
  • #56
Here's one that's more like what you ask for.

GIANCOLI.ch26.p36.jpg


Determine the currents I_1, I_2, and I_3 in the figure. Assume the internal resistance of each battery is 1.0 Ohm.
I1, I2, I3 = __________ A

What is the terminal voltage of the 6.0-V battery?
Vt = _____ V
 
  • #57
That's a great one, thanks! :smile: I'll work on it now!
Sauron mispointed on a small one? I don't recalll that! Was that the book or the movie?
 
  • #58
Sauron misplaced one small ring, that was picked up by one small guy.
He never thought it would matter, especially since not even the great white wizard (Saruman) could withstand his corrupting power. :)
(Both in the books and in the movies.)
 
  • #60
I like Serena said:
Sauron misplaced one small ring, that was picked up by one small guy.
He never thought it would matter, especially since not even the great white wizard (Saruman) could withstand his corrupting power. :)
(Both in the books and in the movies.)


Oh, from some reason I thought you meant mispointed on a hobbit. nvm! Yes, I've seen the movies and read the books a million times, I thought I missed something
 
  • #61
Femme_physics said:
Is the last term correct? I took this loop and I'm just unsure about the last term

Rather than simply answering, let me ask.
What are you unsure of?
Or rather, what do you need to be sure?
 
  • #62
I'll take it as "you're right" :wink:

No, I'm not "unsure" anymore, I looked at it for a while and it makes sense now! :approve:
 
  • #63
Darn! I'll have to cloak my responses more carefully! :eek:
 
  • #65
You have a mistake in your second equation.

And I would suggest doing the math, because I think you have not practiced it enough.

Moreover, you have made the math harder for yourself because you took the entire enclosing loop instead of just the lower loop.
 
  • #66
I agree there is a mistake in the 2nd equation, but it may not be entirely mathematical.
It could be a conceptual one concerning voltage sources.

Taking the larger loop only adds one additional term versus the bottom loop.
Not a large enough difference in math to be concerned.
Either way results in 3 equations with 3 unknowns, which may be solved using
standard technique.
 
  • #67
The mistake is because I copied/pasted the first equation to the second equation to make the process of typing down the second equations easier, then I started deleting terms but I accidentally didn't delete the +12. It's just a typo type of mistake again, really
 
  • #69
Hey Fp! :smile:

I was wondering if I'd see you today.
And I can actually see you! :cool:

I see you worked the problem out carefully now.
And yes! It worked! You have the right answer! :smile:
How does it feel?

That leaves the second part of the problem...
 
  • #70
w00t! Thanks :smile:

Hmm my hair looks kinda too frizzly! :eek:


Wait before we get to the second part I want to ask a question from the test I had... I finally see I have a copy of the test so I can post it.. I think I'll start a different topic for that
 
  • #71
Well, the second part of the problem is almost trivial now.

I mainly attracted your attention to it, since in a real test, you should make sure that you do not forget to answer all the questions, just because you think you're done. :smile:
 
  • #72
I never heard the term "terminal voltage" before, though...can you tell me what it means?
 
  • #73
What do you think it means?
What could they mean?
Can you think of more than 1 interpretation?

(Often enough you'll get questions like this one where you more or less have to deduce what it is that they mean. :wink:)
 
  • #74
After a bit of googling :-p I know the answer!

Vt = 6 - I3r

:smile:

Right?
 
  • #75
I see you've used the tools at hand to "deduce" the meaning. Very good! :rolleyes:

Yes, that's the right formula.
Although on a test I hope you'll also give a numerical result. :smile:
 
  • #76
I will :smile: I'd be also very upset if during the test they wrote us something we hadn't studied like terminal voltage!
 
  • #77
Well, taking terminal voltage as an example.

Isn't it possible that they did teach it, but you missed it, forgot it, or perhaps you weren't attending when it was taught?
(Not that something like that would happen to you! )
Or perhaps the teacher making the test didn't realize that he was using a word that had not been taught.Still, if you can deduce its meaning on your own, you'll be able to answer correctly regardless of what happened!
(Much better imho than complaining that it wasn't taught! :smile:)
 
  • #78
Isn't it possible that they did teach it, but you missed it, forgot it, or perhaps you weren't attending when it was taught?

Why how dare you?!? Humphfh! Why I never!

Okay maybe I've ever'ed.

Or perhaps the teacher making the test didn't realize that he was using a word that had not been taught.

He hasn't use the word "internal voltage" in the problem! Just internal resistance, which is a whole other (SIMPLE) matter
 
  • #79
Pick two points and let me find the potential difference! :smile:
 
  • #80
GIANCOLI.ch26.p36.jpg


The 2 junctions. :smile:
 
  • #81
Vab = 12 -I2 x r -I2 x 11;
12 - 0.50842 x 1 - 0.50842 x 11

= 5.899 [V]

:smile:
 
  • #82
Right! :approve:
 
  • #84
*Protecting my eyes*

What? I can't C! :smile:

Show me yours and I'll show you mine. ;)
 
  • #85
Is this writing in codewords that I am correct? *grins*
 
  • #86
Yes, you're correct. :)

But, assuming the resistors blow up at the same amount of power, how much power must the resistors be able to withstand without blowing up?
 
  • #87
But, assuming the resistors blow up at the same amount of power, how much power must the resistors be able to withstand without blowing up?


:bugeye::bugeye::bugeye::bugeye::bugeye:

Uhhh...

Is that similar to the "max power" question? I really am lost as to how approach it.

Don't scare me before the test lol! :eek:

PS


Is the potential between A and B equals

Vab = E2 - I1 x R2 = 9 - 10.692 x 1 = -1.692 [V]
 
  • #88
Femme_physics said:
:bugeye::bugeye::bugeye::bugeye::bugeye:

Uhhh...

Is that similar to the "max power" question? I really am lost as to how approach it.

Don't scare me before the test lol! :eek:

Just kidding. :devil:
(And no, this one is much, much easier.)


Femme_physics said:
PS


Is the potential between A and B equals

Vab = E2 - I1 x R2 = 9 - 10.692 x 1 = -1.692 [V]

Yep! :smile:
 
  • #89
Phew! :smile:

Yep, think am ready^^
 
  • #90
Going back to the original circuit that started this topic,

If the directions of the currents weren't marked to me, how could I possibly have known it?
 
  • #91
You don't.

It's like mechanics. You don't know beforehand which directions the internal forces will point.
So you choose an arbitrary direction and mark them carefully in your FBD's.
They may turn out negative.

In electronics it's the same, you carefully mark the choose directions for the currents.
(Actually, you still tend to neglect doing that.)

If you do not mark the chosen direction of the current, typically that will result in a mistake in a sign somewhere.
 
  • #92
Thanks, comparing stuff to mechanics helps tons to my understanding :smile:
 
  • #93
I would recommend trying this one after all.

[PLAIN]http://img25.imageshack.us/img25/4407/circuitj.jpg

You don't have to do the dreary math, but just set up the equations.
Just to make sure you do not make mistakes with the signs.
 
Last edited by a moderator:
  • #94
Just a thought: This circuit might be an excellent vehicle for introducing the mesh current method. It'll keep the number of equations to be solved down to the bare minimum (three in this case). The methodology is almost identical to the KVL that you've been doing, but doesn't required introducing a lot of "unknown" current variables and their attendant node current sum equations.
 
  • #95
Hey guys! :smile:

The test is in 8 hours. Figured it's best I spend all my morning/noon time here^^

This is by far the most complex parallel V parallel R circuit I've seen!

Before I'll get to that, I want to go back to this problem



http://img819.imageshack.us/img819/3534/brioot.jpg

Turns out I's result is in MINUS (I think so anyway can you please check me up on that?). At least that what I got with the presumption that current goes like in drawing (a), where in actually it's going like circuit (b) then. True?

http://img143.imageshack.us/img143/5222/draqwingssss.jpg
 
Last edited by a moderator:
  • #96
No, the current goes like in drawing (a).
If you solve it like drawing (b) all the currents wiil be negative.

Consider your visual cues!
Current wants to flow from + to -.
And it will do so, unless there is a stronger battery forcing it the other way.

In drawing (a) all the currents go from + to -.
In drawing (b) all the currents go from - to +.
That should give you a hint (a hint only!).
 
  • #97
No, the current goes like in drawing (a).
If you solve it like drawing (b) all the currents wiil be negative.

Ah... so I was wrong. In my solution before I just rewrote all the results in plus because I figured it doesn't matter, only the value does, but it does kinda matter to know where really current flows.

I'll tell you why I made the mistake. In my calculator I have this charthttp://img804.imageshack.us/img804/4177/abcdt.jpg

If I type D as minuses, I get a positive result. If I type in D as positive, I get negative result. So I guess I should always look at D as being at the other part of the equation!

Funny.

In drawing (a) all the currents go from + to -.
In drawing (b) all the currents go from - to +.
That should give you a hint (a hint only!).

True, that's why I originally thought it flows like in (a), the erroneous results in terms of the signs confused me and made me retract! But, as you've correctly pointed out, I should always trust my visual cues. You have no ideas how many times I did it in mechanics, and it encouraged me to recalculate, change my result and get the correct one. :smile:So the fact that the potential difference here turned out negative is okay right?

Vab = E2 - I1 x R2 = 9 - 10.692 x 1 = -1.692 [V]
 
Last edited by a moderator:
  • #98
Femme_physics said:
Ah... so I was wrong. In my solution before I just rewrote all the results in plus because I figured it doesn't matter, only the value does, but it does kinda matter to know where really current flows.

I'll tell you why I made the mistake. In my calculator I have this chart

http://img804.imageshack.us/img804/4177/abcdt.jpg

If I type D as minuses, I get a positive result. If I type in D as positive, I get negative result. So I guess I should always look at D as being at the other part of the equation!

Funny.

Yeah, well, that's because your calculator expects the system of equations like this:
[PLAIN]http://upload.wikimedia.org/math/9/8/e/98e26fd3a6b65fa8fab337495b46bf81.png[/INDENT]
That is, all the variables left, and all the constants right.
Your D column contains the constants, which should be right (of the equal signs).
Since you have them left (of the equal signs), you need to apply minuses (which they would have if they were on the right side of the equal signs).


Femme_physics said:
True, that's why I originally thought it flows like in (a), the erroneous results in terms of the signs confused me and made me retract! But, as you've correctly pointed out, I should always trust my visual cues. You have no ideas how many times I did it in mechanics, and it encouraged me to recalculate, change my result and get the correct one. :smile:

So something good came from my misspelled queues. :smile:


Femme_physics said:
So the fact that the potential difference here turned out negative is okay right?

Yes. Indeed it is negative.​
 
Last edited by a moderator:
  • #99
I like Serena said:
Yeah, well, that's because your calculator expects the system of equations like this:
98e26fd3a6b65fa8fab337495b46bf81.png
That is, all the variables left, and all the constants right.
Your D column contains the constants, which should be right (of the equal signs).
Since you have them left (of the equal signs), you need to apply minuses (which they would have if they were on the right side of the equal signs).

Duly noted! Thanks for the detailed confirmation!

So something good came from my misspelled queues. :smile:

I can't imagine something bad coming from something *you* do!

Yes. Indeed it is negative.


*victory dance!*
 

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