Parallel voltage sources circuit

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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.
  • #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.
 
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  • #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.
 
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  • #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
 
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  • #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]
 
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  • #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.​
 
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  • #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!*
 
  • #101
Femme_physics said:
So now I want to find Vcd

So!

Vc -> 0

(I can do that, right? Set Vc as my zero point, since it makes things easier)


Vcd = I2 x 0.5 - I3 x 4 = 8.615 x 0.5 - 2.077 x 4 = -4 V

Yes? :smile:

Wow! You're starting to make up your own exercises! Cool! :cool:

Well, let's see...
You followed the wire down and back up.

Can you also follow the wire up and then back down?

(As for setting Vc to zero. I guess that's fine, although I wouldn't write it down.)
 
  • #102
Wow! You're starting to make up your own exercises! Cool!

Well, let's see...
You followed the wire down and back up.

Can you also follow the wire up and then back down?

Vcd = 6 - 10 = -4Vw00t :D

(As for setting Vc to zero. I guess that's fine, although I wouldn't write it down.)
Why wouldn't you?
 
  • #103
Femme_physics said:
Why wouldn't you?

Well, suppose you set Vc to zero because you want to calculate Vcd.

And then you set Va to zero because you want to calculate Vab.

But now... you have set both Va and Vc to zero, which can't both be true!


You can use it as a trick in your mind to find Vcd, but Vc does not really have to be 0 [V].
Especially not if there is a ground attached to some other point of the circuit.
 
  • #104
Well, suppose you set Vc to zero because you want to calculate Vcd.

And then you set Va to zero because you want to calculate Vab.

But now... you have set both Va and Vc to zero, which can't both be true!

Well I don't have to use the fact that Vc = 0 anymore when I look for ab.

I can write IF Vc = 0 then Vcd equals...

And then at the other clause I write

IF Va = 0 then Vab equals...
Especially not if there is a ground attached to some other point of the circuit.

If there is a ground, I have to use the ground they give me, right? As of now, it's kind of a groundless circuit
 
  • #105
Femme_physics said:
Well I don't have to use the fact that Vc = 0 anymore when I look for ab.

I can write IF Vc = 0 then Vcd equals...

And then at the other clause I write

IF Va = 0 then Vab equals...

The problem with those statements is, that Vcd will still equal whatever you calculated, even if Vc is not zero (since Vcd is a "difference").



Femme_physics said:
If there is a ground, I have to use the ground they give me, right? As of now, it's kind of a groundless circuit

Yes. It's groundless.
 
  • #106
Yes. It's groundless.

Is this like our "circuit flowing in space" idea? Not connected to ANYTHING!

The problem with those statements is, that Vcd will still equal whatever you calculated, even if Vc is not zero (since Vcd is a "difference").

So in other words I don't have to define anything as my "zero" point. :smile: Duly noted!
 

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