Parallel voltage sources circuit

[I like Serena]

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.

 

[Femme_physics]

I never heard the term "terminal voltage" before, though...can you tell me what it means?

 

[I like Serena]

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. )

 

[Femme_physics]

After a bit of googling I know the answer!

Vt = 6 - I3r



Right?

 

[I like Serena]

I see you've used the tools at hand to "deduce" the meaning. Very good!

Yes, that's the right formula.
Although on a test I hope you'll also give a numerical result.

 

[Femme_physics]

I will I'd be also very upset if during the test they wrote us something we hadn't studied like terminal voltage!

 

[I like Serena]

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! )

 

[Femme_physics]

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

 

[Femme_physics]

Pick two points and let me find the potential difference!

 

[I like Serena]

The 2 junctions.

 

[Femme_physics]

Vab = 12 -I2 x r -I2 x 11;
12 - 0.50842 x 1 - 0.50842 x 11

= 5.899 [V]

 

[I like Serena]

Right!

 

[Femme_physics]

Excellent!

I want to solve the actual question that was on the test now *mean-faced*

Oh I'm tackling you baby


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



I'll take your I's out!

http://img849.imageshack.us/img849/9387/el1x.jpg

http://img691.imageshack.us/img691/1650/el2r.jpg

http://img89.imageshack.us/img89/416/el3u.jpg

Is that it?

 

[I like Serena]

*Protecting my eyes*

What? I can't C!

Show me yours and I'll show you mine. ;)

 

[Femme_physics]

Is this writing in codewords that I am correct? *grins*

 

[I like Serena]

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?

 

[Femme_physics]

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?

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!

PS


Is the potential between A and B equals

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

 

[I like Serena]

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!

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

PS


Is the potential between A and B equals

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

Yep!

 

[Femme_physics]

Phew!

Yep, think am ready^^

 

[Femme_physics]

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?

 

[I like Serena]

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.

 

[Femme_physics]

Thanks, comparing stuff to mechanics helps tons to my understanding

 

[I like Serena]

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.

 

[gneill]

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.

 

[Femme_physics]

Hey guys!

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

 

[I like Serena]

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!).

 

[Femme_physics]

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. 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]

 

[I like Serena]

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).

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.

So something good came from my misspelled queues.

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

Yes. Indeed it is negative.​

 

[Femme_physics]

Yeah, well, that's because your calculator expects the system of equations like this:

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.

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

Yes. Indeed it is negative.

*victory dance!*

 

[Femme_physics]

So now I want to find Vcd

http://img827.imageshack.us/img827/7514/vcdx.jpg

And I know that

http://img89.imageshack.us/img89/416/el3u.jpg

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?

 

[I like Serena]

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?

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?

(As for setting Vc to zero. I guess that's fine, although I wouldn't write it down.)

 

[Femme_physics]

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?

 

[I like Serena]

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.

 

[Femme_physics]

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

 

[I like Serena]

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").

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.