Parallel voltage sources circuit

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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:
 
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.
 
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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gneill said:
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.

Very true! :blushing:

I intended to ask for the power dissipation in the 4 ohm resistor that is between A and B.



Femme_physics said:
:biggrin:


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 ))

This looks *almost* like a Norton equivalent circuit! :biggrin:

I give you the smurfs. All of them are yours, you *evil* you!
Total power and domination!

Btw, the total power is the sum of the powers dissipated in each of the 3 resistors (I get 259 W).
 
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I intended to ask for the power dissipation in the 4 ohm resistor that is between A and B.

Easy peasy!

P2 = 2 x 5.211^2 = 54.31 [W]

I give you the smurfs. All of them are yours, you *evil* you!

Muwahhaahahhah!

Btw, the total power is the sum of the powers dissipated in each of the 3 resistors.

You mean
Pt = Power dissipated in resistor 1 +Power dissipated in resistor 2 + Power dissipated in resistor 3?

In which case, I was aware of that! :approve:

And I did get the correct result, yes?
 
Femme_physics said:
Easy peasy!

P2 = 2 x 5.211^2 = 54.31 [W]



Muwahhaahahhah!



You mean
Pt = Power dissipated in resistor 1 +Power dissipated in resistor 2 + Power dissipated in resistor 3?

In which case, I was aware of that! :approve:

And I did get the correct result, yes?

Let me see...

We got a total power of 258.65 W.
Subtract 199.46 W from your previous result.
Subtract another 54.31 W that you just got.

That leaves... 4.88 W for the 4 ohm resistor between A and B.
Yes! That's it! You got it! :smile:
 
w00000000000000000t :biggrin:

You rock!I'll probably do the other one tomorrow^^ You're incredible, ILS! This is pretty fun for me, I don't know about you :wink: You're a life saver. I'm pretty stressed about term B in electronics, it's my last chance at "redemption" for this course.
 
You did a nice job ! (!עבודה טובה)
There are typically several approaches to the same question.
Some take fewer steps than others. :smile:

Here is the method I was suggesting you try.
(It may help save you some time on an exam)

You found currents for the initial question to be I1=1.105, I2=5.211A and I3= I0 = 6.316A

Next was find VAB and PAB (with respect to R1).

V= IR ; VAB = I1R1 = (1.105)(4) = 4.42V
P = IV; PAB = I1VAB = (1.105)(4.42) = 4.88W
 
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Duly noted Quabache :smile:

Next time I will certainly be wiser :approve:

I will try me hands on another exercise this morning (it's 4:30 AM here. I'm an early riser!)
 
I like Serena said:
And here's a second one.
Perhaps you can start a separate thread for this problem?For the circuit shown below, determine the voltage for each of the
resistors and label the values on the diagram.
attachment.php?attachmentid=36992&stc=1&d=1310029649.jpg

That was pretty easy. I didn't find P for each just for lack of patience. I know I know it!

http://img18.imageshack.us/img18/8364/otherquestion.jpg
 
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Yea I know finding the P in each of them...but com'on you just apply a formula it's soooooooooooooo easy! Lemme skip that pretty please? :smile:
 
I like Serena said:
How about this one?

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

First question--

How come they get to pick for me the direction of I? Shouldn't I be the one determining it?
 
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Femme_physics said:
First question--

How come they get to pick for me the direction of I? Shouldn't I be the one determining it?

Well...

Look at the drawing in your first post in this thread.
You did not mark the directions of the currents (and as a consequence you made a mistake with it).
So I thought I'd better mark them for you! :devil:

(Just kidding, I just picked the first exercise that fitted your description. :wink:)

The real reason would be that the people who made the exercise would want the same answer from all students, so it's easier for them to check the answers. This means naming the currents and preselecting the directions.
 
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I like Serena said:
Well...

Look at the drawing in your first post in this thread.
You did not mark the directions of the currents (and as a consequence you made a mistake with it).
So I thought I'd better mark them for you! :devil:

(Just kidding, I just picked the first exercise that fitted your description. :wink:)

The real reason would be that the people who made the exercise would want the same answer from all students, so it's easier for them to check the answers. This means naming the currents and preselecting the directions.

Interesting. So you get different answers if you pick different direction for I? I didn't know that!
 
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?
 
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
 
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:
 
I've really decided it's too complex since we're not going to have something THAT difficult on the test!
 
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.
 
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
 
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?
 
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.)
 
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
 

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