Electronic parallel voltage sources exercises?

AI Thread Summary
The discussion focuses on finding exercises related to circuits with parallel voltage sources, with participants sharing problems and seeking solutions. One user expresses difficulty in locating suitable exercises online, prompting others to provide circuit problems for practice. The conversation delves into the differences between voltage and current sources, emphasizing the importance of understanding these concepts for solving circuit problems. Participants discuss strategies for applying Kirchhoff's laws and calculating currents in various circuit configurations. Overall, the thread highlights the collaborative effort to enhance understanding of electrical circuits and problem-solving techniques.
Femme_physics
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I'm trying to google for some but results often lead me to forum questions, theory articles, etc. Basically I'm looking for circuits questions with parallel voltage sources answers, where I am supposed to do the work to find out the solution for the current.
 
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Hi Fp! :smile:

I'll give you a problem (found on PF).


[PLAIN]http://img297.imageshack.us/img297/7053/96217079.jpg

given this circuit find the current on R1,R2 and R3.


Can you solve it?
 
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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
 

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Thanks, ILS! I'll take a crack at it soon :smile: ^^
 
attachment.php?attachmentid=37008&stc=1&d=1310103105.png


attachment.php?attachmentid=37009&stc=1&d=1310103105.png
 

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I like Serena said:
attachment.php?attachmentid=37008&stc=1&d=1310103105.png


attachment.php?attachmentid=37009&stc=1&d=1310103105.png

Also too easy :smile: I think I'll skip it. I'm really looking for one with parallel voltage sources. I've been googling but it's hard to find an exercise with answers. Do you happen to remember or have another one I can work on?
 
Too easy eh? I guess I have not been anticipating your phenomenal progress! :smile:

Ok.
How about this one?

[PLAIN]http://img25.imageshack.us/img25/4407/circuitj.jpg
 
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Eep.Yikes.
Looks intimidating.

*touches her lucky item to imbue self with ILS-like powers*

I'll give it a shot!
 
Once the item has inspired you enough to finish the problem, I have another one here.
I hope it's not too easy! :devil:

attachment.php?attachmentid=20570&d=1253080477.jpg
 
  • #10
Uh, where are the voltage sources?
 
  • #11
Femme_physics said:
Uh, where are the voltage sources?

Who needs voltage sources, when you can get current sources?
Or are you certain you won't get those?

It seemed smart to me, to not only practice exactly the same problem over and over, but to vary it a bit and make sure you're not thrown off by some detail that should not matter.
The method is exactly the same...
 
  • #12
Yes, but I don't know where are they located! Shouldn't you tell me at least where are they located if not their value?
 
  • #13
There are no voltage sources.
Instead there are current sources that are similar.

An (ideal) voltage source provides a constant voltage.
An (ideal) current source provides a constant current (the voltage drop is zero).
 
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  • #14
Oh, well we haven't studied about that, I'm pretty sure it's not going to be on the test! Should I really be trying to solve that?
 
  • #15
Well, yeah!

To get a perfect score, you need to be able to play with the material.
It does not suffice that you have drilled one specific exercise over and over.
You need to have the confidence that whatever they throw at you, you simply know it won't matter, because you know your stuff, and are not thrown off by surprises!

That is, unless you're content with doing just well enough to pass the test.
(But then you'll have these guys looming over you that you don't dare contradict, because you're not sure whether you know better. :wink:)

(And if you don't do these exercises, you'll not be spending time with me! Sniff! :cry:)
 
  • #16
Femme_physics said:
Yes, but I don't know where are they located! Shouldn't you tell me at least where are they located if not their value?

You can transform voltage and current sources interchangeably with resistors.

If you want a big list of circuits problems I would get the Schaum's outline of Electric Circuits btw.
 
  • #17
I like Serena said:
There are no voltage sources.
Instead there are current sources that are similar.

An (ideal) voltage source provides a constant voltage.
An (ideal) current source provides a constant current (the voltage drop is zero).


Well, what's the physical difference between a current source and a voltage source? Voltage source is plus and minus. A place with lower amount electrons connected to a place with greater amount of electrons...Is a current source the same?
 
  • #18
Femme_physics said:
Well, what's the physical difference between a current source and a voltage source? Voltage source is plus and minus. A place with lower amount electrons connected to a place with greater amount of electrons...Is a current source the same?

Uh yes, in the sense that you mention it is the same.
The physical difference is that a voltage source has a fixed voltage difference and a variable current, depending on the circuit.
Whereas a current source generates a fixed current and a variable voltage difference, depending on the circuit.

(Btw, I have to retract my statement that the voltage drop across a current source is zero. It isn't.)
 
  • #19
Hmm...then do I treat a voltage source like a current source when I build my equations? Or do I discount V and build the equations?
 
  • #20
Can I get a good morning from you? :smile:

Start with building the equations that you think you can build.
You should be able to apply KCL.
Use KVL only on loops that do not contain a current source (since you won't get useful information from that).
 
  • #22
Good morning! :smile:

I did say "heya" at the other thread! :approve:Can I minimize this last circuit to this?

http://img580.imageshack.us/img580/4282/minimize.jpg

Start with building the equations that you think you can build.
You should be able to apply KCL.
Use KVL only on loops that do not contain a current source (since you won't get useful information from that).

Seems simple enough. I might just do it, although I don't think we studied about current sources. I mean, I've never seen such a circuit in class. Then again, it seems kinda simple.
 
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  • #23
Femme_physics said:
Good morning! :smile:

I did say "heya" at the other thread! :approve:

You said "heya guys".
I'd like my greeting to be a bit more personal. :smile:

(I wouldn't mind hugs or kisses. :redface:)
Femme_physics said:
Can I minimize this last circuit to this?

Yep!

I was wondering if you got that physical concept and it appears that you do! :approve:
Femme_physics said:
Seems simple enough. I might just do it, although I don't think we studied about current sources. I mean, I've never seen such a circuit in class. Then again, it seems kinda simple.

As far as I'm concerned, it's practice so you don't get stopped by something you didn't see before.
But that you simply look at it, say "Okay" and solve the problem.
 
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  • #24
You said "heya guys".
I'd like my greeting to be a bit more personal.

(I wouldn't mind hugs or kisses. )

Aww in that case *BEAR HHHHHHUGS!* *SMOOCHES*

Yep!

I was wondering if got that physical concept and it appears that you do!

:smile:

Thanks! Seems very easy now!

As far as I'm concerned, it's practice so you don't get stopped by something you didn't see before.
But that you simply look at it, say "Okay" and solve the problem.

Gotcha!

Just a moment
 
  • #25
I like Serena said:
Once the item has inspired you enough to finish the problem, I have another one here.
I hope it's not too easy! :devil:

attachment.php?attachmentid=20570&d=1253080477.jpg

There are so many confusing things about this. Starting with this:

What do the circle and square signs mean? Are these current sources?
 
  • #26
Yes, the circle and diamond signs are current sources. :)

Circle for constant current specified next to it.
Diamond for a dependent current source that provides the current specified next to it.

What's so confusing about it? :confused:
Or is it just that you didn't see it before?Edit: Oh, and here's a more detailed explanation from someone you already know:
https://www.physicsforums.com/showpost.php?p=2351140&postcount=7
 
  • #27
Femme_physics said:
Aww in that case *BEAR HHHHHHUGS!* *SMOOCHES*

Aaaah! I feel much better now! :redface:
 
  • #28
Aaaah! I feel much better now!

Aww o:):blushing:


I like Serena said:
Yes, the circle and diamond signs are current sources. :)

Circle for constant current specified next to it.
Diamond for a dependent current source that provides the current specified next to it.

What's so confusing about it? :confused:
Or is it just that you didn't see it before?


Edit: Oh, and here's a more detailed explanation from someone you already know:
https://www.physicsforums.com/showpost.php?p=2351140&postcount=7

But where it says "mA" -- how am I supposed to know where mA is Ix or IL?
 
  • #29
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  • #30
Femme_physics said:
But where it says "mA" -- how am I supposed to know where mA is Ix or IL?

mA means milliampere.
It's the unit of the corresponding current, which is neither Ix nor IL.
 
  • #31
Femme_physics said:
Going back to this:

Io is in fact the current that flows in all across the last circuit I made, right?

Though I recognize it splits here


when it gets back to point a, does it kinda recombine and spits again? I'm not sure how to explain it...

Do you know what I mean?

When you merge the resistors, you can't "see" Io anymore.
You can calculate the voltage drop and the total current though.
Afterward you have to go back to the original circuit to find Io.
 
  • #32
Forgot to answer
What's so confusing about it?
Or is it just that you didn't see it before?

Yes, it's because I've never seen it before.

I do know that mA means miliampere! I just wanted to know to which it belongs, and now you say it belongs to neither..

so those blue arrows are ANOTHER 3 unknowns?

http://img832.imageshack.us/img832/2659/bluecw.jpg

Uploaded with ImageShack.us
 
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  • #33
I like Serena said:
When you merge the resistors, you can't "see" Io anymore.
You can calculate the voltage drop and the total current though.
Afterward you have to go back to the original circuit to find Io.
I see, so after I find Itotal, I find V. And after I find V I just doV/R1 to find IoYes?
 
  • #34
Well, they are not all really "unknown" are they?
I think the leftmost one is 6 mA, don't you?
That doesn't sound "unknown" to me.
 
  • #35
Femme_physics said:
I see, so after I find Itotal, I find V. And after I find V I just do


V/R1 to find Io


Yes?

No, you will have to use KCL to find Io.
 
  • #36
I like Serena said:
Well, they are not all really "unknown" are they?
I think the leftmost one is 6 mA, don't you?
That doesn't sound "unknown" to me.

Oh, right *smacks forehead*


I like Serena said:
No, you will have to use KCL to find Io.


Alright, I see, give me a few moments to try and solve it...be right back with a scan
 
  • #37
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  • #38
I like Serena said:
Well, they are not all really "unknown" are they?
I think the leftmost one is 6 mA, don't you?
That doesn't sound "unknown" to me.

Good point! lol... that was kinda silly of me.

Okay then... should be easy ...hope I'm not oversimplifying it (which I tend to do)...

http://img714.imageshack.us/img714/6847/600il.jpg
 
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  • #39
Femme_physics said:
Well I'd hate to start solving this 5 eq 5 unknown thing, but this is it right? I mean, this is what it boils down to?

PS Maybe a mod can move this topic to the homework section?

You can do it this way, but I think you would prefer the "smooth ride".

After calculating I_total, don't go back right away to your original circuit, but take the top diagram you drew in your scan, where you have 2 merged resistors.

There you can determine the voltage drop to the middle.

Only then, go back to the original circuit and calculate all the currents with Ohm's law.

Then, finally apply KCL to find Io.
 
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  • #40
I like Serena said:
You can do it this way, but I think you would prefer the "smooth ride".

After calculating I_total, don't go back right away to your original circuit, but take the top diagram you drew in your scan, where you have 2 merged resistors.

There you can determine the voltage drop to the middle.

Only then, go back to the original circuit and calculate all the currents with Ohm's law.

Much easier indeed! Well, I doubt we'd have somethign that difficult on the test thankfuly, and speak of the devil... I'm off to college in a jiffy :smile: wish me luck

XoX

Edit: Actually I don't want too early, I'll try and solve it
 
  • #41
Femme_physics said:
Good point! lol... that was kinda silly of me.

Okay then... should be easy ...hope I'm not oversimplifying it (which I tend to do)...

No, that won't work.

In your loop (a) you treat the current source of 3 mA as a voltage source of 3 V.
But that is not right.
The corresponding voltage drop is unknown.
 
  • #42
I like Serena said:
You can do it this way, but I think you would prefer the "smooth ride".

After calculating I_total, don't go back right away to your original circuit, but take the top diagram you drew in your scan, where you have 2 merged resistors.

There you can determine the voltage drop to the middle.

Only then, go back to the original circuit and calculate all the currents with Ohm's law.

Then, finally apply KCL to find Io.


http://img834.imageshack.us/img834/9089/wttt.jpg


Uhh...okay what am I doing wrong? O_o

btw got less than 30 minutes:rolleyes:
 
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  • #43
I wish you luck! :smile:
 
  • #44
Femme_physics said:
Uhh...okay what am I doing wrong? O_o

btw got less than 30 minutes:rolleyes:

Go back to the original circuit.

Now you have the voltage drop between a and b, so you can calculate the currents through the left and right resistor.

Same for b and the bottom.
 
  • #46
Femme_physics said:
Well this is my last attempt due to time shortage but I don't see how I find Io in any of this:

Good!
Do the same between a and b to find I1 and I2.


Aftward apply KCL at point B to find Io.
 
  • #47
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  • #49
:smile:

It seems that thread didn't have such a smooth ride!
 

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