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Circuit problem with ammeter and unknown resistance

by Femme_physics
Tags: ammeter, circuit, resistance, unknown
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Ouabache
#55
May18-11, 12:13 PM
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Quote Quote by Femme_physics View Post
Heh ;) Do you know Hebrew then?
Doesn't everyone?
I recognize the characters and have rudimentary vocabulary from self-study אולפן.
But in this case, google translate came to the rescue.

(caveat: for those wanting to try this, it requires setting up your keyboard for Hebrew. However, do this at your own peril , an accidental keystroke can suddenly change your keys from Latin to Hebrew and you'll spend the next hour trying to change it back.

Quote Quote by Mindscrape View Post
Sorry if I stole anyone's thunder, but the amount of complexity in this problem was getting a bit absurd, beyond being practical for learning I felt.
What is obvious to one person is often challenging to someone who may not be familiar, in this case, with linear circuit theory. So there is merit in discussing a problem from more than one angle, to learn how to arrive at the solution. This is similar to teaching a concept, by explaining it in different ways.

In fact, it is a good idea to solve these exercises using at least one more method. If you arrive at the same answer, there is a high likelihood your solution is correct. (ILS also alludes to this in step 4 of his last post).
Femme_physics
#56
May18-11, 11:34 PM
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Irht*Rx+Irght*R2+Vs=0
How can this equation be entirely in pluses and equal zero? Unless they're all zero.


Sorry if I stole anyone's thunder, but the amount of complexity in this problem was getting a bit absurd, beyond being practical for learning I felt.
I was wondering about that. I was really getting exhausted trying this 4 eq 4 unknown thing. Not that it's hard, I've just been procrastinating because I felt a bit lost with all the equations and not seeing the light. ILS is wonderful in his approach, though, but if you say I only need 3 equations I rather solve it using just 3 equations and then once I get the final answer I'll start comparing stuff.

Point C really makes the most sense to ground, and give zero potential.

Hmm...does there have to be a point in a circuit that has 0 V potential?
I like Serena
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May18-11, 11:44 PM
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Quote Quote by Femme_physics View Post
but if you say I only need 3 equations I rather solve it using just 3 equations and then once I get the final answer I'll start comparing stuff.
*grumbling* ;)

You seem to have missed my last reply (2 equations!):
http://www.physicsforums.com/showpos...4&postcount=50
Femme_physics
#58
May18-11, 11:45 PM
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Sorry, I lost the thread of this thread!

My worst pun to date ;)

I'll get to it and post back
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#59
May18-11, 11:48 PM
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Equation #4: +I2R2 -I1R1 = 0
Equation #5: -I2Rx +I1R3 = 0


Can you try and eliminate I1 and I2?
I see 3 unknowns. Rx, I1 and I2. Not 2 unknowns!
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May18-11, 11:49 PM
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Quote Quote by Femme_physics View Post
I see 3 unknowns. Rx, I1 and I2. Not 2 unknowns!
*Sigh* ;)

Just try it!

[edit] (You procrastinator you! ) [/edit]
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#61
May19-11, 12:16 AM
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I'm stuck. I have 3 unknowns, and 2 equations. What can I try? I can try moving things around. I don't see how it helps me. I move I2R2 to equal I1r1

Great. So, now I can divide everything by...


Ah I'll just upload it. Hope you can see it clearly!





But now I still have too many unknowns. 2 unknowns 1 equation! I1 and Rx in this case are the unknown.
Femme_physics
#62
May19-11, 12:17 AM
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Quote Quote by I like Serena View Post

[edit] (You procrastinator you! ) [/edit]
I'll work to crack it today :)
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#63
May19-11, 12:18 AM
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Quote Quote by Femme_physics View Post
But now I still have too many unknowns. 2 unknowns 1 equation! I1 and Rx in this case are the unknown.
Divide everything by I1?
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May19-11, 12:30 AM
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Femme_physics
#65
May19-11, 12:37 AM
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LOL is this a mock at my procrastination? Such as I'm overthinking the theory and not doing the work? Hah. Well, I may have exaggerated in this case :)


Here! Doing!






Is that right?
This is bizarre. I never knew you can solve a 3 unknown 2 eq thing. Is this really true? If so, what just happened here?!?


PS
I think I'll use red ink so it's clearer from now on...or something more thick...
ehild
#66
May19-11, 12:39 AM
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Why minus??????????!!!!!!!!!!!!

ehild
Femme_physics
#67
May19-11, 12:40 AM
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My bad, discount the minus. I was excited I was getting somewhere :P

Rx = R3R2/R1

Is that right?
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May19-11, 12:45 AM
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Quote Quote by Femme_physics View Post
LOL is this a mock at my procrastination? Such as I'm overthinking the theory and not doing the work? Hah. Well, I may have exaggerated in this case :)


Here! Doing!
Nope! I just thought it was a funny comic!

Wait! Sorry! No it is a mock!



I was just wondering how Rx could be negative.
What could it mean?
Did you just discover negative resistors?
But that's great! We could generate voltage out of nothing!!!


Quote Quote by Femme_physics View Post
Is that right?
This is bizarre. I never knew you can solve a 3 unknown 2 eq thing. Is this really true? If so, what just happened here?!?
It's a special circuit that is designed to "balance" stuff.
With the method we used, we did introduce 1 unknown to many, but it didn't matter, because it would cancel out.
Femme_physics
#69
May19-11, 12:48 AM
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Wait! Sorry! No it is a mock!
Fair enough heh, I had that coming

But that's great! We could generate voltage out of nothing!!!
LOL

Fixed. (as in my post above ehild!)
Is that right?

It's a special circuit that is designed to "balance" stuff.
With the method we used, we did introduce 1 unknown to many, but it didn't matter, because it would cancel out.
Incredible.
Mindscrape
#70
May19-11, 12:51 AM
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How can this equation be entirely in pluses and equal zero? Unless they're all zero.
En contraire. Irht could, and is in fact, negative! Right? Because I told that you that I was going to pick Irht to be in the wrong direction (CCW), when in actuality, and from the equations we will find that it is negative the direction of I picked which is CW. All is well in the world of arbitrarily assigning currents directions to loops, alright!

Hmm...does there have to be a point in a circuit that has 0 V potential?
No, you don't have to have a point with a 0V potential. Usually you want to though, if you don't have coherent references to ground, you may end up with a circuit like this http://xkcd.com/730/

Of course, you're free to take my approach or not. In my mind the great thinking comes from realizing why you get the answer you do, why the circuit behaves as it does, and not in actually getting the answer. Hey, congrats on getting the problem right though!

Now, try using the voltage divider at the left and rightmost points, setting them equal, and solving for Rx. Once you do, you will see how wonderful keeping it simple really is. Using the voltage divider ought to take 4 minutes at most :)
ehild
#71
May19-11, 01:02 AM
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Quote Quote by Femme_physics View Post
Is that right?
This is bizarre. I never knew you can solve a 3 unknown 2 eq thing. Is this really true? If so, what just happened here?!?
You did not solve the system of equations really, I1 and I2 are still unknown. And one solution can be I1=0 and I2 =0. In this case Rx could be anything. What you got means that unless Rx=R3R2/R1 , both I1 and I2 should be zero.

ehild
Femme_physics
#72
May19-11, 01:57 AM
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En contraire. Irht could, and is in fact, negative! Right? Because I told that you that I was going to pick Irht to be in the wrong direction (CCW), when in actuality, and from the equations we will find that it is negative the direction of I picked which is CW. All is well in the world of arbitrarily assigning currents directions to loops, alright!
Oh yea, you do the same in mechanics. I do it arbitrary when I have 1 unknown with 1 equation. I don't like risking stuff when I have more equations or more unknown, because I figured that it may yield the wrong answer. But, if you say it doesn't in this case, then I take your word for it.

No, you don't have to have a point with a 0V potential. Usually you want to though, if you don't have coherent references to ground, you may end up with a circuit like this http://xkcd.com/730/
*chuckles* is that supposed to be a reference to role-playing maps/mazes?

But I get it, it's important to make a ground point.

So, if I have a simple circuit like this:



Since the voltage drop after the 5 ohm resistor is 10, all the points AFTER the 5 ohm resistor is 0, yes?

Now, try using the voltage divider at the left and rightmost points, setting them equal, and solving for Rx. Once you do, you will see how wonderful keeping it simple really is. Using the voltage divider ought to take 4 minutes at most :)
I would be happy to do it, even if it took 44 minutes :) , but I'm not sure how to use the "voltage divider", this is the first time I hear that term. Googling it, it appears to "cut" a circuit at a point. I never realized you can do that!


You did not solve the system of equations really, I1 and I2 are still unknown. And one solution can be I1=0 and I2 =0. In this case Rx could be anything. What you got means that unless Rx=R3R2/R1 , both I1 and I2 should be zero.

ehild
Is there more for me to learn from this exercise while I can be exercising other problems, Can I just that if indeed I1 and I2 = 0 then Rx = R3R2/R1 and write that as the answer and bob's your uncle?

I can also now answer

Is the value of Rx depends on the ammeter resistance A? Explain.
No, the value of a resistor is a constant.


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