Circuit problem with ammeter and unknown resistance

[ehild]

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]

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:

http://img153.imageshack.us/img153/333/510bw.jpg

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.

 

[ehild]

I wrote that to people who read this thread and like to see in depth of problems.

ehild

 

[I like Serena]

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

Ehild's comment shows his extensive knowledge and experience.
I think it also shows the answer to a trick question that he did not ask.

However, for your answer to the exercise, I think you can suffice with saying that:
Rx = R3R2/R1

Is the value of Rx depends on the ammeter resistance A? Explain.

No, the value of a resistor is a constant.

A little sharper would be that we made no assumptions about the resistance of the ammeter, but that its resistance does not show in the final solution for Rx.
So Rx is indeed independent of the ammeter resistance.

Note that this question is intended to highlight the value of the Wheatstone Bridge.

 

[Femme_physics]

Fair enough ehild. :) I want too! As evidence by 5 pages of what you call a "simple exercise"!

 

[Femme_physics]

Ehild's comment shows his extensive knowledge and experience.
I think it also shows the answer to a trick question that he did not ask.

Woah, woah, have I been misinterpreted? Did I complain? I hope not! Never! Knowledge is power, it's valuable! I'll take what I can get.

I just asked if I should be investing so much in one exercise or now that I have the answer, I'll let go, for the sake my sanity, to continue other exercises :)

@ ehild. You should never have to excuse yourself for writing extra knowledge stuff. And never esp. in my threads! Am very happy you do that :)

However, for your answer to the exercise, I think you can suffice with saying that:
Rx = R3R2/R1

*does victory dance!*

A little sharper would be that we made no assumptions about the resistance of the ammeter, but that its resistance does not show in the final solution for Rx.
So Rx is indeed independent of the ammeter resistance.

Note that this question is intended to highlight the value of the Wheatstone Bridge.

Awesome :) Thank you soooooooooo much everyone! ehild, Mark, ILS, gneil, quabache... hope I didn't forget anyone. Thank you! :)PS I may come back to this thread at some point as I continue to learn.

 

[Mindscrape]

You must have just seen circuits written in a way you're not used to. Here's a voltage divider for a circuit written the way you're used to.

The formula would be Vmiddle=R2/(R1+R2)Vbat. So the voltage across R2 will be VR2 = 2/3*12 = 8V

So really what you have in your balanced wheatstone bridge is two independent voltage dividers.