Circuit problem with ammeter and unknown resistance

Ouabache

Very cool notation you have in the center for "Power Source" (מקְור מתְח).

Femme_physics

I agree that equation 1 with I₀, I₁ and I₂ is unneeded. It appears to just overcomplicates things to add I₀ to the equation.

I am now considering just 4 equations then. It looks like it takes a bit of work to solve it. I'll have some college time this evening so I'll post back later my results. Sorry I am procrastinating with this, but I am really enjoying myself working on this exercise and I wanna take my time and do it right and get to the right conclusions. Had a bit of insomnia lately due to my new job bumping in with school and helping other students, so couldn't get to it. :(


Heh ;) Do you know Hebrew then?

Mindscrape

What I mean about "grounding a point" is that you'll later find that when people design circuits they often times have one reference point from which all other voltages are referenced, and this point is said to have 0 volts (0 is really arbitrary if you think about it because we are really just doing a potential difference, but it makes sense because you don't 5-0 is more trivially 5 than if we gave ground 1V and were subtracting one all the time; e.g. 6-1). This point is usually tied to earth, the earth ground is kind of the ultimate reservoir of electrons, but is sometimes simply tied to chassis of a box (or is sometimes just a "signal ground" on the plane of a PCB). I didn't really want to go searching for a reference that had just the right information, so you'll have to take it from me, a lowly physicists who think he knows circuits well :p. In your circuit you can pick any point from which all other points would be referenced. Point C really makes the most sense to ground, and give zero potential.

Oh right, and the garden hose analogy. Since current is a flow, and water is a flow that's where the analogy begins. You can control the water flow by turning a spigot, analogous to voltage. Certain things, like gravel, may impede water flow just like resistors. The only thing is that I haven't really come up with an example of a water capacitor.

As for gneill's initial advice, which I was soon to echo after having read through the topic, let's address it now:

As others have previously noted, you can assign whatever direction you want to your loops. If you went CCW around the right loop (indicating a CCW current) you would find out later, when you solved for the current, that you get a negative current (meaning that the actual current flow is opposite what you initially picked and really goes CW as we'd expect).

ILS, I agree that there a couple ways to solve this problem, and I fully embrace eventually tackling an alternate solution. I do, however, disagree with starting with the hardest. :p

Mind if I quickly outline our a more practical approach? Also, didn't there used to be letters on your diagram?

Let's do some KVL since that's how you started. Make both loops CCW, just to demonstrate that it will work.

Right loop:
Irht*Rx+Irght*R2+Vs=0
Left loop:
Ilft*R1+Ilft*R3-Vs=0
Now for the constraint from the problem:
We know that there is no current flowing through the ammeter, which means that the leftmost point and the rightmost point are at the same potential
-Irht*R2=Ilft*R1
(you could have also used the voltage loop if what I did was confusing, going CW Irht*R2+Ilft*R1=0)

So, now there are 3 eqns and 3 unknowns, wahoo! Note: for the constraint you could also do the other voltages,
-Irht*Rx=Ilft*R3.

There's other tricks that you could use to solve this problem really quickly. If you treat the leftmost point and rightmost point as voltage dividers (look it up if you haven't seen it), you just equate them and you get the answer presto, virtually no work involved!

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.

Ouabache

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.

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

How can this equation be entirely in pluses and equal zero? Unless they're all zero.


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.


Hmm...does there have to be a point in a circuit that has 0 V potential?

I like Serena

*grumbling* ;)

You seem to have missed my last reply (2 equations!):
http://www.physicsforums.com/showpos...4&postcount=50

Femme_physics

Sorry, I lost the thread of this thread!

My worst pun to date ;)

I'll get to it and post back

Femme_physics

I see 3 unknowns. Rx, I1 and I2. Not 2 unknowns!

I like Serena

*Sigh* ;)

Just try it!

[edit] (You procrastinator you! ) [/edit]

Femme_physics

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

I'll work to crack it today :)

I like Serena

Divide everything by I1?

I like Serena

Femme_physics

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

Why minus??????????!!!!!!!!!!!!

ehild

Femme_physics

My bad, discount the minus. I was excited I was getting somewhere :P

Rx = R3R2/R1

Is that right?

I like Serena

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!!!


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.