Circuit problem with ammeter and unknown resistance

In summary: I got it now! I'm getting it now! Thanks! :DAre you familiar with the concept of the voltage divider where there are two resistors in series placed across a potential difference (say a battery) and the potential at the 'tap point' where the resistors meet is a portion of the applied potential that depends upon ratios of the resistor values? Yes :) But I got it now. Thanks!If so, can you recognize two such voltage dividers in your Wheatstone Bridge circuit?Yes! I can!So, I'm looking at I1. I got this far: [tex]\text{Potential through the first voltage divider} = \frac{I_1}{I_
  • #36
Look at this problem as a set of simultaneous equations. You wouldn't decide on the sign of the answers until the end, would you? It doesn't matter what sign (arrow direction) you start with. The solution to the equations will put it right in the end.
 
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  • #37
Well, the result comes from the fourth and fifth equations (if they are written correctly). Find I1/I2 from both and make them equal: It is the equation for Rx.

ehild.
 
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  • #38
ehild said:
Well, the result comes from the fourth and fifth equation (if they are written correctly). Find I1/I2 from both and make them equal: It is the equation for Rx.

ehild.

Missed that! Good call! :smile:
 
  • #39
You won't need the 5th equation, but please note that you mixed up the currents in this equation. How?

Well if you're referring to the drawing it comes from earlier posts.. I do realize that those currents are opposite.

And the only reason I gave a + to I2 in the 5th equation is because you told me so, I didn't bother to check this time (I thought it was the same idea why I changed the sign in the 4th equation, with the current going from high to low)

Although I'm not sure why I should bother about the sign reading sophie's reply:

Look at this problem as a set of simultaneous equations. You wouldn't decide on the sign of the answers until the end, would you? It doesn't matter what sign (arrow direction) you start with. The solution to the equations will put it right in the end.
 
  • #40
Femme_physics said:
Well if you're referring to the drawing it comes from earlier posts.. I do realize that those currents are opposite.

And the only reason I gave a + to I2 in the 5th equation is because you told me so, I didn't bother to check this time (I thought it was the same idea why I changed the sign in the 4th equation, with the current going from high to low)

Although I'm not sure why I should bother about the sign reading sophie's reply:

It's about which current goes where.

I1 is the current that goes from the top of the battery, through R1. Where does it go after that?
As for the signs, you definitely need the right signs in the right places, or you'll find a different (and wrong) solution for the equations.

What sophiecentaur meant, is that when you solve the equations, you might find that some current is negative. That's not a problem. It's fine. It just means the current actually goes in the opposite direction from what you initially thought.
Note that in this case you won't find a negative current.
And what ehild meant, is that you can discard the first 3 equations, because you can find what you need from the 4th and 5th equation (assuming you have the 5th equation in order).
This will yield the same result as what I suggested before, it's just easier to calculate.
[edit] And it may provide some additional insight in a Wheatstone Bridge! :wink: [/edit]
 
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  • #41
My thoughts:

First, there are too many unknowns. To get the currents I0, I1, and I2, I need to know one more current or voltage. All the non-zero currents will scale with the voltage source.

You may solve for the resistance R, though.

Solving for Rx is simple and shouldn't involve current loops. The voltages on either side of the ammeter are the same, because the ammeter is close to a short circuit anyway, and because it's drawing no current. This means the voltage must divide in the same ratios in both of the two paths. So, R1/R3 = R2/Rx.

This is known as a balanced Wheatstone bridge.
 
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  • #42
I1 is the current that goes from the top of the battery, through R1. Where does it go after that?

It goes to R3

So it's

Sigma V = 0; R3I1 - RxI2 = 0

Is that it?
 
  • #43
Femme_physics said:
It goes to R3

So it's

Sigma V = 0; R3I1 - RxI2 = 0

Is that it?

Yep! :smile:

(Sigh! :wink:)
 
  • #44
So now I have the correct 5 equations if I just replace the last correction with the 5th equation?

First, there are too many unknowns. To get the currents I0, I1, and I2, I need to know one more current or voltage. All the non-zero currents will scale with the voltage source.

You may solve for the resistance R, though.

Solving for Rx is simple and shouldn't involve current loops. The voltages on either side of the ammeter are the same, because the ammeter is close to a short circuit anyway, and because it's drawing no current. This means the voltage must divide in the same ratios in both of the two paths. So, R1/R3 = R2/Rx.

This is known as a balanced Wheatstone bridge.

Well, I'm not sure how to get to it yet. Step by step :)
 
  • #45
I have to say, using the "ammeter" equation seems kinda made-up to me, fake or untrue. It's kinda weird to use a loop that has no current flowing through parts of it. It's as though I've "made up" a loop and really am just looking at 2 arms of this loop. Can I make up loops like that anywhere, or only when I have an ammeter?Also, still looking for confirmation please

So now I have the correct 5 equations if I just replace the last correction with the 5th equation?
 
  • #46
Wow! Five equations for such an innocent looking circuit.

It might of of interest to note that for a complete solution to a given circuit, it is sufficient to have a set of loops where each component of the circuit belongs to at least one loop.

For this circuit, three loops (and thus three equations) suffice to completely solve the circuit for any given values of the its components. And in indeed, there are particular conditions imposed upon this example that allow you to eliminate one of the loops entirely, leaving just two loops and thus two equations and two unknowns to worry about. I refer, of course, to the fact that the problem specifies that the current in the ammeter loop is zero. It is also the case that the only voltage source in the circuit belongs to both remaining loops, the remainders of which consist of two resistors in series. So you end up with two equations that don't even need to be solved simultaneously!

In the attached figure, note that the potential difference between the labeled points c and d is fixed by the battery. We are told that i3 is zero. This means that the currents i1 and i2 are entirely determined by the battery potential and the resistors of the given loop, and that these currents are entirely independent of each other -- the fixed battery voltage isolates one loop from the other, since no voltage variations can occur across an ideal voltage source.

So it remains to write the two equations for i1 and i2, and using i1 and i2, determine the potential at points a and b (with respect to d, say), and equate the potentials.
 

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  • #47
Thank you gneill for bringing some sanity to this topic!

As you can see, I agree whole heartedly with gneill. You really need to not create so much work and confusion for yourself! One, you are always free to ground any point in the circuit (point C seems to make the most sense). Two, there is no current through the ammeter which means don't worry about it. If I had a system of garden hoses and I told you to find out how much gravel was in each garden hose (making a particular flow rate), and if one of these garden hose sections happened to have its pressure completely balanced so there was no flow, you'd leave it alone and just call it good (i.e. ignore the ammeter because it makes no sense to try to include it). Also, you'd not care about whether one particular part of the same flow section had more gravel than another, just the total gravel (i.e. put the resistors in series to make things even better!).

If you take gneill's suggestion, you will solve this circuit hassle free and time free.
 
  • #48
Femme_physics said:
I have to say, using the "ammeter" equation seems kinda made-up to me, fake or untrue. It's kinda weird to use a loop that has no current flowing through parts of it. It's as though I've "made up" a loop and really am just looking at 2 arms of this loop. Can I make up loops like that anywhere, or only when I have an ammeter?

Yes. You can make up loops like that anywhere. :)

Note that if you have a circuit without voltage source, there will be no current, but Kirchhoff's voltage law will still hold, although in that case it's rather trivial. In this case it is not trivial.


Femme_physics said:
Also, still looking for confirmation please
So now I have the correct 5 equations if I just replace the last correction with the 5th equation?

Yes. You have the correct 5 equations.

Note that you only need to solve equations 4 and 5 (as ehild noted) to find Rx.


Let me summarize:

Equation #4: +I2R2 -I1R1 = 0
Equation #5: -I2Rx +I1R3 = 0


Can you try and eliminate I1 and I2?
 
  • #49
Wow! Five equations for such an innocent looking circuit.
Thank you gneill for bringing some sanity to this topic!

Well, ILS has been guiding me through to get to these equations so I thought that makes sense and that he's probably making a point to me knowing him :)

It might of of interest to note that for a complete solution to a given circuit, it is sufficient to have a set of loops where each component of the circuit belongs to at least one loop.

That is indeed an interesting fact to note!

For this circuit, three loops (and thus three equations) suffice to completely solve the circuit for any given values of the its components. And in indeed, there are particular conditions imposed upon this example that allow you to eliminate one of the loops entirely, leaving just two loops and thus two equations and two unknowns to worry about. I refer, of course, to the fact that the problem specifies that the current in the ammeter loop is zero. It is also the case that the only voltage source in the circuit belongs to both remaining loops, the remainders of which consist of two resistors in series. So you end up with two equations that don't even need to be solved simultaneously!

Right! So if I solve the equation I made for the ammeter, it would naturally equal zero! I'd get zero = zero!

In the attached figure, note that the potential difference between the labeled points c and d is fixed by the battery. We are told that i3 is zero. This means that the currents i1 and i2 are entirely determined by the battery potential and the resistors of the given loop, and that these currents are entirely independent of each other -- the fixed battery voltage isolates one loop from the other, since no voltage variations can occur across an ideal voltage source.

We're dealing with basic electricity/circuits, I presume we're initially only dealing with ideal voltage sources?

Although, I have to say, I didn't know that it's because there is a fixed battery voltage that one loop can be isolated from the other.

So it remains to write the two equations for i1 and i2, and using i1 and i2, determine the potential at points a and b (with respect to d, say), and equate the potentials.

Wait a second! I agree with your i2 loop, but your i1 loop is opposite to what it should be, no? It goes from minus to plus!
One, you are always free to ground any point in the circuit

I'm not sure what you mean by "grounding any point in the circuit". Can you explain or provide me with at least a referrence link otherwise?


Two, there is no current through the ammeter which means don't worry about it.
If I had a system of garden hoses and I told you to find out how much gravel was in each garden hose (making a particular flow rate), and if one of these garden hose sections happened to have its pressure completely balanced so there was no flow, you'd leave it alone and just call it good (i.e. ignore the ammeter because it makes no sense to try to include it)

I had thought so initially! :) That's why in my initial suggestion where I started this topic I made 3 equations without the ammeter.

*I'm not sure if I follow your garden hose analogy, but I'll think about it.


I'm kinda rushing this reply because I have to go and would probably only reply back tomorrow.

Thanks everyone! :)
 
  • #50
I like Serena said:
Yes. You can make up loops like that anywhere. :)

Note that if you have a circuit without voltage source, there will be no current, but Kirchhoff's voltage law will still hold, although in that case it's rather trivial. In this case it is not trivial.





Yes. You have the correct 5 equations.

Note that you only need to solve equations 4 and 5 (as ehild noted) to find Rx.


Let me summarize:

Equation #4: +I2R2 -I1R1 = 0
Equation #5: -I2Rx +I1R3 = 0


Can you try and eliminate I1 and I2?


I got to run-- just saw this! Will reply to this tomorrow! :)
 
  • #51
Well, ILS has been guiding me through to get to these equations so I thought that makes sense and that he's probably making a point to me knowing him :)

Let me outline my methodology.
Feel free to agree or disagree. o:)


Step 1: Use a sure fire method to create a set of equations.
In particular one that the OP started with, and is comfortable with (KVL).
On the way clarifying any issues that come along. (I think we resolved a couple of those! :smile:)

Step 2: Solve the set of equations yielding the answer (which should be good for a first Ooooh!).

Step 3: Look back at the problem, see what happened, and see how else it might have been done.

Step 4: Use an alternate and perhaps easier method and verify the result (coming perhaps to an Oooooooooooooh! :cool:)


As yet, I'm still at step 1, and I think the OP's understanding has increased significantly about how circuits and Kirchhoff's laws work. :smile:

I'll have to admit it took longer than I at first expected.
However, I prefer not to rush as long as there is no deadline. :wink:
 
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  • #53
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 want to 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. :(


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

Heh ;) Do you know Hebrew then?
 
  • #54
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:

Wait a second! I agree with your i2 loop, but your i1 loop is opposite to what it should be, no? It goes from minus to plus!
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.
 
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  • #55
Femme_physics said:
Heh ;) Do you know Hebrew then?
Doesn't everyone? :smile:
I recognize the characters and have rudimentary vocabulary from self-study אולפן.
But in this case, http://translate.google.com/#" came to the rescue. :blushing:

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

Mindscrape said:
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).
 
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  • #56
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?
 
  • #58
Sorry, I lost the thread of this thread!

My worst pun to date ;)

I'll get to it and post back
 
  • #59
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!
 
  • #60
Femme_physics said:
I see 3 unknowns. Rx, I1 and I2. Not 2 unknowns!

*Sigh* ;)

Just try it!

[edit] (You procrastinator you! :wink:) [/edit]
 
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  • #61
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!



http://img221.imageshack.us/img221/9679/hopuui.jpg [Broken]

But now I still have too many unknowns. 2 unknowns 1 equation! I1 and Rx in this case are the unknown.
 
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  • #62
I like Serena said:
[edit] (You procrastinator you! :wink:) [/edit]

I'll work to crack it today :)
 
  • #63
Femme_physics said:
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?
 
  • #64
dilbert20021117procrastination.jpg
 
  • #65
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!

http://img862.imageshack.us/img862/3010/12634938.jpg [Broken]http://img69.imageshack.us/img69/8012/95121367.jpg [Broken]

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...
 
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  • #66
Why minus?!

ehild
 
  • #67
My bad, discount the minus. I was excited I was getting somewhere :P

Rx = R3R2/R1

Is that right?
 
  • #68
Femme_physics said:
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! :smile:

Wait! Sorry! No it is a mock! :wink:
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! :smile:
Femme_physics said:
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.
 
  • #69
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.
 
  • #70
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 :)
 
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<h2>1. What is an ammeter and how does it work?</h2><p>An ammeter is a device used to measure the flow of electric current in a circuit. It works by connecting it in series with the circuit, which means the current passes through the ammeter and the ammeter measures the amount of current flowing through it.</p><h2>2. How do you calculate the unknown resistance in a circuit problem?</h2><p>To calculate the unknown resistance in a circuit problem, you can use Ohm's Law which states that resistance (R) is equal to voltage (V) divided by current (I). So, R = V/I. You can also use the formula R = (V1 - V2)/I, where V1 and V2 are the voltages measured across the unknown resistance.</p><h2>3. Can an ammeter be connected in parallel in a circuit?</h2><p>No, an ammeter should always be connected in series in a circuit. Connecting it in parallel can damage the ammeter and give incorrect readings.</p><h2>4. What is the purpose of using an ammeter in a circuit?</h2><p>The main purpose of using an ammeter in a circuit is to measure the amount of current flowing through the circuit. This is important for understanding the behavior of the circuit and for troubleshooting any issues.</p><h2>5. How can you ensure accuracy when using an ammeter in a circuit?</h2><p>To ensure accuracy when using an ammeter in a circuit, it is important to make sure that the ammeter is connected in series and that the correct range is selected. It is also important to check for any loose connections and to use a high-quality ammeter with a calibration certificate.</p>

1. What is an ammeter and how does it work?

An ammeter is a device used to measure the flow of electric current in a circuit. It works by connecting it in series with the circuit, which means the current passes through the ammeter and the ammeter measures the amount of current flowing through it.

2. How do you calculate the unknown resistance in a circuit problem?

To calculate the unknown resistance in a circuit problem, you can use Ohm's Law which states that resistance (R) is equal to voltage (V) divided by current (I). So, R = V/I. You can also use the formula R = (V1 - V2)/I, where V1 and V2 are the voltages measured across the unknown resistance.

3. Can an ammeter be connected in parallel in a circuit?

No, an ammeter should always be connected in series in a circuit. Connecting it in parallel can damage the ammeter and give incorrect readings.

4. What is the purpose of using an ammeter in a circuit?

The main purpose of using an ammeter in a circuit is to measure the amount of current flowing through the circuit. This is important for understanding the behavior of the circuit and for troubleshooting any issues.

5. How can you ensure accuracy when using an ammeter in a circuit?

To ensure accuracy when using an ammeter in a circuit, it is important to make sure that the ammeter is connected in series and that the correct range is selected. It is also important to check for any loose connections and to use a high-quality ammeter with a calibration certificate.

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