Circuit problem with ammeter and unknown resistance

[Femme_physics]

Homework Statement

http://img811.imageshack.us/img811/6902/circuittosolve.jpg

This circuit includes 4 resistors whose value of 3 of them is known and the forth resistor, R_x is unknown. They connected an ammeter A to the circuit and the current through it is 0.

1) Calculate R_x.
Is the value of R_x depends on the ammeter resistance A? Explain.
2) Write the name of the circuit

The Attempt at a Solution

I seem to be getting way too many unknowns!

R_x = unknown
I₀ unknown
I₁ = unknown
I₂ = unknown
V = unknown

I'm not sure how to derive more equations. Or how to use the ammeter. If I'm told it has 0 current, than it must have 0 voltage, and must have 0/0 resistance! So, how do I even treat it? And why do they mean by the "name" of the circuit? I personally named him Electrobob before I started the exercise and I refuse to change it!
http://img810.imageshack.us/img810/7382/toomanyunknowns.jpg

 

[ehild]

Well, it is a very well known "bridge" circuit, used to measure resistance, try to find its real name.
There is a mistake in your third equation. You wrote Rx instead of R3.
Zero current across the ammeter means zero potential difference across its terminals. How are the potential differences across R1 and R2 related?

ehild

 

[I like Serena]

I seem to be getting way too many unknowns!

R_x = unknown
I₀ unknown
I₁ = unknown
I₂ = unknown
V = unknown

I'm not sure how to derive more equations.

Well you didn't draw a loop yet at the top along the ammeter...
That would give you the extra equation you need.

Or how to use the ammeter. If I'm told it has 0 current, than it must have 0 voltage, and must have 0/0 resistance! So, how do I even treat it?

Indeed, from this you can not deduce the resistance.

However, an ammeter is designed to measure the current through a wire.
To do that it's not supposed to alter the current, so its resistance must be very low.
You can treat it as a short circuit, that is, having a resistance of zero.

And why do they mean by the "name" of the circuit? I personally named him Electrobob before I started the exercise and I refuse to change it!

I like your name!
Electrobob it is!

 

[ehild]

However, an ammeter is designed to measure the current through a wire.
To do that it's not supposed to alter the current, so its resistance must be very low.
You can treat it as a short circuit, that is, having a resistance of zero.

Those ammeters which are designed to read very low currents and detect balance at zero current, usually do not have very low resistance. It can be a few hundred ohms.

ehild

 

[I like Serena]

Those ammeters which are designed to read very low currents and detect balance at zero current, usually do not have very low resistance. It can be a few hundred ohms.

ehild

@ehild: Thanks :)
I didn't know that yet.

@FP: Either way, you do not need to know the resistance.
Since the current is zero, the voltage across it is zero, so its contribution to the voltage law is zero.

 

[Femme_physics]

Well, it is a very well known "bridge" circuit, used to measure resistance, try to find its real name.

Oh! I see. Wiki's been a dear :) It's Wheatstone bridge!

There is a mistake in your third equation. You wrote Rx instead of R3.

That's what you get when you copy-paste :-/ Thanks!

Zero current across the ammeter means zero potential difference across its terminals. How are the potential differences across R1 and R2 related?

You mean that they're parallel and I should combine them like parallel resistors?

Well you didn't draw a loop yet at the top along the ammeter...
That would give you the extra equation you need.

But what's the point of drawing a loop through the ammeter of no current goes through it?

I like your name!
Electrobob it is!

*chuckles*

 

[I like Serena]

You mean that they're parallel and I should combine them like parallel resistors?

I'm afraid that won't help you.

But what's the point of drawing a loop through the ammeter of no current goes through it?

It's not about the ammeter, it's about the voltage law equation with I1, R1, I2 and R2.

 

[Femme_physics]

I'm afraid that won't help you.

Yea, I didn't think it would. I was just wondering what ehild means.

It's not about the ammeter, it's about the voltage law equation with I1, R1, I2 and R2.

That would give me a 4th equation. But, I have 5 unknowns. I should need 5 equations, right?

 

[I like Serena]

That would give me a 4th equation. But, I have 5 unknowns. I should need 5 equations, right?

Not in this case. You'll see...

 

[ehild]

Oh! I see. Wiki's been a dear :) It's Wheatstone bridge!

Nice job! Old people used to learn about it and doing lab measurement as students "Measuring resistance with Wheatstone Bridge".

That's what you get when you copy-paste :-/ Thanks!

:) I do the same mistake quite often.

You mean that they're parallel and I should combine them like parallel resistors?


But what's the point of drawing a loop through the ammeter of no current goes through it?

If you want to calculate the current through the battery you can consider R1 parallel to R2, Rx parallel to R3. But you do not need that. You have to calculate Rx. Use the information given, that the current is zero through the ammeter. Write the loop equation for the ammeter circuit.

ehild

 

[gneill]

Are 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?

If so, can you recognize two such voltage dividers in your Wheatstone Bridge circuit?

 

[Femme_physics]

I meant to reply sooner but I got cut off internet connection for a bit.

If you want to calculate the current through the battery you can consider R1 parallel to R2, Rx parallel to R3. But you do not need that. You have to calculate Rx. Use the information given, that the current is zero through the ammeter. Write the loop equation for the ammeter circuit.

ehild

Fair enough.

Not in this case. You'll see...

Okay.

The red arrows are suppose to indicate the loop to which I'm considering KVL-- I'm perfectly aware there's no current flowing through the ammeter section!

http://img269.imageshack.us/img269/6941/cirmh.jpg

Is that right?

Nice job! Old people used to learn about it and doing lab measurement as students "Measuring resistance with Wheatstone Bridge".

That's how "you" learned about that? Not that I'm saying you're old.. you're saying...said..er... *embarrassed*

Are 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?

Makes perfect sense. Doesn't that come from parallel connection of resistors?

If so, can you recognize two such voltage dividers in your Wheatstone Bridge circuit?

Well, yes, R₁ and R₂! I believe.

 

[sophiecentaur]

And two other resistors?

 

[I like Serena]

The red arrows are suppose to indicate the loop to which I'm considering KVL-- I'm perfectly aware there's no current flowing through the ammeter section!

Is that right?

That's fine, the arrows do not have to match the actual current.
They're only defining which way you have chosen to be +.

But no, it is not right.
With your choice for the loop you have forgotten the voltage source.

I meant to reply sooner but I got cut off internet connection for a bit.

That's fine. I've got cut off as well.

 

[Femme_physics]

But no, it is not right.
With your choice for the loop you have forgotten the voltage source.

Right. My bad. But, other than that I got it right. Now I can only use the counterclockwise loop with the ammeter (I presume that's what sophiecentaur meant when he asked "and the two other resistors?").
That would get me yet another equation. This is the most equations with the most unknowns I ever had to solve in my life!

Is it always like that in electronics? And, is that the only way to solve this problem? Ohm's law appears to be completely useless in getting the answers.

That's fine. I've got cut off as well.

Wow, you really are a Jew.

LOL sorry...^^ :)

 

[I like Serena]

Right. My bad. But, other than that I got it right. Now I can only use the counterclockwise loop with the ammeter (I presume that's what sophiecentaur meant when he asked "and the two other resistors?").
That would get me yet another equation. This is the most equations with the most unknowns I ever had to solve in my life!

Well, you could have done with 1 equation less, but oh well. ;)

And no, gneill and sophiecentaur meant something else.
It's another way of reasoning out the answer.

I didn't mention that way yet, but was planning to come to it, when you were done with Kirchhoff. I just didn't want to confuse you...

Either way, it's good practice for setting up equations and solving them.

Is it always like that in electronics? And, is that the only way to solve this problem? Ohm's law appears to be completely useless in getting the answers.

Well, you're applying Ohm's law each and every time you're applying Kirchhoff's voltage law.

And we'll get back to Ohm's law in a minute with an easier way to find the answer.

Wow, you really are a Jew.

LOL sorry...^^ :)

Uhh? Is that a reference to the Jewish birth ritual?

 

[ehild]

The red arrows are suppose to indicate the loop to which I'm considering KVL-- I'm perfectly aware there's no current flowing through the ammeter section!
Is that right?

There is no current through the ammeter so no voltage across it, but you included the battery in the circuit. Why don't you use the much simpler red loop?

That's how "you" learned about that? Not that I'm saying you're old.. you're saying...said..er... *embarrassed*

I learned about Wheatstone bridge at high school first, more than fifty years ago, and later at the university, and later I taught it... but I am not at all old

ehild

 

[Femme_physics]

Well, you could have done with 1 equation less, but oh well. ;)

And no, gneill and sophiecentaur meant something else.
It's another way of reasoning out the answer.

I didn't mention that way yet, but was planning to come to it, when you were done with Kirchhoff. I just didn't want to confuse you...

Either way, it's good practice for setting up equations and solving them.

:) I see.
I'll go at it step by step then.
I'll start with the equations.

There is no current through the ammeter so no voltage across it, but you included the battery in the circuit. Why don't you use the much simpler red loop?

How can I use this loop if it has no voltage source?

KVL states: "The directed sum of the electrical potential differences (voltage) around any closed circuit is zero."

Voltage! There must be voltage source in a KLV loop, no?

Well, you're applying Ohm's law each and every time you're applying Kirchhoff's voltage law.

And we'll get back to Ohm's law in a minute with an easier way to find the answer.

Right, but I didn't directly solve it with the voltage drop methodology (what's the official name for this solving tactic? it can't be "ohm's law" since "ohm's law" just represent the formula, not a solving methodology).

Uhh? Is that a reference to the Jewish birth ritual?

LOL was it silly? I really should silly slap myself over that. Seriously, every time I make a pun a kitten dies.

I learned about Wheatstone bridge at high school first, more than fifty years ago, and later at the university, and later I taught it... but I am not at all old

:) Old is how you feel, anyway!

 

[I like Serena]

:) I see.
I'll go at it step by step then.
I'll start with the equations.

Finally!

Don't make me *command* you again!

How can I use this loop if it has no voltage source?

KVL states: "The directed sum of the electrical potential differences (voltage) around any closed circuit is zero."

Voltage! There must be voltage source in a KLV loop, no?

Hmm, I don't see anything about the KLV loop needing to have a voltage source?
It only talks about voltage differences in general I think...

Right, but I didn't directly solve it with the voltage drop methodology (what's the official name for this solving tactic? it can't be "ohm's law" since "ohm's law" just represent the formula, not a solving methodology).

Uhh... I dunno.
I have to admit I just invented the term "Voltage drop methodology" myself, I thought it sounded sort of cool.

Basically they're all variations of applying Ohm's law. I'm not aware of any names for methodologies here...

LOL was it silly? I really should silly slap myself over that. Seriously, every time I make a pun a kitten dies.

Well? Did you? Silly slap yourself I mean?
Now do you understand what I mean with RPGSM?

 

[ehild]

How can I use this loop if it has no voltage source?

KVL states: "The directed sum of the electrical potential differences (voltage) around any closed circuit is zero."

Voltage! There must be voltage source in a KLV loop, no?

No.
You are just applying Kirchhoff's circuit laws. see: http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws.

ehild
.

 

[Femme_physics]

Hmm, I don't see anything about the KLV loop needing to have a voltage source?
It only talks about voltage differences in general I think...

No.
You are just applying Kirchhoff's circuit laws. see: http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws.

ehild

But where is the voltage difference from? I'm used to seeing a plus and a minus when I use Kirchhoff's circuit laws. I don't see a plus and a minus designated in ehild's red-marked loop. That's confusing.

Uhh... I dunno.
I have to admit I just invented "Voltage drop methodology" myself, I thought it sounded sort of cool.

Basically they're all variations of applying Ohm's law. I'm not aware of any names for methodologies here...

Then I'll call it that too until someone correct me ;) Stuff "you" invent are always cool! And if you give it you ILS seal of approval, it's in my mind as though the science counsel approved of it officially :)

Well? Did you? Silly slap yourself I mean?

Are you enjoying my self-RPGSM?^^

*smacky-slappity-slap-slap*

 

[I like Serena]

But where is the voltage difference from? I'm used to seeing a plus and a minus when I use Kirchhoff's circuit laws. I don't see a plus and a minus designated in ehild's red-marked loop. That's confusing.

The fact that you're looking at an isolated loop doesn't mean the rest of the circuit isn't there anymore. It's just that for any loop, with or without voltage source, KVL must hold.

You don't need a plus or a minus.

What you need is a drawing of the entire circuit, marked with the currents and the direction you have chosen for those currents.
Note that in a loop the currents do not even have to point in the same direction.

With this you can choose any loop and apply KVL, taking into account the proper plus or minus depending on the chosen direction of the currents.

Then I'll call it that too until someone correct me ;) Stuff "you" invent are always cool! And if you give it you ILS seal of approval, it's in my mind as though the science counsel approved of it officially :)

(Be careful! You're making me feel like an expert, and I haven't even made it to homework helper yet!)

Are you enjoying my self-RPGSM?^^

*smacky-slappity-slap-slap*

Yep!
You had that one coming!

 

[gneill]

Something that may simplify your life a bit is to recognize that, anytime that you know that a branch of a network is passing zero current, that branch can be ignored (simply remove it from the diagram!) as far as loop equations are concerned (This assumed that the branch contains finite valued components -- no open switches or infinite resistors!). Further, if the branch is passing zero current then you know that the the two nodes where the branch connects must be at the same potential (so that there's no potential difference that would make current flow in the branch).

In this problem you are told that the ammeter is passing zero amps; there;s no current. So, removing the ammeter and labeling the points where it connected Va and Vb, we conclude that Va must equal Vb. The trick then is to find Rx that makes this happen.

There are a few possible approaches to this. In one approach you write loop equations for the two obvious loops (Battery--R1--R3--Battery; Battery--R2--Rx--Battery), solve them for currents and determine expressions for Va and Vb. Equate Va and Vb, solve for Rx.

In another approach you spot right away that there are two independent voltage dividers operating and that the calculation of Va and Vb can be written directly by inspection.

Moreover, once you recognize that it's voltage dividers at play and that they should have the same potential, and knowing how the 'tap' potential of a voltage divider depends only on the ratio of the resistor values forming the divider, then you can write the expression for Rx immediately, too!

Here's the circuit diagram redrawn to accentuate the loops and dividers I mention. The ammeter has been "deleted" because it draws no current when Va = Vb.

 

[Femme_physics]

I'm going to class soon, but I will write your replies on paper (I wish I had a printer at work) to read it on the bus and I'll probably get back to you probably only tomorrow morning since I'll be too tired to reply tonight. Thanks for your time and effort, gneill, ehild, ILS (also sophiecentaur). You're all amazing in helping me through this. Electronics is a fascinating and beautiful science.

Kodus! And BBL :)

 

[Femme_physics]

I eventually fell asleep on the bus, but I did copy your entire reply to paper. :) [this is what you do when you have no iphone! :( ]

(proof!)

http://img5.imageshack.us/img5/7664/proofsk.jpg Now it's morning so I'm ready to take a deep breath and look at this exercise again.

The fact that you're looking at an isolated loop doesn't mean the rest of the circuit isn't there anymore. It's just that for any loop, with or without voltage source, KVL must hold.

You don't need a plus or a minus.

Really? Okay, then I've just drawn a circuit. If I use KVL on this drawn loop, does it mean the V I should get in the sum of all V is in fact the V after R₁ voltage drop. Its plus end is B, its minus end is C. Hmm... if I'm right then things are beginning to make a bit more sense.

http://img29.imageshack.us/img29/2674/10vyc.jpg

http://img832.imageshack.us/img832/7233/sigmav.jpg

Here's the circuit diagram redrawn to accentuate the loops and dividers I mention. The ammeter has been "deleted" because it draws no current when Va = Vb.

But now we've reduced the problem to 2 equations, 2 unknowns. I'm confused, you and ILS are saying two different things it seems to me. I should keep the ammeter and get enough equations, or I should remove the ammeter and not get enough equations. What ILS says seems to make a bit more sense to me. Isn't it best if I use the ammeter so that at least gives me the number of equations I need?

In one approach you write loop equations for the two obvious loops (Battery--R1--R3--Battery; Battery--R2--Rx--Battery), solve them for currents and determine expressions for Va and Vb. Equate Va and Vb, solve for Rx.

2 equations are not enough equations!

R_x = unknown
I₀ unknown
I₁ = unknown
I₂ = unknown
V = unknownhttp://img818.imageshack.us/img818/2139/correctedequ.jpg

 

[I like Serena]

I eventually fell asleep on the bus, but I did copy your entire reply to paper. :) [this is what you do when you have no iphone! :( ]

(proof!)

I really like these pictures!

Now it's morning so I'm ready to take a deep breath and look at this exercise again.

Really? Okay, then I've just drawn a circuit. If I use KVL on this drawn loop, does it mean the V I should get in the sum of all V is in fact the V after R₁ voltage drop. Its plus end is B, its minus end is C. Hmm... if I'm right then things are beginning to make a bit more sense.

The V should not be part of your equation, since the voltage source is not in the loop.

And could you draw the currents I2 and I3 too?
That is, labeled arrowheads in the wires?

I ask because I think the signs of the currents mismatch what you intended.

But now we've reduced the problem to 2 equations, 2 unknowns. I'm confused, you and ILS are saying two different things it seems to me. I should keep the ammeter and get enough equations, or I should remove the ammeter and not get enough equations. What ILS says seems to make a bit more sense to me. Isn't it best if I use the ammeter so that at least gives me the number of equations I need?

I'd suggest to keep the ammeter in for now.

Then you should have 4 equations (of which 1 is unnecessary).
And you should have 5 unknowns (of which 1 is unnecessary).

Normally, you wouldn't be ables to solve this, but you'll find that 1 unknown cancels out when you try to solve it.

Could you make a list of all the relevant equations?

 

[Femme_physics]

Looks right to me. looks right to you?

http://img853.imageshack.us/img853/5336/5eq.jpg

Uploaded with ImageShack.us

 

[I like Serena]

Looks right to me. looks right to you?

Almost. ;)

I'm afraid that in equation #4 you've got the sign of I2 wrong.

And I see you've included a 5th equation.
All right.
Well, I'm afraid in this one you've mixed up the resistors and the sign of I2 is wrong.

Could you label the resistors and the currents in each wire?
As far as I'm concerned there's no need to show all the drawings now.
I already know them.
Just 1 drawing will suffice with resistors and currents drawn and labeled (currents with the chosen direction as with an arrowhead).
Note that I do not mean the arrows that you've drawn to indicate the loop.

 

[Femme_physics]

Hmm...are you telling me that in a loop, a current can go in opposite directions? Such as in here (current indicated in red, loop in blue)

http://img850.imageshack.us/img850/6974/oncemore222.jpg

 

[I like Serena]

Hmm...are you telling me that in a loop, a current can go in opposite directions? Such as in here (current indicated in red, loop in blue)

Yep! You've nailed it!

 

[Femme_physics]

Thanks, once again I'll have to get back to it later :) (Need to go help 3 students with mechanics). You rock!

 

[Femme_physics]

Another question with respect to this:

http://img850.imageshack.us/img850/6974/oncemore222.jpg [/QUOTE]Am I marking I2 a plus because when I look at the way the loop is going, it is going AGAINST the current at the point of I2, and therefor I2 is a plus?

 

[I like Serena]

Am I marking I2 a plus because when I look at the way the loop is going, it is going AGAINST the current at the point of I2, and therefor I2 is a plus?

Yes.

What you're calculating is the change in potential in the direction of the loop.
Since current flows from a high potential to a low potential, that means that if you walk against the current, the potential goes up (that is, you have a positive sign).

 

[Femme_physics]

Hmm...inteteresting, so the loop representing how you walk through the current. Duly noted..

Now can I solve these 5 eq?

http://img42.imageshack.us/img42/1457/nowt.jpg

Uploaded with ImageShack.us

 

[I like Serena]

Now can I solve these 5 eq?

You can discard the first equation.
It would give you I0, but you don't need that one, and I0 is not in any of the other equations.

From the second equation you can find V.
Combining that with the third equation you can eliminate V, which is good, since you don't need V.

That leaves combining the result with the fourth equation...

You won't need the 5th equation, but please note that you mixed up the currents in this equation. How?

 

[sophiecentaur]

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.

 

[ehild]

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.

 

[I like Serena]

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!

 

[Femme_physics]

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.

 

[I like Serena]

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! [/edit]

 

[MisterX]

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.

 

[Femme_physics]

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?

 

[I like Serena]

It goes to R3

So it's

Sigma V = 0; R3I1 - RxI2 = 0

Is that it?

Yep!

(Sigh! )

 

[Femme_physics]

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 :)

 

[Femme_physics]

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?

 

[gneill]

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.

 

[Mindscrape]

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.

 

[I like Serena]

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.

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: -I₂R_x +I₁R₃ = 0


Can you try and eliminate I₁ and I₂?

 

[Femme_physics]

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! :)

 

[Femme_physics]

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: -I₂R_x +I₁R₃ = 0


Can you try and eliminate I₁ and I₂?

I got to run-- just saw this! Will reply to this tomorrow! :)