# Circuit problem with ammeter and unknown resistance

1. May 14, 2011

### Femme_physics

1. The problem statement, all variables and given/known data

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

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

3. The attempt at a solution

I seem to be getting way too many unknowns!

Rx = 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!

Last edited: May 15, 2011
2. May 15, 2011

### 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

Last edited: May 15, 2011
3. May 15, 2011

### I like Serena

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

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.

Electrobob it is!

4. May 15, 2011

### ehild

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

5. May 15, 2011

### I like Serena

@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.

6. May 15, 2011

### Femme_physics

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

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

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?

*chuckles*

7. May 15, 2011

### I like Serena

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

8. May 15, 2011

### Femme_physics

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

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

9. May 15, 2011

### I like Serena

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

10. May 15, 2011

### ehild

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

:) I do the same mistake quite often.

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

11. May 15, 2011

### Staff: Mentor

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?

12. May 15, 2011

### Femme_physics

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

Fair enough.
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!

Is that right?

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

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

Well, yes, R1 and R2! I believe.

13. May 15, 2011

### sophiecentaur

And two other resistors?

14. May 15, 2011

### I like Serena

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.

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

Last edited: May 15, 2011
15. May 15, 2011

### Femme_physics

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.

Wow, you really are a Jew.

LOL sorry...^^ :)

16. May 15, 2011

### I like Serena

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.

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.

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

17. May 15, 2011

### ehild

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?

I learnt 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 :rofl:

ehild

#### Attached Files:

• ###### bridge.JPG
File size:
15.1 KB
Views:
122
18. May 15, 2011

### Femme_physics

:) I see.
I'll go at it step by step then.

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?

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).

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

:) Old is how you feel, anyway!

19. May 15, 2011

### I like Serena

Finally!

Don't make me *command* you again!

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.... :uhh:

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...

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

Last edited: May 15, 2011
20. May 15, 2011

### ehild

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

ehild
.