Equivalent resistance of a section of a graphite ring

[A13235378]

Homework Statement

The figure shows an ideal electromotive force generator E equal to 12V, a voltmeter V, also ideal, and a graphite ring that has a constant section and a non-zero resistivity, such that a segment corresponding to an arc of pi / 2 radians has a resistance equal to 10 ohms. This ring is connected to the other elements of the circuit through four connection terminals separated angularly from pi / 2 radians. The equivalent resistance value between A and B of the graphite ring.

Relevant Equations

U=RI
R= pL / A , p = resistivity

I considered that the lengths of pi/4 and 3pi/4 would be in parallel:

1/R = 1/30 + 1/10
R = 7,5 ohm

But the answer is 30 ohm. Where am I missing and how do I calculate the potential of the voltmeter.

 

[berkeman]

Homework Statement:: The figure shows an ideal electromotive force generator E equal to 12V, a voltmeter V, also ideal, and a graphite ring that has a constant section and a non-zero resistivity, such that a segment corresponding to an arc of pi / 2 radians has a resistance equal to 10 ohms. This ring is connected to the other elements of the circuit through four connection terminals separated angularly from pi / 2 radians. The equivalent resistance value between A and B of the graphite ring.
Relevant Equations:: U=RI
R= pL / A , p = resistivity

View attachment 268658

I considered that the lengths of pi/4 and 3pi/4 would be in parallel:

1/R = 1/30 + 1/10
R = 7,5 ohm

But the answer is 30 ohm. Where am I missing and how do I calculate the potential of the voltmeter.

I agree with your answer, at least as far as the problem is stated here. Are you sure you copied it down correctly? And the connections to the ring don't change its resistance. The current that is induced depends just on the resistance of the ring in this case. If the voltmetet were replaced with a current meter, that would short out that section of the ring, but an ideal voltmeter has infinite input impedance...

 

[berkeman]

such that a segment corresponding to an arc of pi / 2 radians has a resistance equal to 10 ohms.

Is there any chance that instead of 10 Ohms it says 40 Ohms?

 

[A13235378]

Is there any chance that instead of 10 Ohms it says 40 Ohms?

No. I believe I copied it right. It's from an Olympics in my country

 

[berkeman]

Can you post an image of the whole problem? Or link to the document? Thanks.

 

[DaveE]

I agree your reasoning makes sense to me. I suspect that the problem is with poor phrasing of the question.

Perhaps they meant the resistance between A-B only around the longer path. Of course, this makes no physical sense unless you cut the ring between A-B. Resistance is a property of the ring (composition, dimensions, etc.); it doesn't matter if you connect voltage or current sources, meters, etc. You may change the resistance of the total circuit, but the ring is still the ring, and it has a stable, well defined resistance between any two points.

 

[DaveE]

Another way to interpret the question (OK, really a different question) is to consider the entire circuit. Perfect voltage sources have an intrinsic resistance of 0 ohms. No matter how much current you make flow in the ring, the voltage between A-B is E volts. Since the voltage doesn't change if you add current (which, BTW, will all flow through the battery). Then the resistance of the CIRCUIT (not the ring) between A-B is 0 ohms.

 

[DaveE]

Also, one way to deal with ambiguity in the wording of questions is to carefully, explicitly explain you reasoning and results. This way you can demonstrate that you really do understand all of the physics, you just don't necessarily understand the way the question was asked. Of course, this only helps if a human is reviewing your work and cares more about the process/knowledge than a numerical answer.

You might say something like "If you cut the ring between A and B, the resistance is 30 ohms. If you cut it anywhere else, the resistance is 10 ohms. But since the ring is complete, both paths in parallel yield an equivalent resistance of 7.5 ohms."

 

[etotheipi]

The natural interpretation of the question is to calculate the equivalent resistance between A and B of:

Another way to interpret the question (OK, really a different question) is to consider the entire circuit. Perfect voltage sources have an intrinsic resistance of 0 ohms. No matter how much current you make flow in the ring, the voltage between A-B is E volts. Since the voltage doesn't change if you add current (which, BTW, will all flow through the battery). Then the resistance of the CIRCUIT (not the ring) between A-B is 0 ohms.

I don't follow this part. Why would the resistance of the circuit be zero? Wouldn't that imply an infinite current through the voltage source?

 

[DaveE]

The natural interpretation of the question is to calculate the equivalent resistance between A and B of:
https://www.physicsforums.com/attachments/268663
I don't follow this part. Why would the resistance of the circuit be zero?

The resistance of the voltage source is 0 ohms. It is in parallel with 7.5 ohms in the ring (from A-B). 0||7.5=0.

Consider what happens to a voltage source of E volts (even E(t), if you like, any independent well defined voltage) when you make a current pass through it. The voltage doesn't change dV/dI =0. This is the definition of 0 ohms, the current creates no change in the voltage.

BTW, this isn't analysis, it is the definition of an independent voltage source. There is a dual argument that shows that independent current sources have infinite impedance.

That would imply an infinite current through the voltage source.

Sorry, I don't follow this. That would require an additional source.

However, you are correct if you mean that if you connect a 1V source in parallel with a 2V source, then the current will be infinite. You have to be careful with ideal lumped element component definitions because of zeros, infinities, and degenerate/contradictory equations. This is why your circuit simulator won't accept a model that has loops of only voltage sources and capacitors, or a node with only inductors and current sources.

edit: "if you connect a 1V source in parallel with a 2V source, then the current will be infinite" No, undefined, really. The configuration just doesn't make sense.

 

[etotheipi]

The resistance of the voltage source is 0 ohms. It is in parallel with 7.5 ohms in the ring (from A-B). 0||7.5=0.

Consider what happens to a voltage source of E volts (even E(t), if you like, any independent well defined voltage) when you make a current pass through it. The voltage doesn't change dV/dI =0. This is the definition of 0 ohms, the current creates no change in the voltage.

But the resistance associated with the voltage source (zero) is in series with the resistance of the ring (7.5 ohms), so you would just have a voltage source connected up to a resistance of 7.5 ohms, giving an effective resistance of 7.5 ohms, not zero?

 

[A13235378]

Can you post an image of the whole problem? Or link to the document? Thanks.

http://olimpiada.ifsc.usp.br/06/OBF2006_F2_3a.pdf
I don't know if it will help because it is in Portuguese

 

[DaveE]

But the resistance associated with the voltage source (zero) is in series with the resistance of the ring (7.5 ohms), so you would just have a voltage source connected up to a resistance of 7.5 ohms, giving an effective resistance of 7.5 ohms, not zero?

I think you are confusing my definitions of "ring resistance (=7.5 ohms)" and the "circuit resistance", which includes the current path through the voltage source.

 

[etotheipi]

I think you are confusing my definitions of "ring resistance (=7.5 ohms)" and the "circuit resistance", which includes the current path through the voltage source.

Ah, okay, you meant "get rid of the part of the bit of the ring from A to B, and find the equivalent resistance between A and B (in which case the voltage source and the rest of the ring are in parallel)".

Sorry I misunderstood, I had trouble visualising what you were talking about and was thinking of something completely different .