Finding resistance and its change from a V-I graph

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Homework Help Overview

The discussion revolves around determining the electrical resistance of a component based on a voltage-current (V-I) graph. Participants are examining how resistance is defined and calculated, particularly at specific voltage points V1 and V2, and how it may change between these points.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are exploring the definition of resistance, questioning whether it should be calculated as the slope of the V-I graph or through other means. There are inquiries about the implications of an open circuit at V1 and how that affects resistance calculations at V2.

Discussion Status

The discussion is active with various interpretations being explored. Some participants are seeking clarity on the definitions of static versus dynamic resistance, while others are debating the implications of the graph's characteristics on resistance values. There is no explicit consensus, but several lines of reasoning have been presented.

Contextual Notes

Participants note that the question lacks clarity regarding the type of resistance being asked for, leading to conflicting interpretations. The nature of the circuit at V1 is also under scrutiny, with discussions about whether it can be considered open or not.

Raymond
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Homework Statement


The question reads:
The following graph shows the relationship between voltage V and current I, through a particular electrical component. (The line in graph is straight, not showing properly in the photo)

What is the electrical resistance of this component when the potential difference across it is V2 volts, and how does this resistance change between V1 and V2?

Homework Equations

The Attempt at a Solution


This questions was in an exam and there are conflicting answers. I'd like to know what others think the answer is and why.
 

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In terms of the graph, what is the resistance (i.e. slope, intercept, inverse slope, etc.)?
 
Raymond said:
This questions was in an exam and there are conflicting answers. I'd like to know what others think the answer is and why.
What are those answers and the reasoning behind them? More importantly, what is your answer and your reasoning behind it?
 
gneill said:
What are those answers and the reasoning behind them? More importantly, what is your answer and your reasoning behind it?
I personally think the answer should be C because to work out the resistance you just read it off the graph for voltage V2 which is I2. So the resistance should be V2/I2 and for the second part I believe it decreases because at point V1 the resistance is infinite (open circuit) and then it is conducting so resistance has decreased. However the teachers at the school are saying the correct answer is B and I could not follow their reasoning.
 
Raymond said:
I personally think the answer should be C because to work out the resistance you just read it off the graph for voltage V2 which is I2. So the resistance should be V2/I2 and for the second part I believe it decreases because at point V1 the resistance is infinite (open circuit) and then it is conducting so resistance has decreased. However the teachers at the school are saying the correct answer is B and I could not follow their reasoning.
Under that reasoning, what is the resistance at V1? Does that make any sense?
 
The teachers' answer is going by the definition of resistance, which is not the same as what you might think from a casual application of Ohm's Law for simple ohmic materials (such as typical resistors). Take a look here:

https://www.av8n.com/physics/resistance.htm

One problem I have with the question as it is given is that the resistance at the point on the curve at V1 is not well defined since the curve does not have a unique derivative (slope) at that point.
 
Isaac0427 said:
Under that reasoning, what is the resistance at V1? Does that make any sense?
There's no current going through at that point. so the circuit is open and hence the resistance is infinite.
 
gneill said:
The teachers' answer is going by the definition of resistance, which is not the same as what you might think from a casual application of Ohm's Law for simple ohmic materials (such as typical resistors). Take a look here:

https://www.av8n.com/physics/resistance.htm

One problem I have with the question as it is given is that the resistance at the point on the curve at V1 is not well defined since the curve does not have a unique derivative (slope) at that point.
The definition you referred to is more for differential or dynamic resistance not the static one which I believe what the question asks for.

https://en.wikipedia.org/wiki/Elect...onductance#Static_and_differential_resistance

If we just assume this graph was produced by plotting the results of an experiment. Then wouldn't the ammeter be reading I2 when V2 voltage is applied and wouldn't that logically imply that the resistance of the component at the time of applying V2 is V2/I2?
Also in regards to V1, the current going through the component is zero so the it acts as a open switch (just like a diode) so the resistance is infinite
 
It depends upon what definition is applied for resistance. Your answer will depend upon which definition you assume applies. Since the question is not explicit in this, you need to turn to your course text or notes to determine the definition that is intended.

Usually, the definition based on the differential change in potential difference w.r.t. current is assumed.
 
  • #10
Raymond said:
There's no current going through at that point. so the circuit is open and hence the resistance is infinite.
Reread the question. How I am reading it, it is implied that the circuit is not open. If the element were open, how would it have been possible for there to be current across it at V2? Remember, the graph is only of one electrical element. You may just be thinking too much into the question. If I had seen the question on a test, I would have recognized that the resistance is the reciprocal of the slope of that line, and the slope is constant throughout the graph. What level physics is this? By the question, it seems like an algebra-based physics course, in which that approach is what they would be looking for.
 
  • #11
gneill said:
It depends upon what definition is applied for resistance. Your answer will depend upon which definition you assume applies. Since the question is not explicit in this, you need to turn to your course text or notes to determine the definition that is intended.

Usually, the definition based on the differential change in potential difference w.r.t. current is assumed.
I agree , they should have clarified which type of resistance they were after. At this school they only define resistance as voltage over current.
 
  • #12
Isaac0427 said:
Reread the question. How I am reading it, it is implied that the circuit is not open. If the element were open, how would it have been possible for there to be current across it at V2? Remember, the graph is only of one electrical element. You may just be thinking too much into the question. If I had seen the question on a test, I would have recognized that the resistance is the reciprocal of the slope of that line, and the slope is constant throughout the graph. What level physics is this? By the question, it seems like an algebra-based physics course, in which that approach is what they would be looking for.
The circuit is not open but it acts as if open. This characteristic can be seen in diodes where a certain amount of voltage is needed for the current to go through. If the voltage is lower than that no current goes through.
Resistance being reciprocal of slope is not totally correct and doesn't work at all times. It only gives you the dynamic resistance not the static one.
 
  • #13
Perhaps consider what happens if you have a DC voltage (=V2) with a small AC voltage superimposed on it (eg the voltage fluctuates about V2 by a small amount. That will cause a DC current I2 with a small AC current superimposed on it. The small signal resistance will be the slope at V2.

Then do the same thing at (or rather just above) V1.
 

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