Was there a mistake in our final exam question about resistance?

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  • #1
Eitan Levy
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Homework Statement


This graph was given, which describes the current through a resistor, as a function of the voltage.
Then, we were supposed to find the resistance of the resistor when 3V<V<5V.
But it changes throughout this range, doesn't it? Because in the answers (Mind you, this is the FINAL physics exam in the country), it says we were supposed to do: R=ΔV/ΔI, which I believe isn't correct when the line doesn't go through (0,0).
What do you think?

Homework Equations


V=IR

The Attempt at a Solution

 

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  • #2
Ray Vickson
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Homework Statement


This graph was given, which describes the current through a resistor, as a function of the voltage.
Then, we were supposed to find the resistance of the resistor when 3V<V<5V.
But it changes throughout this range, doesn't it? Because in the answers (Mind you, this is the FINAL physics exam in the country), it says we were supposed to do: R=ΔV/ΔI, which I believe isn't correct when the line doesn't go through (0,0).
What do you think?

Homework Equations


V=IR

The Attempt at a Solution


The formula ##R = \Delta V / \Delta I## is correct; just because the tangent line of the graph does not pass through the origin is no reason to reject it. When resistance varies with voltage (or current), the equation ##V = IR## is no longer true.

See, eg., https://physics.stackexchange.com/q...have-to-be-constant-for-ohms-law-to-be-obeyed
 
  • #3
Eitan Levy
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The formula ##R = \Delta V / \Delta I## is correct; just because the tangent line of the graph does not pass through the origin is no reason to reject it. When resistance varies with voltage (or current), the equation ##V = IR## is no longer true.

See, eg., https://physics.stackexchange.com/q...have-to-be-constant-for-ohms-law-to-be-obeyed
I thought that the definition of resistance is literally V/I, what makes it incorrect to use here?
And also my book says the opposite. Here is an example (it's in Hebrew but look at the numbers, it says the resistance varies in the second part). Is the book really straight up wrong?
 

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  • #4
Eitan Levy
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  • #5
Ray Vickson
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Also, using this definition would give us two different resistances in points where the slope changes, how does that make sense?

No, it does not make a lot of sense, but that is likely due to making a crude, piecewise-linear approximation to a curved graph. For a smooth curve ##I = f(V)## the "local" resistance at ##(V_0,I_0)## is ##R = 1/(df /dV)|_{V = V_0}##, and that will typically depend continously on ##V_0##. It will be the same resistance for small increases or small decreases of voltage near the base value ##V_0##.

In your exam's case, the resistance at some voltages is a bit different for increases than for decreases, but as long as you specify the value of ##\Delta V## the answer will be unique.
 
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  • #6
Ray Vickson
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I thought that the definition of resistance is literally V/I, what makes it incorrect to use here?
And also my book says the opposite. Here is an example (it's in Hebrew but look at the numbers, it says the resistance varies in the second part). Is the book really straight up wrong?

I don't read Hewbrew, so I cannot say. However, the link I included tells you the actual situation as it applies to non-ohmic cirduits.

Sometimes introductory textbooks simplify situations, which is no bad thing because Ohm's Law applies very widely to the vast majority of cases; there are exceptions, however.
 
  • #7
Eitan Levy
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I don't read Hewbrew, so I cannot say. However, the link I included tells you the actual situation as it applies to non-ohmic cirduits.

Sometimes introductory textbooks simplify situations, which is no bad thing because Ohm's Law applies very widely to the vast majority of cases; there are exceptions, however.
I can tell you what is says. It says that throughout the second part, the resistance starts at 0.25Ω and ends at 0.3333Ω, and that in this part 0.25Ω≤R≤0.333Ω, which would be nothing but straight up wrong, I believe, according to what you say. I just find it odd.
 
  • #8
Felipe Lincoln
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The resistance actually can vary over voltage and it's definition is indeed ##\mathrm{d}V/\mathrm{d}I##. If you get the graph of V against I of a non-ohmic component like let's say, a lamp, you can see the graph behaving as a curve instead of a straight line, and at each point of the curve you have a different slope, therefore a different resistance.
 
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  • #9
Felipe Lincoln
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I thought that the definition of resistance is literally V/I
It's the definition of Ohm's law, not resistance. It's only a way to determine the resistance of a resistor that obey Ohm's law.
 
  • #10
Ray Vickson
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I can tell you what is says. It says that throughout the second part, the resistance starts at 0.25Ω and ends at 0.3333Ω, and that in this part 0.25Ω≤R≤0.333Ω, which would be nothing but straight up wrong, I believe, according to what you say. I just find it odd.

Right: it tells you something that seems wrong. Between V = 3 v and V = 5 v, the graph looks very much like a straight line, going from about I = 0.25 A at V = 3 v to about I = 0.33 A at V = 5 v, so a change of about 0.08 A in over a voltage change of 2 v; that gives a resistance ##R = 2/0.08 = 25 \, \Omega## over the whole range.
 
  • #11
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I looked this up on Wiki, and they call V/I the chordal resistance while dV/dI is called the differential resistance. Ethan, I thought that your approach to getting the resistance was totally valid, and in agreement with my experience. But Ray has lots of experience, and his expertise needs to be respected. Is it possible that the exact statement of your problem asked for the differential resistance? What was the exact wording of your problem.
 
  • #12
Eitan Levy
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I looked this up on Wiki, and they call V/I the chordal resistance while dV/dI is called the differential resistance. Ethan, I thought that your approach to getting the resistance was totally valid, and in agreement with my experience. But Ray has lots of experience, and his expertise needs to be respected. Is it possible that the exact statement of your problem asked for the differential resistance? What was the exact wording of your problem.
I am so confused right now. It literally said "calculate the resistance", nothing more.
 
  • #13
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I am so confused right now. It literally said "calculate the resistance", nothing more.
Sorry. If that's what it said, then the question is poorly worded (unless they specifically defined resistance in your course as dV/dI).
 
  • #14
Eitan Levy
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Sorry. If that's what it said, then the question is poorly worded (unless they specifically defined resistance in your course as dV/dI).
They didn't, as you can see in the example I provided above. That's why I am confused.
 
  • #15
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They didn't, as you can see in the example I provided above. That's why I am confused.
Your confusion is not your fault. It is their fault. You have a legitimate complaint.
 
  • #16
Merlin3189
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I'd have thought common sense had a part to play here.
On such a V - I curve, how can you have a single ## R = \frac {V} {I} ## resistance, or an ## R = \frac {dV} {dI} ## resistance, over a range of voltages ?
The only resistance that makes any sense to me here, is the incremental resistance ## R = \frac {ΔV} {ΔI} ##

Though I agree the question should be better expressed.
 
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  • #17
Felipe Lincoln
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Yeah... We can also think of ## R=V/I## as ##R=\mathrm{d}V/\mathrm{d}I## passing through origin, it's exactly the same thing. So in ohmic devices we use ## R=V/I## just to evidence the fact that in this case it will pass through (0,0)
 
  • #18
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I'd have thought common sense had a part to play here.
On such a V - I curve, how can you have a single ## R = \frac {V} {I} ## resistance, or an ## R = \frac {dV} {dI} ## resistance, over a range of voltages ?
The only resistance that makes any sense to me here, is the incremental resistance ## R = \frac {ΔV} {ΔI} ##

Though I agree the question should be better expressed.
For materials exhibiting non-linear stress-strain behavior, the Young's modulus is specifically defined as ##E(\epsilon)=\sigma(\epsilon)/\epsilon##, not as ##d\sigma/d\epsilon##. How does that fit in with your "common sense?"
 
  • #19
CWatters
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I'm an electronic graduate and my kids have also done this in physics at school in the past year.

The resistance at a voltage V is not the slope of the IV curve at V. It's the slope of a line from the origin (0,0) to the point V on the curve.

In this case the resistance varies over the range 3-5V. As far as I can see the problem statement doesn't imply a one value answer.
 
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I'm an electronic graduate and my kids have also done this in physics at school in the past year.

The resistance at a voltage V is not the slope of the IV curve at V. It's the slope of a line from the origin (0,0) to the point V on the curve.

In this case the resistance varies over the range 3-5V. As far as I can see the problem statement doesn't imply a one value answer.
Amen Bro.
 
  • #21
haruspex
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The formula ##R = \Delta V / \Delta I## is correct; just because the tangent line of the graph does not pass through the origin is no reason to reject it. When resistance varies with voltage (or current), the equation ##V = IR## is no longer true.
That does not appear to be the consensus.
Here are the first 8 reasonably authoritative-looking hits I got on the net.
Most either define resistance as V/I and dV/dI as "<some qualifier> resistance" or qualify the terms of both V/I and dV/dI.
Only one defines unqualified "resistance" as dV/dI.

http://www.physics.csbsju.edu/trace/CC.html:
"For any two-terminal device we can rearrange this relationship to define a "voltage dependent resistance": R = V/I
although more often the "dynamic resistance" r = dV/dI is the useful concept."

https://www.physicsforums.com/threads/which-is-the-resistance-1-slope-or-v-i-in-i-v-graph.73312/:
Jeff273 "The resistance of anything is R = V/I"
James R: "The resistance at any point is defined to be V/I at that point."

https://en.wikipedia.org/wiki/Electrical_resistance_and_conductance:
"The resistance (R) of an object is defined as the ratio of voltage across it (V) to current through it (I)..
In other situations, the derivative ##\frac {dV}{dI}## may be most useful; this is called the "differential resistance".
...
  • Static resistance (also called chordal or DC resistance) – This corresponds to the usual definition of resistance; the voltage divided by the current"
  • Differential resistance (also called dynamic, incremental or small signal resistance) – Differential resistance is the derivative of the voltage with respect to the current;
https://www.researchgate.net/post/Resistance_measurement_from_IV_curves:
Christian Binek: "For a non-linear I-V curve the resistance is still defined as R=V/I. ... it is useful to defined the differential resistance R=dV/dI for non-linear I-V characteristic. "

https://www.physics.purdue.edu/~clarkt/Courses/Physics271L/Exp3/Exp3.html:
"static resistance R is defined by
(1) R=V/I.
R is not necessarily a constant if I and V change. In dealing with components for which R varies, it is sometimes useful to work with the dynamic resistance r, defined by :
(2) r=dV/dI "

https://www.quora.com/Can-we-measure-the-resistance-of-a-non-linear-load
Uneet Kumar Singh: "you can calculate resistance for a given set of instantaneous voltage and current simply by dividing voltage by current. "
Gonzalo Perez: "R=dV/dI ; i.e. Resistance is the slope of the V/I curve at a given point."
 
  • #23
Tom.G
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Which version is more useful in practice?
Depends on the practice.
Consider a Zener diode voltage regulator circuit.

To find the value of the series dropping resistor you consider from the origin to the operating point of the Zener.
To find the load regulation you consider the dynamic resistance of the Zener around the operating point.

It seems that both approaches are "Correct."

Cheers,
Tom
 
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  • #24
haruspex
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Depends on the practice.
Consider a Zener diode voltage regulator circuit.

To find the value of the series dropping resistor you consider from the origin to the operating point of the Zener.
To find the load regulation you consider the dynamic resistance of the Zener around the operating point.

It seems that both approaches are "Correct."

Cheers,
Tom
No, it shows that both measures have value. The question in this thread is which is meant by the use of the unqualified term “resistance". It seems that most authorities would qualfy the dV/dI version as "differential", or "dynamic" or whatever. (I would have called it "marginal".)
 
  • #25
CWatters
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Skip to 7min50..

 
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  • #26
CWatters
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No, it shows that both measures have value. The question in this thread is which is meant by the use of the unqualified term “resistance". It seems that most authorities would qualfy the dV/dI version as "differential", or "dynamic" or whatever. (I would have called it "marginal".)
Perhaps the "small signal resistance".
 
  • #27
Merlin3189
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For materials exhibiting non-linear stress-strain behavior, the Young's modulus is specifically defined as ##E(\epsilon)=\sigma(\epsilon)/\epsilon##, not as ##d\sigma/d\epsilon##. How does that fit in with your "common sense?"
Common sense should have told me not to poke a stick into a hornets' nest! But common sense is not so common.

So I take it by analogy with your definition of E, that ## R(i) = \frac {V(i)} {i} \ \ \ and\ not \ \ \frac {dv} {di} \ at \ i ##
So it leaves me with the same problem of what meaning to give to ## R( 3 ≤ v ≤ 5) ## or to ## E(3 ≤ \epsilon ≤ 5) ##
Please forgive my style here. I don't know the mathematical notation, if any, which expresses the question properly.

My naive observation was simply that, if we can't calculate a single value using ## R= \frac {V} {I} ## nor using ## R= \frac {dv} {di} ## , but we can calculate a single value for ## R( a ≤ v ≤ b) = \frac {Δi(v)} {Δv} = \frac {i(b) - i(a)} {b - a} ## then the definition that works could be the one that was intended.
Since the value it gives seems a useful one, I feel it ought to be available to us to use. Whether it needs another name, or indeed already has one, is another question.

I don't know if a value of ## E(\epsilon ) ## calculated in that way would be as useful as the electrical equivalent: I'd think it would mean something like, the increment of stress required to effect extension (or compression) between the values specified.

As far as I can see the problem statement doesn't imply a one value answer.
Yes.
I'd taken the OP's complaint that
we were supposed to find the resistance of the resistor when 3V<V<5V.
But it changes throughout this range, doesn't it?
to imply that he thought he needed a single value.
 
  • #28
Eitan Levy
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Common sense should have told me not to poke a stick into a hornets' nest! But common sense is not so common.

So I take it by analogy with your definition of E, that ## R(i) = \frac {V(i)} {i} \ \ \ and\ not \ \ \frac {dv} {di} \ at \ i ##
So it leaves me with the same problem of what meaning to give to ## R( 3 ≤ v ≤ 5) ## or to ## E(3 ≤ \epsilon ≤ 5) ##
Please forgive my style here. I don't know the mathematical notation, if any, which expresses the question properly.

My naive observation was simply that, if we can't calculate a single value using ## R= \frac {V} {I} ## nor using ## R= \frac {dv} {di} ## , but we can calculate a single value for ## R( a ≤ v ≤ b) = \frac {Δi(v)} {Δv} = \frac {i(b) - i(a)} {b - a} ## then the definition that works could be the one that was intended.
Since the value it gives seems a useful one, I feel it ought to be available to us to use. Whether it needs another name, or indeed already has one, is another question.

I don't know if a value of ## E(\epsilon ) ## calculated in that way would be as useful as the electrical equivalent: I'd think it would mean something like, the increment of stress required to effect extension (or compression) between the values specified.


Yes.
I'd taken the OP's complaint that to imply that he thought he needed a single value.
In the answers only one answer was given.
 
  • #29
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In the answers only one answer was given.
You talk about it as if the definition of resistance is supposed to be carved in stone. There are obviously more than one definition that people subscribe to, and neither of them is generally accepted by all as the convention. Live with it. This is not the only time you are going to encounter ambiguity in you science studies.
 
  • #30
jtbell
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(Mind you, this is the FINAL physics exam in the country)
So this is a standard exam that every student in your country takes? (e.g. for exit from high school or for a standardized university introductory physics course)

If yes, then there must be a standard curriculum that your instructor/professor is supposed to follow, which defines resistance as:

it says we were supposed to do: R=ΔV/ΔI

If your instructor did not do that, then your problem is with the instructor. If he mentioned R = V/I, he should have discussed the considerations that were mentioned in this thread, and made clear which definition you should expect to use on the exam.
 
  • #31
CWatters
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In the answers only one answer was given.
In which case I'd want to see the exact wording of the question. At the very least it's unclear in the OP.

You could try writing to the examining board but most won't enter into correspondence. I'd be tempted to find a forum for professional electrical/electronic engineers in your country and post the question there. Dont post the answer initially, just see what replies you get. Perhaps see if you can't interest them in writing to the board to complain about the quality of the question.

Here in the UK the correct answer for this in a GCSE Plysics exam (taken at age 16) would be the range of values given by R=V/I.
 
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