Which is the resistance? 1/Slope or V/I? (in I-V graph)

[VirtualPhysicist]

I've just done the experiment on investigating the resistance of a piece of filament. As expected, the dots on the I-V graph shows a curve trend. But what is the resistance of each point on the curve? the 1/slope? or just the x coordinate/y coordinate? What's the difference? Thanks.

 

[faust9]

Ohm's law says that E=R(I). R is ususlly said to be constant (temp effects aside), so changing the current through a resistor causs a likewise change in the voltage...

How does that equate to the good old y=mx+b? (what are the constants, the independent variable and the dependent variable)

 

[VirtualPhysicist]

Thx

I agree. But the filament of the bulb doesn't obey ohm's law, i.e. the relationship (rs) is not linear. It is a coincidence that the slope equals to the quotient of any coordinates on the I-V graph when the rs is linear and the line passes through the origin.

So, in the particular case of a curve, which better states the resistance? inverse of the slope or inverse of I/V? By the way, they have different values.

 

[Jeff273]

The resistance of anything is R = V/I. The resistance of the filament at a given point is the slope (derivative) at that point. There are two ways to get this. You can manually (with ruler and pencil) approximate the slope at a point, or you can curve fit over a protion of the curve around the point of interest and take the derivative at that point. Excel will curve fit (add trend line) reasonably well.

 

[faust9]

Thx

I agree. But the filament of the bulb doesn't obey ohm's law, i.e. the relationship (rs) is not linear. It is a coincidence that the slope equals to the quotient of any coordinates on the I-V graph when the rs is linear and the line passes through the origin.

So, in the particular case of a curve, which better states the resistance? inverse of the slope or inverse of I/V? By the way, they have different values.

The filament does follow Ohm's law; however, one cannot disregard the effect of temperature upon resistance when the device in question is heated to hundreds of degrees.

 

[VirtualPhysicist]

The resistance of anything is R = V/I. The resistance of the filament at a given point is the slope (derivative) at that point. There are two ways to get this. You can manually (with ruler and pencil) approximate the slope at a point, or you can curve fit over a protion of the curve around the point of interest and take the derivative at that point. Excel will curve fit (add trend line) reasonably well.

Thx ag
The resistance of anything is R=V/I and the reistance of the filament at a given point is the slope? Don't they seem to be contradictory? Because the inverse of the slope doen't equal to V/l anyway. Can you tell me why? thanks

 

[VirtualPhysicist]

The filament does follow Ohm's law; however, one cannot disregard the effect of temperature upon resistance when the device in question is heated to hundreds of degrees.

Thx
Does it? I think Ohm's law states that I is proportional to V, that is V/I equals a constant. I hope R = V/I in this case as well

If so, what is the meaning of the inverse of the slope? It has the resistance's unit as well. Or it is the resistance? Then, what is V/I?

 

[James R]

The resistance at any point is defined to be V/I at that point. The resistance of a non-ohmic device is NOT defined by the gradient of the voltage vs. current graph.

 

[VirtualPhysicist]

So what is the meaning of the slope? it has the resistance's unit as well...

 

[faust9]

Yes, the slope of the I/V graph is the resistance at a given value of I. If you were to hold the filament at the highest temp it sees under normal operating conditions the the slope would be linear from I_0 to I_max. The problem you are encountering here is that the filament needs to be heated(thousands of degrees) in order for it to emit light which causes a corresponding--noticable change-- in filament resistance ( http://www.tungsten.com/mtstung.html scroll down the the resistivity section). Regular old 1/4 watt resistors undergo this very same thing and that is one of the reasons for the power ratings on resistors. If you exceed the power rating of a resistor then there is no guarantee that said resistor will be within the resistance band the lines say it should be. If you really exceed the power rating the resistor will take is ball home with it when it goes pop.

Well, hope this helped a little
Good luck.

 

[VirtualPhysicist]

Thx, Helps a lot.

I went to the site which you told. hmm...
Electrical Resistivity @ 20 ºC (microhm-cm) 5.5
Electrical Resistivity @ 227 ºC (microhm-cm) 10.5
Electrical Resistivity @ 727 ºC (microhm-cm) 24.3
Electrical Resistivity @ 1727 ºC (microhm-cm) 55.7
Electrical Resistivity @ 2727 ºC (microhm-cm) 90.4
Electrical Resistivity @ 3227 ºC (microhm-cm) 108.5

How it difines the resistivity for tungsten at a certain temperature? The p.d. across it over the current flowing? or the slope? I think they give different values

 

[James R]

Yes, the slope of the I/V graph is the resistance at a given value of I

Let's be clear about this. The slope DOES NOT give you the resistance. The resistance is defined to be V/I, not dV/dI, although these two things happen to be the same for an ohmic resistor.

So what is the meaning of the slope? it has the resistance's unit as well...

Yes, it does have units of resistance, but it doesn't have any defined physical meaning.

 

[faust9]

Let's be clear about this. The slope DOES NOT give you the resistance. The resistance is defined to be V/I, not dV/dI, although these two things happen to be the same for an ohmic resistor.

Yes, thanks for clarifying.

 

[VirtualPhysicist]

Thx for help

 

[joe tomei]

Since the slope is not constant, the resistance is not constant. You need to separate small signal effects and large signal effects. For instance if you put current thru the resistor such that the slope significantly changes, then the Small signal resistance is the slope at that point. If you have an ac coupled small signal, it would see the lower resistance that is equal to the slope at that point of the curve.
On the other hand if you increase the nominal current (say double) then you need to ask: what is the large signal resistance. It will be the total voltage divided by the total current. Note that that is not the slope at thehat point.

 

[SGT]

Resistance is a property defined for linear resistors. The filament is a nonlinear resistor, so there is no sense in talking about its resistance.
As joe tomei mentioned, you can linearize the characteristic around an operating point and define the resistance for small signals. This is dV/dI at the operating point.