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#19
Dec311, 06:24 PM

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PF Gold
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I don't see that any of this relates to Ohm's Law at all. There's nothing inherently non linear about any of these effects and that's all Ohm's Law tells you. It deals specifically with resistance and doesn't concern any transforming effects of distributed components. In all these examples, doubling the Volts will double the Current. So where's any violation?



#20
Dec311, 06:51 PM

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If fact, I cringe when I hear talk of "nonohmic" resistors. Just because the resistance of a resistor changes (due to temperature, for example) does not mean Ohm's Law no longer applies. A small change in voltage still produces a small change in current and the ratio of these two still gives the resistance of the resistor. 


#21
Dec311, 07:00 PM

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PF Gold
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It would have been better if they had called it 'Ohm's Behaviour', perhaps.



#22
Dec611, 11:20 AM

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ohms law fails when the resistance of the conductor increases as it gets hot by electron friction



#23
Dec611, 11:24 AM

PF Gold
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#25
Dec611, 10:45 PM

P: 24

in Ohm's law as resistance is proportional to voltage, then: temperature is a limitation, resistance varies with temperature, and depending on materials and room temp, it can vary a lot. unless we state that temp is constant we can secure the lineality of the ecuation



#26
Dec711, 03:23 AM

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It depends on the definition of Ohm's Law.
There is the one we all saw on day 1 of Electricity at school. Graph voltage vs current and if it is a straight line then you can work out the resistance. According to that version, even a rheostat would be regarded as non ohmic because it can change. It leads to the strange situation where a lamp filament is "nonohmic" if you measure its voltage and current curve slowly (and give it time for the temperature to change) but "ohmic" if you change the voltage rapidly. The current in a lamp filament with 60 Hz AC on it is close to sinusoidal, meaning the resistance is close to constant because the temperature is fairly constant and the resistance is also stable. Much more useful is the incremental one which assumes that Ohm's Law always applies and you can work out the resistance at any point on a V vs I curve by taking a small change in voltage and observing the small change in current that results. This small signal resistance is very real and it can be measured as the actual impedance of a circuit. It may be quite different to the ratio of DC voltage to DC current that applies at that point. 


#27
Dec711, 08:39 AM

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PF Gold
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Ohm's Law is its own definition. The condition for V/I being constant is for the temperature to be constant. Measurement with a 'probe', low level AC of a filaments lamp will reduce the temperature fluctuation and give a value of dV/dI that is not resistance. The resistance is still V/I for the particular point on the temperature dependent VI curve that you will get using DC.
The only time that Ohm's Law truly fails for a metal is for enormous current flux when there are just no more conduction electrons available and the conductor becomes .non linear. 


#28
Dec811, 01:25 AM

P: 324

I wonder though: It is Ohm's Law that is failing or that the device (or circuit) in question is no longer well modeled by an ideal resistor in the examples given? I would argue Ohm's law isn't breaking, the ideal resistor model is.
If one replaced the ideal model in Ohm's law with a new model that also took temp, time, etc., into consideration, wouldn't Ohm's law still hold? I think it does. I agree with Sophie and the only thing that really breaks it is current density and possibly arc'ing across the geometry of the element. Because if the function for R had to take current or voltage as an independent variable then equation we would be left with would be self referencing and I think this would count as a new beast. 


#29
Dec811, 04:02 AM

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PF Gold
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I would say that the only time a metal stops obeying Ohm's law is when it is under such extreme conditions that it can almost not be classed as a metal because its very structure will have changed (metallic bonding won't even be the same). George O. can't really be blamed for not having put a codacil on his Law to take account of this. 


#30
Dec811, 06:32 AM

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#31
Dec3111, 08:33 AM

P: 390

Can anyone quote the exact statement given by Ohm with web reference. Just being curious.
Given a constant temperature current will be equal to voltage multiplied by resistance.  correct Given a constant temperature current will be proportional to voltage.  wrong 


#32
Dec3111, 10:07 AM

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#33
Dec3111, 12:57 PM

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PF Gold
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If you get your history right, you will realise that Ohm only introduced Resistance as the constant of proportionality for a linear relationshipbetween two quantities that he could actually MEASURE. They didn't sell Ohmmeters at the time! 


#34
Dec3111, 05:00 PM

P: 24

I= V R: WRONG 


#35
Dec3111, 09:57 PM

P: 390

"Given a constant temperature current through a material will be proportional to the voltage drop across it (in steady state)." Now, Ohm's law is indeed true if we do not take into consideration the quantum effects. In fact if you are able to measure IV curve for a long conductor with very high accuracy or say you measured it exactly, you will discover that the IV curve is never linear. The variation in resistance is laughably tiny, but it exists and due to QM effects. In nanoscale structures the variation is tremendous. In those cases the resistance depends on the applied bias. Now, as Ohm's law is only a model to predict the behavior of real materials (just like QM is a model to predict the behavior of electrons  we don't even understand what electrons actually are, only know their mass, charge, observables etc.), the idea that Ohm's law only predicts the behavior of ideal conductors is bogus. If it did that we wont be needing it cause there is no ideal conductor. My conclusion is that when it comes to accuracy, the IV relationship derived from QM beats Ohm's law (both in large and small materials) proving Ohm's law does have its limits. 


#36
Dec3111, 10:06 PM

P: 390

I = [itex]\frac{V}{R(V)}[/itex] But, [itex]\frac{I_{2}}{I_{1}}[/itex] [itex]\neq[/itex] [itex]\frac{V_{2}}{V_{1}}[/itex] 


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