Does a diode really follow Ohm's Law?

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

Introducing the constant of proportionality, the resistance,one arrives at the usual mathematical equation that describes this relationship: I = V/R, where I is the current through the conductor in units of amperes, V is the voltage measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms. More specifically, Ohm's law states that the R in this relation is constant, independent of the current."


However, I had a fellow electrical engineer tell me that a diode does follow Ohm's Law, V = IR, except it has a varying resistance which automatically varies in order to keep a relatively constant voltage drop for any current.

Is this true?

Does it or does it not follow Ohm's Law?

Furthermore, if you put a diode at the end of a power supply, with the anode to + and the cathode not connected, you still see a voltage drop with no current flow. Explain this.

Here's a diagram to show voltage drop with respect to current on an HER508 diode:

graph of I-V
Source: http://www.rectron.com/data_sheets/her501-508.pdf

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    \$\begingroup\$ So the question is not really about ohm's law, but how things can have varying resistances, right? \$\endgroup\$ – Harry Svensson Nov 9 '17 at 23:16
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    \$\begingroup\$ I suppose I seek understanding on both. When you don't know what you don't know it's hard to pinpoint what you're trying to find out. \$\endgroup\$ – Gabriel Staples Nov 9 '17 at 23:20
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    \$\begingroup\$ it looks like you probably want upload.wikimedia.org/wikipedia/commons/a/a5/Diode-IV-Curve.svg or something, instead of your image, which looks like its about pulse currents \$\endgroup\$ – BeB00 Nov 9 '17 at 23:29
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    \$\begingroup\$ On that graph the Y axis is logarithmic but the X axis is linear. If both axes were the same the curves would be quite different - and not at all like a resistor! \$\endgroup\$ – Bruce Abbott Nov 9 '17 at 23:40
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    \$\begingroup\$ Gabriel, it depends on how you define Ohm's law. Equations based upon finite variables (common algebra) use averages. To compute how far a car drove, you might say \$D=S\cdot t\$. But you'd be using the total time times the average speed. Note average! However, you could instead write: \$\textrm{d} D = S \cdot \textrm{d} t\$. And now you are using infinitesimal algebra (aka calculus.) In the sense of calculus, most everything obeys the infinitesimal versions of Ohm's law. \$\endgroup\$ – jonk Nov 10 '17 at 0:15

12 Answers 12


This really is not a black and white question and many folks will argue it does not follow "Ohm's Law", and depending how you argue it, they can be right.

However, the truth is the resistance of a diode changes depending on the applied current or voltage. As such, you can not simply look up the resistance of a diode and use "Ohm's Law" to determine the relationship between voltage and current by the good old V=IR formula like you can with a resistor. From that argument, no a diode, or more accurately, semiconductor, does not seem to follow Ohm's Law.

However, if you have a circuit with a diode in it, biased at voltage V or with a bias current of I, the resistance of the diode under those conditions is still a constant. That is, Ohm's formula still applies when the diode is in a steady state. If you are trying to calculate the output impedance of your circuit in that state, that is important to know, while acknowledging the impedance will be different when the circuit is in a different state.

In fact, I would go as far as to argue that a diode always follows Ohm's formula. Yes V=IR. However, in the case of the diode R follows a rather complex equation that includes V or I as variables..

That is for a diode

\$V = I.R_D\$ Where
\$R_D = F(I,V)\$
\$V = I.F(I,V)\$

So yes, mathematically, it does follow Ohm's formula, just not in a form that is much use to you except under very specific static conditions.

For those that argue "Ohm's Law does not apply if the resistance is not constant" I am afraid that is a misquote by Maxwell. Ohm's intent with that was that the resistance should be constant with time under stable excitation conditions. That is, the resistance can't change spontaneously with no change in the applied voltage and current. The truth is, nothing has a fixed resistance. Even your humble quarter watt resistor will change resistance when it warms up and as it ages.

If you think this is just he opinion of one man, you would be right, his name is
Georg Simon Ohm

enter image description here

Chances are you have never actually read his work, or if you read German, the original version. If you ever do, and, at 281 pages or antiquated English and electrical terminology, I warn you, it is a very hard thing to read, you will discover that he indeed covered non-linear devices and, as such, they should be included in Ohm's Law. In fact there is a whole Appendix, some 35 pages, devoted entirely to the subject. He even acknowledges there were things to still be discovered there and leaves it open for further investigation.

Ohms Law states.. according to Maxwell..

"The electromotive force acting between the extremities of any part of a circuit is the product of the strength of the current, and the resistance of that part of the circuit."

That however is only part of Ohm's thesis and is qualified in Ohm's words by the statement, "a voltaic circuit... which has acquired it's permanent state" which is defined in the paper, and I paraphrase, as any element whose resistance is dependent on the applied voltage or current or anything else must be allowed to settle into it's balanced condition. Further, after any change in the excitation of the circuit as a whole, a rebalance must occur before the formula is effective. Maxwell, on the other hand qualified it as, R must not change with V or I.

That may not be what your were taught in school, or even what you have heard quoted or read from many reputable sources, but it is from Ohm himself. The real issue is many people perceive or understand only a very simplified interpretation of Ohm's thesis, penned by Maxwell, that has been, possibly mistakenly, propagated over the decades since the great man actually performed his work as "Ohm's Law".

Which of course leaves you with a paradox.

The fact is Ohm simply stated, once it settles into a stable state the voltage across the circuit is the sum of the current times the resistances of the parts.


simulate this circuit – Schematic created using CircuitLab

\$E = I.R1 + I.R2 + I.R3\$

Where R3 is whatever resistance the diode settles into. As such, it does not matter whether R3 is a diode or not. Which of course is correct. Maxwell, on the other hand, implies that since the circuit contains a non-linear element, the formula does not apply, which of course, is wrong.

So do we believe what Maxwell wrote was an error in oversimplification and go with what Ohm really said, or do we throw away what Ohm really said and go with Maxwell's simplification which leaves non-linear parts out in the cold?

If you believe a diode does not fit your mental model of Ohm's Law, then your model of Ohm's Law is actually Maxwell's Law. Something that needs to be qualified as being a subset of Ohm's thesis. If you believe a diode does fit the model then you are really quoting Ohm's thesis.

As I said, it is not black and white. In the end, it does not really matter since it changes nothing.

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    \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$ – Trevor_G Nov 10 '17 at 1:19
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    \$\begingroup\$ "Even your humble quarter watt resistor will change resistance when you room warms up." And even if the temperature remains constant, the resistance will vary as the resistor ages. \$\endgroup\$ – WhatRoughBeast Nov 10 '17 at 17:26
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    \$\begingroup\$ Trevor, please edit the Wikipedia page with this information and your clarifications--if it's not already there. I think you've done a great job and believe your understanding is the most correct. \$\endgroup\$ – Gabriel Staples Nov 10 '17 at 21:20
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    \$\begingroup\$ All the guys who say diodes don't follow the ohms law, should stop using the formula Rd = ∆V/∆I . Much irony ! :D \$\endgroup\$ – Mitu Raj Dec 9 '17 at 19:42

Diodes do not follow Ohm's Law. But. At any given current level, you can measure the change in voltage (\$\Delta V\$) for small changes in current (\$\Delta i\$), and get a local equivalent resistance called dynamic resistance. Graphically, this is simply the slope of the voltage/current curve for the diode, or \$Rd=\frac{\Delta V}{\Delta i}\$. This is often useful for describing how a diode in a circuit will behave at a given current level.

Your friend is simply describing the behavior of a standard (silicon, non-Schottky) diode, whose v-i curve is an exponential which is essentially essentially zero (for a graph which uses mA as the current axis) and which starts visibly rising at about 0.6 volts and which will normally hit very high currents by about 0.7 volts. That is, the dynamic resistance is very high at low currents and after (about) 0.6 volts rapidly drops. This means that, if you have a forward-biased diode driven by a variable voltage and fixed resistor, over quite a range of voltages the diode forward voltage will be pretty close to 0.6 or 0.7 volts.

  • \$\begingroup\$ Is the dynamic resistance actually useful? For most conditions I would imagine that this number is extremely high, as the diodes forward voltage will not be changing much at all. \$\endgroup\$ – BeB00 Nov 10 '17 at 0:12
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    \$\begingroup\$ @BeB00 - Actually, dynamic resistance as you envision it is very low. Small changes in voltage require large current changes. Dynamic resistance is useful in predicting zener diode performance and RF applications using diodes as mixers. \$\endgroup\$ – WhatRoughBeast Nov 10 '17 at 1:12
  • \$\begingroup\$ I think you put I and v in the wrong places in your definition \$\endgroup\$ – BeB00 Nov 10 '17 at 1:24

Diodes do not follow ohms law. As you can see in your quoted passage, Ohm's law specifically states that R remains constant. If you try to calculate R from V/I while looking at a diodes IV curve, you will see that as you increase the voltage, "R" will change.

Your electrical engineer friend is incorrect. Saying that "the resistance varies to keep a constant Vdrop" is completely meaningless. In this case, the "resistance" is literally just V/I, which is changing. If you allow R to have any value in V=IR, the equation becomes useless because you cannot predict anything.

In your situation, you would not see a voltage drop. Both sides of the device would be at the same positive voltage (relative to the - terminal of the power supply)

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    \$\begingroup\$ what is the internal resistance of your multimeter? Is it passing enough current to explain the drop? \$\endgroup\$ – MAB Nov 9 '17 at 23:24
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    \$\begingroup\$ @GabrielStaples because you're conntecting a resistor between the - leg and the power supply. As MAB hints, the internal resistance of your multimeter is probably around 1-10Mohm. If you measure the current while your multimeter was connected (using another multimeter), you would probably measure a few uA \$\endgroup\$ – BeB00 Nov 9 '17 at 23:26
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    \$\begingroup\$ I calculate that for a typical diode, at 1uA current, you would expect a vdrop of around 0.35V, so you would measure (V+)-0.35 on your multimeter \$\endgroup\$ – BeB00 Nov 9 '17 at 23:39
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    \$\begingroup\$ If you go by "Ohm's law specifically states that R remains constant" then Ohm's law does not apply to anything... \$\endgroup\$ – Trevor_G Nov 9 '17 at 23:57
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    \$\begingroup\$ I would say that that's probably true, in that no device is truly "ohmic". Instead, many device follow ohms law up to a point. If the region that they follow ohms law in is the typical operating region, I would say it's reasonable to call these devices "ohmic", although of course there will be conditions under which they do not follow ohms law. Diodes never follow ohms law, so I would say they are not ohmic. \$\endgroup\$ – BeB00 Nov 10 '17 at 0:12

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

  1. A diode is not a conductor.

  2. '... directly proportional to ...' means a linear relationship between voltage and current over a substantial operating range, which is, clearly, not the case.

So, no; a diode does not follow Ohm's law.

  • \$\begingroup\$ Thanks. I scrolled down to the answer box to write a very similar answer, now I did not have to. :) \$\endgroup\$ – pipe Nov 10 '17 at 15:07
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    \$\begingroup\$ @Chu a diode is a conductor; everything that has a finite resistance is a conductor per definition (a substance, body, or device that conducts, a conductor is an object or type of material that allows the flow of an electrical current in one or more directions etc.); it's not a conductive material, it's not a metal, but it is a conductor. If it conducts, it's a conductor. \$\endgroup\$ – user20088 Nov 10 '17 at 22:37
  • \$\begingroup\$ @vaxquis So why do we call it a semi conductor? \$\endgroup\$ – pipe Nov 11 '17 at 9:14
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    \$\begingroup\$ @Chu, I agree with you. The word "proportional" is important as it is the keyword in our discussion. If I am not wrong, this word does not appear in all the comments listed below Trevors cobntribution. The rule we call "Ohms law" is valid for parts with a linear V-I characteristics only. \$\endgroup\$ – LvW Nov 11 '17 at 17:32
  • \$\begingroup\$ @pipe so why do we call a semi-trailer semi-trailer? The same logic follows here. Re: because it either conducts, or it doesn't, depending on the operating conditions. A diode is mostly a conductor in one direction (assuming low Vd), and an insulator or conductor in the other. The word "conductor" usually means "conductive metal" - it's not limited to it, though; note that, given high enough voltage, diode will always conduct. Still, it doesn't conduct in a range of voltages, temperatures etc. \$\endgroup\$ – user20088 Nov 11 '17 at 17:34

A diode is a diode and does not follow nor care anything we think, write or imagine about it.

So the question could be turned upside down into something like
"Does a diode I/V characteristic can be modelled using Ohm's law?"

In this case answer could be:
"Yes, within certain constraints Ohm's law can be used though it is definetely not the best nor the first option"

Having \$v=R\,i\$ but with varying \$R=f(i)\$ is indeed a big headache when numbers have really to be crunched.

In fact many many models can be pushed to fit diode behaviour, pointing out the right one for your very applications is the job.

The diode could also be modelled as it were a capacitor:

\$v=\frac{1}{C}\int i\,{\mathrm{dt}}\$ with \$\frac{1}{C}=f(v,i,t)\$ popping up and down from zero with appropriate Dirac's deltas to accomplish diode's I/V characteristic.

This is obviously a totally crazy idea and nobody sane would even think of using it.

I just wish to make clear models are just models. They have nothing to do with "reality" -whatever it means- and they are right as long as they give the "right" answers. Then, some of them are better suited to the purpose.

So recapping, depending on what we are after, the more appropriate model shall be found:
constant drop/threshold, constant drop and fixed resistance, exponential models and various differential ones are for sure far better then trying to push unwilling Ohm's law.

  • \$\begingroup\$ Your point about capacitor is interesting though, since we do use the diode capacitance in circuits. \$\endgroup\$ – pipe Nov 11 '17 at 18:04
  • \$\begingroup\$ @pipe Oh yes that's right I am only partially insane then :D. Anyway building the whole DC characteristic up from that is far beyond from refining another "nearly good" model with some voltage dependante capacitances. \$\endgroup\$ – carloc Nov 12 '17 at 13:16

... I had a fellow electrical engineer tell me that a diode does follow Ohm's Law, V=IR, except it has a varying resistance which automatically varies in order to keep a relatively constant voltage drop for any current. Is this true?


  • but only for incremental voltage when saturated and the fixed value of resistance has a wide tolerance, but you may consider nominal VI curve.

What is saturated? When the dynamic logarithmic resistance becomes less than the fixed bulk resistance so that ESR is almost constant and Ohm's Law applies.

  • Note the following def'n is false !!
    A diode that is passing the maximum possible current, so further increases in applied voltage have no effect on current. McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.

What is ESR? Effective Series Resistance usually measured by the tangent of the VI curve or \$ESR=\frac{\Delta V}{ \Delta I}\$ this can be used to measure Cap ESR with step pulse or transistors Vce(sat) or anything with losses over some limited range.

So what current is needed to measure ESR?

  • It becomes more linear and fixed near rated Vf @ current If and may be predicted in general for most diodes using this
  • Since If(max) depends on power rating Pd (max) and chip size ESR is always inversely related to Pd and no longer is logarithmic but rather almost constant. - ESR tolerance may be +/-50% over entire production but < 5% in a batch.
  • For Zener diodes ESR is called \$Z_{zt}\$ @ some If (mA) and is same thing and Ohm's Law applies


enter image description here

\$V_f= V_{th} + I_f*ESR ~~~~~ \$
- Vth is knee of curve like Zener threshold ( LED, Ge, Si etc)

Verify my assertions

Toshiba LED TL1-L3-xxx specs

  • 2.85V (typ) @ 350mA, 1A max (pulse) so measure ESR > 0.1A
  • Pd (typ) = 2.85 * 350mA = 1W
  • ( my rule) ESR = k/Pd for k= 0.5 (good) to 1 (fair)

From spreadsheet above ( generated from datasheet) see how ESR ( dark green) flattens out above Vf = 2.85V

  • ESR @ If
    • (left Y-axis vs right Y-axis)
 1.5 Ω @ 100mA
 1.0 Ω @ 175mA
 0.5 Ω @ 350 mA ( 2.85V )
 0.25Ω @ 1000 mA  ( absolute max)

enter image description here Since above means ESR k factor = 0.5 this is an excellent efficient LED ( more than just good) Low power LEDs like 5mm tend to have k=1 e.g. 65mW , ESR = 16 Ω . Generally the better the product quality and bigger the size, lower k is better, a useful Figure of Merit (FoM). and remember the tolerance on specs is wide, but your results depend on supplier.

Misc (ticky tacky) Info

Diodes are inherently logarithmic over 4 decades when ideal. This is a large power diode so the linear bulk resistance is quite small compare with the logarithmic natural response.

I have often talked about how the incremental linear resistance of diodes follows the inverse Pd rating +/-25% for k=0.5 to 1 for ESR=k/Pd. This is my own discovery , not taught yet consistent with most diodes and transistors. although this part has no Pd rating it’s 5A@1.1~1.7 @60’C implies an avg. of 7W or an ESR of 0.07 to 0.14 ohms or an avg. of 0.1V rise per Amp. This gives a ballpark estimate of the curve in the 1 to 10A range above which becomes linear as shown by the curve in the log-lin graph of the fig 4 in http://www.eicsemi.com/DataSheet/HER501_8.pdf

But this curve you show is only for narrow pulse where the junction temperature is regulated at a constant 25’C.

But for ESR, it does follow a somewhat linear curve between 10% and 100% of max rated current. Below this it the incremental R is logarithmic.

So yes and no are your answers. It depends on ESR.


They don't follow Ohm's law, but that doesn't make the comparison useless.

First off, consider that if I have any two values, such as voltage and current, I can define some function R which is a "resistance" equating the two. In this case, the R of a diode (the "resistance" of a diode) is highly nonlinear. Given that I can create such a relationship for basically any device I please, claiming diodes follow Ohm's Law is akin to saying "anything is air-droppable at least once." (Rule 11)

However, this relationship can be very useful for small signal models. Let's take the basic exponential region of a diode's behavior: \$I=I_0e^{kV}\$, where $k is some constant for that particular diode. If I take the derivative, I get \$\frac{dI}{dV}=kI_0e^{kV}\$. I can use this to construct a small signal model for a diode biased with a certain voltage. As long as the small signal voltage is small enough, it wont create too many non-linear effects, and I can do some circuit design as though the diode was a resistor.


Ohms law works for lots of things besides current and voltage through resistors. But wherever you try to apply it, it will eventually fail. For a resistor, breakdown happens when the current and voltage are high enough to make the resistor go up in smoke. For magnetic circuits, ohm's law fails when part of the circuit is saturated. It can also apply to fluid flow through pipes, models of illegal immigration, and much more.

For ordinary diodes, there is the DIODE EQUATION, developed IIRC by Shockley. It is I = Io (e^(Vd/nVt)-1). A diode does not follow ohm's law. See https://en.wikipedia.org/wiki/Diode_modelling for more details. Of course this model, like all others, has limits beyond which it fails.

In ordinary circuit modeling, I use a voltage controlled switch in series with a voltage source of about 0.6 volts. Less than 0.6 volts, the switch is open and no current flows. Above 0.6 volts, the switch closes and the voltage drop is limited by the voltage source to 0.6, no matter the current. This works well enough in most circuits.

Note that the WP-34s calculator includes the Lambert W function that you can use to solve the diode equation immediately without any iteration, but that is beyond the scope of your question.

At high frequencies, diodes have inductance and capacitance that will have to be modeled, so take care if you encounter such a situation.


Your friend is confusing "Ohm's law" which states a linear relation between voltage and current with the ability to specify differential resistance, the local relation between voltage and current at a given operating point. The former is an actual law making a prescriptive statement, the latter is basically more or less descriptive and only assumes the existing of a relation between voltage and current.

Note that the operating point cannot even uniquely be described by current: a tunnel diode, for example, has a phase of negative differential resistance as the tunneling effect is replaced by normal diode behavior, where current decreases as voltage increases. This makes it viable for driving oscillators.


Diodes are non-linear (whether they emit light or not).

"Non-linear" means they do not follow Ohm's Law in the usual fashion as do resistors, heaters, long wires, etc.

 E=IR              E (volts) = I (amps) x R (ohms).  

At any instant there's a value for E and I, so an effective R can be computed.

But Ohm's Law gives a sense that R stays constant if E or I changes: if E doubles I must double too. That is not true for non-linear things like diodes.


Ohms law is a linear equation and all other thing being held constant results in a straight line plot. A diode is classified as a non-linear device and to claim otherwise is the same as saying the definition of linear is wrong. Would you seriously use the same analogy with a square or cubic plots. Saying a diode follows ohms law sounds like a quote from a politician - and as believable.

  • \$\begingroup\$ Ohm's Law is a linear equation, with respect to what? A diode is classified as a non-linear device, WITH RESPECT TO WHAT? Would you seriously use the same analogy with a square or cubic plots.Perhaps we shouldn't ask questions in our answers. Secondly, I think you have the wrong definition of nonlinear here. Saying a diode follows ohms law sounds like a quote from a politician.... um okay. \$\endgroup\$ – user103380 Nov 11 '17 at 17:06

The actual general idea of resistance is \$R = \frac{dI}{dV}\$.

With passive circuits everything is linear and resistance is also \$R = \frac{dI}{dV} = \frac{V}{I}\$ — the derivative is a constant, linear.

It's this linear resistance (a constant), that people think first when talking about resistance. They're "resistors". It's convenient also that the other active components do not have to be expressed in terms of pure resistance. Even with parasitic resistances (forward parasitic diode resistance, FET \$R_{OUT}\$, etc), we deal with them as resistors. So this solidifies the idea that resistance are only for resistors.

But really if we take \$R = \frac{dI}{dV}\$, nearly any component that has a voltage drop across them and allows current to flow to or from at least one terminal can be expressed as having "resistance".

I'll probably regret saying "to flow to or from at least one terminal", as there is no such practical component (probably an antenna, but I'm not sure)

ALSO, don't buy from Lees's Electronics, they might mistake to give you parts set aside for me and you might end up with faulty components.

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    \$\begingroup\$ I would say that there is a difference between "Ohms law" and just having a quantity R that is equal to V/I. Ohms law is specifically that R=V/I and that R is constant. If R isn't constant, it's not Ohm's law, it's just an equation with some variables in it. \$\endgroup\$ – BeB00 Nov 10 '17 at 0:15
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    \$\begingroup\$ @BeB00 You could argue that. But frankly I'd disagree, but it also depends on context. Once your brain is properly set up for thinking entirely in terms of infinitesimals, you never see a finite equation the same again. They are all immediately re-expressed mentally into infinitesimal form where they are far, far more accurately expressed. \$\endgroup\$ – jonk Nov 10 '17 at 0:21
  • \$\begingroup\$ @jonk Well, I would again disagree with that. R=dI/dV is useful in some respects, but if you're not sure that R is staying constant for large dV, it becomes a lot less useful. \$\endgroup\$ – BeB00 Nov 10 '17 at 0:34
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    \$\begingroup\$ @BeB00: You're right. Really, resistor are just constant up to a certain wattage. \$\endgroup\$ – Dehbop Nov 10 '17 at 1:01
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    \$\begingroup\$ @BeB00 Nothing is ever really constant, except for a few physical constants. But as I said, it's context dependent. So I'm going to leave you to your opinion and end the discussion about this with a semi-disagreement. I think Trevor nailed my own thoughts. \$\endgroup\$ – jonk Nov 10 '17 at 1:29

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