11
\$\begingroup\$

I often see people saying "Ohm's law doesn't apply here", usually in relation to AC circuits and diodes. They describe certain situations as being "non-Ohmic". My understanding of Ohm's law is that it always applies all the time, only it's sometimes not useful to think about it that way.

I'm the kind of person that needs to understand from the most "technically-correct" point and work upwards from there. For instance, AC isn't some special form of electricity, it's just DC with a frequently-inverted voltage. For practical engineering in the real world, it's useful to treat the two as unique, but technically speaking, it's all just electricity.

So with that in mind, does Ohm's law always apply in all situations so long as you sample an infinitely narrow point in time?

Note: When I talk about Ohm's law, I'm talking about the relationship between voltage, resistance and current: given any two you can calculate the third. I've seen some people try to describe Ohm's law as defining that a linear increase in voltage would result in a linear increase in current (which might not be true because increasing current may generate heat and also increase resistance, for instance).

\$\endgroup\$
9
  • 2
    \$\begingroup\$ Your "infinitely-narrow point in time" may be too restrictive. Current is related to time: coulombs-per-second. I suppose you could measure current by measuring the time for one charge carrier to pass a point, but you'd have to do that many times to get an accurate estimation of current. \$\endgroup\$
    – glen_geek
    Commented Apr 12, 2023 at 23:19
  • 2
    \$\begingroup\$ "My understanding of Ohm's law is that it always applies all the time" No it doesn't. Unfortunately, professors don't know that or don't think about it, so they incorrectly teach students that Ohm's law applies to all components. It doesn't. It only applies to resistors and a few other resistive components. \$\endgroup\$ Commented Apr 13, 2023 at 0:38
  • 2
    \$\begingroup\$ @DavideAndrea Professors? Ha, I'm entirely self-taught. Maybe I'd actually know better if I had gone to University haha :) \$\endgroup\$
    – Clonkex
    Commented Apr 13, 2023 at 0:41
  • 4
    \$\begingroup\$ The grain of salt to take with resistors and Ohm's law is noise. \$\endgroup\$
    – greybeard
    Commented Apr 13, 2023 at 2:13
  • 1
    \$\begingroup\$ Consider a circuit with reactive components such that voltage and current are out of phase. Ohm's law may be found to apply moment by moment but the results change continually. While the basis of the changes are well understood any given instantaneous result in isolation is meaningless (maybe "useless") when the system is considered over a time period of an integral number of AC cycles. The fact that Ohm's law may apply at any given point, or at all points in isolation may be true but of no practical value in understanding circuit operation. [That's not as clear as I'd like. Comment welcome]. \$\endgroup\$
    – Russell McMahon
    Commented Apr 13, 2023 at 11:31

8 Answers 8

29
\$\begingroup\$

No, Ohm’s law only applies when considering constant-value resistive elements in a lumped-element circuit model.

Maxwell’s equations apply always in all situations. But that requires vector calculus, and it’s unwieldy for normal use. Ohm’s law is a simpler model that assumes there is a linear (affine) relationship between voltage and current for resistors.

For practical engineering work, first-order analysis assumes resistors have constant resistance. Sometimes we have to account for the way this constant resistance changes over temperature, so it’s not really linear. But it’s close enough.

Diodes, PN junctions, transistors, saturated magnetic, etc, do not have a linear relationship between voltage and current, so we cannot use the simple linear Ohm’s law relationship to describe those elements. Diodes and transistors often require exponential equations instead of linear.

All of these are lumped-element models, where we assume ideal components connected by ideal wires. Again, this is done because the math is much simpler than trying to apply Maxwell’s equations to a complicated design. We can use lumped-element model as long as the signal bandwidth is not too high, so the wires and components are much smaller than the smallest wavelength we care about.

\$\endgroup\$
22
  • 2
    \$\begingroup\$ Note that this is the opinion of 99% of professional engineers. There are some outliers who will argue, but this is certainly the definition that most people find most useful. \$\endgroup\$
    – TimWescott
    Commented Apr 12, 2023 at 23:22
  • 3
    \$\begingroup\$ @TimWescott what is the source of your statistic? Also, do you think that the 99% of engineers would disagree about what law they apply to compute the current in a 12 V, 12W incandescent bulb? \$\endgroup\$ Commented Apr 12, 2023 at 23:37
  • 4
    \$\begingroup\$ I should have mentioned, too, but this is the sensible and useful definition for Ohm's law. It isn't what Ohm wrote down, because when he was doing his investigation, it was not clear what the current versus voltage relationship was in a wire. We have just found it useful in linear circuit analysis to define an ideal resistor as being a perfectly linear component, so we name that usage after Ohm. \$\endgroup\$
    – TimWescott
    Commented Apr 12, 2023 at 23:42
  • 2
    \$\begingroup\$ @Clonkex: here. electronics.stackexchange.com/questions/339172/… \$\endgroup\$
    – TimWescott
    Commented Apr 12, 2023 at 23:46
  • 5
    \$\begingroup\$ @MarkU: Ohm's law has nothing to do with lumped element circuit model. There is also a version of Ohm's law that is compatible with field based models: \$\vec{j} = \sigma \vec{E}\$, where \$\vec{j}\$ (current density) and \$\vec{E}\$ (electric field) are vectors of vector fields and \$\sigma\$ (conductivity) is a scalar of a scalar field. Note also that Maxwell's equations alone are not be sufficient to derive Ohm's law. It requires an additional theory about scattering of charge carriers (electrons) in a conductor (metals). \$\endgroup\$
    – Curd
    Commented Apr 13, 2023 at 8:21
9
\$\begingroup\$

For instance, AC isn't some special form of electricity, it's just DC with a frequently-inverted voltage.

You would be better off thinking of DC as a special case of AC, where the rate of change of the signals goes to zero.

When I talk about Ohm's law, I'm talking about the relationship between voltage, resistance and current: given any two you can calculate the third.

If you choose to define the resistance as the ratio between voltage and current, then of course you can say that Ohm's law applies anywhere at any time.

But that isn't the "most tehcnically-correct" way to define the resistance, and it isn't the most useful for understanding how circuits work or predicting how new circuits you haven't built yet are going to work.

The most technically correct definition of Ohm's law is the one originally written by Ohm. It applies only to metallic conductors (wires, essentially), and it was an observation that (if the temperature of the wire is kept constant) the voltage increases proportionally with the current, or vice versa. Any statement about a device other than a metallic conductor isn't "technically" Ohm's law.

The more useful definition of Ohm's law is that it describes many types of devices (including, but not limited to, metallic conductors kept at constant temperature) that behave similarly: the voltage varies proportionally to the current (or vice versa). If you have such a device, then you can use Ohm's Law to predict the circuit's behavior, using methods such as nodal and mesh analysis.

On the other hand there are many devices that whose I-V characteristics aren't linear, and therefore can't be considered as following the more useful form of Ohm's law. In those cases we need to use more involved methods of predicting their behavior. But those methods might still involve modeling the device as a combination of an ideal (linear, "Ohmic") resistor and some other device such as a voltage source or capacitor, at each step of the analysis (which can be done so as to be equivalent to using Newton's method to solve the nonlinear equation describing a nonlinear circuit).

\$\endgroup\$
10
  • 1
    \$\begingroup\$ @Clonkex, I can't read your mind, I can only respond to what you write in your question post, which is where you wrote that AC is "just DC with a frequently-inverted voltage." If I misunderstood, feel free to ignore that part of my answer and focus on the other part. \$\endgroup\$
    – The Photon
    Commented Apr 13, 2023 at 1:23
  • 2
    \$\begingroup\$ @Clonkex, As for why it's better to look at it one way than the other: The methods used to solve AC circuits can be used to solve linear DC circuits but not vice versa. Similarly, the methods used to solve nonlinear circuits can be used to solve linear circuits but not vice versa. So if you understand AC circuits then you already understand DC circuits. But not vice versa. \$\endgroup\$
    – The Photon
    Commented Apr 13, 2023 at 2:54
  • 3
    \$\begingroup\$ "If you choose to define the resistance as the ratio between voltage and current ..." that is exactly what we do. Many non-linear devices are characterised by a VI curve, and as soon as we starting discussing that it won't be long before someone talks about slope, which is resistance. Even L and C have reactance, also in Ohms, even though they don't obey "classic" Ohms Law - they do once we account for phase. It's time to stop confusing people like the OP by trying to use a too-narrow definition. \$\endgroup\$
    – danmcb
    Commented Apr 13, 2023 at 9:17
  • 2
    \$\begingroup\$ @danmcb, First, I explicitly mentioned the issue of thermal dependence of resistors in my answer. Second, it's easier to teach someone to understand a simple model of a system and then expand that knowledge to include more subtle aspects of the system behavior than to teach them to understand everything about the system and then point out how it can be simplified in specific circumstances. \$\endgroup\$
    – The Photon
    Commented Apr 13, 2023 at 15:13
  • 3
    \$\begingroup\$ @danmcb: Effective series resistance is the radio of dV/dI, which cannot be ascertained merely by looking at the instantaneous voltage and current. \$\endgroup\$
    – supercat
    Commented Apr 13, 2023 at 15:38
5
\$\begingroup\$

It seems to me that you are looking at this backwards.

Given a certain voltage and current, the ratio of the two at any point in time (and I am not going to get bogged down in your "infinitely small" statement - you have to sample across a finite period in order to even quantify current) the ratio of the two is known as resistance, unit being ohms.

It so happens that some types of devices have (more or less) constant resistance (if we don't worry too much about the effects of temperature and so on).

Others don't.

In other words - resistance is a useful ratio, devised by engineers, to help us classify types of material or devices, and make numerical predictions about how they will behave.

So in any situation where you can measure a current and voltage across a defined period of time, you can define a resistance (at that time). Thus, Ohms Law applies - because we say it does.

\$\endgroup\$
4
\$\begingroup\$

Ohm's law does not apply to other materials and devices, including insulators, capacitors, inductors, switches, transistors, vacuum, voltage sources, current sources, dielectrics, semiconductors, and many others. All of these devices and materials violate Ohm's law

Source: https://physics.stackexchange.com/questions/562633/why-ohm-s-law-doesn-t-work-in-these-scenarios-inside-ideal-battery-and-in-vacuu

So that's usually the case where ohms law doesn't apply, in those materials, there is no simple relationship. Ohmic losses are usually from random interactions in the material, a crude way of thinking about is that the electrons are bumping into atoms and imparting energy which is measured as heat.

Materials with ordered electronic states have nonlinear profiles and ohms law doesn't work to approximate their V-I curve.

For instance, AC isn't some special form of electricity, it's just DC with a frequently-inverted voltage.

Actually DC is AC with no frequency according to Fourier. DC would be a sine wave with infinite wavelength

\$\endgroup\$
3
  • 1
    \$\begingroup\$ "Actually DC is AC with no frequency according to Fourier. DC would be a sine wave with infinite wavelength" That sounds like a way to describe DC mathematically. I like this, but with that definition, AC and DC are the same. You just see more or less of the AC-like effects depending on the rate, frequency and amplitude of change. I guess I just struggle with people trying to use clearly-defined boxes to say "this is AC and this is DC and they are different things" when it's actually an infinitely sliding scale. \$\endgroup\$
    – Clonkex
    Commented Apr 13, 2023 at 1:14
  • 1
    \$\begingroup\$ @Clonkex Hmmmmh yes and no. Frequency can not be infinite, as this would mean wavelength is zero. So the infinite slider, from a frequency perspective, becomes finite. The defintion of DC is: No changes over time. So frequency = 0. Therefore, there are "two boxes". The one box beeing every electrical effect with f not 0 (AC) and every effect with f = 0 (DC). The difference beeing: Are electro/magneto- DYNAMIC effects at play or not. Sure, you can say: AC is DC with these effects inseatd of saying DC is AC without these effects. But: There are many effects at play which favor option 2. \$\endgroup\$ Commented Apr 13, 2023 at 2:13
  • 1
    \$\begingroup\$ @ElectronicsStudent I guess I was imagining representing the wave as length and amplitude. From that perspective, there's nothing stopping length being infinite and thus frequency being 0. It probably stops representing a wave, but that's accurate because there is no longer a wave. I guess you could also lower the amplitude to 0 for the same effect, except then the frequency no longer has to be 0 and it could probably still be considered a wave. So imagine that inductance is the slope of the wave at some point. If the wave is flat, the slope is flat and so there is no inductance. \$\endgroup\$
    – Clonkex
    Commented Apr 13, 2023 at 2:28
3
\$\begingroup\$

Hmmmh, you have some misconception here.

AC is not just DC with a frequently-inverted voltage .

The deeper you go (Waves, Signal-Theory and so on) the more they differ!

In fact, DC is a special form of AC and not the other way around. Many conceptions do not apply for DC, which have to be taken into account with AC.

To make it all worse, DC-Signals like DC Square-Waves are a fascinating topics of AC-Analysis.

So with that in mind, does Ohm's law always apply in all situations so long as you sample an infinitely narrow point in time?

No. Not at all. Maxwell Equations apply. DC Ohms Law is a special case (strongly defined) of these equations

\$\endgroup\$
6
  • 1
    \$\begingroup\$ I find it hard to believe that AC and DC could be considered so different. That's like saying a stationary wheel is a special case of a turning one. Yes, the turning wheel exhibits effects not visible on the stationary wheel (vibration, resonance, maybe the tyre expands if it's turning fast enough) but that's because it's turning. It's still a wheel, but depending on the rate of rotation you see more or less of the effects. Electricity is still electricity, but depending on the rate of change you see more or less of the effects commonly associated with AC. (I'm uneducated, please correct me.) \$\endgroup\$
    – Clonkex
    Commented Apr 13, 2023 at 0:55
  • 1
    \$\begingroup\$ @Clonkex If you consider electricity beeing electrons flowing through conductors - then yes! i can see your point. As soon as you start seeing electricity as fields with gradients in respect to time and position in space, one dimension (the dimension of time) is dropped for pure DC considerations. So DC is a much more well constraint case of AC, as less effects can take place. \$\endgroup\$ Commented Apr 13, 2023 at 1:30
  • 1
    \$\begingroup\$ @Clonkex Sure, this is a purely theoretical approach. Think about electricity beeing a form of "light" focused through lenses (cables) made of conductive materials. With light beeing a wave and DC having infinite wavelength, DC makes no sense at all. It is some sort of: Actually its AC, but we make our lives easy (Math and so on) and introduce DC for this. \$\endgroup\$ Commented Apr 13, 2023 at 1:30
  • 1
    \$\begingroup\$ @Clonkex Also, it speaks alot about your mind to state on the internet, that you are "uneducated" (In your words, i dont think they apply here!). I like it. \$\endgroup\$ Commented Apr 13, 2023 at 1:33
  • 1
    \$\begingroup\$ Ah, I get your point. If they are the same (DC is just AC with infinite wavelength), then everything is AC and DC is just a special form of AC. I hated maths in school and never went beyond algebra, trig and some extremely basic calculus, so I have no idea what "gradients in respect to time and position" would be haha. Maybe one day I'll do a maths course to catch up since I actually love maths these days and rely on it as a computer programmer. \$\endgroup\$
    – Clonkex
    Commented Apr 13, 2023 at 2:07
2
\$\begingroup\$

Others have already spoken about Ohm's law with respect to different devices, so I'll address an aspect of Maxwell's equations and AC vs DC.

Something that I did not appreciate for a very long time is that the seemingly basic concept of voltage only really makes sense when your magnetic field is not changing (i.e., when you are dealing with magnetostatics). This is because Faraday's law of induction implies that Kirchoff's loop rule does not need to hold when there is a changing magnetic field. If you have alternating currents, then Ampère's circuital law implies you have changing magnetic fields.

Looking at instantaneous points in time doesn't save you here: when there are alternating currents, there is simply no such thing as "the" voltage difference between two points in a circuit anymore -- there can be current loops! At best, you can hope that your currents are alternating slowly enough that magnetostatics gives good enough of an approximation to what's going on -- that is, you hope your current loops have negligible net current.

For an ideal resistor, it still makes sense to talk about "the" voltage across it despite these complications. However, for physical resistors, for example those with a spiral of conductive material around a ceramic insulator, at high enough frequencies you are going to have to face the fact that they have non-negligible inductance since they internally have a time-varying magnetic field. Ohm's law does not describe the behavior of inductors.

I'll mention that there are some tricks where you take advantage of a mathematical property called linearity and, rather than resistance, you consider a frequency-dependent complex-valued quantity called impedance. You can analyze what a resistor does to each frequency component of a signal independently (how much does it scale it and how much does it time shift it), and then to see what a resistor will do to a particular signal you bring in Fourier transforms. There are linear relationships in the analysis, but I think it's safe to say we're outside the realm of Ohm's law.

\$\endgroup\$
7
  • 1
    \$\begingroup\$ In a circuit containing only linear elements driven by a periodic sinusoidal waveform, magnetic and electric field effects exhibit themselves in the form of inductance and capacitance. If the voltage is viewed as a complex number, with the instantaneous portion being the real component and the 90-degrees-out-of-phase portion being the imaginary component, and if passive linear elements are viewed as having complex impedance, ohms law will work when applied to any individual frequency. For waveforms containing the sum of multiple frequency components, ohm's law may be applied... \$\endgroup\$
    – supercat
    Commented Apr 13, 2023 at 15:34
  • 1
    \$\begingroup\$ ...to individual components and the results summed together. I don't know that Ohm learned in his lifetime that his Law could be applied this way, but it does work for any device that can be modeled as a combination of ideal resistors, capacitors, and inductors. \$\endgroup\$
    – supercat
    Commented Apr 13, 2023 at 15:35
  • \$\begingroup\$ @supercat Aw this is where the gaps in my knowledge of maths start to show through. I can google the definition of complex numbers but I'd have to work with them for a while to really grok/understand them. I have deep understanding of some things but giant gaps for others. I guess I should find a good maths course to fill in some of the gaps and then come back. \$\endgroup\$
    – Clonkex
    Commented Apr 13, 2023 at 22:25
  • \$\begingroup\$ @Clonkex: At any particular frequency, perfect inductors and capacitors both behave with respect to Ohm's Law as though their resistance is an imaginary number (inductors and capacitors have opposite signs). If one has an inductor in series with a sinusoidal voltage source with an entirely-real component, the current will be an imaginary value, indicating that its phase lags the voltage by 90 degrees. A capacitor would have the opposite sign, and yield a current with the opposite sign, indicating that its phase leads the voltage by 90 degrees. \$\endgroup\$
    – supercat
    Commented Apr 13, 2023 at 22:28
  • \$\begingroup\$ @Clonkex: What's amazing is that if one apply's Ohm's Law to any network of series and parallel resistors, capacitors, and inductors, using the rules for multiplying and dividing complex numbers, everything will "work" to model the combined effects of resistors, capacitors, and inductors at any particular frequency. \$\endgroup\$
    – supercat
    Commented Apr 13, 2023 at 22:31
2
\$\begingroup\$

Generally speaking simple linear models of resistance like Ohm's law are making the implicit assumption that the power dissipated by the circuit element (and thus its voltage) depends only on the instantaneous current, and not its history (i.e. time derivatives of current). However, all materials ultimately are composed of microscopic particles, and their resistance is caused by scattering due to these particles.

Therefore, such a model will only hold if the time resolution of our measurements is low compared to the relaxation times of whatever is scattering our current and causing the dissipations. If we measured instead across a time scale this small, we would observe small fluctuations around the Ohm's law voltage, due to the fact that the energy loss and gain of individual scatterers becomes resolvable in such short time scales.

So to summarise, even for an Ohmic material, Ohm's law itself is describing the average of a stochastic process caused by some underlying microscopic process, and at sufficiently high time resolution one can measure the fluctuations of this process directly, and see small random deviations around the Ohm's law prediction.

\$\endgroup\$
1
\$\begingroup\$

If you consider complex values of U and I and impedance instead of resistance, you'll start getting close. Considering real values, no. A coil or capacitor in an AC circuit puts the voltage and current out of phase with one another. Basically, the instantaneous current is proportional to the instantaneous voltage at a different point in time. The moments when the voltage is zero, the current can be close to maximum in alternating directions.

\$\endgroup\$
1
  • \$\begingroup\$ Just a nitpick: since this site language is English and in almost all English documentation about electronics voltage is indicated with V and not U, I suggest you to change that symbol, otherwise your answer could be misinterpreted (U may be seen as potential energy, for example), especially when talking to a newbie. And yes, I know there are countries where U is used to indicate voltage (e.g. Germany). \$\endgroup\$ Commented Apr 15, 2023 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.