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Here is my simple reasoning. We apply a potential difference across a resistor. All the electrons begin responding. Since it takes time for electrons to respond, our current is not yet fully established. In fact, immediately after we apply the voltage, the current is zero! And yet we have a non-zero V and R. So am I wrong here, or does V sometimes not equal IR?

EDIT: I'm having a bit of an issue here. People want to say that an inductance of zero will allow for instantaneous current initiation, but I wonder. Is it ever possible to accelerate an object with mass infinitely quickly with nothing but an electric field? Sounds impossible more than likely. I feel that an instantaneous acceleration can only be caused by an infinite amount of energy given.

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    \$\begingroup\$ If you're going to start modeling the movement of electrons, you have to do it throughout the circuit, including inside the power supply. You can't take a "perfect" power supply and apply it to a "real world" circuit and expect to be able to disprove V=IR. Either use perfect components everywhere and try to disprove it, or practical components everywhere and try to disprove it, but mixing theory and practice will trip you up every time. \$\endgroup\$ – Adam Davis Jul 14 '14 at 18:04
  • \$\begingroup\$ Could you define a perfect power supply? Do you mean that perhaps the power is applied too fast for a real power supply? I'm a bit confused on your logic, but thanks. \$\endgroup\$ – Andres Salas Jul 14 '14 at 18:11
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    \$\begingroup\$ You tell me. You say, "We apply a potential difference" but since you can't apply a potential difference instantaneously in the real world, then you need to model that part of the circuit as well before you can disprove V=IR. So describe what you mean by "We apply a potential difference" and how it actually works in your circuit where you are looking at electrons. \$\endgroup\$ – Adam Davis Jul 14 '14 at 18:15
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    \$\begingroup\$ The basics of Ohm's Law, is an aproximation similar to Newtonian Physics. Not 100% encompassing, but good enough for pretty much all intensive practial uses. \$\endgroup\$ – Passerby Jul 14 '14 at 21:20
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    \$\begingroup\$ Right. A small flaw, in extreme circumstances, does no hurt in the overall usefulness of physical Laws/theories. \$\endgroup\$ – Andres Salas Jul 14 '14 at 21:23
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The only reason the current would be zero is if you have a non-zero inductance (which all circuits do). Once you factor in the inductance, you'll be able to calculate the rise time of the current after you apply a voltage. Only in the ideal world where there's no inductance or capacitance, will V=IR be true at all times over a resistor.

In our non-ideal world, you're right that V=IR only applies at steady state after the influence from capacitance and inductance fall away.

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    \$\begingroup\$ Phew. I was worried about that non-zero inductance comment until you said that all circuits have a non-zero inductance. Thanks for your comment. \$\endgroup\$ – Andres Salas Jul 14 '14 at 17:34
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    \$\begingroup\$ This is the right answer. \$\endgroup\$ – Matt Young Jul 14 '14 at 18:07
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    \$\begingroup\$ +1 Yup. Even a straight wire has a bit of inductance. We're effectively measuring the inductance of a wire a few mm long to a part in a million or so. \$\endgroup\$ – Spehro Pefhany Jul 14 '14 at 18:38
  • \$\begingroup\$ Actually, in one sense your question is valid. When you apply a voltage across a resistor, it takes a finite amount of time for that information to propagate through the resistor, which will happen at approximately the speed of light. So you do get a delay of sorts, measured in picoseconds, probably. Of course, during the propagation time the current will be flowing in part of the resistor, so it's an "interesting question" whether current is flowing through the resistor. \$\endgroup\$ – WhatRoughBeast Jul 15 '14 at 3:16
  • \$\begingroup\$ @WhatRoughBeast, what I'm trying to think here is that not only does it take a while for the electrons to receive that information(electric field), it also takes another amount of time to accelerate to the drift velocity, but I think that resistance is constant so V=IR isn't quite correct for a period of time, including this time you describe. \$\endgroup\$ – Andres Salas Jul 15 '14 at 17:23
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V=IR is only valid at steady state.

You are trying to apply it outside of that condition, so it doesn't always work. Transient Responses are not modeled at all in a 1st order system.

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    \$\begingroup\$ for all values of "steady state" less than the speed of light. \$\endgroup\$ – placeholder Jul 14 '14 at 17:31
  • \$\begingroup\$ Transient response is so modeled for first order systems. \$\endgroup\$ – Matt Young Jul 14 '14 at 17:42
  • \$\begingroup\$ What @MattYoung said. A inductor circuit exhibits first order (transient) response. \$\endgroup\$ – Li-aung Yip Jul 14 '14 at 18:05
  • \$\begingroup\$ The term "steady state" can also include periodic oscillatory behavior, if "R" is defined as being the complex impedance at the oscillating frequency. A truly beautiful example of real-world phenomena following the rules of complex arithmetic. \$\endgroup\$ – supercat Jul 14 '14 at 18:57
  • \$\begingroup\$ @Horta said it right - "In our non-ideal world, you're right that V=IR only applies at steady state after the influence from capacitance and inductance fall away." \$\endgroup\$ – Jeff Wurz Jul 14 '14 at 19:32
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You are assuming that voltage can be applied instantaneously (zero capacitance) but that current cannot rise instantaneously (non-zero inductance). You are also assuming a theoretically perfect resistor that always applies precisely the same resistance. These seem like an unreasonable set of assumptions.

As far as the resistor goes, isn't it true that resistance can be calculated directly from resistivity at a given temperature?

Steady-state, theoretical resistance can. But the actual resistance displayed at a particular instant in time cannot. If you don't see why, imagine if we're sending electrons through one at a time.

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  • \$\begingroup\$ I believe you may be on to something here. But, I have a few statements: firstly, an instantaneous voltage, as I explained to Adam Davis above, isn't a necessary assumption. It only needs to be applied fast enough, which is physically possible. Also, I am not 'assuming' that current cannot rise instantaneously, I feel that I am proving it. Instantaneous acceleration=infinite work. As far as the resistor goes, isn't it true that resistance can be calculated directly from resistivity at a given temperature? If that is true, I'm ok. I need more light though, thank you \$\endgroup\$ – Andres Salas Jul 14 '14 at 19:18
  • \$\begingroup\$ @AndresSalas See update. \$\endgroup\$ – David Schwartz Jul 14 '14 at 19:29
  • \$\begingroup\$ Which brings up the question: why then are resistors labeled as having a particular resistance, say 50 Ohms, even when voltages and current may vary as you say? So, in this idea you have, would the instantaneous resistance be huge or small? Huge right? \$\endgroup\$ – Andres Salas Jul 14 '14 at 19:35
  • \$\begingroup\$ @AndresSalas Everyone understands what those resistance labels mean. \$\endgroup\$ – David Schwartz Jul 14 '14 at 20:22
  • \$\begingroup\$ Uhoh. Seems like I missed that party. I'm confused, becasue initially you seem to state that a resistor has a non-constant value at specific times. Later on, you seem to argue the opposite point, which seems that a 50 Ohm resistor is a 50 Ohm resistor with a 50 Ohm resistance at all times. Apparently I don't understand what those little labels mean haha. I may ask you to define these labels if I'm missing out some point here \$\endgroup\$ – Andres Salas Jul 14 '14 at 20:34
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Some of the other answers mention that there are AC effects and things like inductance and capacitance. But there is a bit more than that.

The thing is, you can't apply the voltage instantly anyway across all parts of the circuit, because you are limited by the speed of light. Assuming a perfect circuit, the moment the voltage is applied you have an electromagnetic wave propagating down the wire at near the speed of light. The actual movement of the electrons is much slower and doesn't come into play here, the energy (and voltage) is propogated in the EM wave. So V=IR isn't being violated at all, because your current is propagating near light speed with the voltage.

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  • \$\begingroup\$ A poor answer. The electrons respond near-instantaneously with the EM wave (although wait a minute here, are you thinking of a straight wire? Cuz a short, bent wire is not driven by the electric field from the battery, but from the electric field propogated from the battery through the individual atoms which create dipoles. The electric field is bent through the creation of dipoles, which is dependent on electron movement, which is slow.) They may respond at the speed of light, but they do not accelerate to the Ohm's Law current speed instantaneously, that is impossible. \$\endgroup\$ – Andres Salas Jul 14 '14 at 22:32
  • \$\begingroup\$ Hope I do not come negatively across here (see what I mean, negatively...electronegatively...haha) I appreciate the example \$\endgroup\$ – Andres Salas Jul 14 '14 at 22:33
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@AndresSalas Its all in the definition. An ohm is equivalent to a volt per ampere. http://en.wikipedia.org/wiki/Resistor

All this applies to DC (direct current) circuits, using AC or changing voltage/current will have additional effects like impedance and capacitance.

If you are doing this quickly (your instantaneous voltage) then you no longer have pure DC characteristics and have to factor in impedance and capacitance to the equation for how it will behave.
Electrical Impedance seems more appropriate to what you are trying to do.

http://en.wikipedia.org/wiki/Electrical_impedance

See if that doesn't help answer your thoughts.

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  • \$\begingroup\$ Ah I see what you're saying, this is basically (but not quite, since the voltage isn't applied in reverse) the first step in an ac sinusoidal voltage application. \$\endgroup\$ – Andres Salas Jul 14 '14 at 20:48

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