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Suppose current runs through a resistor from left to right, and we define the left-to-right direction as positive.

Then from left to right, the potential decreases. So $V,$ the voltage across the resistor (defined as the electric potential at the right of the resistor minus the electric potential at the left of the resistor), is negative, while $I,$ the current through the resistor, is positive. So it would seem $V = -IR$ is the correct relationship, but I don't recall ever seeing Ohm's law written this way. Why not?

(For example, in the introductory textbook I have on hand, Ohm's law is derived using $E = V/\Delta L,$ even though, in the previous chapter, the book writes $E_x = -\Delta V_x/\Delta x.$ [1] The intermediate book I have on hand argues that $V$ and $I$ must be proportional, saying "$V$ is the line integral of $\mathbf{E}$ on a path through the conductor...", but it would seem more appropriate to say "$V$ is the negative of the line integral of $\mathbf{E}$..." The book defines $V$ as "the difference in potential between those terminals" without specifying the direction. [2])

[1] Physics, Resnick, Halliday, and Krane, 5ed. equations 29-12 and 28-37

[2] Electricity and Magnetism, Purcell and Morin, 3ed. text before equation 4.12

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  • $\begingroup$ Are you sure what V represents? If it is a "drop" in potential then Ohm's law is providing the magnitude. $\endgroup$ – ggcg Jan 8 at 18:24
  • $\begingroup$ I'm not sure of anything, but I find the references I've consulted unclear, so I'm asking. $\endgroup$ – Mark Eichenlaub Jan 8 at 18:31
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    $\begingroup$ I thought this was just because $V$ was defined here as the voltage drop, i.e. the voltage on the left minus the voltage on the right. Kirchoff's loop rule is usually written in terms of voltage drops, too. $\endgroup$ – knzhou Jan 8 at 18:45
  • $\begingroup$ I guess that could be, but then voltage drop and voltage would share the same symbol, $V$. How is a student to know whether $V$ means voltage or voltage drop, aside from lots of experience? Sources like en.wikipedia.org/wiki/… don't say they're using voltage drop for $V$, it's actually explicitly voltage if you look at the derivation. $\endgroup$ – Mark Eichenlaub Jan 8 at 18:50
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    $\begingroup$ "So V, the voltage across the resistor (defined as the electric potential at the right of the resistor minus the electric potential at the left of the resistor)" - According to the passive sign convention, the electric current enters the positive labelled terminal of a resistor. You've chosen the opposite convention. Why? $\endgroup$ – Hal Hollis Jan 8 at 20:46
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I think that the answer lies in the fact that resistors are not the only components in an electronic circuit.

Consider the following circuit.

enter image description here

Following the path of the electric current (positive charges) $\int \vec E \cdot d \vec l$ is negative for component $A$ and positive for component $B$.

What does that mean?
It means that when positive charge go through component $A$ work is done on the positive charge and when positive charge goes through component $B$ work is done by the positive charge.
In simple terms this can be interpreted as $A$ being a source of electrical energy and $B$ being a sink of electrical energy.
If $A$ is a photocell the light is being converted into electrical energy and if $B$ is an LED then electrical energy is being converted into light.

For both components the power is $|VI|$ so how does one differentiate between these two components?
One introduces a sign convention which is usually the passive sign convention which then results in the electrical power being produced in component $A$ negative (a source of electrical energy) and that in component $B$ positive (a sink of electrical energy).

There is another convention which is used and that is the active sign convention which makes the power of sources of electrical energy positive and that of sinks of electrical energy negative.
For example, this is used when generators are sold, you buy a $(+)1\,\rm kW$ generator not a $-1\,\rm kW$ one.

The difficulty with running both conventions together, the passive for sinks and the active for sources is that some circuit elements can be a source (battery connected to a bulb) or a sink (battery being recharged).

The passive sign convention is the one most commonly used and then for a resistor $V=IR$ with the current flowing from a high potential to a lower potential and $V$ is the drop in potential (in the direction of the current) across the resistor ie a positive quantity.

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In the comments, Hal Hollis pointed to the "passive sign convention".

The passive sign convention is defined like this:

passive sign convention diagram, from wikipedia

So it is simply taken as a definition that the positive direction for current is the opposite direction as the positive direction for voltage. With this convention, $V=IR$ is the correct equation.

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I think it's all a matter of definition. Electric current's direction is just defined as the direction in which (effective) positive charge carriers move. Positive charges move from the higher voltage points to lower-voltage points, so a flux variable (current) defined in that direction would be positive. You can, for example, define "current"'s direction as the direction in which electrons travel, and write V = -IR.

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  • $\begingroup$ How are you defining $V$? $\endgroup$ – Mark Eichenlaub Jan 8 at 18:58
  • $\begingroup$ @Mark V is the electric potential difference, which equals integral of the Electric field (dl) between the two points. $\endgroup$ – Ali Lavasani Jan 8 at 19:02
  • $\begingroup$ Can you be more specific than "electric potential difference"? What are you subtracting from what? Suppose that at point $A,$ the electric potential is $0\;\mathrm{V}$ and at point $B,$ the electric potential is $1\;\mathrm{V}.$ Then what is the "electric potential difference" from $A$ to $B$? $\endgroup$ – Mark Eichenlaub Jan 8 at 19:05
  • $\begingroup$ @Mark I mean, if you integrate the electric field on any arbitrary path from point A to point B, it gives you V. In your example, integral of E(x,y,z) from A to B (dl) would equal 1. $\endgroup$ – Ali Lavasani Jan 8 at 19:10
  • $\begingroup$ No, that's wrong. If you do a line integral of the electric field from $A$ to $B,$ you get $-1$ times (potential at $B$ minus the potential at $A$). $\endgroup$ – Mark Eichenlaub Jan 8 at 19:13
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You're right. $R$ is a constant, and we have variables $V \in \mathbb{R}$ and $I \in \mathbb{R}$ where $V$ stands for the change in potential between two points and $I$ is the current from the current vector $\vec{I} = I \;\hat{u}$. So $V$ and $I$ are just some numbers, positive or negative. Ohm's law should be stated $V = -IR$ as you say.

Draw a number line on a sheet of paper as you would in algebra class. The positives are pointing in some direction. You could have drawn the positives in another tilted direction. You could even draw a curvilinear number line. A curvilinear number line has no fixed direction (each point on the line has an instantaneous positive direction), but it still has the sense of a single orientation. A curved number line is still one-dimensional from this standpoint.

Because of this (circuits are curvy), my guess is that Ohm's law is written as $|V| = |I|R$ (from my definitions of $V,I$ above) because there is no fixed direction (current is moving in this direction, then this direction, then down, etc).

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  • $\begingroup$ (also, potential doesn't have a direction, nor does voltage. But voltage is measured in a "direction". Once you pick an orientation, let $V$ and $I$ be variables with respect to that orientation and let the math do the work for you) $\endgroup$ – DWade64 Jan 8 at 21:15
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There is a minus sign in $\vec J = \sigma \vec E$ and $\vec E = - \vec \nabla V$: the current goes down the voltage gradient. However, V in Ohm's law is not the potential but the potential drop caused by the resistor. I am all in favour of $\Delta V = -IR$. Yes, I changed my story ...

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  • $\begingroup$ Do you have any reference corroborating that $V, I,$ and $R$ are scalars without direction? In that case, how could a circuit analysis with Ohm's law ever tell someone which direction the current was running through a particular circuit element? $\endgroup$ – Mark Eichenlaub Jan 8 at 19:14
  • $\begingroup$ Negative voltages and negative current flows are quite ordinary when analyzing an electronic circuit. The value of a "resistor" component always is a non-negative constant, but there are electronic components (e.g., a tunnel diode) and combinations of electronic components that behave like a negative-valued resistor over some limited range of currents and voltages. $\endgroup$ – Solomon Slow Jan 8 at 19:53

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