I have read that Ohm's law, $V=IR ,$ just means that $V$ is equivalent to $IR ,$ not that voltage is the cause of current.

This is similar to the interpretation of Newton's second law, $\mathbf{F} = m \mathbf{a} ,$ where force is not the cause of acceleration.

Question: Is it true that Ohm's law is merely a relationship between voltage and current without it implying that voltage causes current?

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    $\begingroup$ I have a battery that says that voltage drives current. $\endgroup$
    – my2cts
    Mar 8, 2019 at 9:36
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    $\begingroup$ It is a bit like $v=f \lambda$, where the velocity is just a property of the medium, not something that is caused by a frequency of a wavelength. Such formulas are mathematical connections between quantities, and they can be rearranged. $\endgroup$
    – user137289
    Mar 8, 2019 at 9:46
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    $\begingroup$ @my2cts and I have a current source that says current drives voltage... your point? $\endgroup$
    – UKMonkey
    Mar 8, 2019 at 11:55
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    $\begingroup$ @UKMonkey The point might be that just reading something doesn't make it true. Or useful. $\endgroup$
    – Mast
    Mar 8, 2019 at 12:00
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    $\begingroup$ How would either answer change your understanding of the behavior of a circuit? $\endgroup$
    – nasu
    Mar 8, 2019 at 14:27

8 Answers 8


As usefully pointed out in the comments, Ohm's law just states that voltage $V$ and current $I$ have a relation. Therefore, Ohm's law per se means equivalence and not causation. This could be not immediate to deduce, there is plenty of Physics laws which imply both a relation and a causation, but this is not the case for neither Ohm's law, nor Newton's second law, as you correctly mentioned.

The causation is established according to the specific circuit and its features.

For example, a battery is a device which tries to always impose a fixed voltage $V_b$ between its terminals $V_+$ and $V_-$. In this example, $V_b$ is the cause and the independent quantity. If you connect a lamp between the battery terminals, it can be represented as a resistance $R_{\mathrm{lamp}}$. The voltage across $R_{\mathrm{lamp}}$ is $V_b$. The current on $R_{\mathrm{lamp}}$ is then the effect of the presence of the battery, and the dependent quantity. This is a little circuit which consists in a voltage generator (the battery), a resistor (the lamp) and a single loop. As an effect of the imposed voltage, a current $I$ will flow through the resistor. $I$ is the dependent quantity: to determine its value, you must know both $V_b$ and $R_{\mathrm{lamp}}$; Ohm's law can be written as

$$I = \frac{V_b}{R_{\mathrm{lamp}}}$$

Of course, the battery may not be able to erogate enough current to create or maintain a $V_b$ across $R_{\mathrm{lamp}}$, or the current may be so high that $R_{\mathrm{lamp}}$ is damaged, and so on, but these real cases are not relevant here.

On the other hand, transistors can be used as current generators in several configurations. Current generators try to always impose the value $I_b$ of current between their terminals $I_+$ and $I_-$. If you connect a resistor $R_{\mathrm{lamp}}$ between these two terminals, you already know the current $I_b$ flowing into the resistor: $I_b$ is the independent quantity and the cause. This is again a simple circuit consisting in a current generator, a resistor and a single loop. As an effect of the current $I_b$, a voltage $V$ will appear across the resistor: in this case, $V$ is the dependent quantity. To determine its value, write Ohm's law as:

$$V = R_{\mathrm{lamp}} I_b$$

There are again some real scenarios when $I_b$ can not be imposed by the current generators, but they can be discarded here.

The same Ohm's law is used in both the examples, but with exchanged roles between $V$ and $I$, according to the circuit.

In any circuit, regardless of its structure, if you know the value of a resistor $R$ and the current $I$ flowing through it, you can determine the voltage between the resistor terminals as $V = RI$. If instead you know the voltage existing between the resistor terminals, you can determine the current as $I = V / R$. In the previous cases, the cause was always the independent quantity and the effect was the dependent quantity. In this case instead, you don't know what is the cause, but you can still use Ohm's law: the dependent quantity is now, for you, the one you still do not know.

As correctly pointed out by Whit3rd's answer, Ohm's law mainly refers to a resistor in a circuit and it involves three quantities: the voltage across it, the current through it and the resistance value. If you know two of them, regardless of what is the cause and what the effect, you can determine the third one.


In circuitry, there are materials that conduct electric current. While that is useful, it is NOT the only way that currents come about: an electron beam in a vacuum, or a charged belt motor-driven in a van de Graaff machine, do not have a material conductor, nor any associated 'R' constant.

If a conductor with one end grounded has voltage source applied to its other end, or has a current source connected to it, the resistance of that conductor does relate the current and voltage according to Ohm's law (with very minor exceptions). That relationship tells us about the resistor, but not about the causation of a current-is-flowing episode. Knowing both the resistor AND one of (voltage-difference, current), we can calculate the third variable.

And, it works the third way, too. Knowing a voltage applied and how much current passes, we can compute a resistance.

Three unknowns, and one equation given by Ohm's law. Two measurements (two more equations) and you can solve for all three variables.

the law of Ohm just means equivalence and that voltage is not the cause

Yes, that seems correct. Does generator voltage drive a resistive light bulb, or generated current? Both are equally valid statements.


The potential difference, or voltage, between two points is defined as the work per unit charge required to move the charge between the points. So the voltage across the resistor is the work required per unit charge to move the charge through the resistor. Since power in a resistor is $i^{2}R$, for a current of 1 ampere (1 coulomb per second) and resistance of 1 ohm, we have 1 watt (1 joule per second) of work. After 1 second, 1 joule of work was done to move 1 coulomb of charge through the resistor. That work represents a drop in potential.

Some voltage source was needed to establish the current in the circuit where the resistor is used. But unless that source was connected directly across the resistor (with no other circuit elements in series), the voltage across the resistor was not the "cause" of the current through the resistor. It is simply, as others said, the relationship between the current and resistance, $V=IR$.

Hope this helps

  • $\begingroup$ Then what's the cause Bob? $\endgroup$
    – TechDroid
    Mar 8, 2019 at 10:45
  • $\begingroup$ @TechDroid The cause is the voltage source (the source of electrical potential energy) that drives the current in the circuit. Let's say it is a battery with voltage E. If the battery is connected directly across the resistor, then E=IR and that E is the cause of the current in the resistor. But suppose there are two resistors in series. The battery voltage E is the "cause" of the current that flows in both resistors. The voltage across each resistor is E/2, not E. E/2 is not the "cause" of the current in the circuit. $\endgroup$
    – Bob D
    Mar 8, 2019 at 10:54
  • $\begingroup$ Oh I get it. But it's just the terminology that complicate things sometimes, there's voltage, then potential difference, then emf,... If I may ask Bob, what is the difference between emf and potential difference and voltage? $\endgroup$
    – TechDroid
    Mar 8, 2019 at 10:59
  • $\begingroup$ So really it just depends on the situation? $\endgroup$ Mar 8, 2019 at 11:31
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    $\begingroup$ @TechDroid There is no "cause" and no "effect". Two things happen simultaneously. End of story. If you insist on trying to believe in "cause" and "effect" think about an electrical machine that can operate as either a motor or a generator. Does electrical energy "cause" it to do mechanical work (a motor) or does mechanical work "cause" it to generate electricity (a generator)? Neither, because asking about which causes which is meaningless. $\endgroup$
    – alephzero
    Mar 8, 2019 at 13:47

I posted this as an answer to another question of yours about a week ago, where I thought you were asking the exact question you're asking now. I'll link it below.

It turned out that you were asking something entirely different!

But as I feel like my previous answer is extremely well suited for this question (and since that previous question got closed), I'm going to post it here.

Hope it helps!!!

First off, it's not voltage (directly) that causes charges to move through a circuit. Voltage does not cause current, at least not directly. It's an electric field that forms within the circuit and pushes charges (electrons) through the circuit.

(And if you want to get really specific, it's an electric field due to rings of charge that build on the circumference of the components of the circuit - but that's for another question.)

However, what causes this electric field in the first place?

(What causes the rings of charge that create an electric field throughout the entire circuit)?

Usually, it's a battery inserted somewhere in the circuit. Without getting into the specifics of how a battery works, let's just say there's an overabundance of electrons on one side of the battery and a lack of electrons of the other side, and since electrons repel each other, they want to move from the side with more electrons to the side with less of them.

This cramming of electrons creates an electric field opposite to the direction the electrons want to move (since the electric field points from positive to negative, while electrons move from negative to positive). Again, this isn't REALLY how a battery works, but it's just meant to be a super handwavy explanation of what a battery does

When we insert the battery in a circuit, the battery causes an electric field throughout the circuit that will give the electrons some net velocity (drift velocity) opposite to the direction of the electric field (since, once again, electrons move opposite to the direction of the electric field).

The electrons' velocity is approximately proportional to the strength of the electric field created by the battery. We could go into the details of why that is, but it's really for another question.

The current through a circuit, by definition, is the amount of charge that flows through a cross-sectional area of the circuit per second. Obviously then, the current will be proportional to both the velocity of the electrons moving through the circuit and the cross-sectional area of the wire in the circuit.

$I \propto AV_d$

(the reason for the $_d$ in $V_d$ is because it's the drift velocity of the electrons, which you can reasearch more on - but if that doesn't make sense, you can ignore it for now and simply read: $I \propto AV_e$, where $A$ is the cross-sectional area of wire in the circuit and $V_e$ is the velocity of the electrons)

Now, let's get back to voltage, and your original question. I said we weren't going to get into the specifics of how a battery works, but there is one thing you have to accept: for some (electrochemical) reason or another, when you purchase a battery, it doesn't come with an "electric-field rating" that tells you how big of an electric field it'll create when used to power a circuit. No! Batteries come with their VOLTAGE specified! They come with the potential difference between the positive and negative terminals specified!!!

(I suggest you try and understand the meaning of voltage in the context of electrostatics and electric fields before trying to understand Ohm's law, and the role that voltage plays in a circuit, and stop reading my answer here. But, for after you've understood voltage, I'll continue anyway!)

Let's look at what the fact that batteries come with a set potential difference between the positive and negative terminals implies.

Remember, voltage is the product of the dot product of the electric field pushing the electrons and the distance the electrons move ($V = \int [\vec{E}\cdot dr]$):

  1. Let's say we get some wire and connect the positive and negative terminals of the battery together with it. This wire isn't perfectly conductive so that we don't short our circuit.

    *(At this point you're probably angry about the fact that I said "...isn't perfectly conductive..." to avoid saying "...wire has some resistance..."

    Relax-you can understand the term "not perfectly conductive" without understanding "resistance".

    Me saying that the wire isn't perfectly conductive means that there is something about the material composing the wire which resists the movement of electrons through it - like the way sand in a pipe would resist the flow of water through the pipe.

    Because of this, in order for the wire to conduct electricity, the battery needs to create an electric field throughout that wire. If the wire was perfectly conductive ("super-conductive" is the correct term actually) then even without an electric field pushing them forwards, electrons could move through the wire at REALLY QUICKLY AND FOREVER. Trying to create super-conductive electrical components is one of the things our current generation of physicists is obsessed with doing.)*

    Right, so we connect the positive and negative terminals of the battery with this wire, and the battery creates an electric field within the wire that causes electrons to start moving from the negative terminal of the battery to the positive terminal.

    The electric field points in the direction of the wire (for why this is, why the electric field in a wire points along the direction of the wire, search it up on Stack-Exchange. I remember seeing some really good answers a while back), meaning that the voltage will simply be the product of the magnitude of the electric field through the wire, and the wire's length.

    But remember: the electric field wasn't what the battery came set with as a constant - for some reason or another, the battery comes with a fixed voltage difference between the positive and negative terminals.

    So, let's say we were to double the wire's length. In order for the voltage to stay the same, the electric field through the wire would get divided by $2$. And since the electric field was approximately proportional to the velocity of the electrons in the wire, the velocity of the electrons would also get divided by $2$. Meaning that the current would get divided by $2$ as well.

    We've just discovered our first relationship: for a certain voltage (which I can't emphasize enough is what is FIXED in the circuit due to the way in which a battery works) caused by some battery throughout some circuit, the current is inversely proportional to the length of the wire, but directly proportional to the voltage, since the voltage was directly proportional to the electric field caused within the wire.

    $I \propto \frac{V}{L}$

  2. Now, let's say we were to double the cross-sectional area of this wire. Once again, remember what's fixed is the voltage of the battery, which is given by the electric field the battery generates within the wire multiplied by the length of the wire.

    Since the length of the wire didn't change, the electric field in this wire won't change either. Meaning that the velocity of the electrons moving in this wire won't change either.

    However, going back to the definition of current, we saw it was both proportional to the velocity of the electrons and the cross-sectional area those electrons were traveling through!

    That means that if we double the cross-sectional area of the wire, the current (amount of charge flowing through that area) must've double as well. Aaaand, we've discovered our second relationship: the current is proportional to the cross-sectional area of the wire.

    $I \propto AV$ (where $A$ is the cross-sectional area, and once again the current is proportional to the voltage $V$ because the voltage is proportional to the electric field created within the wire, which is proportional to the velocity of the electrons. I know I'm repeating myself a lot, but I feel like it'll help. If the repetition is annoying, tell me and I'll edit this post).

  3. We've got one more relationship left.

    Remember when I said the material making the wire wasn't "perfectly conductive", so that it required an electric field of some significant magnitude to be created within the wire in order for electrons to move through it? Well, there's actually a way to measure how "perfectly conductive" this wire is - how much it resists the flow of electrons through it. It's called the "resistivity" of the material. We won't go into what causes materials to have more or less resistivity, but the best way to think of it is that the higher the resistivity of the material, the more tightly its atoms hold on to their electrons.

    What that means is that the higher the resistivity, the greater of an electric field will be needed in order to move the electrons at a certain velocity through the wire. If two wires are identical in shape, but the material composing the second wire has twice the resistivity as the material composing the first wire, then in order to cause the same current through the second wire as through the first wire, we'll need an electric field of twice the magnitude through the second wire.

    The resistivity of a material is usually denoted by $\rho$

    And thus, we've discovered our third relationship:

    $I \propto \frac{V}{\rho}$.

Let's now put the three relationships together. We have:

$I \propto \frac{V}{L}$

$I \propto AV$

$I \propto \frac{V}{\rho l}$

Putting them all into one...

$I = \frac{AV}{\rho l}$

Now, all Ohms did (or whoever invented the concept of resistance) was give the term $\frac{\rho l}{A}$ a name. He called it the RESISTANCE of the wire.

$R = \frac{\rho l}{A}$

And thus, the current through the wire is given by:

$I = \frac{V}{R}$

Which agrees with the wording of your original question. It's Ohm's law!!!

Hope that helped :). Please suggest edits and comment if something's confusing!!!! I'll edit this answer as much as need be - after like 40 minutes of writing, I'm really attached to it!!!

Best of luck with your electrical endeavors!

Old question: https://physics.stackexchange.com/a/462720/184540


It depends on how you want to look at it. Voltage is defined to be the energy required to raise charges to a particular potential OR the amount of energy (potential or dynamic) per coulomb of charge in the system, commonly known as the potential difference across the terminals of the electricity source (regardless of source). Current is the amount of charge moving through a conductor and passing a randomly chosen point every second.

Let's assume a piece of wire with length $l$ and diameter $x$, the number of free and mobile electrons in the whole of the conductor is constant. If we set a potential difference $V$ between the two ends of the wire, the current in the wire is defined as the number of total electrons in coulomb moving past a point per second (disregard direction). To have more or less current flowing through the wire, it has to be either the electrons are moving faster or slower because their numbers are limited. So what can make electrons move faster or slower in the wire, it's how much potential difference is set between the two ends of the wire.

If we consider all sources of electricity in existence, we can conclude that the potential difference (~voltage) determines how much current flows through a wire of a particular resistance. I think the relation $V=IR$ isn't just an equivalence, but a causation too.


It is important to note that Ohm’s law, both in the form you’ve written it, and in more fancy forms, is simply an approximation applicable only in a specific range of parameters.

To give some intuition about where this result comes from, consider a conducting material. At any point in time, this material will have some electric field $\textbf{E}$ and some current density $\textbf{J}$ passing through it. Ohm’s law is then simply an approximation, valid in many materials, stating that the electric field and current density are proportional:


where $\rho$ is a constant (or matrix) of proportionality known as the “resistivity” of the material, which is something akin to a “resistance density.” In materials in which the electric field and current density are roughly proportional to each other, this relationship defines resistance. Ohm’s law in the way you’ve written it is simply a hop, skip, and a jump away from this definition.

So in this picture, Ohm’s law, so long as it is valid, can be thought of as a definition for resistance. Non-Ohmic materials obey a nonlinear relationship between the applied voltage and the observed current, and don’t have a simple notion of resistance.

Now, to (what I think is) the heart of your question. The heart of Ohm’s law lies in the relationship between the current in a material and the force applied to it. In the equation above, this corresponds to thinking of the electric field as the “force per unit charge” that “pushes” the corresponding charges to move, creating a current (see section 7.1 of [1] for a discussion). In this sense, Ohm’s law could be thought of as defining a causal relationship: the applied electric field causes a current.

[1] D. Griffiths, Introduction to Electrodynamics, 3rd Edition, 1999


In the dynamical system modeling world, a resistive element can exhibit either type of causality: if an effort source (like a constant-voltage power supply) is applied to a resistor, the output variable is a flow (current). Alternatively, if a flow source (like a constant-current power supply) is applied to a resistor, the output variable is an effort (voltage).

It is worth noting here that dual causality cannot be exhibited by inertances (inductors) or compliances (capacitors). This is because it is not possible to assert an instantaneous effort (voltage) across a compliance nor can you assert an instantaneous flow (current) through an inertance.

For an inductor, then, voltage is the input variable and current is the output; for the capacitor, current is the input variable and voltage is the output.


If by Ohm's Law, you mean the mathematical equation, no, it doesn't imply anything more than the relation between three variables. It is physical context that makes this relation meaningful. Causation is about time and a physical context is needed for attributing time to variables. In the physical context, I think we can say that, yes, a voltage causes a current.

Ohm's Law can be seen as a steady state solution of a kinetic model of charge carriers (Drude model). The Drude model isn't very accurate for metals owing to quantum effects, but I don't think quantum mechanics affects the chain of causation. The Drude model simply emerges Newton's Second Law applied to a system of charge carriers bouncing atom cores. So whether Ohm's Law says that voltage causes current hinges on whether we can say that force (or potential energy) cause acceleration (changes in velocity).

Why do you say that the net force does not cause acceleration in Newton's Second Law? I'm not aware of any problems created by considering it that way. A net force causes an acceleration and the acceleration over time leads to displacement. With Newton's Second Law, we usually proceed this way:

  1. The spatial arrangement of particles at time $t$ determines the force (you can include the velocities of the particles for frictional forces)
  2. The force determines the acceleration at time $t$
  3. The acceleration determines how the velocity changes in time
  4. The velocity determines how the arrangement (positions) of the particles changes in time

Then we go back to (1), calculating a force at time $t + \mathrm{d}t$ for the new arrangement of particles, which determines the acceleration at this time.

This is how Newton's Laws can be implemented numerically (for example, in molecular dynamics). The way of calculating the acceleration, velocity, and position vary between numerical schemes (the velocity is sometimes ignored entirely), but the chain of causation seems fairly clear: force causes acceleration. The cause (the force) and the effect (the displacement) happen at different times in the numerical scheme; however, Newton's Law is continuous and would be the limit of the numerical scheme as the time between steps goes to zero. Philosophically, this limit is strange because the cause and effect would only be infinitesimally separated in time. But for any finite $\mathrm{d}t$, the cause and effect are clear.

Ohm's Law is a little different, since time isn't involved. However, we can see Ohm's Law as the steady state limit of the Drude model. In the Drude model,

  1. The applied voltage and spatial arrangement of charge carriers and ionic cores determines the local electric field
  2. The local electric field determines the acceleration a given charge carrier
  3. The acceleration defines how its velocity changes in time
  4. The velocity defines how its position changes in time

And we can go back to (1), with the new positions determining the new spatial arrangement of charge carriers.

Eventually, an average velocity is established at all points in the circuit (different positions can have different average velocities). This is called the drift velocity and is related to the resistivity. So I would say: the voltage causes a steady-state electric field to develop throughout the circuit and the steady-state electric field causes individual charge carriers to drift, which constitutes a current flow.

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    $\begingroup$ I do not disagree with what you say, but see: we may turn Ohm’s law around and say: current causes a voltage drop to appear over a resistor. This happens in practice, when you connect a current source to a resistor. $\endgroup$
    – flaudemus
    Mar 8, 2019 at 16:40
  • $\begingroup$ @flaudemus That's certainly a valid way to look at it. I suppose that constant voltage sources seem more natural to me, especially from a kinetic theory perspective. $\endgroup$ Mar 8, 2019 at 16:46
  • $\begingroup$ I wanted to point out that your own statement about causality actually implies that Ohm’s law is nothing more than a relation between quantities. There is no reason to talk about causality here, in particular not in the physics context. And this was the OP’s question. $\endgroup$
    – flaudemus
    Mar 8, 2019 at 21:18

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