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According to Ohm's law,$$ \left[\text{voltage}\right]~=~\left[\text{current}\right] \times \left[\text{resistance}\right] .$$Does that mean if there is no resistance, then there can be no voltage? Why is that so? If you make a circuit with no load, and no resistors, why does the voltage have to be zero?


marked as duplicate by sammy gerbil, stafusa, Prahar, M. Enns, Cosmas Zachos Jun 4 '18 at 15:59

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    $\begingroup$ The question seems to answer itself. You know the relations between resistance, current and voltage, so I'm not clear why you don't see the connection between zero resistance and zero voltage. $\endgroup$ – StephenG Jun 3 '18 at 19:37
  • $\begingroup$ Hint: if the voltage is non-zero, what will the current be? $\endgroup$ – PM 2Ring Jun 3 '18 at 19:40
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    $\begingroup$ Possible duplicate of Does voltage drop occur with zero load and zero resistance $\endgroup$ – sammy gerbil Jun 3 '18 at 20:16
  • $\begingroup$ See also How can Ohm's law be correct if superconductors have 0 resistivity? $\endgroup$ – sammy gerbil Jun 3 '18 at 20:23
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    $\begingroup$ It is remarkable how many wrong answers you can get by asking about the Ohm's law. Consider a circuit with a battery V, resistor R, and superconductor in series. The current in the superconductor is $I=\dfrac{V}{R}$, the voltage on the superconductor is zero. The Ohm's law for the superconductor is correct $V_s=R_s I$ or $0=0\cdot I$ $\endgroup$ – safesphere Jun 4 '18 at 6:14

Do not assume that a lack of resistors connecting two points implies that the resistance is zero between those points. In fact, it’s the reverse: The conductivity is zero, the resistivity is infinite, the current is zero, and therefore the voltage difference is undefined. That is, you can apply any voltage you wish between two unconnected points.

In contrast, if you connect a perfect conductor between two points (e.g., an idealized superconductor or circuit wire), then the resistance is zero between the points, the conductivity is infinite, the voltage difference is zero, and the current is undefined. Here, the current is determined by the rest of the circuit.


tl;dr- No, the $V$ in Ohm's law is meant as voltage drop, not absolute voltage; it's better to write it as $\Delta V = IR = 0 ,$ which just means that there's no voltage drop between the terminals of the resistor when $R=0.$

Ohm's law is $\Delta V = IR$

Ohm's law is a concept, not a specific algebraic equation, so it can be written different ways. $V=IR$ is common, but it's kinda lazy.

A better expression would be$$ \underbrace{\Delta V}_{\begin{array}{c} V_\text{in} - V_\text{out} \end{array}} ~~=~~ IR ~,$$i.e. that the change in voltage from one end of the resistor to the other is $\Delta V$. This better reflects the concept,$$ {\def\place#1#2#3{\smash{\rlap{\hskip{#1px}\raise{#2px}{#3}}}}} {\def\hline#1#2#3{\place{#1}{#2}{\rule{#3px}{2px}}}} {\def\thline#1#2#3{\place{#1}{#2}{\rule{#3px}{0.5px}}}} {\def\vline#1#2#3{\place{#1}{#2}{\rule{2px}{#3px}}}} {\def\tvline#1#2#3{\place{#1}{#2}{\rule{0.5px}{#3px}}}} % \hline{0}{75}{70} \hline{230}{75}{70} \place{25}{70}{\Large{\bullet}} \place{0}{77}{\begin{array}{c} {\phantom{\small{\bullet}}}\sideset{_{\phantom{\text{in}}}}{_{\text{in}}}{V} \\ \hspace{50px} \end{array}} \place{250}{70}{\Large{\bullet}} \place{225}{77}{\begin{array}{c}{\phantom{\small{\bullet}}}\sideset{_{\phantom{\text{out}}}}{_{\text{out}}}{V} \\ \hspace{50px} \end{array}} \tvline{30}{40}{5} \tvline{30}{50}{5} \tvline{30}{60}{5} \tvline{30}{70}{5} \tvline{255}{40}{5} \tvline{255}{50}{5} \tvline{255}{60}{5} \tvline{255}{70}{5} \place{-10}{70}{\llap{\large{I}}} \place{297}{70}{\rlap{{\Large{\blacktriangleright}}}} \place{70}{50}{\rule{160px}{50px}} \place{73}{53}{\color{white}{\rule{154px}{44px}}} \place{30}{40}{\underbrace{\hspace{225px}}_{\lower{10px}{\Large{\Delta V ~~{\equiv}~~V_\text{in}-V_\text{out}}}}} \place{75}{72}{\begin{array}{c} \hspace{150px} \\ \text{Resistor} \\ \small{\text{resistance, } R~{\equiv}~\frac{\Delta V}{I}} \\ \hspace{150px} \end{array}} % \phantom{\rule{300px}{100px}}_{\hspace{25px}\huge{.}} $$Then in this case,$$ \require{cancel} {\def\place#1#2#3{\smash{\rlap{\hskip{#1px}\raise{#2px}{#3}}}}} {\def\hline#1#2#3{\place{#1}{#2}{\rule{#3px}{2px}}}} {\def\thline#1#2#3{\place{#1}{#2}{\rule{#3px}{0.5px}}}} {\def\vline#1#2#3{\place{#1}{#2}{\rule{2px}{#3px}}}} {\def\tvline#1#2#3{\place{#1}{#2}{\rule{0.5px}{#3px}}}} % \hline{0}{75}{70} \hline{230}{75}{70} \place{25}{70}{\Large{\bullet}} \place{0}{77}{\begin{array}{c} {\phantom{\small{\bullet}}}\sideset{_{\phantom{\text{in}}}}{_{\text{in}}}{V} \\ \hspace{50px} \end{array}} \place{250}{70}{\Large{\bullet}} \place{225}{77}{\begin{array}{c}{\phantom{\small{\bullet}}}{\sideset{_{\phantom{\text{out}}}}{}{V_{\text{out}}}} \\ \hspace{50px} \end{array}} \tvline{30}{40}{5} \tvline{30}{50}{5} \tvline{30}{60}{5} \tvline{30}{70}{5} \tvline{255}{40}{5} \tvline{255}{50}{5} \tvline{255}{60}{5} \tvline{255}{70}{5} \place{-10}{70}{\llap{\large{I}}} \place{297}{70}{\rlap{{\Large{\blacktriangleright}}}} \place{70}{50}{\rule{160px}{50px}} \place{73}{53}{\color{white}{\rule{154px}{44px}}} \place{30}{40}{\underbrace{\hspace{225px}}_{\lower{10px}{{\left.V_\text{in}-V_\text{out}~{=}~\left(0\Omega\right)I~{=}~0\right.~~~~{\Longrightarrow}~~~~\boxed{\left.V_\text{out}~{=}V_\text{in}\right.}}}}} \place{75}{72}{\begin{array}{c} \hspace{150px} \\ \frac{\Delta V}{I} ~{=}~ \cancelto{0\,\Omega}{~R~} \\ \hspace{150px} \end{array}} % \phantom{\rule{300px}{100px}}_{\hspace{25px}\Large{,}} $$so zero resistance basically just means zero voltage drop in a passive resistive element, e.g. an ideal resistor.

Where's $V=IR$ come from?

Voltage is a relative thing. For example, say that you have a AA-battery; they're specified to have a $\Delta V=1.5 \, \mathrm{V}$. However, you could just as readily say that one terminal of your battery is at $1000000000\,\mathrm{V}$ while the other is at $1000000001.5\,\mathrm{V}$; so long as you keep the voltage difference consistent, it's accurate, if slightly misleading out-of-context.

$V=IR$ works the same way. Here, they set $V_\text{out}$ to $0\,\mathrm{V}$, such that$$ \require{cancel} \Delta V ~~=~~ V_\text{in} - \cancelto{0\,\mathrm{V}}{V_\text{out}} ~~=~~ V_{\text{in}} ~~=~~ V ~~=~~ IR ~.$$So basically, $V=IR$ assumes a context in which you specify one terminal of a resistor to be at $0\,\mathrm{V}$.

So, if you use Ohm's law only once in a circuit analysis, and you're willing to say that the output terminal of the resistor is at $V_{\text{out}}=0 \, \mathrm{V}$, then you actually can use $V=IR$ directly.

But, in practice, most folks just use it as shorthand for $\Delta V = IR$.

Example circuit analysis

Here's an example of Ohm's law in action:


For $R_1$, Ohm's law predicts$$ \Delta V ~~=~~ I R ~~=~~ \left(1.5\,\mathrm{A}\right) \left(6\,\Omega\right) ~~=~~ 9\,\mathrm{V} \,.$$This leads to the voltage drop of $9\,\mathrm{V}$ shown in the figure.

Then since$$ \underbrace{{\Delta V}_1}_{9\,\mathrm{V}} ~~{\equiv}~~ \underbrace{{V_1}_{\text{in}}}_{12\,\mathrm{V}} - {V_1}_{\text{out}} \,,$$${V_1}_{\text{out}}=3\,\mathrm{V}$, as shown in the figure. And that's Ohm's law!

Using the same math shows that ${\Delta V}_2 = 3\,\mathrm{V}.$ But, note that it's also true that ${V_2}_{\text{in}}=3\,\mathrm{V}.$ As described in the above section, this is a direct consequence of ${V_2}_{\text{out}}=0\,\mathrm{V}.$

Then to address the case of $R=0\,\Omega$ asked about in the question, it should be clear that while $\Delta V = 0 \, \mathrm{V}$ follows from Ohm's law, that just means that $V_{\text{in}}=V_{\text{out}},$ not that they're both zero (unless one of them is selected as their common reference or/and one of them happens to be at $0\,\mathrm{V}$ due to other circumstance).

The shorthand $\cancel{\Delta} V \to V$ has led to widespread confusion

When I originally wrote this answer, I included the section about why Ohm's law is often written as $V=IR$ rather than the more proper $\Delta V = IR$, as this is a known source of frequent confusion. However, after looking around the internet a bit, it seems that the frequent use of the shorthand has led even Wikipedia's article on voltage to get it mixed up in parts! (Other parts of the Wikipedia article are correctly written.)

So to be clear:

  1. Voltage, $V$, is also known as electric potential.

  2. Voltage drop, $\Delta V$, is also known as electric potential difference.

  3. Voltage drop only reduces to voltage when one of the points is selected as the common reference, e.g. as with $V_{\text{out}}=0.$

  4. Voltage drop is often referred to as "voltage" in simple scenarios where one of the points is already the common reference.

    • For example, a AA battery's voltage drop is typically specified as $\Delta V = -1.5\,\mathrm{V}$, meaning that a AA battery should raise the voltage from the input terminal to the output terminal (where in/out are defined by the sign convention on the current, $I$). However, it's more common for people to say that a AA battery has a voltage of $1.5\,\mathrm{V}.$
  5. Despite being a useful shorthand in simple cases, that context-switching seems to cause conceptual confusion and tends to be grossly impractical in larger circuit analyses.

HyperPhysics and much of the response to this SE.Physics question are helpful. Note that HyperPhysics does use the shorthand at places, too, but their use appears appropriate.

  • $\begingroup$ Voltage is a difference of electric potentials, not a difference of voltages. There is no "in" and "out" voltage, just a voltage between input and output. There is no "delta" in the Ohm's law, just voltage. There is no such thing as "absolute voltage", voltage is relative, just like speed. "Voltage drop" is a technical concept that is not helpful here, but only adds more confusion. And yes, the Ohm's law is a "specific algebraic equation" indeed. $\endgroup$ – safesphere Jun 5 '18 at 6:00
  • $\begingroup$ @safesphere "in" and "out" voltages are the voltages at the "in" and "out" terminals, where flow direction is determined by the charge sign convention (i.e., in the direction of positive $I$). Then, what do you mean about "voltage drop" being a "technical concept" that only adds more confusion? It seems incredibly simple to me, plus I think that it's taught to kids in school these days. I mean it's a basic, introductory principle for learning about electronic circuits; folks need to understand voltage drop to, say, understand how batteries work together in a flashlight. $\endgroup$ – Nat Jun 5 '18 at 6:16
  • $\begingroup$ Voltage is measured between two terminals. You could measure the "in" voltage, for example, between the "in" terminal and the ground, but this is misleading, because the ground has nothing to do with the question and only brings an irrelevant complication. In this question you simply measure the voltage between the "in" and "out" terminals. It is one voltage, not two voltages. $\endgroup$ – safesphere Jun 5 '18 at 6:21
  • $\begingroup$ @safesphere Ohhh I think that you're misunderstanding voltage; naw, voltage isn't between two terminals; that's voltage drop. Rather, voltage is kinda like pressure, and voltage drop is kinda like pressure drop (via the hydraulic analogy). But I think that it's commonly confused because voltage drop is often referred to as voltage using the shorthand described in the second half of the answer. $\endgroup$ – Nat Jun 5 '18 at 6:23
  • $\begingroup$ It's actually pretty interesting reading the Wikipedia article on voltage. They get the distinction right in a lot of places, but the first sentence is very misleading, which they only rectify in a parenthetical note in the first sentence. I hadn't realized that this was such a commonly confused topic! Basically, the Wikipedia article points out that they're using a shorthand, and they give the formally correct definition too, but the article on "voltage" still largely conflates it with voltage drop. $\endgroup$ – Nat Jun 5 '18 at 6:27

Voltage is not 0 since the potential difference is determined by the source. In the limit of 0 resistance, it will be like completely turning on a faucet as opposed to letting it trickle. You get an increase in current keeping the p.d. roughly constant (assuming the source has a sufficiently large capacity). The "0 resistance" is what is commonly referred to as a short circuit in everyday circuits. The high levels of current will be evident even from the thermodynamic properties of the wire since there will be significant heating.

  • $\begingroup$ current source? $\endgroup$ – hyportnex Jun 3 '18 at 20:06
  • $\begingroup$ Wrong answer. The voltage is indeed zero and the current can have any value we want. No heating in the superconductors either. $\endgroup$ – safesphere Jun 4 '18 at 6:19
  • $\begingroup$ I specifically stated "in everyday circuits". Superconductors are a beast of their own. If the voltage is 0 then you cannot have a current in the normal circuit because that would imply that the electric field is zero $\endgroup$ – Jepsilon Jun 4 '18 at 9:12
  • $\begingroup$ You should address your comments using the @ sign. Otherwise no notification is sent to whom you respond. $\endgroup$ – safesphere Jun 5 '18 at 6:09

It is important to realize that equations in physics have physical meaning behind them. So mathematically yes, $R=0$ means $V=0$, but what does this equation actually mean?

This equation is for when we apply a voltage to an ohmic material to determine what current will flow through it. Therefore, we are already assuming the object has resistance. If $R=0$ then we do not use this equation, since it no longer applies.

This is what I mean by saying our equations have meaning. The resistance does not determine potential difference. The resistance determines the resulting current due to a potential difference.

  • 2
    $\begingroup$ I disagree with this answer. Ohm's law is valid if $R=0$, you just have to read it backwards: no matter what current you put in, the voltage will always be zero. $\endgroup$ – Javier Jun 3 '18 at 20:12
  • $\begingroup$ @Javier How would we "put in a current" with no voltage difference? $\endgroup$ – Aaron Stevens Jun 3 '18 at 20:19
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    $\begingroup$ by applying a force somewhere else to push electrons along your zero-resistance material. To a very good approximation, this is just what happens in the cables in a circuit. $\endgroup$ – Javier Jun 3 '18 at 20:21
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    $\begingroup$ Ohm's law does apply when $R=0 \, \Omega$; in fact, an ideal wire is basically just a resistor with $R=0 \, \Omega$, while real-life wires fall short of this ideal, having small-but-non-zero resistances. $\endgroup$ – Nat Jun 3 '18 at 22:04
  • $\begingroup$ @Nat I am not debating the usefulness or existence of $R=0$. And as for your other comment, I know the math works out fine, but I guess it just seems weird to me to say that resistance determines the potential difference. I guess it's kind of like if we use Hooke's law with $k=0$, are we even talking about ideal springs anymore? $\endgroup$ – Aaron Stevens Jun 3 '18 at 23:00

No. $V=IR$ is only valid for an ideal resistor. For instance, an ideal inductor has $R=0$, but the voltage across it is $V=L{dI\over dt}\ne0$.

An ideal circuit with no load and no resistors does not have $R=0$, it has $R=\infty$. $I=0$, so $V$ is given by the indeterminate form $0\times\infty$, which means any value is valid.

  • $\begingroup$ -1: The Ohm's law is valid for an ideal resistor, including $R=0$. Consider a superconductor fed by a current source. The Ohm's law stands: $0=0\cdot I$. $\endgroup$ – safesphere Jun 4 '18 at 6:25
  • $\begingroup$ @safesphere How does that disagree with my statement? All I say is that $R=0$ does not imply $V=0$, since not every component with $R=0$ is an ideal resistor. $\endgroup$ – Chris Jun 4 '18 at 6:44
  • $\begingroup$ The question is about the Ohm's law that applies to ideal resistors, so the question is about an ideal resistor with a zero resistance (superconductor). For such a superconductor (DC current), $R=0$ does imply "V=0". I didn't actually downvote. $\endgroup$ – safesphere Jun 4 '18 at 6:52
  • $\begingroup$ I disagree on your reading of the question. Part of OP's question is "If you make a circuit with [...] no resistors, why does the voltage have to be zero?" The resolution to that question is that Ohm's law does not apply in general, but only if you have ideal resistors in the first place. $\endgroup$ – Chris Jun 5 '18 at 2:39
  • $\begingroup$ Perhaps misreading the question is the reason why you are in a negative territory here counting my virtual downvote ;) $\endgroup$ – safesphere Jun 5 '18 at 5:53

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