Why does 0 resistance mean 0 voltage?

Question

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?

Answer 1

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 $R=0.$

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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:

$\hspace{50px}$

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.