# Current across a wire with zero potential difference

If there was a circuit connected with a $$\50 \,\Omega\$$ resistor and a $$\5 \, \rm V\$$ battery and we measured the voltage across two points of the wire that have no resistor or battery, does it mean the voltage is zero? Then, according to $$\V = I R\$$, is the current also zero?

Assume that the wire has negligible resistance.

• Current is through the wire. Voltage is measured across two points.
– JRE
Feb 26, 2019 at 6:37
• In the limit, $I=\frac{V}{R}=\frac{0}{0}$, which is indeterminate.
– Chu
Feb 26, 2019 at 7:19
• NB you might find it difficult to have an ideal wire with no resistance at all in practice, even low temperature superconductors have some. Feb 26, 2019 at 9:31
• What is the smallest voltage you can measure, and how does it compare to 'negligable'? Feb 26, 2019 at 14:23
• @eckes from wikipedia: "Superconductivity is a phenomenon of exactly zero electrical resistance" Feb 27, 2019 at 4:46

[in a plain wire] does it mean the voltage is zero?

Yes, the voltage across both ends of an ideal wire is always zero.

[given U = R * I] is the current also zero?

No, it means that the current can have an arbitrary value. Because in ...

0 V = 0 Ohm * x Ampere


... x can have any value.

• Theoretically, current can be any value but in practice there's melting point of wire's material. So send enough amps and the wire will melt. Feb 26, 2019 at 23:43
• @SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value. Feb 27, 2019 at 3:51
• @user60561 Well, theoretically, yes. But here's 500 Amp vs wrench Feb 27, 2019 at 3:58
• @SergiyKolodyazhnyy 100000A over 0Ω: eurekalert.org/pub_releases/2014-07/nion-mff072514.php. The actual current is passed through those two 5mm by 20mm superconductors. Feb 27, 2019 at 4:45
• @user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks. Feb 27, 2019 at 4:55

If you connect a 5 V voltage source with a 50 Ohm resistor, there will be a current of:

$$I = U/R = 0.1 \ A$$

Even if you measure a voltage of 0 Volt between two spots with a resistance of (almost) 0 Ohm between them, there is still a current of 0.1 A. It's simply a problem of measurement accuracy: As the resistance decreases, so does the voltage you can measure.

Let's say the Ohmic resistance is not zero but 1 nano-Ohm, then you would expect a voltage of:

$$V = R \cdot I = 1 \ n\Omega \cdot 0.1 \ A = 100 \ pV$$

Of course, measuring such a voltage of 100 pico-Volt would be a challenge.

• OP specifically mentioned the wire between the measuring points had a resistance of zero. It is clearly a theoretical question so this answer is therefore incorrect
– MCG
Feb 26, 2019 at 12:28
• @MCG: No, please read the question again! In fact, this answer clarifies what it implies to have a "negligible resistance" without talking about mathematical limits. Feb 26, 2019 at 13:03
• You need to read it again. The measurement is taking place over a bit of wire with zero resistance. The e-cell with the negligible resistance and the 50 ohm resistor are at different places of the circuit. The measurement is over zero ohms. So again. This is incorrect
– MCG
Feb 26, 2019 at 13:06
• @MCG: Where do you read zero? Feb 26, 2019 at 13:08
• I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly
– MCG
Feb 26, 2019 at 13:12