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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.

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

2 Answers 2

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[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.

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  • \$\begingroup\$ 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. \$\endgroup\$
    – Sergiy Kolodyazhnyy
    Feb 26, 2019 at 23:43
  • \$\begingroup\$ @SergiyKolodyazhnyy but it won't: P=R*I^2, 0W=0Ω*(x A)^2, x can have any value. \$\endgroup\$
    – flaviut
    Feb 27, 2019 at 3:51
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    \$\begingroup\$ @user60561 Well, theoretically, yes. But here's 500 Amp vs wrench \$\endgroup\$
    – Sergiy Kolodyazhnyy
    Feb 27, 2019 at 3:58
  • \$\begingroup\$ @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. \$\endgroup\$
    – flaviut
    Feb 27, 2019 at 4:45
  • \$\begingroup\$ @user60561 Well that's just cheating with superconductors :) JK. Interesting article. Thanks. \$\endgroup\$
    – Sergiy Kolodyazhnyy
    Feb 27, 2019 at 4:55
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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.

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  • \$\begingroup\$ 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 \$\endgroup\$
    – MCG
    Feb 26, 2019 at 12:28
  • \$\begingroup\$ @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. \$\endgroup\$
    – Frank from Frankfurt
    Feb 26, 2019 at 13:03
  • \$\begingroup\$ 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 \$\endgroup\$
    – MCG
    Feb 26, 2019 at 13:06
  • \$\begingroup\$ @MCG: Where do you read zero? \$\endgroup\$
    – Frank from Frankfurt
    Feb 26, 2019 at 13:08
  • \$\begingroup\$ I think it's clear what the OP means if you read the question properly. The answer by nikolai answers perfectly \$\endgroup\$
    – MCG
    Feb 26, 2019 at 13:12

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