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I have been doing some thought experiment and came across this interesting problem.

Say I have a standard normal wire that is 1 meter long, now I have a DC battery at some voltage level and I connect the positive and negative on each end of the wire. Of course, because of the low resistance of the wire, a large current will be drawn from the battery, causing the wire to heat up.

But here is the question, if I place a volt-meter between the positive side of the wire and to the middle of the wire, what will the volt-meter read? if we can assume the wire absolutely homogeneous in density across the span of the wire, will it read half of the supplied voltage?

enter image description here

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  • \$\begingroup\$ Can we assume an ideal voltage source instead of a DC battery? Then we can focus on the behavior of the wire instead of the behavior of the battery. \$\endgroup\$ Commented Jul 9, 2020 at 18:17
  • \$\begingroup\$ Yes you can, but how would this be different? \$\endgroup\$ Commented Jul 9, 2020 at 18:19
  • \$\begingroup\$ Batteries have internal resistance. So if the battery had an internal EMF of 1.5 V (for example), then unless your 1-m wire was very very thin, most of the 1.5 V would drop across the internal resistance and only a small fraction would appear across the wire. Since this isn't what I think you're asking about, I'd rather just give an answer that focuses on the behavior of the wire. \$\endgroup\$ Commented Jul 9, 2020 at 18:22

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I will assume

  • Instead of a battery you use an ideal voltage source to provide the potential difference across the wire.

  • The connections from the source terminals to the ends of the 1-meter wire are very short and/or made of much larger-diameter wire than the wire being tested.

  • The voltage produced by the voltage source is \$V\$

Then your voltmeter will read \$V/2\$.

Because the resistance of a section of wire is given by $$R=\rho\frac{l}{A}$$ where \$\rho\$ is the resistivity of the wire, \$l\$ is the length of the wire, and \$A\$ is the cross-sectional area of the wire.

By measuring between one end and the midpoint of the wire, you are effectively dividing your wire into two segments of length \$x/2\$. Thus you create a voltage divider with two equal resistors, and the voltage across either of them is then half the voltage across the series combination.

Actually you could measure the voltage across any section (not just one starting at the end of the wire) of the wire of length \$x/2\$ and you would get \$V/2\$, it would just take a little more math to show why.

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  • \$\begingroup\$ Ahh, the voltage divider you mentioned made a whole lot of sense, the problem can be reduced to a voltage divider circuit, thanks! \$\endgroup\$ Commented Jul 9, 2020 at 18:26
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Interesting Thought!

If taking everything only in ideal condition(resistance uniformity, no internal resistance of the battery and infinite current support capability of the battery and wire). If everything is prepared for even FIRE. Then, lets go ahead and see the situation.

Since wire resistivity is uniform across the length of the device, YES, the voltmeter will read half the voltage.

But, please don't try this at home!

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  • \$\begingroup\$ This reads too calm, you should put more in shouty capitals, bold, italics and exclamation marks... :-) \$\endgroup\$ Commented Jul 9, 2020 at 18:27
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Have a look at this wirewound pot photo (courtesy of our friends at Aliexpress):

enter image description here

The pot element consists of a wire, wound around the white core you see, from the terminal at the left bottom to the terminal at the right bottom. The wiper (currently near top center) can rotate from one side to the other. When it is in the middle, there are (ideally) exactly equal lengths of wire on each side, so it will measure halfway between the two voltages, just like a voltage divider.

The wire used in a pot is something along the lines of Nichrome, an alloy with high resistivity and (ideally) very low temperature coefficient. Manganin is common in high accuracy applications.

If you connect a wire across a battery, the battery voltage will drop due to the internal resistance of the battery. The battery may get very hot or do other unpleasant and possibly dangerous things if the resistance is too low.

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