# Voltage Splitting Demonstration Video

The video linked is about an experiment connecting 60 USB chargers in parallel. At 2:30 minutes into the video, the narrator says that when a metal rod is welded into a loop, and a low voltage, high-amp current is applied, the rod glows red hot, because the resistance is halved. I found this part of the demonstration interesting because I have been studying voltage splitting circuits involving two resistors. This video appears to have something to do with that effect.

I can't watch the video 'cause I'm at work, so I may not understand what he's doing.

If the guy is putting using metal bar (almost zero resistance) to "short out" one or more USB power supplies, that's dumb because the power supplies can be fried. Just putting two or more power supplies in parallel is usually "not advised".

the rod glows red hot, because the resistance is halved.

If the current travels half the distance through the wire/rod (because it's folded/looped), the resistance is halved. If there are two current-paths (around both sides of the "loop") you have two resistances in parallel, and the resistance is halved again.

Also, if the power is dissipated in a shorter-section of the wire, the heat is more concentrated and more "intense".

With half the resistance, you may or may-not get twice the current because the voltage may not be constant with the power supplies shorted. You are certainly not getting 5V across a metal bar, even with 60 power supplies.

The presenter is incorrect. The resistance would be quartered when the connections are rearranged that way, if the resistance was evenly split between the two branches.

The reason is, suppose the wire logo has a total resistance of R. When the connection is readjusted to the midpoint of the sides instead of the end, each branch has R/2 resistance. When they ends are soldered together, the two branches for a parallel combination of resistances that is R/2 || R/2, which calculates to an equivalent resistance of R/4.

In reality, the clamps weren't actually put at the proper midpoint. You can clearly see that the base of the flame logo heated up faster than the top of it, meaning that it was a lower-resistance branch.

Thanks for your reply. I connected two 330 Ω resistors, parallel. My multimeter gives total reading 168-170 Ω, about half of only one. Doesn't that replicate a resistor wire ring, as in the video? Next, I connected the two 330 Ω resistors in series, just to check the theory. Total resistance reads 651 Ω, about double that of one.
The resistors are cheap parts, which explains the discrepancy. Both read less than 330 Ω alone.
The test I would try, if I had the device shown in the video, would be to apply the current to a metal rod at various segments. Is resistance half of the total length at half? I won't assume so without testing that. We do know that a potentiometer is a ring, but not a complete circle.
I am testing a mechanical pencil lead with my multimeter. The full length of the lead gives a reading of about 5 Ω. The reading is about half of that, at half the pencil lead's length. I thought graphite was actually used in resistors, or still is, being a weak conductor of electricity.

brianeo:
Thanks for your reply. I connected two 330 Ω resistors, parallel. My multimeter gives total reading 168-170 Ω, about half of only one. Doesn't that replicate a resistor wire ring, as in the video? Next, I connected the two 330 Ω resistors in series, just to check the theory. Total resistance reads 651 Ω, about double that of one.
The resistors are cheap parts, which explains the discrepancy. Both read less than 330 Ω alone.
The test I would try, if I had the device shown in the video, would be to apply the current to a metal rod at various segments. Is resistance half of the total length at half? I won't assume so without testing that. We do know that a potentiometer is a ring, but not a complete circle.
I am testing a mechanical pencil lead with my multimeter. The full length of the lead gives a reading of about 5 Ω. The reading is about half of that, at half the pencil lead's length. I thought graphite was actually used in resistors, or still is, being a weak conductor of electricity.

Carbon resistors change value over time(years). I have a box of new, assorted value and size resistors that are perhaps 40 years old. None of them are even close to the value indicated in the color bands. and that is why they are not used, or even available any more.
Paul

brianeo:
Thanks for your reply. I connected two 330 Ω resistors, parallel. My multimeter gives total reading 168-170 Ω, about half of only one. Doesn't that replicate a resistor wire ring, as in the video? Next, I connected the two 330 Ω resistors in series, just to check the theory. Total resistance reads 651 Ω, about double that of one.
The resistors are cheap parts, which explains the discrepancy. Both read less than 330 Ω alone.
The test I would try, if I had the device shown in the video, would be to apply the current to a metal rod at various segments. Is resistance half of the total length at half? I won't assume so without testing that. We do know that a potentiometer is a ring, but not a complete circle.
I am testing a mechanical pencil lead with my multimeter. The full length of the lead gives a reading of about 5 Ω. The reading is about half of that, at half the pencil lead's length. I thought graphite was actually used in resistors, or still is, being a weak conductor of electricity.

This post is quite ambiguous. Your experiment confirms what I wrote about in my post, yet you speak of a "discrepancy", implying that something is not what you thought it would be. What's the problem you still have?

Resistance for a wire is inversely proportional to cross-sectional area (thicker = less resistance) and proportional to length (longer = more resistance).

That was a really stupid, poorly executed "experiment". No way were those supplies load sharing they way the brainiac thought they were. Poor plan, poor wiring, no validation. It's your everyday run-of-the-mill video crap that passes as science these days... sad.