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How can I measure the resistance of a hot dog using Ohms Law?

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ScottH

Sat Feb 25 2017, 05:05PM

Registered Member #61373 Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87

I put 2 forks in a hot dog and hooked my ammeter in series to measure the amps drawn by the hot dog. I also hooked up a separate voltmeter on the 2 forks to measure the voltage between the hot dog ends.
When the amperage peaked at 3.29A, the voltage went down to 123v. The outlet puts out 126v, so it was a 3v drop.

I am confused. I know Ohms Law uses the voltage drop of a resistor and the current flowing through the resistor (hotdog). When I plugged the numbers in a Ohms Law calculator using 3v, it said 0.9 ohms and only 9.8w (the hotdog cooked and busted). Using 123v gave me more realistic numbers.

I know the voltage didn't drop 123v down to 3, or else a car battery would cook it faster with 12v. Can somebody educate me on my scenario? Thank you.

Sulaiman

Sat Feb 25 2017, 05:12PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

The 0.9 Ohms is the source impedance of the ac supply looking back from the forks,
a fraction of the resistance will be in the mains wiring
most will be in your plug and wires

Bjørn

Sat Feb 25 2017, 05:14PM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

Make a drawing of your complete circuit and write down all your known values where they are measured. Look at the 123V and 126V, the drop will be between those two. Now you should be able to see what the voltage across your meat resistor is.

Enceladus

Sun Feb 26 2017, 01:22PM

Registered Member #61428 Joined: Sat Jan 14 2017, 12:39PM
Location:
Posts: 50

LOL. You and you hot dogs.

Physikfan

Sun Feb 26 2017, 04:48PM

Registered Member #60240 Joined: Mon May 16 2016, 07:01PM
Location:
Posts: 304

Hi ScottH

"When the amperage peaked at 3.29A, the voltage went down to 123v. "

Therefore the resistance of your hot dog is 123V/3.29A, about 37 Ohm.

And the power, consumed by your hot dog, is 3.29A times 123V, about 400 W.

And, as said by others before, the resistance of your wires, plugs etc. delivering your current of 3.29 A is (126V-123V)/3.29, about 0.9 Ohm.

ScottH

Mon Feb 27 2017, 08:11PM

Registered Member #61373 Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87

Physikfan wrote ...

Hi ScottH

"When the amperage peaked at 3.29A, the voltage went down to 123v. "

Therefore the resistance of your hot dog is 123V/3.29A, about 37 Ohm.

And the power, consumed by your hot dog, is 3.29A times 123V, about 400 W.

And, as said by others before, the resistance of your wires, plugs etc. delivering your current of 3.29 A is (126V-123V)/3.29, about 0.9 Ohm.

Ok, so I don't use the voltage drop in the Ohms Law formula. I use the voltage that the supply drops to?

ScottH

Mon Feb 27 2017, 08:11PM

Registered Member #61373 Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87

Enceladus wrote ...

LOL. You and you hot dogs.

radiotech

Sat Mar 04 2017, 05:28AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

Somehow i doubt that 3.29 amps times 125 volts represents the wattage in a hotdog. The casing would blow apart.

The forks and the salt water probably are acting like a capacitor.

Do your experiment in a glass vessel filled with water, and convert the rise of temperature of the water
into watt seconds and find out.

Then figure out how to convert the volt and amps, and watt seconds, into uFd.

If the experiment passes muster, you may relish the conclusion.

ScottH

Sat Mar 04 2017, 04:32PM

Registered Member #61373 Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87

radiotech wrote ...

Somehow i doubt that 3.29 amps times 125 volts represents the wattage in a hotdog. The casing would blow apart.

The forks and the salt water probably are acting like a capacitor.

Do your experiment in a glass vessel filled with water, and convert the rise of temperature of the water
into watt seconds and find out.

Then figure out how to convert the volt and amps, and watt seconds, into uFd.

If the experiment passes muster, you may relish the conclusion.

Muster and relish sounds good with hotdogs

The casing did blow at peak watts.

KrowBar

Mon Mar 06 2017, 09:52PM

Registered Member #57401 Joined: Sat Sept 19 2015, 08:06PM
Location: Huntsville, AL
Posts: 10

ScottH wrote ...

Ok, so I don't use the voltage drop in the Ohms Law formula. I use the voltage that the supply drops to?

I think you misunderstand the usage of the term voltage drop. Think drop as in dropping something from a high place releases some potential to do work - dropping voltage across a load is reducing the potential as work is done.

Your volt meter always measures a voltage difference. - If in a closed circuit, the difference is a drop between the two points in the circuit you are connected to. The "supply" voltage would be the open circuit voltage difference between the supply wires. When in a closed circuit , the voltage drop across the load should be quite close to the supply voltage - if it's not, then you have inadequately sized wiring, or significant internal resistance in your source.

What you saw as a 3v change in the supply voltage would more commonly be referred to as sag. Using the gravity potential metaphor again - if the high place you want to drop something from is not rigid, but is sagging, then you don't have as much potential and can't do as much work. The sag depends on the load, but also on the properties of whatever is trying to supply the initial potential (flimsy legs on a tower make it sag, as high resistance wiring in your home/extension cord makes the voltage sag)

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