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AFPhys... i never knew that about extension cords. I use the heavy duty (thick orange ) ones mostly . Are they okay or will they also result in damage to power equipement (chainsaw, etc) that one is using

It all depends on how much current (the "I" that V=RI equation) the device draws, and what the resistance (the "R") of the cord is. The cords are generally rated not by their resistance but by the maximum current they can carry. The "V" in that equation is then the voltage drop from the wall plug to the device. Too much drop and the device can be damaged. There are other effects too, but are generally not too important.

BTW, electrical/electronic engineers are taught that equation a bit differently, but it's the same equation.

E=IR, where I and R are the same, and E is the same as "V", but stands for "electromotive force" (the term goes way back). The use of "I" for current I'm not sure about, but a quick search says it is "current intensity" It's a measure of how many electrons per second are passing through the wire. (1 ampere, shortened to "amps", is 6,241,509,479,607,717,888 electrons per second) The unit was defined before they knew the sign of the charge on the moving charge carriers. They of course got it wrong, so to keep things straight, engineers say the current flows from positive to negative, but actually the electrons go the other way. (In your power cord they change directions 60 times a second (in the US, YMMV in other locations).

So there is no one answer to your question, but the longer the cord, the bigger the wires need to be. This is because, like water through a pipe, their is less resistance the bigger the wire. So to get the same voltage drop for a longer cord, (The wires have so many Ohms (the unit of resistance) per unit of length), you have to have a lower resistance per unit of length, and thus a fatter wire).

Ohm, Ampere and Volta were some of the earlier pioneers in electricity, so their names are applied to the units of measure of those three quantities.

El Gato, MSEE, 1977.

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It all depends on how much current (the "I" that V=RI equation) the device draws, and what the resistance (the "R") of the cord is. The cords are generally rated not by their resistance but by the maximum current they can carry. The "V" in that equation is then the voltage drop from the wall plug to the device. Too much drop and the device can be damaged. There are other effects too, but are generally not too important.

BTW, electrical/electronic engineers are taught that equation a bit differently, but it's the same equation.

E=IR, where I and R are the same, and E is the same as "V", but stands for "electromotive force" (the term goes way back). The use of "I" for current I'm not sure about, but a quick search says it is "current intensity" It's a measure of how many electrons per second are passing through the wire. (1 ampere, shortened to "amps", is 6,241,509,479,607,717,888 electrons per second) The unit was defined before they knew the sign of the charge on the moving charge carriers. They of course got it wrong, so to keep things straight, engineers say the current flows from positive to negative, but actually the electrons go the other way. (In your power cord they change directions 60 times a second (in the US, YMMV in other locations).

So there is no one answer to your question, but the longer the cord, the bigger the wires need to be. This is because, like water through a pipe, their is less resistance the bigger the wire. So to get the same voltage drop for a longer cord, (The wires have so many Ohms (the unit of resistance) per unit of length), you have to have a lower resistance per unit of length, and thus a fatter wire).

Ohm, Ampere and Volta were some of the earlier pioneers in electricity, so their names are applied to the units of measure of those three quantities.

El Gato, MSEE, 1977.

Thank you!!! I have read this twice. I am SURE that three is a charm.! SO often I realize how really niave (and lucky) I am in using equipement that I have no clue about.