# Wanted: Free physics lesson **URL:** \ **Category:** Factual Questions\ **Created:** [June 22, 1999, 11:32pm UTC](https://boards.straightdope.com/t/wanted-free-physics-lesson/1452 "1999-06-22T23:32:00Z")\ **Posts on this page:** 6\ **Page:** 1
**Author:** ![system](https://global.discourse-cdn.com/straightdope/original/2X/e/e489c3b7d8fce19c4b355dd4fc3f88cc39c34b87.png) [@system](https://boards.straightdope.com/u/system)\ **Post date:** [June 22, 1999, 11:32pm UTC](https://boards.straightdope.com/t/wanted-free-physics-lesson/1452/1 "1999-06-22T23:32:00Z")
I learned this once, but for the life o’ me I can’t remember… How do you measure the cost of electrical power that is consumed by a household appliance? I know that you need to know the $/kwH and the ohm rating (or is that amperage?), but what do you do with them? There was also something about potential and resistance (that’s voltage, right?). Thanks for your help… * * * “Anything is peaceful from one thousand, three hundred and fifty-three feet.” ---
**Author:** ![system](https://global.discourse-cdn.com/straightdope/original/2X/e/e489c3b7d8fce19c4b355dd4fc3f88cc39c34b87.png) [@system](https://boards.straightdope.com/u/system)\ **Post date:** [June 23, 1999, 1:29am UTC](https://boards.straightdope.com/t/wanted-free-physics-lesson/1452/2 "1999-06-23T01:29:00Z")
It seems to be simply plugging the right values into the equation. You need to know the $/kWh the power company charges, the wattage of the appliance, and the time it runs. Say you are charged $0.06/kWh. If you run a 100-watt light bulb one hour, you’re using 0.1 kWh of power. Multiply that by $0.06 and you get less than a cent. ---
**Author:** ![system](https://global.discourse-cdn.com/straightdope/original/2X/e/e489c3b7d8fce19c4b355dd4fc3f88cc39c34b87.png) [@system](https://boards.straightdope.com/u/system)\ **Post date:** [June 23, 1999, 2:32am UTC](https://boards.straightdope.com/t/wanted-free-physics-lesson/1452/3 "1999-06-23T02:32:00Z")
Makes sense… thanks I guess the next question is, how do I figure the wattage of a coffee maker? That might be where the ohm rating comes into play, but I don’t remember how. * * * “Anything is peaceful from one thousand, three hundred and fifty-three feet.” ---
**Author:** ![system](https://global.discourse-cdn.com/straightdope/original/2X/e/e489c3b7d8fce19c4b355dd4fc3f88cc39c34b87.png) [@system](https://boards.straightdope.com/u/system)\ **Post date:** [June 23, 1999, 2:57am UTC](https://boards.straightdope.com/t/wanted-free-physics-lesson/1452/4 "1999-06-23T02:57:00Z")
If you know the resistance of a circuit (in ohms), and want to figure out its power draw (in watts), here’s how. First, because you asked for a lesson, here’s the definitions. One watt is the power produced by a current of one ampere, across a potential difference of one volt. One ohm is the resistance of a circuit in which a potential difference of one volt, causes a current of one ampere. Potential, resistance, and current are related in the equation: V=IR. V = potential (or voltage) I = current R = resistance (or impedance) So to find the wattage of your appliance, the procedure is simple: Check the label on the back. =b^) ---
**Author:** ![system](https://global.discourse-cdn.com/straightdope/original/2X/e/e489c3b7d8fce19c4b355dd4fc3f88cc39c34b87.png) [@system](https://boards.straightdope.com/u/system)\ **Post date:** [June 23, 1999, 2:38pm UTC](https://boards.straightdope.com/t/wanted-free-physics-lesson/1452/5 "1999-06-23T14:38:00Z")
You turn off everything else electrical in your house and only run that one appliance for an entire month. When you get your electric bill, you will have determined the cost of electrical power for that appliance. [Ow! Ow! Hey, stop hitting me!] ---
**Author:** ![system](https://global.discourse-cdn.com/straightdope/original/2X/e/e489c3b7d8fce19c4b355dd4fc3f88cc39c34b87.png) [@system](https://boards.straightdope.com/u/system)\ **Post date:** [June 23, 1999, 11:03pm UTC](https://boards.straightdope.com/t/wanted-free-physics-lesson/1452/6 "1999-06-23T23:03:00Z")
In case you havent gotten it yet here it is spelled out for you. P=IV P=Wattage/Power (Watts) I=Current(Amps) V=Voltage (Volts) V=IR V=Voltage (volts) I=current (Amps) R=Resistance (Ohms) P=V^2/R You know the voltage supplied (120V in the US), and the resistance is on the appliace usually. To really figure it out the wattage exactly needs to be done with an oscilliscope, but these formulas work well for close approximation. Now to find the cost, just multiply the wattage times the time used and times by the $/kWH. Hope this is complete enough * * * The facts expressed here belong to everybody, the opinions to me. The distinction is yours to draw… Omniscient; BAG