Can you calculate power dissipation using only Volts and Resistance? - Page 1
Forum section: electronics / beginners

author: cannontrodder
author_rank: Newbie
timestamp: August 13, 2015, 01:48:33 pm
content: Hi all!I'm a beginner and working my way through The Art of Electronics 3rd edition and I'm glad I didn't just skim over the exercises because I'm getting a lot more insight into things. Although with this one, I'm coming up a little short...The exercise is 1.6 which talks about how much power would be lost if you supplied New York City with a copper cable at 110 V.Voltage = 110Resistance of copper cable per foot is 5x10-8OhmsAmount of power used per day for NYC estimated at 1010WattsThe exercise asks for:1) "The power lost per foot from I2*R losses."2) "Length of cable over which you will loose all 1010watts"The bit that has confused me is the power dissipation is P=IV and Ohm's law gives me the equivalents of P=I2R and P = V2/R. Other people have said to choose the formula that has the the values you already have so I picked P=V2/R which gives me P=2.4211Watts. Supposedly that is the power lost per foot so dividing into 1010shows me it would use all power after 0.413 feet. Seems wrong!The question hints at I2*R losses so I went back and used that:I = V/R = 2.29Amps..then P = I2*R = 2.4211- AGAINAm I working this out correctly? It feels like I should be able to use the resistance over 1ft and the voltage to work out power dissipation but other answers dotted around the web seem to suggest I should be getting an answer around 24 feet. Am I correctly using P=V2/R to work out dissipation?

author: grumpydoc
author_rank: Super Contributor
timestamp: August 13, 2015, 02:20:06 pm
content: Your mistake is that you don't have a drop of 110V every foot of cable.Work out the current using voltage & power, and then put that into I2R to get the power loss per foot.Oh, and learning how to use thesuperscriptandsubscriptfunctions can help writing formulae.

author: andtfoot
author_rank: Unknown
timestamp: August 13, 2015, 02:31:32 pm
content: It's probably not the most efficient, but this is how I would initially attempt to work it out:I think of the cable and the city is effectively 2 resistors in series creating a voltage divider. We need to isolate just the cable side to figure out #1.First I divide the amount of power by the voltage to get the current through the system.10E10 / 110 = 90,909,091 ampsThe resistances, being in series, will have the same current through both. We can now either figure out the power lost in the cable with I^2*R.90,909,091^2 * 5x10E-8 = 413,223,141 wattsWe can then divide the 10E10 by the above, to get number of feet = 24.2

author: grumpydoc
author_rank: Super Contributor
timestamp: August 13, 2015, 03:07:33 pm
content: Quote from: andtfoot on August 13, 2015, 02:31:32 pmOr, to put it another way, you can't feed 1x1010watts down a cable at 110V even if it does have only 50 nano ohms resistance per foot.Which is what I suspect the exercise is trying to demonstrate.

author: cannontrodder
author_rank: Newbie
timestamp: August 13, 2015, 03:27:05 pm
content: Ok, the voltage drop over 1 foot is the obvious problem - ie, it doesn't drop 110V over 1 foot!!So am I saying that the Voltage *does* drop at the point where the last of the power dissipates?Anyway, I = P/V = 1010/110 = 9.09x107ASo now I think of the cable as lots of resistors in series, each 1 foot long so I can use the resistance of 1 foot to work out power dissipation:P = I2R = 4.132 x 108WAnd divide the total power loss by that to get the number of feet: 24.2 feet!!!Thank you for your help grumpydoc, and andtfoot thanks for laying it out like that, I was still making silly maths mistakes once I had the right methodGrumpydoc, yes that seems to be the purpose of the exercise - it goes on to ask "what is the solution to this problem?". So if I=P/V and I increase the voltage, I can lower the current. Lower the current and the power dissipation specified by P=I2R will drop dramatically.Thanks again!

author: PSR B1257
author_rank: Frequent Contributor
timestamp: August 13, 2015, 04:50:42 pm
content: QuoteThis is true for the total length of the conductor.But not for the cable, which consists of at least two conductors