Can you calculate power dissipation using only Volts and Resistance?

[cannontrodder]

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 = 110
Resistance of copper cable per foot is 5x10^-8 Ohms
Amount of power used per day for NYC estimated at 10¹⁰ Watts

The exercise asks for:

1) "The power lost per foot from I²*R losses."
2) "Length of cable over which you will loose all 10¹⁰ watts"

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=2.42¹¹ Watts. Supposedly that is the power lost per foot so dividing into 10¹⁰ shows 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 = 2.2⁹ Amps

..then P = I²*R = 2.42¹¹ - AGAIN

Am 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?

[grumpydoc]

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 the ^superscript and _subscript functions can help writing formulae.

[andtfoot]

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 amps
The 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 watts
We can then divide the 10E10 by the above, to get number of feet = 24.2

[grumpydoc]

We can then divide the 10E10 by the above, to get number of feet = 24.2

Or, to put it another way, you can't feed 1x10¹⁰ watts 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.

[cannontrodder]

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 = 9.09x10⁷A

So 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 10⁸W

And 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 method 


Grumpydoc, 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!

[PSR B1257]

And divide the total power loss by that to get the number of feet: 24.2 feet!!!

This is true for the total length of the conductor.
But not for the cable, which consists of at least two conductors