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I am currently studying The Art of Electronics, third edition, by Horowitz and Hill. Exercise 1.6 says the following:

C. Power in resistors
The power dissipated by a resistor (or any other device) is \$P = IV\$. Using Ohm’s law, you can get the equivalent forms \$P = I^2 R\$ and \$P = V^2 / R\$.
Exercise 1.6. Optional exercise: New York City requires about \$10^{10}\$ watts of electrical power, at 115 volts (this is plausible: 10 million people averaging 1 kilowatt each). A heavy power cable might be an inch in diameter. Let’s calculate what will happen if we try to supply the power through a cable 1 foot in diameter made of pure copper. Its resistance is \$0.05 \ \mu \Omega\$ (\$5 \times 10^{−8}\$ ohms) per foot. Calculate (a) the power lost per foot from “\$I^2R\$ losses," ...

So we have that \$P = \dfrac{I^2}{5 \times 10^{-8} \ \Omega} \$, which means that we need to find the current \$ I \$. I sought to use Ohm's law: \$I = \dfrac{115 \ \text{V}}{5 \times 10^{-8} \ \Omega} = 2.3 \times 10^{9}\$. Have I done this correctly? If not, then why is this incorrect, and what is the correct way to do this?

The Pointer
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    Make a mental note that there is something called "skin depth" that affects AC but not DC current. Skin depth is a reason we would never use a 1 foot diameter solid copper wire even for a low frequency like 60 Hz. Copper is expensive and the current won't flow in the middle of the wire anyway. – user57037 Dec 30 '20 at 10:10
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    P = I^2 R, not I^2 / R. – StainlessSteelRat Dec 30 '20 at 13:53
  • @StainlessSteelRat Thanks for spotting that. – The Pointer Dec 30 '20 at 13:55

2 Answers2

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What you have done is applied 115 volts across the 1 foot of copper cable. This isn't what we are doing here. What you need to find is the amount of current through the cable, but you obtain that by calculating from the power and the voltage. Since P=EI, I will be P/E. When you get done, you can also find I*R, and that will be the voltage dropped across the 1 foot of cable.

gbarry
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  • I don't understand what you mean from "but you obtain that by calculating from the power and the voltage" onwards. Please clarify. – The Pointer Dec 30 '20 at 07:31
  • Your cable is part of a circuit, with a generator on one end, and New York City on the other. Consider New York City as a load that is dissipating 10^10 watts. With 115 volts across it. Now you know enough to calculate the current. – gbarry Dec 30 '20 at 07:36
  • I don't know what to do with this information. – The Pointer Dec 30 '20 at 08:35
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It is important here that you differentiate between the power dissipation (losses) in the wire and the power that is still available for the consumer. The power that is supposed to be available for the consumer is provided: $$P_{load} = 10^{10}\; \text{W}$$ You are also provided with the load voltage for the consumer: $$V_{load} = 115\; \text{V}$$ From this information you can calculate the current that is required to achieve the stated power at the given voltage: $$I = \frac{P_{load}}{V_{load}} = \frac{10^{10}\; \text{W}}{115\; \text{V}} = 8.7 \times 10^7 \; \text{A}$$ Now that you know the current you can calculate the resistive losses caused by the wire with the provided resistance per foot: $$P_{loss} = I^2R$$ The results are of course pretty silly. Nobody would think about supplying all of New York via one thick wire at 115V. But generally this kind of calculation can be applied to more realistic scenarios.

Lars Hankeln
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  • Ahh, ok, I understand. I got \$ 3.8 \times 10^8\$ A\$^2\$ \$\Omega\$ (per foot), where A\$^2\$ \$\Omega\$ is equal to watts? – The Pointer Dec 30 '20 at 09:30
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    Yes. This is basically the reason why power lines are high voltage, to reduce the required current and thus the resistive losses in the power lines. – Lars Hankeln Dec 30 '20 at 09:35