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## Homework Statement

A small city requires about ##17 \mbox{MW}## of power. Suppose that instead of using high-voltage lines to supply the power, the power is delivered at ##120 V##.

Assuming a two-wire line of 0.60cm -diameter copper wire, estimate the cost of the energy lost to heat per hour per meter. Assume the cost of electricity is about ##9.5\, \mbox{ cents}/\mbox{kWh} ##

## Homework Equations

- ##R = \dfrac{\rho \ell}{A}##
- ##P = I^2R##
- ##W = IV##

## The Attempt at a Solution

Given that a small city requires that amount of watts ##W##

##W = IV##

##I = \dfrac{W}{V}##

By the resistance formula, rewrite it in terms of ohms per meter, which is

##R = \dfrac{\rho}{A}##

But we also know that the "face" of the wire is circle, so

##R = \dfrac{\rho}{\pi r^2}##

which is equivalent to

##R = \dfrac{\rho}{\dfrac{d^2\pi}{4}}##

where ##d## is the diameter of the wire. For the power rate, we have double the power rate of one wire, so

##P = 2I^2R##

##P = 2\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

The units of that ##P## are now ##\mbox{W}/\mbox{s}##. By unit conversion, we obtain

##P = 7.2\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

Given that the cost of electricity is around 9.5 cents per kilowatts hour,

##\text{Cost of lost energy}=7.2\cdot 9.5\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

##\text{Cost of lost energy} = 68.4\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

So converting cents into dollars, we finally obtain

##\text{Cost of lost energy} = .684\left( \dfrac{W}{V}\right)^2 \left(\dfrac{\rho}{\dfrac{d^2\pi}{4}} \right)##

By substitution, I got around $8156585 per hour per meter, but Mastering Physics marked it incorrect. I don't see where I went wrong.

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