WEBVTT
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This is College Physics Answers
with Shaun Dychko.
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We want to find the resistance
of a piece of wire
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such that the cost
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of the energy lost due to heat
in this regular wire
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equals the cooling cost needed
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to cool a superconducting
wire that loses no heat.
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So there is 1 liter of liquid nitrogen used
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to cool the superconducting wire
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and the cooling cost is 30 cents per liter
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of coolant per hour
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and the wire is meant to carry 100 amps.
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And so the power lost in a regular wire
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is the current through it times
its resistance
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or I should say the current squared
times its resistance—
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that's equation 30 in chapter 20—
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and the cost of the energy lost
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in this regular wire is gonna be the power
loss multiplied by time
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multiplied by this energy cost rate
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which we have to express in
cost per watt hours
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because this power has units of watts
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and so we need our cost in
dollars per watt hour.
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And this power loss we'll express as
*I* squared *R*
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since we know what the current is.
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And then the cost of the coolant is
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30 cents per liter per hour
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multiplied by time multiplied by
the volume of coolant
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and we are told that the costs
are meant to be equal;
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this is a break even point
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for the different costs.
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So we can equate this
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*I* squared times resistance times
time times the energy-cost rate
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equate that to the coolant cost rate
times time times volume
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and the times will cancel and we'll
divide both sides by
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the current squared and divide by
this energy cost rate
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And so we have 30 cents per liter per hour
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times the volume divided by 1 times 10 to
the minus 4 dollars per watt hour
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times the current squared.
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And we substitute 1 liter in for the volume
and 100 amps in for the current
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and we end up with a resistance of
0.30 *Ω**'s
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for the regular wire.*