# Art of Electronics exercise1.6

Discussion in 'Homework Help' started by dacart, Jun 9, 2010.

1. ### dacart Thread Starter New Member

Jun 7, 2010
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As I mentioned elsewhere, I am trying to teach myself EE primarily using the Horowitz and Hill book "The Art of Electronics." My math is VERY rusty right now so I wind up doubting my results on some things. Case in point is Exercise 1.6. It has to do with calculating a: the power lost per foot from "P=IIR" losses (II meaning "I squared") b:the length of cable over which you will lose all 10 to the 10th power watts of power. and C: how hot the cable will get using the formula σ=6 X 10 to the minus 12thW/K to the 4th X cm squared. The parameters are 10 to the 10th power Watts at 110 volts through a 1 foot diameter pure copper cable with a resistance of 5 X 10 to the -8th Ohms/foot

To get the power loss per foot I've used the formula P = Vsquared/R and come up with the answer of 242,000,000,000 (!?) I then divide 10 to the 10th Watts by that answer and get .041322314... As far as C: is concerned I THINK I would put K on the left side and multiply the Watts time the cm squared. Is my methodology correct? or am I making fatal algebraic errors somewhere

Dan Carter

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Is this the question?

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3. ### dacart Thread Starter New Member

Jun 7, 2010
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Precisely... I don't have access to a digital copy otherwise I would have spared you the verbosity of my question

Last edited: Jun 9, 2010
4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It's the old adage about one picture being worth a thousand words.

You actually have to use I^2R to work out the power loss per foot.

The current at 10^10 Watts and 110V is 9.091x10^7 amps.

So the power loss per foot is

P=[(9.091x10^7)^2]x[5x10^-8]=4.132x10^8 Watts

To consume the total 10^10 Watts would take

Length = (10^10)/(4.132x10^8) = 24.2 feet

Which is a rather modest length to chew up all that energy.

I suspect the copper cable would probably melt well before the current reached anywhere near 90.91 million amps .....

BTW you can express large numbers using a useful shorthand notation - called scientific notation.

So

10^10 may be written as 1E10

or

5x10^-8 as 5E-8

It's a standard function on a scientific calculator and it will make sense to your readers.

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5. ### dacart Thread Starter New Member

Jun 7, 2010
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Thanks! I AM aware of scientific notation but I was unaware of how to express it as you demonstrated. I'll try not to bump into too much furniture while I'm learning the ropes.
BTW is my algebra correct for the last part of the question i.e.
K^4 = 6 x 10^-12Wcm^2 ?
Thanks again
Dan

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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$[Radiant \ Power \ Loss] \ P_{loss}=\sigma T^{4}A$

from which you get

$T=(\frac{P_{loss}}{\sigma A})^{\frac{1}{4}}$

You are given

$\sigma = 6E-12 \ W K^{-4}cm^{-2}$

$P_{loss}=10^{10} \ Watts$

You calculate the radiant area A as the cylindrical area of the 1 foot diameter copper conductor of length 24.2 feet. You have to convert this area to square cm since emissivity value σ is quoted for square cm.

You'll end up with a ludicrous temperature - far beyond the melting point of copper.

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