How Do You Solve a Problem Involving Wire and Power Dissipation?

[james11223]

Homework Statement

I have attached the problem and all equations.

Homework Equations

The Attempt at a Solution

I have no idea where to even start this problem, and our teacher said that our test would have a similar problem, so please any help would be GREATLY appreciated!

 

[gneill]

Start by working out the end-to-end resistance of the copper cylinder.

 

[james11223]

Ok, I got 149,599.65 J
and 4.675*10^22 electrons

Can someone confirm this?

 

[gneill]

Ok, I got 149,599.65 J
and 4.675*10^22 electrons

Can someone confirm this?

The values are good. Well done.

You might consider being a little less enthusiastic about the number of digits The values given in the problem statement didn't have nearly so many significant figures!

 

[james11223]

Ok thank you very much !

I used P = V^2/R

R = e(l/a) = (1.6*10^-8)(0.5/0.001^2 pi)
p = 20^2/0.00267 = 1.50*10^5 watts

V = IR
I = V/R = (20/0.00267) = 7479 amps = C/s
(7479 C/s)*(1e-/ 1.69*10^-19 C) = 4.68*10^22 electrons

Is my reasoning here correct?

 

[gneill]

Ok thank you very much !

I used P = V^2/R

R = e(l/a) = (1.6*10^-8)(0.5/0.001^2 pi)
p = 20^2/0.00267 = 1.50*10^5 watts

V = IR
I = V/R = (20/0.00267) = 7479 amps = C/s
(7479 C/s)*(1e-/ 1.69*10^-19 C) = 4.68*10^22 electrons

Is my reasoning here correct?

Looks good. So you're providing the value for continuous power (Watts, or J/s) rather than the energy dissipated in one second (Joules); The question wording was a bit vague and open to either interpretation. You've likely chosen the correct one.

 

[james11223]

Wouldnt you just divide the answer by one? thus, giving me the same answer...

 

[gneill]

Wouldnt you just divide the answer by one? thus, giving me the same answer...

Well, multiply by one second to get the power dissipated in one second: (J/s)*1s = J .