What is the energy dissipation through the wire?

[november1992]

Homework Statement

A 3 m long copper wire that has a diameter of 6 mm is connected to a 9 V battery.

a) What is the current through the wire?
b) If wire a was connected to a battery for 1 second, how much energy would the wire dissipate?

Homework Equations

A = [itex]\pi[/itex][itex]r^{2}[/itex]
R = [itex]\rho[/itex][itex]\frac{L}{A}[/itex]
i = [itex]\frac{V}{R}[/itex]
[itex]\rho[/itex] = 1.72 x [itex]10^{-8}[/itex]
P = i * V
W = P * t

The Attempt at a Solution

I got:
R = 1.82mΩ
i = 4945 A
p = 44505.45

I'm unsure of my answers. The values seem to be too high.

 

[rude man]

Your equations are all correct. Your R is too low by 2 orders of magnitude! Recompute A!

 

[november1992]

Well, I tried to recalculate A and I got the same answer. I even used an online calculator and I got the same answer

A = [itex]\pi[/itex][itex]r^{2}[/itex]
A = [itex]\pi[/itex] [itex](3 \times 10^{-3}) ^{2}[/itex]
A = [itex]\pi[/itex] (9 [itex]\times[/itex] [itex]10^{-6}[/itex])
A = 2.83 [itex]\times[/itex] [itex]10^{-5}[/itex]

 

[rude man]

3mm = 3 x 10^-3 m?

That's news to me!

 

[november1992]

I thought the prefix milli had a magnitude of [itex]10^{-3}[/itex]?

 

[rude man]

I thought the prefix milli had a magnitude of [itex]10^{-3}[/itex]?

Er, oops, senior moment on my part. Of course that's right. And so is your original answer. Pardon my goof.