What is the energy dissipation through the wire?

AI Thread Summary
The discussion revolves around calculating the current and energy dissipation in a 3 m long copper wire connected to a 9 V battery. The resistance was initially calculated as 1.82 mΩ, leading to an implausibly high current of 4945 A. Participants identified that the area calculation for the wire's cross-section was incorrect due to a misunderstanding of the diameter's conversion from millimeters to meters. After clarification, it was confirmed that the original calculations were indeed correct, and the confusion stemmed from a misinterpretation of the wire's dimensions. The thread highlights the importance of accurate unit conversion in electrical calculations.
november1992
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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 = \pir^{2}
R = \rho\frac{L}{A}
i = \frac{V}{R}
\rho = 1.72 x 10^{-8}
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.
 
Last edited:
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Your equations are all correct. Your R is too low by 2 orders of magnitude! Recompute A!
 
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 = \pir^{2}
A = \pi (3 \times 10^{-3}) ^{2}
A = \pi (9 \times 10^{-6})
A = 2.83 \times 10^{-5}
 
3mm = 3 x 10-3 m?

That's news to me! :smile:
 
I thought the prefix milli had a magnitude of 10^{-3}?
 
november1992 said:
I thought the prefix milli had a magnitude of 10^{-3}?

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

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