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Ohm's Law: Help please and thank you!

  • Thread starter Malaya
  • Start date
  • #1
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Homework Statement



A cylindrical copper cable carries a current of 1300 A. There is a potential difference of 1.6 10-2 V between two points on the cable that are 0.38 m apart. What is the radius of the cable?

I=1300A
V=1.6e-2 V
l=0.38m
rho=1.7e-8(resistivity of copper)
r=?

Homework Equations


V=IR
Where: V=voltage, I=Current, R=Resistance

R=rho* L/r^2pie
Where: R= resistance, rho=1.7e-8(resistivity of copper), Length in meters, r=radius

The Attempt at a Solution



1. Find Resistance.
V=IR
1.6e-2V=1300A*R
R=1.23E-5 ohms

then I used:
R=rho* L/rpie^2
1.23E-5 ohms=1.7e-8*(0.38m/r^2pie)
2.09e-13=0.38m/r^2pie
6.57e-13=0.38m/r^2
6.57e-13*r^2=0.38m
5.78e11=r^2
r=760333.43m

what do I keep doing wrong??
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
1.23E-5 ohms=1.7e-8*(0.38m/r^2pie)
2.09e-13=0.38m/r^2pie
That step is woefully wrong.

Perhaps if you rewrote it

r2= 1.7*10-8*(.38)/(π*1.23*10-5)
 

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