A cube of side L=2 m is made of a cooper cable with 2mm^2

AI Thread Summary
The discussion focuses on calculating the electric resistance of a cube made of copper cable with a side length of 2 meters and a cross-sectional area of 2 mm². The resistivity of copper is given as 1.72 x 10^-8 Ω*m, leading to a calculated resistance of approximately 0.0172 Ω for one side of the cube. For the equivalent resistance between points A and B, the user is advised to consider the symmetry of the cube and how current distributes across the edges. The conversation emphasizes understanding current flow through the cube's edges when voltage is applied. The user remains confused about the second part of the problem, indicating a need for further clarification on equivalent resistance.
GaussianSurface

Homework Statement


A cube of side L=2m is made of a cooper cable with 2mm^2 of cross section area.
1) Find the electric resistance of one side of the cube.
2) Find the equivalent resistance in between the points A and B.
 

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Actualization: I've done the first statement. By doing this:
The electrical resistivity is defined with the formula R=ρL/A where ρ is the resistivity of the material measured in Ω*m, L is the length in meters and A is the Area measured in m^2.
So that it is:
L=2m
R=?
A=2mm²=2*10^-6m²
ρ= 1.72*10^-8
Replacing these terms in the formula is equal to ≈ 0.0172Ω
I got the first one but the second one I still confused.
 
GaussianSurface said:
the second one I still confused.
You can make good use of the symmetry.
Suppose you apply some voltage across A and B and a current I flows in one of the edges out of A. What current flows along the other edges out of A? What about the six edges those three edges lead to?
 
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