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ollegr [7]
3 years ago
11

Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to

that between the large faces (D x L.) 2W. What is the ratio of the resistance as measured between the small faces (WxW) to that between the large faces (DxL)?
Physics
1 answer:
Ludmilka [50]3 years ago
6 0

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

        R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

         A = W W = W²

        R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

    R₂ / R₁ = 2W²/L

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Question:

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