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16 October, 03:59

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) ?

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  1. 16 October, 04:13
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    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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