(a) should be, respectively, L and A

(b) should be, respectively, 2L and A/2

(c) should be, respectively, L/2 and 2A

(d) do not matter, SINCE the volume of copper

remains the same.

R = ρ L /A

case a R = ρ L /A

case b = R = ρ 2L /(A/2) = 4 ρ L /A

case c = R = (ρ L/2) /2A = (1/4) ρ L /A

option C has minimum resistance

L = L/2

A = 2A

2A = πR²

=> R = √(2A/π)

=> Diameter = 2R = 2√(2A/π)

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