If a certain piece of copper is to be shaped into a conductor of minimum resistance, its length (L) and cross sectional area (A) shall be respectively
Answers
(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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