Three copper wires have lengths and cross-sectional areas as (l, A), (2l, A/2), and
(l/2,2A). resistance is minimum in
A. wire of cross-sectional area A/2
B. wire of cross-sectional area A
C. wire of cross-sectional area 2A
D. same in all three cases.

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Solution

$R e s s i s \tan c e i n W i r e, R = \frac{ρ l}{A} F o r F i r s t W i r e, R_{1} = \frac{ρ ι}{A} F o r S e c o n d W i r e, R_{2} = \frac{ρ 2 l}{A / 2} = \frac{4 ρ l}{A} F o r T h i r d W i r e, R_{3} = \frac{ρ l / 2}{2 A} = \frac{ρ l}{4 A} R e s s i s \tan c e w i l l b e m i n i m u m i n T h i r d W i r e h a v i n g C r o s s s e c t i o n 2 A$

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