Four cylindrical wires of different sizes are made of the same material. Which of the following combinations of length and cross-sectional area of one of the wires will result in the smallest resistance? a. Length Area
3L 3a
b. Length Area
3L 6a
c. Length Area
6L 3a
d. Length Area
6L 6a

Explanation:

For resistance of a wire , the formula is as follows .

R = ρ L/S

where ρ is specific resistance , L is length and S is cross sectional area of wire .

for first wire resistance

R₁ =  ρ 3L/3a = ρ L/a

for second wire , resistance

R₂ = ρ 3L/6a

= .5 ρ L/a

For 3 rd wire resistance

R₃ = ρ 6L/3a

= 2ρ L/a

For fourth wire , resistance

R₄ = ρ 6L/6a

=  ρ L/a

So the smallest resistance is of second wire .

Its resistance is .5 ρ L/a

i would say

higher potential

explanation:

because it is in a high speed so it must move into a of high potential