Question

Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1.0 mm. Conductor B is a hollow tube of outside diameter 2.7 mm and inside diameter 2.0 mm. What is the resistance ratio RA/RB, measured between their ends?

Answer 1

Answer:

The resistance ratio is 3.23

Explanation:

Since the conductors are made of the same material and length the product of the resistivity and length of the conductor are the same for both conductors and is constant.

R = ρL/A = constant / A

So RA = constant

And RaAa = RbAb

Ra/Rb= Ab/Aa

D = 1mm = 1×10‐³m

Aa = πD²/4 = π(1×10-³)²/4 = 7.85×10-⁷ m²

D1 = inner diameter of conductor B = 2.0mm = 2×10-³m

D2 = outer diameter of conductor B = 2.7mm = 2.7×10-³m

Ab = π(D2²-D1²)/4 = π((2.7×10-³)²-(2.0×10-³)²)/4 = 2.58×10-⁶m²

So,

Ra/Rb= Ab/Aa = 2.58×10-⁶/7.85×10-⁷ = 3.23

The resistance ratio is 3.23. This makes sense because a larger cross sectional area means more electrons can flow through the conductor per time and so lesser resistance. Conductor B has a larger cross sectional area and so lesser resistance when compared to conductor A.