Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

# A 23 −m long cable consists of a solid-inner, cylindrical, nickel core 10 cm in diameter surrounded by a solid... Can you help me?

A 23 −m long cable consists of a solid-inner, cylindrical, nickel core 10 cm in diameter surrounded by a solid-outer cylindrical shell of copper 12 cm in inside diameter and 20 cm in outside diameter. The resistivity of nickel is 7×10^-8 Ωm

A) Calculate the resistance of this cable? So R = ? (unit is μΩ)

B) If we think of this cable as a single material, what is its equivalent resistivity? So ρeff = ? (unit is Ω⋅m)

Relevance
• 1 month ago

The longitudinal resistance for a conductor of constant cross-sectional area is

R = ρL / A

For the inner cylinder, A = pi * (0.05 m)^2

R_nickel = 205 μΩ

For the outer cylindrical shell, A = pi * [ (0.1 m)^2 - (0.06 m)^2 ]

Use ρ = 1.7 × 10^-8 Ωm for copper in case you weren't provided it.

R_copper = 19.4 μΩ

A)

When you connect one end of the cable to one potential and the other to another, the two conductors are connected in parallel.

The resulting resistance is

R = R_nickel * R_copper / (R_nickel + R_copper)

B)

The combination as a whole has the cross-sectional area

A = pi * (0.05 m)^2 + pi * [ (0.1 m)^2 - (0.06 m)^2 ]

So ρeff = R * A / L

• 1 month ago

A) For Solid nickel cylinder, By ρ = RA/L

=>R = ρL/A=>R(nickel) = [7 x 10^-8 x 23]/[πr^2]=>R(nickel) = [7 x 10^-8 x 23]/[3.14 x (5 x 10^-2)^2]=>R(nikel) = 2.05 x 10^-4 ΩFor Hollow shell of copper, by R = ρ/2πL x ln(b/a) {where b & a are the radius of outer & inner shell}=>R(copper) = [1.71 x 10^-2/{2 x 3.14 x 23}] x ln(10/6)=>R(coppper) = 60.48 x 10^-4 Ω

B) The equivalent resistence of theses two by, 1/R = 1/R(nikel) + 1/R(copper)

=>1/R = 1/(2.05 x 10^-4) + 1/(60.48 x 10^-4)

=>R = 5.04 x 10^3 Ω

Thus equivalent resistivity (ρ) = RA/L = [5.04 x 10^3 x 3.14 x (10 x 10^-2)^2]/23

=>ρ = 6.89 Ωm