A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the

1.68 x 10⁻⁸Ωm

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A               (i)

Where;

A = πd² / 4              [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4      

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

Thin, aluminium and buried underground.

Explanation:

When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.

i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.

ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.

iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.