## A cylindrical cable with 2.38 x 10-8 m resistivity carries a current of 1100 A. There is a potential difference of 1.5 x 10-2 V between two

Question

A cylindrical cable with 2.38 x 10-8 m resistivity carries a current of 1100 A. There is a potential difference of 1.5 x 10-2 V between two points on the cable that are 0.22 m apart. What is the radius of the cable

in progress 0
2 months 2021-07-30T13:06:06+00:00 1 Answers 3 views 0

11.1 mm

Explanation:

The resistance of the cable R = ρl/A where ρ = resistivity of cable = 2.38 × 10⁻⁸ Ωm, l = distance between the two points on the cable = 0.22 m and A = cross-sectional area of cable = πr² where r = radius of cable. Also, from Ohms’ law, R = V/I where V = potential difference between the two points = 1.5 × 10⁻² V and I = current in cable = 1100 A

So,

ρl/A = V/I

A = ρlI/V

πr² = ρlI/V

r² = ρlI/πV

taking square root of both sides, we have

r = √(ρlI/πV)

substituting the values of the variables into the equation, we have

r = √(ρlI/πV)

r = √(2.38 × 10⁻⁸ Ωm × 0.22 m × 1100 A/[π × 1.5 × 10⁻² V)

r = √(575.96 × 10⁻⁸ Ωm²A/4.712 × 10⁻² V)

r = √(122.2225 × 10⁻⁶ m²)

r = 11.06 × 10⁻³ m

r = 11.06 mm

r ≅ 11.1 mm