A copper transmission cable 170 km long and 10.0 cm in diameter carries a current of 100 A .

a. What is the potential drop across the cable?
b. How much electrical energy is dissipated as thermal energy every hour?

a = = 37.2V

b = 13.39MJ

Explanation:

Given that

L = 170 × 10³

r = d/2

= 10cm / 2 = 5 cm

current I = 100A

we are to find the potential drop across the cable

so, we can use ohm’s law

V = IR = I (ρL/A)

ρ = resistivity of the copper

= 1.72 × 10⁻⁸ Ω.m

A = πr²

V = I(ρL/πr²)

= 100 (( 1.72 × 10⁻⁸ * 170 × 10³) / ( π * 0.05²))

= 37.2V

(b)

Energy (loss) = Pt

Enery (loss) = IVt

 3600s per hour

= (100A)(37.2V)(3600s)

= 13.39MJ

the answer is b ) square root of 5

50 inches should be the resultant vector