Copper and aluminum are being considered for a high-voltagetransmission line that must carry a current of 50.0 A. The resistance per unit length is to be0.200 /km. The densities of copper and aluminum are 8960 and 2700kg/m3, respectively.

Compute:
(a) the magnitude J of the averagecurrent density for a copper cable
1 A/m2
(b) the mass per unit length λ for a coppercable.
2 kg/m
(c) the magnitude J of the average current density for analuminum cable
3 A/m2
(d) the mass per unit length λ for an alumniumcable.
4 kg/m

(a) Jc = 5.81×10⁵A/m²

(b) λ = mass per unit length = 8960×8.60×10-⁵ = 0.771kg/m

(c) Ja = 3.62×10⁵A/m²

(d) λ = mass per unit length = 2700××1.38×10-⁴= 0.373kg/m

Explanation:

Given:

the current to be carried in the conductors I = 50.0A

Densities of copper and aluminum, 8960kg/m³ and 2700kg/m³ respectively.

R/L = 0.2Ω/km = 0.2×10-³Ω/m

Required to find

(a) J the current density in A/m² for copper

We don't know the Cross sectional area of the conductors but we know the current flowing through them and the current density is J = I/A. So we need to find the cross sectional area first. A relation that can help us is that of the resistance of a conductor which is

R = ρL/A

Where R is the resistance of the conductor in Ohms (Ω)

ρ is the resistivity of the (a property) conductor in (Ωm)

L is the length of the conductor

A is the cross sectional area of the conductor

From the formula, the resistance per unit length R/L = ρ/A

So A = ρ ÷ (R/L)

For copper, resistivity is ρ = 1.72×10-⁸Ωm

So Ac = cross sectional area for copper = 1.72×10-⁸/(0.2×10‐³)= 8.6×10-⁵ m²

Ac = 8.6×10-⁵m²

I = 50.0A

Jc = I/Ac = current density

Jc = 50.0/(8.6×10-⁵) = 5.81×10⁵A/m²

(b) λ = mass per unit length = density × Area

Copper: density = 8960kg/m³, Ac = 8.6×10-⁵ m²

λ = mass per unit length = 8960×8.6×10-⁵ = 0.771kg/m

(c) For aluminium, ρ = 2.75×10-⁸Ωm

So Aa = 2.75×10-⁸/0.200×10‐³ = 1.38×10-⁴ m²

Aa = 1.38×10-⁴ m²

Ja = I/Aa= current density

Ja = 50.0/(1.38×10-⁴) = 3.62×10⁵A/m²

(d) Aluminium: density = 2700kg/m³, Ac = 1.38×10-⁴m²

λ = mass per unit length = 2700×1.38×10-⁴ = 0.373kg/m

Jc = i/Ac = 50/8.45 x 10⁻⁵ = 5.91 x 10⁵ A.m⁻²

λc = density x area = 8960 x 8.45 x 10⁻⁵ = 0.75kg.m⁻¹

Ja = i/Aa = 50/13.7 x 10⁻⁵ A.m⁻² =3.65 x 10⁻⁵

λa = density x area = 2700 x 13.7 x 10⁻⁵ = 0.37kg.m⁻¹

Explanation:

The resistance of the wire is defined as

R = ρL/A

where ρ is the resistivity of the material, L is the length of the wire and A is the cross-section area of the wire

so the resistance per length is given by

R/L = ρ/A

The value of the resistance per unit length has to be 0.200Ω/km = 0.20 x 10⁻³Ω/m

so the cross sectional area of the wire is given by

A = ρ/0.20 x 10⁻³ = ρ/0.20 x 10³m³

for copper ρc = 1.69 x 10⁻⁸Ωm

and the cross sectional area of the copper wire has to be

Ac = 1.69 x 10⁻⁸/0.20 x 10³ = 8.45 x 10⁻⁵m²

a) The current density Jc = i/Ac = 50/8.45 x 10⁻⁵ = 5.91 x 10⁵ A.m⁻²

b) Mass per unit length of the copper wire is given by

λc = density x area = 8960 x 8.45 x 10⁻⁵ = 0.75kg.m⁻¹

For aluminium ρa = 2.75 x 10⁻⁸Ω.m

so the cross sectional area of the aluminium wire has to be Aa = 2.75 x 10⁻⁸/0.20 x 10³ = 13.7 x 10⁻⁵m²

c) The current density Ja = i/Aa = 50/13.7 x 10⁻⁵  = 3.65 x 10⁻⁵A.m⁻²

d) Mass per unit length of the aluminium wire is given by

λa = density x area = 2700 x 13.7 x 10⁻⁵ =  0.37kg.m⁻¹

it would be c. predictability