Respuesta :
Answer:
a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²
b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m
c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²
d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m
Explanation:
The expression for electric field of conductor is,
[tex]E = \frac{V}{L}[/tex]
The general equation of voltage is,
V = iR
The expression for current density in term of electric field is,
[tex]J = \frac{E}{p}[/tex]
Substitute (V/L) for E in the above equation of current density.
[tex]J = \frac{V}{pL} ------(1)[/tex]
Substitute iR for V in equation (1)
[tex]J = \frac{iR}{pL} ------(2)[/tex]
Substitute 1.69 × 10⁸ Ω .m for p
50A for i
0.200Ω.km⁻¹ for (R/L) in eqn (2)
[tex]J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2[/tex]
The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²
b) The expression for resistivity of the conductor is,
[tex]p = \frac{RA}{L}[/tex]
[tex]A = \frac{pL}{R}[/tex]
The expression for mass density of copper is,
m = dV
where, V is the density of the copper.
Substitute AL for V in equation of the mass density of copper.
m=d(AL)
m/L = dA
λ is use for (m/L)
substitute,
pL/R for A and λ is use for (m/L) in the eqn above
[tex]\lambda = d\frac{p}{\frac{R}{L} } ------(3)[/tex]
Substitute 0.200Ω.km⁻¹ for (R/L)
8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m
[tex]\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1[/tex]
c) Using the equation (2) current density for aluminum cable is,
[tex]J = \frac{iR}{pL}[/tex]
p is the resistivity of the aluminum cable.
Substitute 2.82 × 10⁻⁸Ω.m for p ,
50A for i and 0.200Ω.km⁻¹ for (R/L)
[tex]J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2[/tex]
The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²
d) Using the equation (3) mass per unit length for aluminum cable is,
[tex]\lambda = d\frac{p}{\frac{R}{L} }[/tex]
p is the resistivity and is the density of the aluminum cable.
Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p
[tex]\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m[/tex]
The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m
Answer:
(a) Jc = 6.88×10⁵A/m²
(b) λ = mass per unit length = 8960×8.82×10-⁵ = 0.790kg/
(c) Ja = 4.30×10⁵A/m²
(d) λ = mass per unit length = 8960×8.82×10-⁵ = 0.7367kg/m
Explanation:
Given:
the current to be carried in the conductors I = 60.7A
Densities of copper and aluminum, 8960kg/m³ and 2600kg/m³ respectively.
R/L = 0.195Ω/km = 0.195×10-³Ω/m
Required to find
(a) J the current density in A/m²
To do this we would need to know what the cross sectional area is. 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 = 1.72×10-⁸/0.195×10‐³ = 8.82×10-⁵ m²
Ac = 8.82×10-⁵ m²
I = 60.7A
Jc = I/Ac = current density
Jc = 60.7/(8.82×10-⁵) = 6.88×10⁵A/m²
(b) λ = mass per unit length = density × Area
Copper: density = 8960kg/m³, Ac = 8.82×10-⁵ m²
λ = mass per unit length = 8960×8.82×10-⁵ = 0.790kg/m
(c) For aluminium, ρ = 2.75×10-⁸Ωm
So Aa = 2.75×10-⁸/0.195×10‐³ = 1.41×10-⁴ m²
Aa = 1.41×10-⁴ m²
Ja = I/Aa= current density
Ja = 60.7/(1.41×10-⁴) = 4.30×10⁵A/m²
(d) Aluminium: density = 2600kg/m³, Ac = 1.41×10-⁴m²
λ = mass per unit length = 2600×1.41×10-⁴ = 0.367kg/m
