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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,
The general equation of voltage is,
V = iR
The expression for current density in term of electric field is,
Substitute (V/L) for E in the above equation of current density.
Substitute iR for V in equation (1)
Substitute 1.69 × 10⁸ Ω .m for p
50A for i
0.200Ω.km⁻¹ for (R/L) in eqn (2)
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,
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
Substitute 0.200Ω.km⁻¹ for (R/L)
8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m
c) Using the equation (2) current density for aluminum cable is,
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)
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,
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
The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m
Answer:
Explanation:
The concept of elastic and inelastic demand is applied.
for an elastic demand, the elasticity must be greater than 1 and for an Inelastic demand, the elasticity must be less than 1.
The steps and appropriate calculation is as shown in the attached file.
