(a) How long must a 0.60 mm diameter aluminum wire be to have a 0.48 A current when connected to the terminal?

(a) current I = V/R => R= V/I = 1.5V/.48 A = 3.13 Ohm, So the AL wire needs to have 3.13 Ohm as resistance. Find the resistivity for Aluminum 9they should have given you this in the problem, but you can google it), R = rho L/A = 2.13 Ohm, use pi r^2 = A, => L= RA/(rho L).

(b)half the length will have half resistance =3.13/2 =1.56Ohm, current will be double.

Resistivity of aluminum ρ = 2.82E-8 Ω∙m

Resistance required: R = V/i = 1.5/0.48 = 3.13 Ω

a) L = πd²R/(4ρ) = π*.0006²*3.13/(4*2.82E-8) = 31.38 m

b) i = 2*0.48 = 0.96 amp