A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thick, so the total radius of the wire is 2.0 mm. The resistivity of copper is 1.7×10−8 Ω·m and that of aluminum is 2.8×10−8 Ω·m. What is the total resistance of the wire? (Hint, are the two diﬀerent materials in series or in parallel with each other?)

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, [tex]R_{Total}[/tex]

[tex]\frac{1}{R_{Total}} =\frac{1}{R_{cu} }+\frac{1}{R_{al}}[/tex]

Parameters given:

Length of wire = 1 m

Cross sectional area of copper [tex]A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3} )^{2} =3.142\times10^{-6} m^{2}[/tex]

Cross sectional area of aluminium wire

[tex]A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3} )^{2}-(1\times 10^{-3} )^{2}] =9.42\times10^{-6} m^{2}\\[/tex]

Resistivity of copper [tex]\rho _{cu}=1.7\times 10^{-8} \Omega .m[/tex]

Resistivity of Aluminium [tex]\rho _{al}=2.8\times 10^{-8} \Omega .m[/tex]

Resistance of copper [tex]R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} } =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega[/tex]

Resistance of aluminium [tex]R_{al}= \frac{\rho_{al} \times l}{A_{al} } =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega[/tex]

The total resistance of the wire can be obtained as follows;

[tex]\frac{1}{R_{Total}} =\frac{1}{5.41\times10^{-3} }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}[/tex]

[tex]R_{Total}= 1.917\times 10^{-3}\Omega[/tex]

∴ The total resistance of the wire = [tex]1.917\times 10^{-3}\Omega[/tex]

nwandukelechi nwandukelechi · Answer 2 · 2025-04-11T04:21:11-04:00

Answer:

Explanation:

Given:

Length, L = 1 m

radius, rc = 1.0 mm

Area of inner copper, Ac = pi × (0.001)^2

= 3.142 × 10^-6 m^2

Thickness, t = 1.0 mm

Total radius of the wire, rt = 2.0 mm

Area of outer aluminum sheathe, Aa = area of total wire, At - area of copper core, Ac

Area of total wire = pi × (0.002)^2

= 1.26 × 10^-5 m^2

Aa = 1.26 × 10^-5 - 3.142 × 10^-6

= 9.42 × 10^-6 m^2

Resistivity of copper, Dc = 1.7×10−8 Ω·m

Resistivity of aluminum, Da = 2.8×10−8 Ω·m

D = (R × A)/L

Rc = (Dc × L)/Ac

= (1.7×10−8 × 1)/3.142 × 10^-6

= 5.41 × 10^-3 Ω

Ra = (2.8×10−8 × 1)/9.42 × 10^-6

= 2.97 × 10^-3 Ω

Since both wires are connected at the same time to the voltage supply, therefore,

1/Rt = 1/Ra + 1/Rc

= 1/2.97 × 10^-3 + 1/5.41 × 10^-3

= 521.54

Rt = 1.92 × 10^-3 Ω