Question

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?)

Answer 1

Step-by-step 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

Answer 2

The total resistance of the wire is =
`1.917*10^(-3)`

Step-by-step explanation:

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,
`R_(Total)`

`(1)/(R_(Total)) =(1)/(R_(cu) )+(1)/(R_(al))`

Parameters given:

Length of wire = 1 m

Cross sectional area of copper
`A_(cu)= \pi r^(2)= \pi * (1* 10^(-3) )^(2) =3.142*10^(-6) m^(2)`

Cross sectional area of aluminium wire

`A_(al)= \pi( R^(2)-r^(2))\\\\ = \pi * [ (2* 10^(-3) )^(2)-(1* 10^(-3) )^(2)] =9.42*10^(-6) m^(2)\\`

Resistivity of copper
`\rho _(cu)=1.7* 10^(-8) \Omega .m`

Resistivity of Aluminium
`\rho _(al)=2.8* 10^(-8) \Omega .m`

Resistance of copper
`R_(cu)= (\rho_(cu) * l)/(A_(cu) ) =(1.7* 10^(-8) * 1)/(3.142*10^(-6) ) =5.41* 10^(-3)\Omega`

Resistance of aluminium
`R_(al)= (\rho_(al) * l)/(A_(al) ) =(2.8* 10^(-8) * 1)/(9.42*10^(-6) ) =2.97* 10^(-3)\Omega`

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

`(1)/(R_(Total)) =(1)/(5.41*10^(-3) )+(1)/(2.97*10^(-3))=521.52(1)/(\Omega)`

`R_(Total)= 1.917* 10^(-3)\Omega`

∴ The total resistance of the wire =
`1.917* 10^(-3)\Omega`