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

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

Answer 2

Answer:

The total resistance of the wire is = $1.917\times10^{-3}$

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, $R_{Total}$

$\frac{1}{R_{Total}} =\frac{1}{R_{cu} }+\frac{1}{R_{al}}$

Parameters given:

Length of wire = 1 m

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

Cross sectional area of aluminium wire  

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

Resistivity of copper $\rho _{cu}=1.7\times 10^{-8} \Omega .m$

Resistivity of Aluminium $\rho _{al}=2.8\times 10^{-8} \Omega .m$

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

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

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

$\frac{1}{R_{Total}} =\frac{1}{5.41\times10^{-3} }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}$

$R_{Total}= 1.917\times 10^{-3}\Omega$

∴ The total resistance of the wire = $1.917\times 10^{-3}\Omega$