Question

A cylindrical aluminum pipe of length 1.42 m has an inner radius of 1.60 ×10^-3 m and an outer radius of 3.39 ×10^-3 m. The interior of the pipe is completely filled with copper. What is the resistance of this unit? (Hint: Imagine that the pipe is connected between the terminals of a battery and decide whether the aluminum and copper parts of the pipe are in series or in parallel.)

Answer 1

In this case potential difference across both wires is same, So both wires are connected in parallel with each other

Now we know that in parallel circuit:

1/Req = 1/R1 + 1/R2

Req = R1*R2/(R1 + R2)

R1 = resistance of copper part = rho_Cu*L/A

rho_Cu = resistivity of copper wire = 1.72*10^-8 ohm-m

L = length of wire = 1.42 m

A = Cross-sectional area of copper wire = pi*r1^2

r1 = radius of inner part = 1.60*10^-3 m, So

R1 = rho_Cu*L/(pi*r1^2)

R1 = 1.72*10^-8*1.42/(pi*(1.60*10^-3)^2) = 3.04*10^-3 ohm = Resistance of copper part

R2 = resistance of Aluminum part = rho_Al*L/A

rho_Cu = resistivity of Aluminum wire = 2.82*10^-8 ohm-m

L = length of wire = 1.42 m

A = Cross-sectional area of Aluminum wire = pi*(r2^2 - r1^2)

r1 = radius of inner part = 1.60*10^-3 m,

r2 = radius of outer part of wire = 3.39*10^-3 m, So

R2 = rho_Al*L/(pi*(r2^2 - r1^2))

R1 = 2.82*10^-8*1.42/(pi*[(3.39*10^-3)^2 - (1.60*10^-3)^2]) = 1.43*10^-3 ohm = Resistance of Aluminum part

Now equilibrium resistance will be:

Req = 3.04*10^-3*1.43*10^-3/(3.04*10^-3 + 1.43*10^-3)

Req = 9.73*10^-4 ohm = resistance of complete wire

Let me know if you've any query.