A cylindrical aluminum pipe of length 1.52 m has an inner radius of 2.20 10-3 m and an outer radius of 3.00 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 in Ohms
To find the resistance of the unit, we need to determine whether the aluminum and copper parts of the pipe are in series or in parallel.
To analyze this, let's consider the electrical conductivity of each material. Copper is a conductor with high conductivity, while aluminum has lower conductivity. When two conductors are connected in parallel, the total resistance decreases, as each path offers a lower resistance. On the other hand, when two conductors are connected in series, the total resistance increases, as the resistances of each path add up.
In this case, the cylindrical aluminum pipe is filled with copper. The electrical current flowing through the pipe will have two paths—one through the aluminum and another through the copper. Since the aluminum and copper are connected end to end (from the outer surface of the aluminum to the inner surface of the copper), the electrical current divides between the two materials.
Therefore, the aluminum and copper parts of the pipe are in parallel. The total resistance of two resistors connected in parallel can be calculated using the formula:
1 / Rt = 1 / R1 + 1 / R2
where Rt is the total resistance and R1, R2 are the resistances of the two parts.
Now, let's find the resistance of each part separately:
The resistance of the aluminum part can be calculated using the formula:
R_aluminum = (ρ_aluminum * L) / A
where ρ_aluminum is the resistivity of aluminum, L is the length of the aluminum part, and A is the cross-sectional area of the aluminum part.
The resistivity of aluminum, ρ_aluminum, is approximately 2.82 x 10-8 Ω * m.
The length of the aluminum part, L, is given as 1.52 m.
The cross-sectional area of the aluminum part, A, can be calculated using the formula:
A = π * (r_outer^2 - r_inner^2)
where r_outer is the outer radius and r_inner is the inner radius of the aluminum part.
Substituting the given values:
A = π * ((3.00 x 10^-3 m)^2 - (2.20 x 10^-3 m)^2)
Now, calculate the resistance of the aluminum part using the obtained values.
Next, let's calculate the resistance of the copper part. Since it completely fills the interior of the pipe, its resistance will be the same as if it were a solid copper cylinder. The resistance of a solid cylindrical conductor can be calculated using the formula:
R_copper = (ρ_copper * L) / A_copper
where ρ_copper is the resistivity of copper, L is the length of the copper part, and A_copper is the cross-sectional area of the copper part.
The resistivity of copper, ρ_copper, is approximately 1.68 x 10^-8 Ω * m.
The length of the copper part, L, is given as 1.52 m.
The cross-sectional area of the copper part, A_copper, can be calculated using the formula:
A_copper = π * r_inner^2
Substituting the given values:
A_copper = π * (2.20 x 10^-3 m)^2
Now, calculate the resistance of the copper part using the obtained values.
Finally, calculate the total resistance by substituting the calculated resistance values into the parallel resistance formula mentioned earlier:
1 / Rt = 1 / R_aluminum + 1 / R_copper
Now, solve for Rt to find the total resistance of the unit.