Two copper wires have equal lengths, but the diameter of one is three times that of the other.
a) The resistance of the thinner wire is _____ times that of the thicker wire: (Prove your work)
1) 3
2) 1/3
3) 9
4) 1/4
1/3
To determine the resistance of the thinner wire relative to the thicker wire, we need to consider the relationship between resistance, length, and cross-sectional area.
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Mathematically, this relationship can be expressed using the formula:
R = ρ * (L/A)
Where R is the resistance, ρ (rho) is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
Since the two wires have the same length and are made of the same material (copper), the only difference lies in their cross-sectional areas. Let's denote the diameter of the thicker wire as D and the diameter of the thinner wire as d, where D = 3d.
The cross-sectional area of a wire is given by the formula:
A = π * (diameter/2)^2
Substituting the values for the diameters:
A (thicker) = π * (D/2)^2
A (thinner) = π * (d/2)^2
Now, we can calculate the ratio of the cross-sectional areas:
A (thinner) / A (thicker) = (π * (d/2)^2) / (π * (D/2)^2)
= (d^2/4) / (D^2/4)
= d^2 / D^2
Since D = 3d, we can substitute the value:
=d^2 / (3d)^2
=d^2 / 9d^2
= 1/9
Therefore, the ratio of the cross-sectional areas is 1/9.
Now, let's go back to the resistance formula:
R (thinner) = ρ * (L/A (thinner))
R (thicker) = ρ * (L/A (thicker))
Substituting the ratio of cross-sectional areas:
R (thinner) = ρ * (L / (1/9)) (since A (thinner) / A (thicker) = 1/9)
R (thicker) = ρ * (L / 1)
Simplifying:
R (thinner) = 9 * ρ * L
R (thicker) = ρ * L
Therefore, the resistance of the thinner wire is 9 times that of the thicker wire.
So the correct answer is:
3) 9