Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L1 and wire 2 has length L2, how do L1 and L2 compare if wire 1 is made from copper and wire 2 is made from aluminum? The resistivity of copper is 1.7 × 10-5Ω·m and the resistivity of aluminum is 2.82× 10-5Ω·m.
L1 = 0.60 L2
L1 = 1.7 L2
L1 = 0.36 L2
L1 = 2.8 L2
A wire has resistance R. A second wire has twice the length, twice the diameter, and twice the resistivity of the first wire. What is its resistance?
8 R
The resistance is not given.
R
R/4
To compare the lengths of wire 1 (copper) and wire 2 (aluminum), we need to consider their resistances. The resistance of a wire is given by the formula:
R = (resistivity * length) / cross-sectional area
Since both wires have the same resistance, we can set up the following equation:
(resistivity of copper * L1) / (area of copper wire) = (resistivity of aluminum * L2) / (area of aluminum wire)
The area of a wire is proportional to its diameter squared, so if the wires have the same diameter, their areas will be equal. Let's assume that the diameter is denoted as D.
Therefore, the equation becomes:
(resistivity of copper * L1) = (resistivity of aluminum * L2)
Substituting the given values, we have:
(1.7 × 10-5 Ω·m * L1) = (2.82 × 10-5 Ω·m * L2)
We can simplify this equation further:
L1 = (2.82 × 10-5 Ω·m / 1.7 × 10-5 Ω·m) * L2
L1 = 1.66 * L2
So the answer is that L1 is approximately 1.66 times the length of L2.
For the second part of the question, let's denote the resistance of the first wire as R. The second wire has twice the length, twice the diameter, and twice the resistivity of the first wire.
Since resistance is directly proportional to both length and resistivity, and inversely proportional to the cross-sectional area, we can use the following formula:
R2 = R * (2 * length) / (2 * diameter) * (2 * resistivity)
Simplifying the equation, we get:
R2 = R * length / diameter * resistivity
Since the second wire has twice the length, twice the diameter, and twice the resistivity of the first wire, we can substitute these values:
R2 = R * 2 * length / 2 * diameter * 2 * resistivity
R2 = R * length / diameter * resistivity
Therefore, the resistance of the second wire is equal to R, which is the resistance of the first wire.
To compare the lengths of wire 1 and wire 2, we can use the formula for resistance:
R = ρ * (L / A)
Where R is resistance, ρ is resistivity, L is length, and A is cross-sectional area.
For wire 1 (copper), let the length be L1 and the cross-sectional area be A1. The resistance of wire 1 is R1.
R1 = ρcopper * (L1 / A1)
For wire 2 (aluminum), let the length be L2 and the cross-sectional area be A2. The resistance of wire 2 is R2.
R2 = ρaluminum * (L2 / A2)
Since wire 1 and wire 2 have the same resistance, we can set R1 equal to R2:
R1 = R2
ρcopper * (L1 / A1) = ρaluminum * (L2 / A2)
Since the cross-sectional area is the same for both wires (same diameter), we can cancel it out:
ρcopper * L1 = ρaluminum * L2
Now we can compare the lengths:
L1 / L2 = ρaluminum / ρcopper
Using the given resistivity values, we have:
L1 / L2 = (2.82 * 10^-5 Ω·m) / (1.7 * 10^-5 Ω·m)
Simplifying this expression, we find:
L1 / L2 = 1.66
Therefore, the correct comparison is:
L1 = 1.66 * L2
For the second question:
Let the resistance of the first wire be R.
The second wire has twice the length (2L1), twice the diameter (2d), and twice the resistivity (2ρ).
The resistance of the second wire, R2, can be calculated using the resistance formula:
R2 = ρ * (2L1) / (π * (2d/2)^2) = ρ * (2L1) / (π * d^2)
Dividing R2 by R:
R2 / R = (ρ * (2L1) / (π * d^2)) / R = (2ρ * L1) / (π * d^2 * R)
Since R = ρ * (L1 / (π * (d/2)^2)):
R2 / R = (2ρ * L1) / (π * d^2 * ρ * (L1 / (π * (d/2)^2)))
Cancellation of terms:
R2 / R = 2 / 1 = 2
Therefore, the resistance of the second wire is:
R2 = 2R