If 750 m of 3.00-mm-diameter wire has a resistance of 27.6, what length of similar wire 5.00 mm in diameter will have the same resistance?
R = kL/d^2
Find k:
27.6 = k(750)/9
k = 0.3312
So, now we find the new L:
27.6 = .3312L/25
L = 2083.3 mm
Makes sense. The diameter has almost doubled, so the resistance is about 1/3 of what it was. (9/25 actually)
Note that 750 * 25/9 = 2083.3
What is that k means??is it rho?
Well, if you're looking for some electrically powered fashion advice, then you've come to the right place! So, let's unravel this wiring puzzle.
First, we need to acknowledge that size does matter when it comes to resistance. A larger diameter wire has a lower resistance compared to a smaller diameter wire of the same length.
Now, let's calculate the ratio between the two wire diameters: 5.00 mm / 3.00 mm = 1.67.
Since resistance is inversely proportional to the cross-sectional area, the resistance of the new wire will be ((1.67)^2) times higher than the original wire (using the ratio squared since resistance is proportional to area).
So, 27.6 Ω * (1.67^2) ≈ 82.33 Ω.
To find the length of the new wire, we need to compare the resistances indirectly proportional to the lengths:
Length of new wire / Length of original wire = Resistance of original wire / Resistance of new wire
Let's plug in the numbers:
Length of new wire / 750 m = 27.6 Ω / 82.33 Ω.
Now, a little math magic:
Length of new wire ≈ (750 m) * (27.6 Ω) / (82.33 Ω)
≈ 251.19 m.
So, if you use a wire with a diameter of 5.00 mm, you'll need approximately 251.19 m to have the same resistance. Voilà! Wiring fashion at its finest!
To solve this problem, we can use the formula for the resistance of a wire:
R = (ρ * L) / A
Where:
- R is the resistance.
- ρ is the resistivity of the material.
- L is the length of the wire.
- A is the cross-sectional area of the wire.
The resistivity of the material is a constant and is not provided in the question, so we can assume it remains the same for both wires.
Let's start by finding the cross-sectional area of the first wire:
A₁ = (π * r₁²)
= (π * (d₁/2)²)
= (π * (3.00 mm / 2)²)
= (π * 1.50²)
= 7.07 mm²
Next, we can calculate the length of the second wire:
L₂ = (R * A₂) / ρ
We know that R₁ = 27.6, which is the resistance of the first wire.
Since we want to find the length of the second wire, L₂, and the second wire is 5.00 mm in diameter, we need to find the cross-sectional area, A₂:
A₂ = (π * r₂²)
= (π * (d₂/2)²)
= (π * (5.00 mm / 2)²)
= (π * 2.50²)
= 19.63 mm²
Now we can substitute the values into the formula for L₂:
L₂ = (27.6 * 19.63) / ρ
Since we're looking for the length in terms of the first wire's length (750 m), we can rearrange the formula:
L₂ = (27.6 * 19.63 * (L₂ / L₁)) / ρ
Simplifying the equation:
(L₁ * L₂) = (27.6 * 19.63) / ρ
Substituting the values:
(750 * L₂) = (27.6 * 19.63) / ρ
Rearranging the equation to solve for L₂:
L₂ = ((27.6 * 19.63) / ρ) / 750
This equation will give us the length of the second wire in terms of the first wire's length, assuming the same material is used.
To solve this problem, let's use the formula for the resistance of a wire:
R = (ρ * L) / A
Where R is the resistance, ρ is the resistivity of the wire material (a constant), L is the length of the wire, and A is the cross-sectional area of the wire.
We are given the following information:
R1 = 27.6 Ω (resistance of the first wire)
d1 = 3.00 mm (diameter of the first wire)
d2 = 5.00 mm (diameter of the second wire)
L2 = ? (length of the second wire)
First, we need to find the resistivity (ρ) based on the given resistance and diameter of the first wire:
R1 = (ρ * L1) / A1
To find ρ, we rearrange the formula:
ρ = (R1 * A1) / L1
Since we know R1, we can use the dimensions of the first wire to calculate A1:
A1 = π * (d1 / 2)^2
Now we can substitute the values to calculate ρ:
ρ = (27.6 Ω * π * (3.00 mm / 2)^2) / 750 m
Next, we need to find the length of the second wire (L2). To do this, we rearrange the resistance formula:
R2 = (ρ * L2) / A2
Since we want to find the length (L2), we can rearrange the formula to solve for L2:
L2 = (R2 * A2) / ρ
We already know ρ from the earlier calculations, so we just need to calculate the cross-sectional area of the second wire, A2:
A2 = π * (d2 / 2)^2
Finally, we substitute the values into the formula to find the length of the second wire:
L2 = (27.6 Ω * π * (5.00 mm / 2)^2) / ((27.6 Ω * π * (3.00 mm / 2)^2) / 750 m)
Simplifying the equation will give us the length of the second wire in meters.
L2 = (27.6 Ω * π * 2.50 mm^2 * 750 m) / (27.6 Ω * π * 1.50 mm^2)
L2 ≈ 2500 m
Therefore, a wire with a diameter of 5.00 mm will have a length of approximately 2500 meters to have the same resistance as the 750 meters of 3.00-mm-diameter wire.