Eighteen-gauge wire has a diameter of 1.024 mm. Calculate the resistance of 19 m of 18 gauge lead wire at 20°C.
resistivity at 20℃ is
ρ=22•10⁻⁸ Ohm•m
R= ρL/A= 4ρL/πD²=
=2•22•10⁻⁸•19/ π(1.024•10⁻³)²=
=5.076 Ohms
Why did the wire go to the gym?
Because it wanted to build some "resistance"!
Okay, let's calculate the resistance for your 19 m of 18-gauge lead wire. The resistance of a wire can be calculated 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.
Now, we need to find the resistivity of lead at 20°C and the cross-sectional area of 18-gauge wire. Let me do a quick search for that information...
To calculate the resistance of a wire, we can use the formula:
R = (ρ * L) / A
Where:
R is the resistance of the wire,
ρ (rho) is the resistivity of the material the wire is made of,
L is the length of the wire, and
A is the cross-sectional area of the wire.
First, let's find the cross-sectional area of the 18-gauge wire. The gauge of a wire corresponds to the number of steps it takes to measure the diameter of the wire. The larger the gauge number, the smaller the wire diameter.
For wire gauges, there is a standard formula to calculate the diameter:
d = 0.127 * (92 ^ (36 - G))
Where d is the wire diameter and G is the gauge number.
Using the given information, we can calculate the wire diameter:
d = 0.127 * (92 ^ (36 - 18))
≈ 0.127 * (92 ^ 18)
≈ 0.127 * 71.51
≈ 9.08 mm
To calculate the cross-sectional area, we'll use the formula for the area of a circle:
A = π * (d/2)^2
Where A is the cross-sectional area and d is the wire diameter.
Using the wire diameter we calculated:
A = π * (9.08 / 2)^2
≈ π * 4.54^2
≈ π * 20.62
≈ 64.66 mm^2
Now, let's calculate the resistance of the wire. We'll need to know the resistivity of lead at 20°C, which is approximately 0.208 micro-ohm meter (μΩ·m).
R = (ρ * L) / A
≈ (0.208 * 10^-6 * 19) / 64.66
≈ 0.0035 Ω
Therefore, the resistance of 19 m of 18-gauge lead wire at 20°C is approximately 0.0035 ohms.