You have a piece of copper with a volume of 2 cm3, and wish to form a wire having a resistance of
0.15 ƒ¶. How long should the wire be if the resistivity of copper is 1.72 �~ 10-8 ƒ¶. m?
V=A•L
A=V/L
R=ρL/A = ρL²/V
L=sqrt{RV/ρ} =
=sqrt{ 0.15•2•10⁻⁶/1.72•10⁻⁸} = 4.17 m
To determine the length of the wire, we can use the formula for resistance:
R = (ρ * L) / A
Where:
R = resistance (0.15 Ω)
ρ = resistivity of copper (1.72 x 10^-8 Ω·m)
L = length of the wire (unknown)
A = cross-sectional area of the wire
We also have the volume of the copper (2 cm³), which we can convert to the cross-sectional area of the wire using the formula:
A = V / L
Where:
V = volume of the wire (2 cm³)
L = length of the wire (unknown)
Let's calculate the cross-sectional area first:
A = 2 cm³ / L
Next, substitute this value into the resistance formula:
0.15 Ω = (1.72 x 10^-8 Ω·m * L) / (2 cm³ / L)
Now, let's convert the units to be consistent. Since the resistivity is given in Ω·m, we need to convert centimeters to meters:
A = (2 cm³ / L) * (1 m / 100 cm)³
Simplifying the expression:
0.15 Ω = (1.72 x 10^-8 Ω·m * L) / ((2 cm³ / L) * (1 m / 100 cm)³)
Now, rearrange the equation to solve for L:
L = sqrt((0.15 Ω * 2 cm³ * (1 m / 100 cm)³) / (1.72 x 10^-8 Ω·m))
L ≈ 0.1462 m (or approximately 14.62 cm)
Therefore, the length of the wire should be approximately 0.1462 meters (or 14.62 centimeters).