Answer: 0.258
Explanation:
The resistance [tex]R[/tex] of a wire is calculated by the following formula:
[tex]R=\rho\frac{l}{s}[/tex] (1)
Where:
[tex]\rho[/tex] is the resistivity of the material the wire is made of. For aluminium is [tex]\rho_{Al}=2.65(10)^{-8}m\Omega[/tex] and for copper is [tex]\rho_{Cu}=1.68(10)^{-8}m\Omega[/tex]
[tex]l[/tex] is the length of the wire, which in the case of aluminium is [tex]l_{Al}=12m[/tex], and in the case of copper is [tex]l_{Cu}=30m[/tex]
[tex]s[/tex] is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:
[tex]s=\pi{(\frac{d}{2})}^{2}[/tex] (2) Where [tex]d[/tex] is the diameter of the circumference.
For aluminium wire the diameter is [tex]d_{Al}=2.5mm=0.0025m[/tex] and for copper is [tex]d_{Cu}=1.6mm=0.0016m[/tex]
So, in this problem we have two transversal areas:
For aluminium:
[tex]s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}[/tex]
[tex]s_{Al}=0.000004908m^{2}[/tex] (3)
For copper:
[tex]s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}[/tex]
[tex]s_{Cu}=0.00000201m^{2}[/tex] (4)
Now we have to calculate the resistance for each wire:
Aluminium wire:
[tex]R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}[/tex] (5)
[tex]R_{Al}=0.0647\Omega[/tex] (6) Resistance of aluminium wire
Copper wire:
[tex]R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}[/tex] (6)
[tex]R_{Cu}=0.250\Omega[/tex] (7) Resistance of copper wire
At this point we are able to calculate the ratio of the resistance of both wires:
[tex]Ratio=\frac{R_{Al}}{R_{Cu}}[/tex] (8)
[tex]\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}[/tex] (9)
Finally:
[tex]\frac{R_{Al}}{R_{Cu}}=0.258[/tex] This is the ratio
