Answer:
a) E = 4.26 W
b) E' = 6.724 W
c) copper wire is the safer option to use.
Explanation:
Given:
Diameter of the 12 gauge copper wire, d = 0.2053 cm
Thus, Radius of the 12 gauge copper wire, r = 0.2053 cm
/ 2 = 0.10265 cm = 0.10265 × 10⁻² m
Now.
the area (A) comes out as
A = π × (0.10265 × 10⁻²)²
A = 3.3103 × 10⁻⁶ m²
Length of the copper wire, L = 2.10 m
a) The resisitivity (ρ) of copper = 1.68 × 10⁻⁸ ohm m
Now,
the resistance of the copper , R = ρL/A
or
R = (1.68 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)
or
R = 0.01065 ohm
The Energy (E) is given as,
E = I²R
where, I is the current
I = 20.0 A
on substituting the values, we get
E = 20.0² × 0.01065
E = 4.26 W
(b) For the aluminium
Resisitivity, ρ' = 2.65 × 10⁻⁸ ohm m
Now, the resistance of the aluminium wire, R' = (ρ' × L) / A
Since the cross-section of the aluminium wire is same as the copper wire
thus,
R = (2.65 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)
or
R = 0.0168 ohm
Therefore,
The Rate of energy produced by the aluminium wire, E' = I²R'
or
E' = 20.0² × 0.0168
or
E' = 6.724 W
(c) From the above results, we can conclude that the power consumed or the rate of energy produced by the aluminium wire is more.
Hence, copper wire is the safer option to use.
