Answer:
[tex]V=2.177*10^{-4}V[/tex]
Explanation:
Given data
Diameter d=1.628 mm=1.628×10⁻³m
Current I=13.5 mA=13.5×10⁻³A
The Cross-sectional Area is given as:
[tex]A=\frac{\pi }{4}d^{2} \\A=\frac{\pi }{4}(1.628*10^{-3} )^{2} \\A=2.08*10^{-6}m^{2}[/tex]
The potential difference across a 1.95 m length of 14-gauge copper wire
Length L=1.95 m
Copper Resistivity p=1.72×10⁻⁸Ω.m
The Resistance of copper wire is given by:
[tex]R=\frac{pL}{A} \\R=\frac{(1.72*10^{-8} )(1.95)}{2.08*10^{-6} } \\R=0.0161[/tex]
R=0.01612Ω
The Potential difference across the copper wire is:
[tex]V=IR\\V=(13.5*10^{-3} )(0.01612)\\V=2.177*10^{-4}V[/tex]
2.126 x 10⁻⁴ V
First, let's calculate the resistance (R) of the wire as follows;
The resistance (R) of a current carrying wire with a length (L), a cross-sectional area (A) and resistivity (ρ) is given by;
R = ρL / A -------------------(i)
Where;
A = π d² / 4 [where d = diameter of the wire]
From the question, let's calculate the area (A) as follows;
d = 1.628mm = 0.001628m
=> A = π (0.001628)² / 4 [Take π = 3.142]
=> A = (3.142) (0.001628)² / 4
=> A = (3.142) (0.001628)² / 4
∴ A = 2.08 x 10⁻⁶ m²
Where;
L = length of the copper wire = 1.95m and
ρ = resistivity of copper wire = 1.68 x 10⁻⁸Ωm
A = 2.08 x 10⁻⁶ m² [as calculated above]
Substitute these values into equation (i) as follows;
R = 1.68 x 10⁻⁸ x 1.95 / (2.08 x 10⁻⁶)
R = 1.575 x 10⁻²Ω
The resistance of the wire is therefore, 1.575 x 10⁻² Ω
Now, let's calculate the potential difference across the wire.
The potential difference (V) across a wire of resistance (R) carrying a current (I) is given by Ohm's law as follows;
V = I x R ---------------------(ii)
Where;
I = 13.5mA = 13.5 x 10⁻³A
R = 1.575 x 10⁻²Ω
Substitute these values into equation (ii) as follows;
V = 13.5 x 10⁻³ x 1.575 x 10⁻²
V = 21.26 x 10⁻⁵ V
V = 2.126 x 10⁻⁴ V
Therefore, the potential difference across the wire is 2.126 x 10⁻⁴ V
