Answer:
V = 3.6 volts
Explanation:
From Ohm's Law, we know that:
V = IR
but,
R = ρL/A
Therefore,
V = IρL/A
where,
V = Potential Difference = ?
I = Current = 4 A
ρ = resistivity of copper = 1.68 x 10⁻⁸ Ω.m
L = Length = 70 m
A = Cross-sectional Area = πd²/4 = π(1.29 x 10⁻³ m)²/4 [16 gauge wire has a diameter of 1.29 mm]
A = 1.31 x 10⁻⁶ m²
V = (4 A)(1.68 x 10⁻⁸ Ω.m)(70 m)/(1.31 x 10⁻⁶ m²)
V = 3.6 volts
The potential difference across the given length of copper wire is obtained to be 3.6 V.
The resistance of the wire depends on the length, area of cross-section of the wire and resistivity of the material.
Given the length of the wire, [tex]L = 70\,m[/tex].
The diameter of a 16 gauge copper wire is given by, [tex]d = 1.29\times 10^{-3}\,m[/tex].
Therefore, the radius of the cross-section is;
[tex]r=\frac{d}{2} = \frac{1.29\times 10^{-3}\,m}{2} =6.45\times 10^{-4}\,m[/tex]
So, the resistance of the given length of wire can be written as,
[tex]R= \rho\, \frac{L}{A} =(1.68 \times 10^{-8} \, \Omega .m)\times \frac{70\,m}{\pi \times (6.45\times 10^{-4}m)^2} = 0.9\, \Omega[/tex]
According to Ohm's Law, we can write;
[tex]V=IR[/tex]
Substituting the known values, we get the voltage drop is;
[tex]V = 4A \times 0.9\Omega = 3.6\,V[/tex]
Learn more about Ohm's law here:
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