Answer :
To develop the problem we will apply Ohm's law that relates the voltage, current and resistance. Previously, it will be necessary to bring the resistance data per unit length to the net resistance, since we count the length. We will also consider that the potential difference across the bird’s feet will be the same as potential difference across the wire beneath its feet because both are parallel.
First,
Resistance of wire will be,
[tex]R = R/m * l[/tex]
[tex]R = 14.4*10^{-5}\Omega/m (0.15m)[/tex]
[tex]R = 2.16*10^{-6} \Omega[/tex]
Now applying Ohm's law,
[tex]V = IR[/tex]
[tex]V = (85.8)(2.16*10^{-6})[/tex]
[tex]V = 1.89 * 10^{-4} V[/tex]
Therefore the potential difference across the bird's feet is [tex]1.89*10^{-4}V[/tex]
Answer:
1.853×10⁻³ V
Explanation:
Potential Difference: This can be defined as the work done when one coulombs of charge moves from one point to another in an electric Field. The S.I unit is volt (V)
From Ohm's Law,
V = IR................ Equation 1
Where V = Voltage, I = Current, R = Resistance.
From the question,
R = R'd.............. Equation 2
Where R' = Resistance per length, d = distance
Substitute equation 2 into equation 1
V = IR'd............ Equation 3
Given: I = 85.8 A, R' = 14.4×10⁻⁵ Ω/m, d = 15 cm = 0.15 m
Substitute into equation 3
V = 85.8×14.4×10⁻⁵×0.15
V = 1.853×10⁻³ V.