To solve this problem we will apply the relation of Ohm's law, at the same time we will use the concept of resistance in a cable, resistivity and potential difference.
According to Ohm's law we have to
[tex]V= IR[/tex]
Here,
V = Voltage
I = Current
R = Resistance
At the same time resistance can be described as
[tex]R = \frac{\rho l}{A}[/tex]
Here,
[tex]\rho[/tex]= Resistivity of the material
l = Length of the specimen
A = Cross-sectional area
From the above expression we can write the current as,
[tex]I = \frac{V}{R}[/tex]
[tex]I = \frac{V}{\frac{\rho l}{A}}[/tex]
[tex]I =\frac{VA}{\rho l}[/tex]
Replacing we have that,
[tex]I = \frac{(0.8V)(0.4*10^{-6}m^2)}{(5.6*10^{-8}\Omega \cdot m)(1.5m)}[/tex]
[tex]I = 3.809A[/tex]
Therefore the current in the wire is 3.809A
Note: The value obtained for the resistivity of Tungsten was theoretically obtained and can be consulted online.
