Question

A 0.800-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.400 mm2. What is the current in the wire

Answer 1

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

$V= IR$

Here,

V = Voltage

I = Current

R = Resistance

At the same time resistance can be described as

$R = \frac{\rho l}{A}$

Here,

$\rho$= Resistivity of the material

l = Length of the specimen

A = Cross-sectional area

From the above expression we can write the current as,

$I = \frac{V}{R}$

$I = \frac{V}{\frac{\rho l}{A}}$

$I =\frac{VA}{\rho l}$

Replacing we have that,

$I = \frac{(0.8V)(0.4*10^{-6}m^2)}{(5.6*10^{-8}\Omega \cdot m)(1.5m)}$

$I = 3.809A$

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.