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 = (\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 = (V)/(R)`

`I = (V)/((\rho l)/(A))`

`I =(VA)/(\rho l)`

Replacing we have that,

`I = ((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.