Question

A 1.50 V potential difference is maintained across a 1.50 m wire that has a cross-sectional area of 0.600 mm2. How much power is dissipated in the wire if its resistivity is 5.25x10-8Ωm?Group of answer choices8.00 W17.1 W12.5 W4.50 W15.0 W

Answer 1

Given that the potential difference is V = 1.5 V.

The length of the wire is l = 1.5 m.

The cross-sectional area is

`\begin{gathered} A\text{ = 0.6 }mm^2 \\ =0.6*10^(-6)m^2 \end{gathered}`

The resistivity of the wire is

`\rho\text{ = 5.25}*10^(-8)\Omega\text{ m }`

We have to find the power dissipated in the wire.

First, we need to calculate resistance.

The resistance can be calculated as

`\begin{gathered} R\text{ = }\rho(l)/(A) \\ =5.25*10^(-8)*(1.5)/(0.6*10^(-6)) \\ =0.13125\Omega \end{gathered}`

The formula to calculate power is

`P\text{ =}(V^2)/(R)`

Substituting the values, the power will be

`\begin{gathered} P\text{ = }((1.5)^2)/(0.13125) \\ =17.1\text{ W} \end{gathered}`

Thus, the power dissipated in the wire is 17.1 W