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

$\begin{gathered} A\text{ = 0.6 }mm^2 \\ =0.6\times10^{-6}m^2 \end{gathered}$

The resistivity of the wire is

$\rho\text{ = 5.25}\times10^{-8}\Omega\text{ m }$

We have to find the power dissipated in the wire.

First, we need to calculate resistance.

The resistance can be calculated as

$\begin{gathered} R\text{ = }\rho\frac{l}{A} \\ =5.25\times10^{-8}\times\frac{1.5}{0.6\times10^{-6}} \\ =0.13125\Omega \end{gathered}$

The formula to calculate power is

$P\text{ =}\frac{V^2}{R}$

Substituting the values, the power will be

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

Thus, the power dissipated in the wire is 17.1 W