2.3. An electric motor requires 2 kW of power dissipation when connected to a 250 V
supply. If the radiator coils are made from a wire of 0.15 mm diameter with a
resistivity of , calculate the total length of the wire required.

Ask by Hampton Coleman. in South Africa

Mar 15,2025

Question

2.3. An electric motor requires 2 kW of power dissipation when connected to a 250 V
supply. If the radiator coils are made from a wire of 0.15 mm diameter with a
resistivity of , calculate the total length of the wire required.

Ask by Hampton Coleman. in South Africa

Mar 15,2025

UpStudy AI · Accepted Answer

We start by finding the resistance that the wire coil must have. For a load dissipating power $$ P $$ at a voltage $$ V $$ the resistance is given by

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

Given $$ V=250\,\mathrm{V} $$ and $$ P=2000\,\mathrm{W} $$ we have

$$
R=\frac{250^2}{2000}=\frac{62500}{2000}=31.25\,\Omega.
$$

The resistance of a wire of length $$ L $$ and cross‐sectional area $$ A $$ is given by

$$
R=\rho\,\frac{L}{A}.
$$

We can solve for $$ L $$ as

$$
L=\frac{R\,A}{\rho}.
$$

The wire has a diameter of $$0.15\,\mathrm{mm}$$. Converting to meters,

$$
d=0.15\times10^{-3}\,\mathrm{m},
$$

so the radius is

$$
r=\frac{d}{2}=7.5\times10^{-5}\,\mathrm{m}.
$$

The cross–sectional area is

$$
A=\pi r^2=\pi\,(7.5\times10^{-5})^2.
$$

Calculating the square of the radius:

$$
(7.5\times10^{-5})^2=5.625\times10^{-9}\,\mathrm{m}^2,
$$

so

$$
A=\pi\times5.625\times10^{-9}\approx1.767\times10^{-8}\,\mathrm{m}^2.
$$

The resistivity is given as

$$
\rho=80\,\mu\Omega\cdot\mathrm{m}=80\times10^{-6}\,\Omega\cdot\mathrm{m}.
$$

Now substitute all these into the expression for $$ L $$:

$$
L=\frac{R\,A}{\rho}=\frac{31.25\times1.767\times10^{-8}}{80\times10^{-6}}.
$$

Performing the multiplication in the numerator,

$$
31.25\times1.767\times10^{-8}\approx5.522\times10^{-7},
$$

and then dividing by $$80\times10^{-6}$$,

$$
L=\frac{5.522\times10^{-7}}{80\times10^{-6}}\approx6.9\times10^{-3}\,\mathrm{m}.
$$

Thus, the total length of the wire required is

$$
L\approx6.9\times10^{-3}\,\mathrm{m}\approx6.9\,\mathrm{mm}.
$$
The total length of the wire required is approximately 6.9 mm.