91Ó°ÊÓ

The expression for the power is given by,

\(P = {I^2}R\)

Here\(P\)is the power,\(I\)is the current flowing through the wire and\(R\)is the resistance.

The expression for the energy is given by,

\(\begin{align}E = P \times t\\ = {I^2}Rt\end{align}\)

According to the information provided.

The cooling a superconductor wire costs \({\rm{1 L}}\) of \({{\rm{N}}_{\rm{2}}}\) per hour,

Or else, \({\rm{\$ 0}}{\rm{.30}}\) per hour,

We'd spend an hour together.

\({\rm{price = 0}}{\rm{.30 \$ }}\)

We would extract energy \(E\) from a power source to spend the same amount for electric energy in a typical wire at the price of \({\rm{0}}{\rm{.10 \$ / kWh}}\).

\(\begin{align}E{\rm{ }} &= {\rm{ }}\frac{{{\rm{0}}{\rm{.30 \$ }}}}{{{\rm{0}}{\rm{.10 \$ / kWh}}}}\\ &= {\rm{ 3 kWh}}\end{align}\)

Then, the power in a normal wire is:

\(P = {I^2}R\)

With the released energy as:

\(\begin{align}E = Pt\\ = {I^2}Rt\end{align}\)

So, the value of\({\rm{R}}\)is obtained as:

\(\begin{align}R &= \frac{E}{{{I^2}Rt}}\\ &= \frac{{{\rm{3kWh}}}}{{{{{\rm{(100A)}}}^{\rm{2}}}{\rm{1h}}}}\\ &= \frac{{{\rm{3 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{W}}}}{{{\rm{1}}{{\rm{0}}^{\rm{4}}}{{\rm{A}}^{\rm{2}}}}}\\ &= {\rm{0}}{\rm{.3}}\,{\rm{\Omega }}\end{align}\)

Therefore, the resistance is \({\rm{0}}{\rm{.3 \Omega }}\).