a) What current is needed to transmit\({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{ MW}}\)of power at\({\rm{10}}{\rm{.0 kV}}\)? (b) Find the resistance of\({\rm{1}}{\rm{.00 km}}\)of wire that would cause a\({\rm{0}}{\rm{.0100\% }}\)power loss. (c) What is the diameter of a\({\rm{1}}{\rm{.00 km}}\)long copper wire having this resistance? (d) What is unreasonable about these results? (e) Which assumptions are unreasonable, or which premises are inconsistent?

Power: The power of a process is the amount of some type of energy converted into a different type divided by the time interval\({\rm{\Delta t}}\)in which the process occurred:

\({\rm{P = }}\frac{{{\rm{\Delta E}}}}{{{\rm{\Delta t}}}}\) ….. (1)

The SI unit of power is the watt (W).

Because the potential difference across a resistor is given by\({\rm{\Delta V = IR}}\)we can express the power delivered to a resistor as

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

\({\rm{P = }}\frac{{{{{\rm{(\Delta V)}}}^{\rm{2}}}}}{{\rm{R}}}\) ….. (2)

The energy delivered to a resistor be electrical transmission appears in the form of internal energy in the resistor.

Electrical resistance: The electrical resistance R of a resistive circuit element depends on its geometric structure (its length L and cross-sectional area A) and on the resistivity\({\rm{\rho }}\)of the material of which it is made:

\({\rm{R = \rho }}\frac{{\rm{L}}}{{\rm{A}}}\)

The power output of the transmission line is:

\(\begin{aligned}{}{{\rm{P}}_{{\rm{out }}}} &= \left( {{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{2}}}{\rm{MW}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{6}}}{\rm{\;W}}}}{{{\rm{1MW}}}}} \right)\\ &= {\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;W}}\end{aligned}\)

The voltage at which the power is transmitted is:

\(\begin{aligned}{}{\rm{\Delta V}} &= \left( {{\rm{10}}{\rm{.0kV}}} \right)\left( {\frac{{{\rm{1000\;V}}}}{{{\rm{1kV}}}}} \right)\\ &= {\rm{10}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;V}}\end{aligned}\)

(\({\rm{0}}{\rm{.0100\% }}\)of power is dissipated in the transmission lines.)

The power output of the transmission line is found from equation (1):

\({P_{out}} = I\Delta V\)

Solve the above equation for I:

\(I = \frac{{{P_{out}}}}{{\Delta V}}\)

Substitute numerical values:

\(\begin{aligned}{}I &= \frac{{1.00 \times {{10}^8}\;W}}{{10.0 \times {{10}^3}\;V}}\\ &= 10.0 \times {10^3}\;A\\ &= \left( {10.0 \times {{10}^3}\;A} \right)\left( {\frac{{1{\rm{ }}kA}}{{{{10}^3}\;A}}} \right)\\ &= 10.0{\rm{ }}kA\end{aligned}\)

Therefore, current needed to transmit the given power is \({\rm{10}}{\rm{.0 kA}}\).

The power dissipated\({{\rm{P}}_{{\rm{dissipated }}}}\)in the transmission lines is found from equation (2). But we are given that\({{\rm{P}}_{{\rm{dissipated }}}}\)is equal to\({\rm{0}}{\rm{.0100\% }}\)of the power transmitted\({{\rm{P}}_{{\rm{out }}}}\).

So

\(\left( {0.0100\% \pi } \right){P_{out }} = {I^2}R\)

Solve the above expression for R as below.

\({\rm{R}} = \frac{{\left( {{\rm{0}}{\rm{.0100\% }}} \right){{\rm{P}}_{{\rm{out }}}}}}{{{{\rm{I}}^{\rm{2}}}}}\)

Entering the values for\({{\rm{P}}_{{\rm{out }}}}\)and I, and you obtain:

\(\begin{aligned}{}{\rm{R}} &= \frac{{\left( {{\rm{0}}{\rm{.0100\% }}} \right)\left( {{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;W}}} \right)}}{{\left( {{\rm{10}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{\;A}}} \right)}}\\ &= {\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ \Omega }}\end{aligned}\)

Therefore, resistance of wire is \({\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ \Omega }}\).

The resistance of the wire is found from equation (3):

\({\rm{R = }}\frac{{{\rm{\rho L}}}}{{\rm{A}}}\)

Here, the cross-sectional area of the wire is\(\frac{{{\rm{\pi }}{{\rm{D}}^{\rm{2}}}}}{{\rm{4}}}\). Therefore,

\({\mathop{\rm R}\nolimits} = \frac{{4rL}}{{p{D^2}}}\)

Solve the above equation for D:

\({\rm{D}} = \sqrt {\frac{{{\rm{4\rho L}}}}{{{\rm{\pi R}}}}} \)

Entering the values\({\rm{\rho ,L,}}\)and\({\rm{R}}\), you have

\(\begin{aligned}{}{\rm{D}} &= \sqrt {\frac{{{\rm{4}}\left( {{\rm{1}}{\rm{.72 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{ \Omega }} \cdot {\rm{m}}} \right)}}{{{\rm{\pi }}\left( {{\rm{1}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ \Omega }}} \right)}}} \\ &= {\rm{0}}{\rm{.468\;m}}\end{aligned}\)

Therefore, the diameter of a resistance wire is \({\rm{0}}{\rm{.468\;m}}\).

The diameter of the wire is unreasonably large. A wire with such a diameter would be too heavy for use.

The power loss is extremely low for a transmission line at this voltage.

For example, a \({\rm{160\;km}}\) transmission line at \({\rm{345 kV}}\) (which is much greater than the voltage in our problem) carrying \({\rm{100 MW}}\) of power can have losses of \({\rm{4}}{\rm{.2\% }}\).