If the power output in the previous problem is 1000 MW and line resistance is 2.00 Ω, what were the old and new line losses?

Power is defined as the rate at which one performs work. It's also known as the rate of energy consumption.

The power output is$P_{Loss}=1000MW\left(\frac{{10}^{6}W}{1MW}\right)={10}^{9}W$

The resistance of the line is $R=2\Omega$

Voltage in the secondary coil of the transformer is $V_S^1=335kV\left(\frac{1000V}{1kV}\right)=3.35\times {10}^{5}V$and$V_S^2=750kV\left(\frac{1000V}{1kV}\right)=7.50\times {10}^{5}V$

The power loss in the circuit can be expressed as,

$P_{Loss}=I^{2}R$……………(1)

Where P, I, and V are the power, voltage, and current in the circuit, respectively.

Substituting the given data in equation (1) can be expressed as,

When the value of output voltage is, $V_S^1=3.35\times {10}^{5}V$, then

\(\begin{align}{}{P_{{\rm{Loss }}}} &= {\left( {\frac{P}{{V_S^1}}} \right)^2}R\\ & = {\left( {\frac{{{{10}^9}{\rm{W}}}}{{3.35 \times {{10}^5}\;{\rm{V}}}}} \right)^2} \times 2\;\Omega \\ & = 1.782 \times {10^7}\;{\rm{W}}\left( {\frac{{1\;{\rm{MW}}}}{{{{10}^6}\;{\rm{W}}}}} \right)\\ & = 17.82\;{\rm{MW}}\end{align}\)

When the value of output voltage is, $V_S^2=7.50\times {10}^{5}V$, then

$$\begin{gathered} P_{Loss}={\left(\frac{P}{V_S^2}\right)}^{2}R \\ =\left(\frac{{10}^{9}W}{7.50\times {10}^{5}V}\right)\times 2\Omega \\ =3.56\times {10}^{6}W\left(\frac{1MW}{{10}^{6}W}\right) \\ =3.56MW \end{gathered}$$

Therefore, the old line loss is $17.82MW$for 335 kV voltage output and the new line loss is $3.56MW$for the voltage output of 750 kV.