91Ó°ÊÓ

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 $R=2\mathrm{Î\copyright }$

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

The power loss in the circuit can be expressed as,

$P_{Loss}=I^{2}R$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(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, $V_S^1=3.35×{10}^{5}V$, 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, $V_S^2=7.50×{10}^{5}V$, then

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

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.