91Ó°ÊÓ

The resistance of transmission line is, $R=0.30\mathrm{Î\copyright }\text{/cable}$

The power of the generator is,

$$\begin{gathered} P=250kW \\ =250×{10}^{3}W \end{gathered}$$

Ohm's law states that the current flowing through a conductor between two points is directly proportional to the voltage across both points.

Calculate the total resistance over two lines first. Then use the expression relating power, voltage, and current to determine the desired values.

Formula:

Write the equation for voltage as below.

$V=IR$

Here, V is the voltage, I is the current, R is the resistance.

Write the equation for power as below.

$P=VI$

The resistance R for two cables; hence, the total resistance is as follows:

$$\begin{gathered} R=2×0.3 \\ =0.6\mathrm{Î\copyright } \end{gathered}$$

$$\begin{gathered} I_{rms}=\frac{P}{V_t} \\ =\frac{250×{10}^3}{80×{10}^3} \\ =3.125A \end{gathered}$$

$$\begin{gathered} \mathrm{Δ}V=I_{rms}R \\ =3.125×0.6 \\ =1.9V \end{gathered}$$

The voltage $\mathrm{Δ}V$ along transmission line is 1.9 V.

$$\begin{gathered} P_d=I_{rms}^2×R \\ ={\left(3.125\right)}^2×0.6 \\ =5.9W \end{gathered}$$

The rate $P_d$ is 5.9 W.

$$\begin{gathered} I_{rms}=\frac{P}{V_t} \\ =\frac{250×{10}^3}{8×{10}^3} \\ =31.25A \end{gathered}$$

$$\begin{gathered} \mathrm{Δ}V=I_{rms}R \\ =31.25×0.6 \\ =19V \end{gathered}$$

The voltage $\mathrm{Δ}V$at $V_t=8.0kV$is 19 V.

$$\begin{gathered} P_d=I_{rms}^2×R \\ ={\left(31.25\right)}^2×0.6 \\ =590W \end{gathered}$$

The power ${\mathit{P}}_{\mathbf{d}}$ at $V_t=8.0kV$ is 590 W .

$$\begin{gathered} I_{rms}=\frac{P}{V_t} \\ =\frac{250×{10}^3}{0.8×{10}^3} \\ =312.5A \end{gathered}$$

$$\begin{gathered} \mathrm{Δ}V=I_{rms}R \\ =312.5×0.6 \\ =190V \end{gathered}$$

The voltage $\mathrm{Δ}V$ at $V_t=0.80kV$ is 190 V.

$$\begin{gathered} P_d=I_{rms}^2×R \\ =1.{875}^2×0.6 \\ =59000W \end{gathered}$$

The power $P_d$ at 0.8 kV is 59000 W.