An ac generator provides emf to a resistive load in a remote factory over a two-cable transmission line. At the factory a stepdown transformer reduces the voltage from its (rms) transmission value ${\mathit{V}}_{\mathbf{t}}$to a much lower value that is safe and convenient for use in the factory. The transmission line resistance is $\mathit{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathit{\Omega }\mathbf{\text{/cable}}$, and the power of the generator is $\mathit{P}\mathbf{=}\mathbf{250}\mathbf{}\mathit{k}\mathit{W}$. If ${\mathit{V}}_{\mathbf{t}}\mathbf{=}\mathbf{80}\mathbf{}\mathit{k}\mathit{V}$, what are (a) the voltage decreases V along the transmission line and (b) the rate $\mathit{\Delta }\mathit{V}$ at which energy is dissipated in the line as thermal energy? If ${\mathit{V}}_{\mathbf{t}}\mathbf{=}\mathbf{80}\mathbf{}\mathit{k}\mathit{V}$, what are (c) V and (d)$\mathit{\Delta }\mathit{V}$ ? If ${\mathit{V}}_{\mathbf{t}}\mathbf{=}\mathbf{80}\mathbf{}\mathit{k}\mathit{V}$, what are (e) V and (f) ${\mathit{P}}_{\mathbf{d}}$?

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

The power of the generator is,

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

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:

$$\begin{gathered} R=2\times 0.3 \\ =0.6\Omega \end{gathered}$$

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

$$\begin{gathered} \Delta V=I_{rms}R \\ =3.125\times 0.6 \\ =1.9V \end{gathered}$$

The voltage $\Delta V$ along transmission line is 1.9 V.

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

The rate $P_d$ is 5.9 W.

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

$$\begin{gathered} \Delta V=I_{rms}R \\ =31.25\times 0.6 \\ =19V \end{gathered}$$

The voltage $\Delta V$at $V_t=8.0kV$is 19 V.

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

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

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

$$\begin{gathered} \Delta V=I_{rms}R \\ =312.5\times 0.6 \\ =190V \end{gathered}$$

The voltage $\Delta V$ at $V_t=0.80kV$ is 190 V.

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

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