Question

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) ${\mathit{P}}_{\mathbf{d}}$?

Short answer

1. The voltage $\Delta V$ decreases along the transmission lineis 1.9 V.
2. The rate$P_d$is$5.9W$
3. The voltage $\Delta V$at${\mathrm{V}}_1=80\mathrm{KV}$is 19 V.
4. The power $P_d$at$V_t=8.0kV$is 590 W.
5. The voltage $\Delta V$ at$V_t=0.80kV$is 190 V.
6. The power $P_d$ at $0.8kV$ is $59000W$

Step-by-step solution

Listing the given quantities:

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

The power of the generator is,

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

Understanding the concepts of power:

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:

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

(a) Calculations of the ΔV along transmission line:

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

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

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

(b) Calculations of the rate of  Pd:

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

The rate $P_d$ is 5.9 W.

(c) Calculations of the  ΔV at Vt=8.00 kV:

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

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

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

(d) Calculations of the  Pd at  Vt=8.00  kV

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

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

(e) Calculations of the  ΔV at  Vt=0.80 kV:

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

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

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

(f) Calculations of the  Pd at  0.8 kV

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

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