Question

Electric power from a 100 MW power station is to be transmitted to a distant load through long and thin cables. Should it be transmitted at 20,000 V or at 200 V, to minimise power wastage?

#NoSpam Answers​

Answer 1

Question:-

➣

Electric power from a 100 MW power station is to be transmitted to a distant load through long and thin cables. Should it be transmitted at 20,000 V or at 200 V, to minimise power wastage?

Answer:-

➣

We know:-

➣ Power [P] =$100 MW = {100 × 10}^6 W$

Solution:-

➣

Let R be the resistance of the cables. The current in the cables, when power is transmitted at $\bf{20,000 V}$ will be

${i}_1 = {\frac{P}{V}}$ =${\frac{100 × {10}^6 W}{20,000 V}}$

$= 5 × {10}^3 A$

and the current at $\bf{200V}$ transmission will be

${i}_2 = {\frac{100 × {10}^6}{200}}$

$= 5 × {10}^5 A$

The power dissipations as heat in the cable in the two cases are :

${{i}_1}^2 R = (5 × {10}^3)^2 × R \\ = 25 × {10}^6 watt$

${{i}_2}^2 R = (5 × {10}^5)^2 × R$

$= 25 × {10}^{10} watt$

The power of dissipation is much less in the first case, that is ,at $\bf{20,000 V}$ transmission.

Answer 2

$\huge\tt\red{Answer}$

As we know

1 Mw = 1000000

That is$10^6$ watt

Convert power from Megawatt to watt.

$100 \times {10}^{6} W$

=} Let the resistance of the cables be R.

Calculate the current transmitted in the cables with both the voltages.

Take the current as $i_1$ and $i_2$ respectively .

______________________

Power (P) = $100 × 10^6$

Voltage (V) = 20,000 V

$i_1 = \frac{P}{V}$

$\frac{100 \times {10}^{6} }{20000}$

Cut all the zeros.

And then divide

=} $5×10^3 A$

_______________________

Power (P) = $100 × 10^6$

Voltage (V) = 200 V

$i_2 = \frac{P}{V}$

$\frac{100 \times {10}^{6} }{200}$

divide the terms.

=}

$5 \times {10}^{5}A$

The power of dissipations is $5×10^3 A$ and $5 \times {10}^{5}A$

$\begin{lgathered}{{i}_1}^2 R = (5 × {10}^3)^2 × R \\ = 25 × {10}^6 watt\end{lgathered}$

${{i}_2}^2 R = (5 × {10}^5)^2 × R$

$= 25 × {10}^{10} watt$

The power of dissipation in the first case is lesser.