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How does high voltages of electricity coming from power plants also helps in saving energy?

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The power cables have some resistance. Power lost in the wires can be calculated as P=R*I^2 with R as the resistance of the wires and I as the current that passes through them. Power at the load is P=V*I. From this you can see that if you increase the voltage by 2x, you need only half the current to deliver the same power. However, if you pass half the current on the same wires, you will lose only a quarter of the power.

An example with made up exaggerated numbers:

Wire resistance is 10 Ohm (not very long power line), power required is 100kW. At 220V, the required current will be 100kW/220V=454A. The power lost at the wire will be 10Ohm*454A^2=2.061MW. Voltage lost at the wire will be 10Ohm*454A=4540V. So, the power plant will have to generate 4760V so that the load can receive 220V. Power generated at the plant - 2.061MW+0.1MW=2.161MW. Efficiency is 0.1MW/2.161MW*100%=4.6%

Now suppose the plant generates 330kV (and the voltage is stepped down to 220V very close to the load). Now the current needed is only 100kW/330kV=0.3A. Power lost on wires is 10Ohm*0.3A^2=0.9W, so the efficiency (ignoring power lost in the transformer) is 100000W/100000.9W*100%=99.9991%. Transformers are very efficient, about 98% for the big ones used in power distribution.

Now scale that up to, say, a power plant that generates 1GW and you will see why Tesla won the current wars (hint: there were no efficient DC-DC converters back then).

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Summarised solution:

  • Resistive power losses both in absolute terms AND as a percentage of energy carried drop as voltage increases.

  • The reduction in loss and/or material (copper and/or Aluminium conductor) is so vast that use of high voltage is very desirable.


Power loss in transmission of power over distance is mainly (but not wholly) by resistive losses.

Resistive power loss = \$I^2 \cdot R\$ (current squared x resistance).

Power transferred in a circuit = Voltage x Current = V x I

So power lost as a proportion of power carried:

$$ {P_{loss} \over P_{carried}} = {(I^{2} \cdot R_{line}) \over {(V \cdot I)}} = {I \cdot R \over V} $$

Note that this has units of Voltage/Voltage which cancel to give a dimensionless ratio (as you'd hope). The "top line" Voltage is the I x R voltage drop in the line and the bottom line Voltage is the transmission voltage. So loss ratio is effectively resistive_voltage_drop / line_Voltage.

So, for a given conductor of resistance R the percentage of power lost will increase as current increases and will decrease as voltage increases.

But as Power carried = V x I, if we double V we halve I.
If we multiply V x 10 we reduce I by 10.
This is a win-win situation for high voltage.

This can be put two ways.

  • For a given transmission line, carrying the same power at higher voltage will reduce loss%.

and

  • For a given desired loss of power (say 1% end to end) we can use increasingly higher resistance lines (and so less metal) as voltage increases.

A simple calculation shows that loss % decreases as the inverse SQUARE of Voltage !. Increase voltage by 10 x and for the same line resistance losses will drop by a factor of 100!. Murphy was asleep that day!

The gains are so great that if this was the only factor then as high a voltage as possible would make sense.

There are other factors such as losses due to corona and the need to provide substantially increased insulation and clearances and tower sizes as voltage foes up BUT economically, it all leads towards big tall ugly very high voltage towers.


Simplistic example:

1000 Watt power transfer. Rline = 1 ohm.

(1) V= 100V, I = 10A. Power transmitted = 100 x 10 = 1000 W.
Power loss = I^2 x R = 10^2 x 1 = 100 W.
100/1000 = 10%.
10% of power is lost.

(2) V= 1000V, I = 1A. Power transmitted = 1000 x 1 = 1000 W.
Power loss = I^2 x R = 1^2 x 1 = 1 W.
1/1000 = 0.1%.
0.1% of power is lost.

(3) Increase line resistance to 10R = use 10% of original material. Still use 1000V. V= 1000V, I = 1A. Power transmitted = 1000 x 1 = 1000 W.
Power loss = I^2 x R = 1^2 x 10 = 10 W.
10/1000 = 1%.
1% of power is lost. Even when the line uses 10% of the conductor material of the original circuit, an increase in voltage by 10x decreases power loss by a factor of 10 !!

!!!!!!!!!!!!!!!!!!!!!!!!

The reduction in loss and/or material (copper and/or Aluminium conductor) is so vast that use of high voltage is obviously very desirable.

There are other factors that make the gains less great in practice BUT the existence of large and larger power pylons demonstrates that reality is still well served by higher voltages.

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Power lost as heat due to the wire's resistance given by:

P = I^2 * R (equation 1)

Hence, if current I, small, power lost is small.

Power intended, or needed is given by:

P = I*V (equation 2)

For larger voltage, smaller current needed to deliver same power. Hence, larger voltage means smaller current; smaller current means less power lost (equation 1)

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