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2021-12-18

Digah Company

45

**High voltages vs. Energy loss**

Summarised solution: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 ISo 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 would 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.andA 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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**230 and 240 transformer leads**

You may have a transformer that was built for use during the time when Europe was standardizing the line voltage. Up until 1995, various countries in Europe used different line voltages for the outlets. For example, in Germany it was 220V, but in the UK it was 240V. In 1995 (Mains voltage wikipedia ), they all began switching to 230V ,and were supposed to have completed the switchover by 2008. So, you may have a transformer that was built to be installed into a device and used when the line voltage was 240V, but could then be switched (switch or jumper) to use the lower voltage after the changeover.Here in Germany in the early 1990's I used a big test rig for two way radios that had a switchable supply voltage - it was adjustable from like 210V up to 250V. I also own an oscilloscope that was built in 1967 that has a jumper on the back to switch between 110V and 220V - with various jumper positions for voltages from 220V up to 250V (or something like that I do not have it memorized.)

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**having different voltages from 2 batteries in serie (3 voltagesâ€¦)**

It's quite simple, you need 3 batteries - it's just a question of where ground is:simulate this circuit - Schematic created using CircuitLabBe warned though that CR2032 batteries can only supply very small currents

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