dave_dj88

Hey all.
I'm going through a senior thesis and it mentioned that in electrical power systems voltage is stepped up so that current is lowered in order to reduce power losses.
It all looks alright when considering that power is conserved i.e. VI is equal on both primary and secondary sides.
But when you consider ohm's law V=IR, isnt current INCREASED when voltage is increased?

I looked at it this way:

consider an AC power source of voltage V connected to a load of resistance R. So the current is I = V/R. Now, if a transformer is used in between to step up the voltage to 2*V. Then isnt current to the load also increased to 2*V/R = 2*I?
Is this correct or am I missing something here?

ericgibbs

hi,
If you had a fixed LOAD in the secondary of the 1:1 transformer and you applied 100Vrms to the transformer primary
[assuming 100% eff] you would get 100Vrms on the secondary.

If now you connected a 100R across the secondary 1Amps would flow in the primary and secondary windings.

If you now replaced the 1:1 transformer with a 1:2 step up, you would get 200Vrms across the 100R load
and 2Amps would flow in the secondary winding and 4Amps in the primary.

The reason that high voltage transmission line are used is so that lower currents are required for the same power,
so the wire can be a lower gauge. The idea is to reduce losses in the transmission cables.

Do you follow OK.?

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

stevez

Another point of view -

If you power a 240 watt incandescant light from a 240 volt source the current is 1 amp - the assumption being that this light is designed to operate at 240 volts.

If you were to power a 240 watt incandescant light from a 120 volt source the current is 2 amps - on the assumption that the light is designed to operate at 120 volts.

The flow of 2 amps in a conductor would result in greater losses than the flow of 1 amp - all other things being equal. Yes, you could make the conductor carrying 2 amps larger. The transformer provides an opportunity to reduce the current flow - you just have to adjust your thinking to account for the fact that the load must be modified to account for the change in voltage. In the example one would have to select a lamp for the intended operating voltage - same power, same light output (presumed) but different voltage.

__________________
stevez

rjvh

In an ideal transformer (100% efficient and thus not existing) the power you put in is the power you get out

electrical power is potential difference times current P=UxI

so if you increase the voltage (potential differance) the current will decrease

in case of a the power cable

U=IxR

the resistance is fixed deu to the usage of the matterial of the wire

if you have a powercable that is 240Ω and you want to transport 240 Watt of power at a voltage of 240 volt than the current is 1 Amp and the potential will be used all so at the end of the line there is 0 volt left over

now we want to transport the same amount of power (240W) and use the same cable but we transform the voltage in 24000V

we only use than 0.01Amp

the voltage drop over the cable would be 2.4V

now we reverse it back so 24000-2.4=23997.6V x 0.01Amp =239.976Watt

Efficient way of transporting

in the first transport methode we have 0v left over at 1 amp so that's 0 watt of power

no efficient way of transporting

Robert-Jan