How come step-up transformers reduce current in transmission lines?

By applying ohms law in the secondary circuit, I=V(stepped up)/R , a stepped up voltage would result in an increase of current, i am aware that the step up transformer steps down current, but it is the secondary circuit that determines the current draws (ohm's law), and the primary current is multiples of it......So ?!!!!How come step-up transformers reduce current in transmission lines?Transformer coils are electrically isolated, the voltage in the secondary being induced by that in the primary. There is therefore no constant current between. The POWER (W=AV) in the primary has to have an an induced equal counterpart in the secondary. In this case, as the voltage in the secondary is larger, the current drops proportionally. At the end of the transmission line it's stepped down again, and then you get a current.How come step-up transformers reduce current in transmission lines?I can't make too much sense out of your question.



Your use of Ohm's law implies a constant resistance, which is not true.



Power is constant between the input of a transformer and the output. When you increase the voltage you decrease the current, and the power remains constant (with a small lose due to inefficiencies).



.How come step-up transformers reduce current in transmission lines?Here are some basic equations for transformer.

V-high / V-low = N-high / N-low Eq. 1

I-low / I-high = N-high / N-low Eq. 2

where, V-high = voltage at high voltage side

V-low = voltage at low voltage side

I-high = current at high voltage side

I-low = current at low voltage side

N-high = number of winding turns at high side

N-low = number of winding turns at low side

Eq. 2 above could be written as

I-high = (N-low / N-high)I-low %26amp; substitute Eq. 1

I-high = (V-low / V-high)I-low Eq. 3

Let us have an example to give a good answer to your question. Consider a 30 MVA, 13.8-138 kV, 3-phase step up transformer. The 13.8 kV is at the generator side and the 138 kV is at the transmission side.

Low side current is I-low = 30,000,000 / 1.732(13,800) = 1255 A

High side current is I-high = 30,000,000 / 1.732(138,000) = 125.5 A

it shows that I-high is less than I-low. Also, using Eq. 3 to further confirm result, we have

I-high = (V-low / V-high)I-low

I-high = (13,800 / 138,000) 1,255

I-high = 125.5 A