Electric transferring from powerplant

[EN1986]

I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage.

The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me.

I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable consumption. And according to this a higher voltage (while assuming a constand resistance that comprises of the line resistance and the load resistance) leads to a higher current that results in a higher power loss.

What am I missing?

 

[.Scott]

Power lines are resistors. The more current that flows through them, the greater the voltage drop from source to user, and thus the more power is dissipated.
The total power delivered is volts times amps. So, given a certain amount of required power, higher voltage means less current and thus less transmission losses.

 

[Baluncore]

W = I ⋅ V ;
If you double the line voltage, then for the same power, the load current will be halved.
Then consider power lost in the line resistance, R.
Ohm's Law; V = I ⋅ R ;
W = I² ⋅ R ;
By doubling the voltage, power loss in the line is reduced to one quarter.