Ohmic Heating in Wires please could someone tell me why Ohmic losses are always referred to as $I^2 R$ losses? Here is my problem. If the power coming from a power station is fixed then you can either deliver this power as high voltage, low current or high current, low voltage. But isn't $I^2R$ equal to $V^2 / R$, therefore if R is constant doesn't the power depend on the square of the voltage so surely it doesn't matter whether it is high voltage or high current. The only way I can reconcile this is that a high current must cause a greater heating effect than a high voltage. I can't figure why though. If this is the case then is there a reason why a high current causes more heating than a high voltage?
 A: 
But isn't $I^2R$ equal to $V^2/R$, therefore if $R$ is constant doesn't the power depend on the square of the voltage so surely it doesn't matter whether it is high voltage or high current. 

Consider the wires connecting the power plant to the appliance; let the effective amplitude of oscillating current flowing through all wires be $I$ and let $V$ be drop of electric potential across one wire conducting current $I$. The situation can be drawn like this:

              V
  <------------------------>
  o‒‒∧‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒o
     |  
     |U
  o‒‒∨‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒o


  o‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒o

power                     appliance
plant

The Joule loss of energy per unit time in the wires is $P_{loss}=VI$ and since  Ohm's law says $V=I/R$, we have expressions $P_{loss}=V^2/R= I^2R$.
When people say higher voltage means lower losses of energy, by "voltage" they do not mean $V$; that would make, as you realized, no sense since the energy losses are proportional both to $I^2$ and to $V^2$.
By "higher voltage", they mean higher voltage $U$ between two separate wires at the same distance from the power plant. Higher $U$ than generated can be achieved in the power plant using appropriate voltage transformer.
Why is higher $U$ beneficial?
The power utilizable at the end of the power line is $P_{useful}=UI$. The power that is being lost is $RI^2$.
So by making $U$ higher, the same useful power can be transferred with much lower current $I$ and thus much lower energy losses $RI^2$ in the power line. It is easy too see that if we double the voltage $U$, the power loss decreases by factor of four.
A: If the only load is the wire, such as a hot wire cutter, then supply voltage and wire resistance are all that matter, and both formulas are equivalent.
But most of the time, wire size is chosen so that most of the voltage appears on the load. As an example, a toaster uses a power cord with wires which drop much less voltage than the heaters - if this were not done, the power cord wires would get as hot as the toaster.
With most of the voltage being dropped by the load and not the wires, it is the load resistance which sets the current in both the load and the wires. As far as the wires are concerned, the supply voltage becomes pretty much irrelevant, so the power dissipated is viewed in terms of the current only.
For example, if the load resistance is 100 times the wire resistance, a 5% difference in the load resistance will cause a 10% variation in the wire power, but at 5% difference in the wire resistance will cause a .01% variation. 
A: Just adding $0.02 for clarity:
The formulas $I^2R$ and $V^2/R$ describe the power dissipated in a resistor.
If you're considering the power lost in power transmission lines, the "resistor" is the power line, not the appliance in the home where the power is wanted.
The $V$ in the $V^2/R$ formula is the voltage beetween the two ends of the resistor so, in the power transmission problem, $V$ is not the voltage delivered to the home:  It's the voltage difference between one end of the transmission line and the other.
A: Just remember ohms law
$$V=IR$$
If you have a high voltage across a resistor, you automatically get a high current because they're related. If you have high current, you're going to need a high voltage to drive it. It's referred to as $I^2R$ by convention and for convenience; $V^2/R$ or even $VI$ are equally valid.
