Power loss in step-up/down transformer Why is the equation Power Loss = $I^2\times R$ rather than Power Loss = $V\times I$?
What I mean is why use $I\times R$ to represent $V$?
Also if Power Loss is equivalent to $V\times R$, doesn't step up transformers which creating higher voltage also cause Power Loss to increase which contradicts to textbooks stating that power loss decreases if the voltage rises given that power supply is same and Power Loss = $I^2\times R$?
 A: There are two equations that are used to calculate power loss in a circuit.  Those equations are:
$P=IV$ (the power equation) and $V=IR$ (Ohm's Law). If current through a resistor is known, and the voltage across that resistor is also known, the equation $P=IV$ can be used directly to calculate the power generated by the resistor.  If current through the resistor is unknown but the voltage across the resistor is known, the equation $V=IR$ can be solved for current, and that can be substituted into the original equation.  Thus, for this case, $P=(V/R)*V=V^2R$.  For the case where the voltage across a resistor is unknown, but the current through the resistor is known, Ohm's Law can be used to substitute $IR$ into the power equation to obtain $P=I*(IR) = I^2R$.
Regarding power loss in a transformer, that loss comes from the resistance of the wires that the transformer is made of, and from hysteresis involved in the alternating magnetic field that is running through the iron core of the transformer.  The voltage and current that is changing through a transformer, which you referenced in your post, is what is transmitted to a load, and that voltage is different than the voltage drop through the transformer itself, which is due to the transformer's electrical resistance.  Thus, to calculate the power loss in the transformer, you need to measure the voltage drop across either the primary or secondary of the transformer, and multiply that by the current through the primary or the secondary of that transformer.  In addition, consideration must be given to the fact that the transformer is running on AC current, so hysteresis in the iron core of the transformer needs to be accounted for.
