Hey, I'm a bit confused about the power loss formula.
So I know to derive it, you sub V=IR into P=VI, but why does power turn into power loss when you do so?
Also, why can't you sub in I=V/R instead to get Ploss=V2/R? I know that this formula doesn't work because stepping up the voltages significantly reduces power loss, so it wouln't make sense to have Ploss proportional to V2. I just don't understand why subbing in the same formula in a different way doesn't work. Could someone please clarify why this is the case?
Step into my office
Legit, the moment in my first year ELEC lecture when I
finally figured out why this is, best moment ever. Not difficult! Just a little intricate.
Right, so \(P=VI\) gives the power dissipated/lost in a circuit element (EG - the wires of a transmission network), as a product of the current through the element and the voltage
across the element. The formula is identical to \(P=I^2R\) and \(P=\frac{V^2}{R}\). So why does only one work?
Notice what I emphasised in the text above -
Across the element. Meaning, the voltage at one end minus the voltage at the other, the potential
difference. This goes right back to
Electrical Energy in the Home, circuit analysis in Year 11 involved analysing
voltage drops across resistors.
The problem is that we are almost always given the voltage going
into the transmission wires, never the voltage ACROSS those wires. We might put 100,000V into the wires, but that's not the voltage
across them. The voltage across them is the difference between 100,000V and the voltage at the other end, which is calculable as \(V=IR\), Ohm's Law!
If we use the voltage
across the wires, those other formulas work - \(P=VI\) and \(P=\frac{V^2}{R}\). If we just use the input voltage, they break. THIS is why we usually use \(P=I^2R\), because we don't need to look at the other end of the wires. We know how much current goes in, we know how much resistance there is - And that is all we need
This is really hard until you have a click moment, then it is really easy. If it is still a little confusing let me know, in which case I'll do a numerical example using Year 11 techniques to show you the difference