AVOIDING POWER EQUATION CONFUSION
Most pupils learn that P = VI, which is useful. It's then a short hop to substitute in V = IR to get P = I ²R and P = V ²/R, which are also fine. However, it might not take some pupils long to spot what looks like a contradiction and ask:
'P = I ²R says that a higher power appliance would have a lower resistance (because P ∝ R). But doesn't P = V ²/R seem to say the opposite is true (because P ∝ 1/R)?'
To which I would then say something like:
'Yes, it does look that way, doesn't it? But with P = I ²R, P is only proportional to R if I is constant. With P = V ²/R, P is only inversely proportional to R if V is constant.
So, if you're comparing, say, heaters which are connected in parallel with the same power supply (i.e. so that V is the same across each), the lower resistance heater would work at a higher power.
On the other hand, if you had the same heaters connected in series (i.e. so that I was the same through each), the higher resistance heater would operate at a higher power.'
The fact that this sort of thing is ripe for confusion explains why situations like this are often used in Olympiad papers and University interviews.
Related to this is...
WHY DO WE NOT USE P = V ²/R WHEN DISCUSSING THE USE OF TRANSFORMERS IN THE NATIONAL GRID?
The usual justification for stepping V up before long-distance transmission is that it lowers I which therefore reduces the power loss in the wires and P = I ²R is usually (quite correctly) quoted as justification for this. In a similar way to the above discussion, though, some pupils might wonder why P = V ²/R doesn't lead us to the conclusion that you should actually step V down to reduce power losses.
The reason is that, in this case, the relevant V is not the pd between the transmission wire and ground (i.e. the 440kV often quoted on warning signs) but rather the pd across the opposite ends of a given transmission cable due to energy loss in the wire. In other words, it's the potential difference between the power station end of the wire and the consumer end of the wire - which, of course, over a long distance will not be the zero Volts we would assume it to be in a short laboratory wire. It was a long time before I spotted this and I'm only glad I didn't get asked that question in the meantime!