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:
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'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)?'
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To which I would then say something like:
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'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.
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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.
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On the other hand, if you had the same heaters connected in series (i.e. so that I were the same through each), the higher resistance heater would operate at a higher power.'
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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.
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Related to this is...
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WHY DO WE NOT USE P=V ²/R WHEN DISCUSSING THE USE OF TRANSFORMERS IN THE NATIONAL GRID?
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The usual justification for stepping V up before 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, 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. It was a long time before I spotted this and I'm only glad I didn't get asked that question in the meantime!
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