Resistance of elements in kettles and light bulbs

Hi everyone,
For a constant voltage source (such as that provided by the mains), the power dissipated across a resistance is inversely proportional to
the resistance. Low resistance, high power, lots of heat dissipated...
How stuff works states that:
"A thinner conductor heats up more easily than a thicker conductor because it is more resistant to the movement of electrons."
Wouldn't something that has more resistance draw less power, and thus less heat? A colleague suggested that perhaps more power/unit area is dissipated with a thiner conductor, and thus it is hotter.
Similarly, it is often stated that the heating element in an electrical kettle is high resistance so that lots of heat is generated. Again, wouldn't a high resistance decrease the power dissipated for a constant voltage source?
I'm obviously missing something here :)
Thanks
Taras
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Don't know what "How stuff works" is, but that statement is far too simplistic to have any serious validity. What on earth does "heats up more easily" mean? I suppose it would heat up faster in a gas flame, so it's not completely invalid, but far too vague.

Yes, from a constant voltage source.

Not sure what's on the web, but you might want to look up the design of light bulb filaments. For example, it's impossible to produce a tungsten wire which will work as a 100W filament in a 240V lamp (probably true to 120V also). When you've worked out what resistance you require and hence the length of wire, you'll find the wire is far too long to fit in a light bulb, and has so much surface area it won't get hot enough to glow. So you need something which is much shorter and has much less surface area so it actually gets hot enough to glow. This is done by coiling or double coiling the filament so it gets much shorter, and has much less effective surface area (as most of the real surface area radiates back onto the filament itself, it doesn't count anymore).

Yes. The 'high' may have been in relation to the supply cable though.
You are at a level where you need better technical resources than "How stuff works" (whatever that is) to answer your questions.
--
Andrew Gabriel
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Right, if you consider the total resistance of the load which is placed across the mains. So, for example, an electric kettle has a much lower resistance than a 50 W light bulb, and so draws (and dissipates) more power.

That's layman's handwaving for something stated more precisely as "A thinner conductor has more resistance than a thicker one because it has less cross-section area. When a constant current is passed through that conductor, the heat dissipated is proportional to the resistance. So thinner conductors get warmer under constant current conditions".
Note the difference in the two sets of conditions. Wires are not generally the major source of resistance across the line, so changes in wire diameter do not cause significant changes in the total current - the current is primarly determined by the resistance of the kettle (or whatever the intentional load is).
To a first approximation, the heater in the kettle operates under constant-voltage conditions, and power is inversely proportional to resistance. To a first approximation, wires operate under constant current conditions, and the power dissipated *in the wire* is directly proportional to resistance.
If you connect a low-resistance wire across a constant-voltage source like the line, you may dissipate a lot of power for a few milliseconds - until the wire explodes, or a breaker trips. Thus, it has too little resistance to make a good heating element.

That's misleading. A kettle has whatever resistance is needed to produce the desired amount of heat at the desired voltage. Any more or less resistance and you get the wrong amount of heat, and possibly too much load for the outlet it's plugged into.
If you consider a kettle and some wire connected in series, so the current is the same in both, then it make sense to say that the kettle's heating element has much higher resistance *than the wire* and so dissipates much more heat *than the wire*. But this only applies when both are in series, as they are in a typical circuit.
    Dave
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On Apr 22, 4:00 am, snipped-for-privacy@cs.ubc.ca (Dave Martindale) wrote:

Right, that's under constant current conditions. When using the mains as the power source, you would have constant voltage conditions. So would it be correct to say that the resistance of the element in the kettle must be chosen so that it is high enough to prevent excessive power dissipation (and possibly melting/explosions), but low enough so that enough power is dissipated (and that the other descriptions I have provided are too simplistic/incorrect)?

Yes, that makes sense, since the two are in series and thus have the same current.
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You have a constant voltage applied *to the wire and kettle connected in series*. Most of the resistance is in the kettle heating element, so the current (and power) taken from the line is almost entirely determined by the element resistance. The wire resistance has almost no effect in a circuit where the wire is sized adequately for the load. (This is because "adequately sized" is *defined* in terms of low power dissipation in the wire.)
So the assembly of wire and kettle in series is operating under constant voltage. And since almost all the voltage drop is across the kettle, the kettle element itself is effectively operating under constant voltage. But if you look *at the wiring alone, without the kettle*, it is operating under constant-current conditions. And that's the circumstances the "thinner conductor heats up more" statement assumed.

Yes. That's how you design the kettle element.
You design the house wiring so that, with the maximum expected load current applied, the heat dissipated in the wire is not enough to start a fire or degrade the wire insulation, nor is there excessive voltage drop across the wire. That's why the minimum acceptable wire size for a circuit depends on the insulation type, the ambient temperature, what the wire is installed in, and its length - in addition to the amount of current carried.
    Dave
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On Apr 23, 2:19 am, snipped-for-privacy@cs.ubc.ca (Dave Martindale) wrote:

Thanks Dave, that cleared things up nicely :)
Taras
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