Low resistance wire = more power used? wait...

[Enoko]

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This is bothering me, they say light bulb filaments and toasters use high resistance wires since they want to convert electric energy to heat energy.

But then I look at the formulas.

V = IR
P = IV

Wall outlets have 120V, lower Resistance means higher current. Current goes into the power formula.
So [Low Resistance] -> Really high current - > Powerful toaster 😕

Shown in numbers, power used by a 1ohm resistor vs a 120 ohm resistor
**1 ohm filament in a toaster**
V=IR
120V = 1ohm * 120 amps

P= V*I
P= 120amp * 120 Volts
Power used by 1 ohm resistor = 14,400 Watts

**120 ohm filament in a toaster**
V=IR
120V = 120ohm * 1 amp

P= V*I
P= 1amp * 120 Volts
Power used by 1 ohm resistor = 120 Watts

The numbers say a weak resistance in the toaster (or lightbulb) makes for a large power output. But intuition tells us we want our toasters and lightbulb filaments to be high resistance wires... right?😕

 

[unleash500]

Large power output does not mean more energy converted to heat energy.

Resistance is what converts electrical energy to heat energy ( for toasters).

Thus, higher resistance means more heat energy produced.

A larger power can just mean more electrical energy going through the circuit.

 

[lorenzomicron]

large power output does mean more energy converted to heat, as the energy is the time integral of the power.

in the toaster example, the current source must be kept constant (i.e. voltage source must get larger), when a higher resistance is put in in order to get more power. the reason for this is that the voltage source, if kept constant, can only supply some maximum amount of energy.

 

[phltz]

Power dissipated in a resistor can be calculated by
P = V^2 / R

Looking at this, we see that if you hold the voltage across a resistor constant, the power dissipated is inversely proportional to resistance.

This may seem counter-intuitive at first, but it makes a lot of sense if you keep thinking about it. Consider a standard plug socket. There's a voltage of 120V from one side of it to the other. If nothing is plugged into it, there's just air between the two sides. Air can be considered a pretty good insulator, because it doesn't conduct electricity well. Thus it has a high R. And we expect that when nothing is plugged into the socket, there should be almost no energy being dissipated through it.

On the flip side, if you just stick a fork or something into the plug socket in the wall, you'll be putting a bug chunk of conductive metal across it. And it's going to be a big problem, because you'll have a very low R and so a very high energy output. If your house lacked circuit breakers, you could easily start a fire or otherwise end poorly.

At any given voltage the higher the resistance, the less current actually flows through the resistor. And the more current that flows, the more power you can dissipate.


Now, it's worth noting that this analysis only holds if the voltage across the resistor is held constant. If we're talking about one resistor in a larger circuit, or that's hooked up to some sort of constant-current supply or something like that, we'll need to reorganize our equations.

 

[fizzgig]

OP i might get the issue you're having trouble with - if Pdissipated is inversely proportional to R, and Pdisspated is how we see light in our lightbulbs, then we should lower R to get high Power and a brighter light, which implies an increasingly thick bulb filament! but clearly we don't have bulbs with superthick filaments.

so dealing with P=V^2/R since our house has constant Vrms and not constant current, say you have a bulb with some filament. this filament material/thickness has been tuned to be a useful lightbulb, but what if we change it.

i do not know exactly what happens when you play with variables, but here's stuff to at least consider:

if you lower the filament resistance by making it thicker, current does increase (V=IR), and power does too. but, also consider you're changing the physical wire, so the surface area available to radiate to the surroundings is larger, so for all i know you could get a lower blackbody curve. alternatively, this wire may be set to be as thick as it can be already without melting! this material has it's own melting point after all. if you make it thicker, yep, more power, but the heat may just kill your device by melting wires. or maybe a little thicker wire really would make the bulb brighter, but it'd have a shorter lifespan.

if you raise filament resistance by making it thinner, current decreases, so does power. but, your wire has less area to radiate, so maybe it could actually get hotter and glow brighter. or, for a filament's length and assuming it is the only big resistor, maybe you could make it thin enough to lower current enough to lower glow, or maybe it' be so thin at that point that even that amount of current would break it.

the main bullet point is that (as i understand it for right now, corrections/help more than welcome) yes, weird as it seems, power dissipated by your Rs does go down with increasing R. for devices that seem to favor lots of dissipation, when you wonder why not lower resistance a bunch so you'll have huge power dissipation, keep in mind it's a tuned circuit made of specific materials, and they only want it to get SO HOT, because otherwise the device will fail.