I recently read and watched some videos and articles about how resistive heat-generating devices such as incandescent lamps and hairdryers work. These articles (and many other sources) claim that the hairdryers, etc. heat up because the heating elements have a high resistance, but they would not with a lower resistance
But using
P=VI
and
V=IR
Let us assume that we have two resistive elements. One is 100?, and the other is 200?. Both are at mains voltage.
120 V / 100 ? = 1.2 Amps
120 V / 200? = .6 Amps
Now subsisting those values into P=VI:
120 * 1.2 = 144 Watts
120 * 0.6 = 72 Watts
So the lower resistance is generating more heat....
Am I doing my calculations wrong, or are the articles wrong?
Do the lamps and hairdryers actually have a lower resistance? How are they creating heat?
Whoever made the video didn't know zilch about physics. Your calculation is totally correct. Large power devices have lower resistance. You use either P=V^2/R or P=V*I to explain it.
The P you calculate is the power consumption on the resistive heater, that power turns into heat mostly. That is what resistors do, they turn power into heat.
No heat is created by current flowing through.
For a given voltage you get more current with a low resistance than a high one so more heat is generated because there is more current.
You cannot equate dissipated power to resistance value (High versus Low) without also considering one of the other required variables (current or voltage).
Let us say the resistor is 1kohm (what you might call "high")
With 10 volts across it the power dissipated equates to 0.1 watts - cold
With 100 volts the power is 10 watts - warm
with 1000 volts the power is 1000 watts - hot
with 10,000 volts the power is 100kW - !!!!
So "high" resistance (whatever that means) can dissipate appreciable power if the voltage is correspondingly high.
You cannot simply state low resistance dissipates high power whilst high resistance does not - unless of course you are considering the resistors applied across the same voltage.
If however you consider the same current flowing through the two resistors then the power dissipated across the high resistor is greater than that across the low.
jack, everything you say is true but very inappropriate here, you're just BS or BSO. Where do you use your hair dryer and household lamps with 10,000volts, with or without flux capacitor? :0
baum wanted basic calculations checked and all you do is confusing people.
You cannot equate dissipated power to resistance value (High versus Low) without also considering one of the other required variables (current or voltage).
I clearly stated that my device(s) was/were at mains voltage. I live in the United States, just so you know. But maybe in Scotland they have a mains voltage of 10kV. Never been there.
You cannot simply state low resistance dissipates high power whilst high resistance does not - unless of course you are considering the resistors applied across the same voltage.
Sorry if that part was confusing. What I meant to say was that Resistance A (which is greater than resistance b) creates Heat a (which is less than heat b).
If however you consider the same current flowing through the two resistors then the power dissipated across the high resistor is greater than that across the low.
Have you ever seen current-regulated mains? Maybe they have that in Scotland, too.
with or without flux capacitor?
Without. If we add it, it will use 1.21gW of heat, which will melt my breadboard (assuming constant voltage!).
The resistance of a heating element is always "high" compare to the wires that connect it to the power supply, but lower values of "high" resistance result in higher power devices.
Your math is fine, but what you left out is that that the house wiring and power cords and such also have resistance:
Suppose that the house wiring has a resistance of 1/3 ohm, the power cord has a resistance of 2/3 ohm, and the lightbulb has a resistance of 99 ohms. Then the total resistance is 100 ohms, and the current (I=V/R) is one amp everywhere in the loop. That means that the power dissipated in the lightbulb (IIR) is 99 Watts. The power dissipated in the house wiring is 1/3 W, and the power dissipated in the power cord is 2/3 W. No problems there.
If the lightbulb is instead only 9 ohms, the current is 10 A, and the lightbulb puts out 900 W (10109) (more like a hair drier now!) So lower resistance gives you higher power. We now also have 33.3W dissipated in the house wiring and 66.7W in the power cord (as much as a small lightbulb, although over a larger device. It might be noticeably warm.)
Now suppose we get extreme has use a lightbulb with a resistance of only 0.1 ohm. Current is now about 91 Amps, but the lightbulb dissipates only about 830 W. That's because most of the power is now being dissipated in the wires: 2760 W in the house wiring and a whopping 5500 W in the power cord. This would not be good. Hopefully a circuit breaker would trip or a fuse would blow, rather than having the house burn down.
liudr, your comment: "Where do you use your hair dryer and household lamps with 10,000volts, " negates the use of a transformer. Add one of those and you can run lots of high voltage devices at 120V... such as the flyback on a CRT.
Man, someone asking a basic question gets this type of "I know so much more than you can ever ask for" treatment :0. He just needed a straight answer and he got it already.
You're both crazy imaginary!
IF the light bulb is indeed very high in wattage, rare in household (because it causes fire like you described) but just say you have one of those, the socket that connects to the bulb must have thicker wires and has a special plug that won't go in your household outlets unless it's put together by some brain-dead electrician. Your assumption although valid in math, again won't help someone trying to get a straight answer to a simple question and won't go very far in reality. Engineers make sure proper wire gauges are used on devices so that the devices, not their wires, get the most power. How come my car jumper cables are thicker than my breadboard jumper wires? You tell me. You can't assume they are the same size and you attach a huge wattage light bulb on a breadboard jumper wire and see who's getting most of the juice. But that was what you did. My 30W soldering iron has 500ohm resistance, with 120V voltage, it is about 30W. I guess most of the power is on the iron, since the wire is only lukewarm at best when the iron is operating. The engineers designed the right gauge of wires so wires don't count, not in my approximation.
BTW, if you really have 0.1ohm load and run it on 120V then you should prepare to pay for 144KWh each hour, if you get all the wiring properly. Although I don't own a house, I think larger appliances have their special outlets and larger circuit breakers, but nothing household has that large requirement of power.
This type of answers is too intimidating to beginners and becomes silly. Could we stop it? 8)
This type of answers is too intimidating to beginners and becomes silly
I understand what everyone is saying, but I was just asking a simple question. I don't care if it is a lamp or a hairdryer, 110V or 10kV, AC or DC, I just wanted to know whether power is inversely proportional to resistance assuming constant voltage.
