How does a hairdryer produce 1500 watts?

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In summary: I get it now.)The resistor in a night light is physically small but electrically large.So the heat output from the coils in the hair dryer is drastically reduced because of the large resistor?
  • #1
DaveC426913
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I took Electricity in HS and Electronics in Continuing Ed, and I still get tripped up over this very basic question.

A household circuit with no resistance - a short - will allow enough current through to burn it out.
A household circuit with a giant resistor can power a tiny nightlight, by reducing the current flow.

How is a 1500w hairdryer able to "draw" 10 amps if the coils are basically a giant resistor?

Is it a balancing act? The coils let enough current through to heat the coils, but not enough to reduce off the current to a trickle?

If so, how does the giant resistor in the night light circuit not heat up as hot as the coils in the hairdryer circuit?
 
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  • #2
V=IR, so I=V/R

P=VI=RI^2=(V^2)/R

The nightlight is designed to consume about 1W, so given Vrms=120V (or whatever), you can find the filament "hot" resistance that is needed to consume that power. There is no other resistance in series with the filament.

For the 1500W hair dryer, you can calculate what the heater coil resistance would need to be from the above equations, no? :smile:
 
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  • #3
The resistance of the heating elements in each case (filament for the nightlight, heating element for the hair dryer) are chosen to draw the appropriate current for the given application. Ohm's law applies.

Household electric outlets are to a good approximation ideal voltage sources. They will supply any current (within the limits of their circuit breaker's or fuse's limiting values) in order to maintain the specified outlet voltage (Specified voltage may vary with geopolitical location. Typically 120 V rms for North American consumers, but may vary between 110 V and 125 V).

The power developed across a given resistance is proportional to the voltage squared divided by the resistance. P = V2/R . For a given voltage, smaller resistance = larger power, and larger resistance = smaller power.

The nightlight filament will have a larger resistance than the hair dryer heating coil.

Describing the heating coil of a hair dryer as a "giant resistor" is ambiguous. It will have whatever resistance it's designed to have at its operating temperature in order to draw the required current and produce the specified watts of heat flow.. No doubt the resistance will be lower than the resistance of a typical nightlight. I've looked at several nightlights and they usually specify a bulb ≤ 5 W. So for a given nightlight and a given hairdryer, and a given voltage supplied, we could calculate the resistance of each.
 
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  • #4
berkeman said:
V=IR, so I=V/R

P=VI=RI^2=(V^2)/R

The nightlight is designed to consume about 1W, so given Vrms=120V (or whatever), you can find the filament "hot" resistance that is needed to consume that power. There is no other resistance in series with the filament.

For the 1500W hair dryer, you can calculate what the heater coil resistance would need to be from the above equations, no? :smile:
Calculating without understanding won't do me much good :frown:.

How is the night light "designed" to consume only 1w? That's by using a giant resistor, right?

Am I misunderstanding then, that the resistor in the night light is not essentially the same as the coils in the hairdryer?

Or is my original assumption correct, that the coils provide little resistance, thereby drawing 10A?

So if there's little resistance, why do the coils heat up so much?
 
  • #5
DaveC426913 said:
I took Electricity in HS and Electronics in Continuing Ed, and I still get tripped up over this very basic question.

A household circuit with no resistance - a short - will allow enough current through to burn it out.
A household circuit with a giant resistor can power a tiny nightlight, by reducing the current flow.

How is a 1500w hairdryer able to "draw" 10 amps if the coils are basically a giant resistor?

Is it a balancing act? The coils let enough current through to heat the coils, but not enough to reduce off the current to a trickle?

If so, how does the giant resistor in the night light circuit not heat up as hot as the coils in the hairdryer circuit?

I think you might be confusing the physical size of a resistor with it resistance. These aren't necessarily related.

The resistor in a 1500w hair dryer is physically big but electrically small.
The resistor in a night light is physically small but electrically large.

The power is determined by the electrical size.
 
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  • #6
CWatters said:
I think you might be confusing the physical size of a resistor with it resistance. These aren't necessarily related.
No. By 'big' I mean high resistance.
CWatters said:
The resistor in a 1500w hair dryer is physically big but electrically small.
OK, so the coils are low resistance.
Maybe I'm misunderstanding where the heat comes from. I've been assuming high resistance = high heat output.
(Of course, that can't be true, as the short circuit example demonstrates the opposite.)
 
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  • #7
DaveC426913 said:
How is the night light "designed" to consume only 1w? That's by using a giant resistor, right?
For elements like filaments, you use a metal material that has a volume resistivity in the range that you want, and choose the diameter of the filament and the length (usually coiled) to give you the overall resistance you need to generate consume the power that you want. See my equations in my original reply.
DaveC426913 said:
Or is my original assumption correct, that the coils provide little resistance, thereby drawing 10A?
The hair dryer coil has whatever resistance corresponds to 1500W from the 120Vrms wall socket. You choose the material, wire diameter and wire length (in the coil shape) to give you that resistance
DaveC426913 said:
So if there's little resistance, why do the coils heat up so much?
See my equations in my original reply. P=RI^2,P=V^2/R so the lower the R, the higher the power consumed by the coil. BTW, that equation starts to change as the coil resistance starts to drop low enough to become comparable to the source resistance of the AC Mains circuit supplying the power to the coil, so you can't just keep dropping the coil resistance and get infinite power out of the wall socket... :smile:
 
  • #8
OK, so if you ensmallified the resistor in the night light, it would allow more current to flow as well as turning that power into heat.
 
  • #9
DaveC426913 said:
OK, so if you ensmallified the resistor in the night light, it would allow more current to flow as well as turning that power into heat.
Correct, and sorry for not being clearer about light and heat from a light bulb. They go hand in hand for incandescent light bulbs.

If you shorted out half of the filament in a light bulb, that cuts the resistance in half, which doubles the current and doubles the power/light/heat output.
 
  • #10
berkeman said:
If you shorted out half of the filament in a light bulb, that cuts the resistance in half, which doubles the current and doubles the power/light/heat output.
But @DaveC426913 be sure you understand that if you did what berkeman just correctly said, the night light would likely burn out instantly because it is designed for half the current you would be putting through it.

Physical size and the ability to dissipate heat are design considerations for both a night light and a hair dryer along with the V=IR considerations.
 
  • #11
phinds said:
But @DaveC426913 be sure you understand that if you did what berkeman just correctly said, the night light would likely burn out instantly because it is designed for half the current you would be putting through it.
Yeah. Producing heat in the process. Like a very short-lived hairdryer.

I was erroneously thinking the heat was produced by the resistance. It's not; it's produced by the current flow.
 
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  • #12
DaveC426913 said:
I was erroneously thinking the heat was produced by the resistance. It's not; it's produced by the current flow.
It is really a combination of current AND resistance. You can pass 1,000 amps through a very low resistance wire and get less heat than what you would get by passing 1 amp through a high resistance.
 
  • #13
phinds said:
You can pass 1,000 amps through a very low resistance wire and get less heat than what you would get by passing 1 amp through a high resistance.
Then I guess I'm back where I started... :frown:
 
  • #14
Let's try something different.

If I put household current across a variable resistor, going from zero to a bajillion ohms, where would it draw the most current? Where would it generate the most heat?
 
  • #15
DaveC426913 said:
Then I guess I'm back where I started... :frown:
Well, the equations are very simple V=IR (voltage = current x resistance) and P=I^2 x R (power is current squared x resistance).

SO ... if your hair dryer creates 1500 watts from a 110volt wall current, then its resistance must be about 8 ohms

My nightlight (a particularly "strong" one) draws 7 watts, so its resistance must be about 1700 ohms

In the US the standard night light is 4 watts, so would have a resistance of about 3000 ohms.
 
  • #16
DaveC426913 said:
Let's try something different.

If I put household current across a variable resistor, going from zero to a bajillion ohms, where would it draw the most current? Where would it generate the most heat?
At near zero is would be drawing a huge current so the power dissipation would be huge. At a huge resistance it would be drawing next to no current so the power dissipation would be trivial. The key in this experiment is that you are holding the voltage constant (which is reasonable). See the examples in the post directly above
 
  • #17
DaveC426913 said:
Let's try something different.

If I put household current across a variable resistor, going from zero to a bajillion ohms, where would it draw the most current? Where would it generate the most heat?
Don't think of it as household current. Think of it as household voltage: Some essentially fixed voltage supplied at the plug. Electric companies ideally provide a fixed voltage source that you can draw whatever amount of

Heat is the power developed by the load (your variable resistor).

##P = \frac{V^2}{R}##

So, small resistance going to zero implies major power. This is what's called a short circuit. Current limiting fuse/circuit breaker blows, end of experiment.

Large resistance implies lower power (and current), going to zero as the resistance approaches infinite. So, nice warming effect, or glowing incandescent light bulb, or operating whatever device the resistance is meant to represent. All's fine so long as the current drawn is less than the trigger value of the fuse/circuit breaker.
 
  • #18
DaveC426913 said:
Let's try something different.

If I put household current across a variable resistor, going from zero to a bajillion ohms, where would it draw the most current? Where would it generate the most heat?
When the resistance of your resistor matches the impedance of the supply.

Imagine your power source as a perfect battery with a resistor in series with it. This resistor represents the impedance of your source. When you attach a load to this, the load is in series with the impedance.
For example, assume a 10 v supply with a 10 ohm internal resistance. If I attach a 1 ohm resistor across it, the total current will be 10v /(10+1) = 0.90909 amps,
Power used by the the load resistor would be 0.90909^2 x 1 =.826 w.
If I attach a 100 ohm resistor, the total current would be 10/(10+100) = 0.090909 amp and the power used by the load resistor would again be 0.826 w.
However, if I attach a 10 ohm resistor, the total current is 10/ (10+10) = 0.5 amp, and the power used by the load resistor is 2.5 w. More than either of the other resistors.

In case 1 the total power used by the system is 10^2/11 = 1100 w, but most of the power is dissipated by the source resistance.
In case 2 the total power used by the system is 10^2/110 = 0.90909 w, most of the power is dissipated by the load, but the total power used is low.
in case three, the total power is 10^2/20 =5 w. more total power than case 2 and half the power is dissipated by the load.

This impedance matching between source and load is an important factor in electronics. To get maximum sound out of a speaker, I want it's impedance to match the output impedance of my amplifier.
 
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  • #19
DaveC426913 said:
I was erroneously thinking the heat was produced by the resistance. It's not; it's produced by the current flow.
With the given voltage of the outlet, it is more like 'current set by the resistance', and 'heat produced by the current on the resistance'.
 
  • #20
Janus said:
When the resistance of your resistor matches the impedance of the supply.
While your post is certainly true, and a good discussion, I think by moving away from ideal sources you are likely to be just confusing the issue for a beginner like Dave, who is just trying to understand the relationship among voltage, current, resistance, and power at a basic level.
 
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  • #21
DaveC426913 said:
If I put household current across a variable resistor, going from zero to a bajillion ohms, where would it draw the most current? Where would it generate the most heat?
Maximum resistance corresponds to nothing being plugged into the wall. For the sake of your budget, you should be glad that corresponds to zero current, and zero $ on your monthly bill; thank goodness for that.

Janus said:
When the resistance of your resistor matches the impedance of the supply.

That's true when the source is a battery, but it should not be used for the power grid where there are many active devices that try to hold voltage within required limits.

Edison said:
"While this controversy raged in the scientific papers, and criticism and confusion seemed at its height, Edison and Upton discussed this question very thoroughly, and Edison declared he did not intend to build up a system of distribution in which the external resistance would be equal to the internal resistance.

He said he was just about going to do the opposite; he wanted a large external resistance and a low internal one. He said he wanted to sell the energy outside of the station and not waste it in the dynamo and conductors, where it brought no profits... In these later days, when these ideas of Edison are used as common property, and are applied in every modern system of distribution, it is astonishing to remember that when they were propounded they met with most vehement antagonism from the world at large." Edison, familiar with batteries in telegraphy, could not bring himself to believe that any substitute generator of electrical energy could be efficient that used up half its own possible output before doing an equal amount of outside work.
 
  • #22
DaveC426913 said:
If so, how does the giant resistor in the night light circuit not heat up as hot as the coils in the hairdryer circuit?
I had the same thought one day, and I nearly finished an EE BS degree!
I resolved it by knowing, of all things, what happens to power with a centrifugal pump at different throttle settings.
If the throttle valve is closed, no fluid will flow, and the effective power of the system will be zero. (In reality of course, the power consumed by the pump doesn't go to zero.)
With the throttle fully open, fluid flow is maximized, and hence, power is maximized.

I think it strikes us as odd, as this is contrary to what most people experience in real life.

Example: Driving your car really fast increases the wind resistance, and requires an increase in power. Resistance goes up, and power goes up.

With electricity, when voltage is held constant and resistance goes up, power goes down.

So there's some inherent difference in "electrical" and "our real world" concepts of "resistance".
Real world "resistance" would be pushing a bicycle up a hill, or pushing a lawnmower.

I'm afraid I'm not willing to describe how electricity actually works, nor try and describe "electrical resistance", as I've tried it at least three times, and no one seemed to like my explanations.
In fact, I went back and looked at my very first attempt, and it took me forever to figure out what the hell I was talking about.

I think it's mainly because "volts" and "ohms" are REALLY weird creatures, and as far as I can tell, have no real world analogies.
Even "amps" are kind of peculiar, IMHO.
Wiki has some nice equations for "volts" and "ohms" that are fun to look at, and scratch your head about.
For instance, volts = (potential energy)/(charge).
This makes absolutely no sense to me whatsoever. (Outside of electrical theory, of course.)

Ohms are just as bad. Possibly worse, as they have two definitions with "kilograms".

wiki.ohm.defs.png


Amps are kind of funny, as I can only analogize them to: "the number of people or things going past a point per unit time".
Watts are kind of OK. But people argue about power also, so they're not totally innocent. Watts, that is.
 

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  • #23
DaveC426913 said:
No. By 'big' I mean high resistance.
OK, so the coils are low resistance.
Maybe I'm misunderstanding where the heat comes from. I've been assuming high resistance = high heat output.

That would be true if you had a constant current source, rather than a constant voltage one.
With a constant voltage source, current varies according to the load by I=V/R. Power is I x V, so it can also be expressed as P = V2/R
Lower resistance lead to more power used by the load.

If you had a constant current source, it would supply a fixed current value (varying its voltage as the load changes). This gives us P=I2R, and power usage goes up with increased resistance of the load.
 
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  • #24
DaveC426913 said:
Let's try something different.

If I put household current across a variable resistor, going from zero to a bajillion ohms, where would it draw the most current? Where would it generate the most heat?
If you assume an ideal voltage source, then if you plot amperage (red line) and Power (blue line) against increasing resistance, then you get something like this:
power graph.png


With both amperage and power tending towards infinite as R decreases and tending to zero as R increases (resistance increases from left to right).
 

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  • #25
I don't know whether it will contribute to this discussion, but when I read Dave's early posts,this is what I thought.
hairdryer.png

I don't know if this diagram helps at all, but what kills me about PF is the lack of pictures. I always start with some sort of picture.
 

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  • #26
Merlin3189 said:
but what kills me about PF is the lack of pictures.
Probably because a Physics bent and an Artistic bent tend to be disjoint sets... then there are those Mathematicians hangin' around too! :wink:
(besides, pictures are a lot of work)
 
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  • #27
Yes, all that. I don't have much beyond Paint for drawing, since Windows left my versions of Corel and Autosketch behind. But that is often enough to sketch a simple diagram. Some people draw on paper and snap it with their phone, which is less simple than it sounds, but can be done by almost anyone.

In all seriousness I do think it is important though. Many posts require an early response of, " have you drawn a FBD? " and questions about all sorts of topics often become much clearer once a diagram is provided.

Along with requiring people to define their variables, instead of assuming everyone else knows what, for eg., x, s, d, or h stands for (distance), what direction and sense they have and what units are being used, I think requiring a diagram would be a useful addition to the HW template. The diagram with variables marked would catch two rabbits in one bush.
 
  • #28
DaveC426913 said:
OK, so if you ensmallified the resistor in the night light, it would allow more current to flow as well as turning that power into heat.
The word "resistivity" has only ben used once so far in this thread and it's at the centre of the business. For a given resistivity, the resistance of a uniform wire is
R = ρl/A
where ρ is the resistivity of the particular metal, l is the length and A is the cross sectional area. "Ensmallifying' a wire would, I imagine, mean scaling it down in all its dimensions but that means (because Area is proportional to the radius squared) the resistance will go UP and UP as size decreases. Metals differ greatly, from Silver, the lowest resistivity, copper (nearly as good and much cheaper for wiring) to some alloys (e.g. constantan) with very high resistivities.
If you decide you want a number of Watts from your resistor then the length and area would be chosen. But there's another thing - the temperature it will operate at. A bulb filament is made deliberately small (a low total surface area) so that it gets white hot (very visible) before it can get rid of the power. ) A 0.1W bulb is quite visible but you need tens of Watts to light a room. All the filaments will be at around the same temperature because the dimensions are chosen right. (Bright enough but not melting.)
Now take a 1500W hair dryer. You don't want it to glow white hot - just a dull red. You choose a long enough coil of wire to reach equilibrium when it's the wanted temperature. Block the air inlet with your hand fluff and the reduced air flow will allow the element to reach orange heat and burn the device up. (The cut out will normally prevent that).
DaveC426913 said:
I was erroneously thinking the heat was produced by the resistance. It's not; it's produced by the current flow.
Don't beat yourself up about it. As with most Engineering matters, there are two or more factors involved and we don't tend to talk in terms of 'agencies' in Science. You can obtain a given heat / light output with various combinations of current and resistance. But we tend to use a Voltage Source for our electricity supplies. I2R and V2/R will both equally well tell you the Power dissipated. You can have a 20W car bulb (12V) or a 20W household bulb (120V); you just need 100 times more resistance in the mains bulb.
It's a good exercise to try many different combinations of figures and different versions of the basic equations. Getting to terms with the simple maths is definitely worth while and can give you a good Intuition.
 
  • #29
Merlin3189 said:
Some people draw on paper and snap it with their phone, which is less simple than it sounds, but can be done by almost anyone.

I totally agree that more pictures would improve the quality of the discussion. But it is not so easy for many people. It also takes more time. Perhaps 20 seconds to type a 2 sentence reply, versus 30 minutes to create and post a graphic.

Pictures can bring their own problems.

https://www.physicsforums.com/threads/guidelines-for-students-and-helpers.686781/ said:
In addition to the typed version, you may want to attach a scan or electronic copy of the problem statement. Please do not do this in lieu of typing up the statement. Posting images, while convenient for you, can be troublesome for others, especially if they're trying to view the thread on a phone or tablet.
...
Don't post poor images.
When you do use an image in your post, make sure it's in focus, oriented the right way, well lit, etc. It seems like this should be obvious, but experience has shown that people frequently post incredibly poor images. Add images as attachments to the post. Don't host it externally.

So, we can encourage more graphics and better graphics but we can't require it.
 
  • #30
anorlunda said:
Pictures can bring their own problems.
So, we can encourage more graphics and better graphics but we can't require it.
I am very visual and also a graphic artist.

A long time ago, I asked if anyone on PF had any diagrams and stuff they wanted rendered - I would happily do them for free. I also tend to use illos a lot for my own contributions:

http://illustrator.davesbrain.ca/images/pic_parallax.gif
http://illustrator.davesbrain.ca/images/pic_roofing_greenhouse.gif
http://illustrator.davesbrain.ca/images/pic_trig.gif
http://illustrator.davesbrain.ca/images/pic_area_circle.gif
http://illustrator.davesbrain.ca/images/pic_lensfocus.gif
http://illustrator.davesbrain.ca/

If I could figure out a way of easily understanding what people wanted to see in diagrams they needed, I could live a full and happy life just whipping up diagrams like this.
pic_em-waves.gif
 

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  • #31
Merlin3189 said:
I don't have much beyond Paint for drawing, since Windows left my versions of Corel and Autosketch behind.
I'm running 64 bit windows 7 professional which let's you setup and run a virtual 16 bit XP mode machine. See; I can use my Autosketch program inside of windows 7.

autosketch in xpmode.jpg
 

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  • #32
Tom.G said:
Probably because a Physics bent and an Artistic bent tend to be disjoint sets... then there are those Mathematicians hangin' around too! :wink:
(besides, pictures are a lot of work)
Challenge accepted!
1 coil using outputting 1 Watt, total resistance 1 Ohm
watt1.png


Two coils, each outputting 1 watt, total output 2 Watts, total resistance 0.5 Ohms
watt2.png


Three coils outputting 1 Watt each. Total output 3 Watts. Total resistance 0.333... Ohms
watt3.png
 

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1. How does a hairdryer produce 1500 watts?

A hairdryer produces 1500 watts of power through the use of a heating element and a fan. The heating element converts electrical energy into heat, while the fan blows the heated air out of the dryer.

2. What is the purpose of the heating element in a hairdryer?

The heating element in a hairdryer is responsible for converting electrical energy into heat. This heat is what dries and styles the hair.

3. How does the fan in a hairdryer contribute to its power output?

The fan in a hairdryer plays a crucial role in the power output of the device. It blows the heated air out of the dryer at a high speed, increasing the airflow and therefore the drying power of the hairdryer.

4. Is 1500 watts the maximum power output for a hairdryer?

No, 1500 watts is not the maximum power output for a hairdryer. Some hairdryers can have a higher power output, up to 2000 watts, which allows for faster and more efficient drying of hair.

5. How does the wattage of a hairdryer affect its performance?

The wattage of a hairdryer directly affects its performance. The higher the wattage, the more powerful the hairdryer will be, resulting in faster and more efficient drying of hair. However, a higher wattage hairdryer may also consume more energy and be more expensive.

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