# How does a hairdryer produce 1500 watts?

#### DaveC426913

Gold Member
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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#### berkeman

Mentor
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? • russ_watters and CWatters

#### gneill

Mentor
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.

• Klystron

#### DaveC426913

Gold Member
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? Calculating without understanding won't do me much good .

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?

#### CWatters

Science Advisor
Homework Helper
Gold Member
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.

• krater

#### DaveC426913

Gold Member
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.

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 cant be true, as the short circuit example demonstrates the opposite.)

• Windadct

#### berkeman

Mentor
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.
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
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... #### DaveC426913

Gold Member
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.

#### berkeman

Mentor
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.

#### phinds

Gold Member
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.

#### DaveC426913

Gold Member
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.

• Windadct

#### phinds

Gold Member
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.

#### DaveC426913

Gold Member
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... #### DaveC426913

Gold Member
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?

#### phinds

Gold Member
Then I guess I'm back where I started... 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.

#### phinds

Gold Member
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

#### gneill

Mentor
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.

#### Janus

Staff Emeritus
Science Advisor
Gold Member
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.

• Klystron, Asymptotic and davenn

#### Rive

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'.

#### phinds

Gold Member
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.

• Tom.G and Rive

#### anorlunda

Mentor
Gold Member
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.

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.

#### OmCheeto

Gold Member
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". 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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#### Janus

Staff Emeritus
Science Advisor
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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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#### Janus

Staff Emeritus
Science Advisor
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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: 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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#### Merlin3189

Homework Helper
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I don't know whether it will contribute to this discussion, but when I read Dave's early posts,this is what I thought. 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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