Is a Watt a Watt?

The power is the same. The loads releasing this identical power are different.

Thanks! So a 20 watt bulb would see no difference?

What can differ is the lumens per watt, or brightness....then there's the temperature of the light...

http://www.seesmartled.com/kb/choosing_color_temperature/

...then you have radiance and illuminance...

http://www.theledlight.com/lumens.html

So why is it the 5700K bulbs are described as "cool", while the much lower temperature, 2700K are considered "warm"? Confuses the heck out of me every time I go to buy an LED light bulb.

In interior lighting and in photography, strangely enough, hotter color temperature light is more "blue" and is considered "cool" whereas lower color temperature light is typically more rich in red tone, and is considered "warm". Don't ask me how that got started, it is just what I remember from hanging around with a brother-in-law who just happended to be a great professional photographer in industrial photography, advertising stills, etc.

So, basically, the non-scientific types set up the "cool" / "warm" thing based on mood, not reality. I don't trust the various manufacturers to get it right (wrong), so I still check the K rating for the color I want - not the description. Unfortunately, I explained the reality versus description thing to my wife, who has a nursing degree, and now I have to verify which bulb it is we need every time. She's totally confused. She never met Kelvin, and doesn't care to.

Yes, Lord Kelvin would probably not be the central topic in Chemistry for Nurses 101. If you think that is bad, my wife is a penny-pinching retired LMSW, and not only do I have to pay attention to the wattage bulb, but also need to watch the color rating, and how the bulb is feeling about being purchased, and also if the bulb has plans for post-release, i.e. living quarters, income level, etc.

Tell your wife if a patient comes in with a temperature of 310.15 °K, to please not panic, as that is precisely 37 °C and is also 98.6 °F. I hope I don't also get you in trouble with my suggestions.

Won't even try to tell her - My wife just says about all this - "Why do we need three temperature scales? Isn't one enough?" I doubt she'll ever past Fahrenheit.

What you are looking at is the clash of two cultures, artists and engineers.

When looking at the color palette, reds and oranges are considered 'warm' colors, blues and greens are considered 'cool' colors, and yellows and purples are considered 'neutral' colors. When describing lights, a 'warm glow' is always used to describe the reds and orange-yellows of a campfire, while a 'cold glare' would describe the blueish-tinted illumination from a flourescent(1) bulb or the bluish-white light reflected off a snowdrift. These meanings are pure emotion, which is why 'warm' lighting relaxes us (our inner caveman feels like it is relaxing by the campfire) and 'cool' lighting puts us on edge (and why we generally dislike the 'cold, sterile' environment of an examination room). When engineers began describing light, they described it using a tungsten fillament as the reference, naming the light after the heat of the fillament, which is why it is referred to as 'color temperature,' ANY piece of tungstin, in fact any material at all, will glow at the the same color when raised to the same temperature(2).

So because of that culture clash, we have the appearant paradox that 'cooler' temperature's produce 'warmer' colors.

Notes:

1) Yes, I have trouble spelling that word, no, I'm not going to keep fixing it.

2) Assuming the material can reach that temperature without self-destructing.

I agree with your (1) - tough bugger isn't it?

Basically as I said - non-scientific set the rules and they are pretty much a "mood" thing.

Yes, it would change!

The formula for power is P = V • I

P = Power in Watts
V = Voltage in Volts
I = Current in Amps

If you take a 60 Watt light bulb and apply the rated voltage of 120 VAC it will draw I=W/V or 60W/120V = 0.5 Amps, which is 60 Watts of power.

If you drop the Voltage to 60 Volts into that same bulb it will draw less current and produce less power.

We can solve that question with another famous equation, Ohm's Law: I = V/R, where R = Resistance in Ohms.

If the first case of 60 Watts from 120 Volts is true, then the resistance is R = V/I = 120VAC/.5A = 240Ω (Ohms).

Assuming the resistance in our ideal light bulb does not change (it really does, but for simplicity we say it doesn't), then at 60 VAC applied to our 240 Ω bulb we have 60VAC/240Ω = .25 Amps.

Now we can easily calculate the actual power produced from the bulb as: P = 60VAC • .25A = 15 Watts.

Notice the power (in Watts) is not the same at.

In incandescent lighting, wattage is power usage, not power produced.

No, if you use the same load with different voltages then you produce different power outputs. If you use different loads with different voltages you can produce the same power output. This assumes that the loads are linear. There are very cleverly designed non-linear loads that will produce the same power with different voltages. A simple filament based light bulb is not such a clever device. A 20 watt LED or florescent bulb might be a clever non-linear device.

There's no such thing as "a" 20-watt bulb. A 10 volt 20 watt bulb has higher resistance than a 2 volt 20 watt bulb. It's not hard to work out the resistance of each, from the formulas others have posted.

Well, yes, I see your point, but when it is stamped right on the damned thing, sort of hard to identify it as other than a "20W bulb". You and I may know the difference, but not everyone even needs to know the difference. Yes, if you want a resistor to go in series with a low input impedance AC induction single phase shaded pole motor, you cannot substitute a light bulb for the up potential load, as you might may never draw enough current to spin the motor, however if you eliminate this part of the circuit, you just trip breakers. Therefore, do not remove the motor from 1500 W heater it was built into. It is not designed to operate outside that defined circuit.

That is part of the fun: declare what the circuit is supposed to be and the produce it, and test to see if it operates as declared.

I'm not sure I see yours ! Where do the motor and heater come from?

I was just replying to OP's #4. But bulbs are (or should be) stamped with the voltage as well as wattage. If the user doesn't give it the right volts it won't give the right watts.

My stupidity lesson came from taking apart a space heater, and attempting to scavenge the fan motor from the circuit by wiring it direct to attemperate a testing enclosure I needed. After tripping the circuit breaker on the wall I couple of times, I next tried various light bulbs, but alas none were highest enough power rating to provide the needed current for that motor, as it only has about 10 loops on the core.

Just because it is a motor, does not mean it is or was ever intended to be directly coupled to a power source as its own independent circuit.

P = V * A , or

P = A * V

It's not complicated.

Yes a power output of 20 watts is the same, regardless of how you arrived at that. And yes, what you can do with that 20 watts depends upon the application- the voltage has to be suitable for the desired usage .

No, a watt is an HP, Joule/sec, Newton meter per second

Those are all units of power but there is an important scaling factor difference.

It is just like the difference between being paid two hundred dollars, two hundred pennies, two hundred yen and two hundred lire. They are all units of money but they all have different values.

I know how you got your ihs name. Coz your at the first level of defense, Lol.

2 out of 3 correct. Must try harder!

May be next time perfect, when i get to reflect on receptacles and orifices hahaha;)

20 watts is 20 watts however produced. If 20 watts is turned to heat with 100 % efficiency then the work done to produce that heat is 20 watts. The work done by 2 amps at 10 volts will be the same as the work done by 20 amps at one volt, the time taken will be different. Careful when you pass 20 amps at one volt through a conductor. Remember W= I squared R so resistance will be 05 ohms at one volt. You will have sparks. At 2 amps 10 volts the resistance will be .5 ohms. The work that both sources can do will be the same. If the supply is AC then one must use RMS values.

(Taking the question seriously) An analogy using water might be revealing. Imagine lifting a gallon of water through 10 feet every minute - that's a fixed work rate (equivalent to Watts). You might use a small pipe with high flow velocity or a big pipe with low flow velocity. They both achieve the same work rate but have different characteristics which have to be chosen to suit the equipment used.

good comparison

In the steady state, your descriptions produce identical results. However, there are transients involved in turning loads on and off. These important classes of loads (such as welding) are game changers as they involve dynamic loads and impedance matching. These classes of loads follow different rules, and not necessarily ohms law.

For the unwary novice perhaps yes '...These classes of loads follow different rules, and not necessarily ohms law..'

but for the purist, every electrical circuit obeys ohms law - there are no exceptions.

The power capacity for both will be the same 20 watts but not interchangeable!

The battery that can supply the 2 volts-10 Amps will be a Different type as compared to the 10 volts -2 Amps battery...

The load to be connected will be your application's limiting factor...

No the bulb will not light at 2 volts and 10 amps if it is designed to work at 10 volts and 2 amps. If it does light it will be a very dim glow. To understand power (watts) you need to understand OHMS Law first. Think of a pipe and water running through it. The larger the pipe, the more water can run through it. The water is the current (I in ohms law) The pressure or force that is pushing the water through the pipe is representative of Voltage (E or Electromotive force) in Ohms law. If you convert say a 4 inch diameter pipe into a 1 inch diameter pipe, there is less water flowing. This restriction becomes the resistance in Ohms law. This is the explanation I use when instructing people desiring to become amateur (ham) radio operators.

I used to have a cute cartoon that pictured the relationship, but unfortunately can't find it right now.

Therefore E=I x R if E equals voltage

I = current

R= Resistance

A very good analysis of a low voltage filament light bulb. Kudos.

However, it appears from the nature of the question from the OP that Ohm's Law is a new concept. Your "spot on" analysis is so far over their head they probably cannot see the landing gear, let alone buy a ticket for this plane ride.

Hahaha, there is actually 3 idiots giving you GA in here. You seemingly sounds right though. The worst thing is you dont start right, clueless- Dearth Vader How is your Voltage and current dictate your resistance on the circuit. Ohms law is a reversible equation, however in reality. It can not be, to any system I follows with mostly on circuit load or resistance and supplied voltage. V could be anytime equals to I*R but, reality is current is a follower to Resistance. At your tungsten filament anology, isothermal process, using 10V and 5 Volts, R is fixed on what ever voltage you might have as long as its wattage is good it will work. There are cases were you have a current supply though how you may force to supply certain current on the circuit, system resistance is not dependent on it. Resistance is intrinsic in the material or any system resisting current flow. How do you define resistance then?

No, a fixed resistance impedance is actually a very rare device. In fact I do not know of any fixed resistance devices at all. Take the simple 470 ohm 1/4 watt resistor as an example. In many applications this component will vary its resistance such a small amount with nominal operating temperatures that many would consider this a permanent fixed value device. However, apply the common industrial controls voltage of 24 V across this component and the resistance value will change quickly.

I will leave it to the student to figure out the simple mathematics why.

PS. It is wiser to try and figure out why three people felt an answer merited a GA tag than to call them idiots because you do not understand.

Geez... You guys really know how to take a simple question and turn it into a major event.

The OP has probably buried his head in the sand by now.

The question had nothing to do with lamp filaments or any other type of load.

The simple answer to the actual question is that in either example given, the same wattage will apply and it will do the same amount of work.

It may not achieve the same result in any specific load, but that is another exercise.

Where have I heard this before?

To our original poster, as you read through all of these you can easily see the difference in the amount of actual work electricity can do as illustrated in words. Here you will find examples of voltage, amperage, and of course resistance. ;-)!

I think it would be fair to say that we have expended a considerable number of watt-hours on this by now.