Watt's the problem??

[supergeek]

Ok, I know there's an easy explanation for this, hopefully someone can straighten me out....... Since P=IE, I'm thinking a 60 watt lamp on a 120 Volt circuit should draw about 0.5 Amps.
Going a step further, since E=IR, the resistance across the lamp should be roughly 240 Ohms. But, I took a new lamp out of the box and the resistance measures 16 Ohms. Is the resistance reading affected by heat or something?
I'd love to find out what I'm not understanding about this! Thanks.

 

[hmspe]

The resistance of the filament in an incandescent lamp increases as the filament heats up. This is normal.

 

[W6SJK]

Yes, resistance will vary. Resistance is futile. If you have an autotransformer type dimmer (not SCR) you could try varying the voltage.

I remember a high school experiment where we had to prove it and show a graph... Oh those were the days...

 

[al hildenbrand]

Geek,

Try something like this.

Put your 60 W lamp in a table luminaire, leave it unplugged but turned on, and take a resistance measurement at the prongs of the plug on the luminaire cord. The best lamp position for retaining heat is base down.

Plug the cord in and let the lamp burn for five minutes or so.

Unplug the cord and immediately take a resistance measurement at the plug prongs. Repeat the resistance measurement at the plug prongs every five seconds for the first minute, then every ten seconds for the next minute or so.

 

[vmtrob]

Ptc

Ptc

Yes this is known as a PTC-positive temperature coefficient. As the temperature increases so does the resistance

 

[jwelectric]

Might also be of help to the original post to explain that true wattage can only be obtained in a DC circuit.

 

[rattus]

Huh?

Huh?

Might also be of help to the original post to explain that true wattage can only be obtained in a DC circuit.

I don't believe that. Explain yourself please.

 

[jwelectric]

I don't believe that. Explain yourself please.

True ?power? or ?wattage? can only be obtained in an AC circuit when the voltage and current are both in exact phase with each other.

The filament of an incandescent light bulb has a twist to it thus giving a little inductance therefore the current and voltage is not in exact phase with each other.

There is never a time that the current and voltage is in exact phase with each other in the real world although we can make them look as though they are in a lab environment or math problem.

True or false?

 

[rattus]

True ?power? or ?wattage? can only be obtained in an AC circuit when the voltage and current are both in exact phase with each other.

The filament of an incandescent light bulb has a twist to it thus giving a little inductance therefore the current and voltage is not in exact phase with each other.

There is never a time that the current and voltage is in exact phase with each other in the real world although we can make them look as though they are in a lab environment or math problem.

True or false?

FALSE! That is what power factor is all about.

Pavg = Irms x Vrms x PF

Furthermore, the PF in this case is so close to unity that it is not a consideration.

 

[jwelectric]

Wait just a minute,

What you are talking about is ?close? not pure.

Furthermore, the PF in this case is so close to unity that it is not a consideration.

What I was talking about is pure or true.

Might also be of help to the original post to explain that true wattage can only be obtained in a DC circuit.

What you are doing is trying to do is clean up the wattage by

FALSE! That is what power factor is all about. Pavg = Irms x Vrms x PF

Now you explain what is meant by the term ?Pavg?
Does this mean ?power average??
If so then it can not be true or pure wattage or power can it?

I take the words pure or true to mean that it is exact not an average.

Yes I understand that in the world of engineering that something close count as exact and that the third place to the right of the decimal is dropped but in an exact world this could not be accepted.

When this method is used then the words true and pure are set aside and an average is accepted as the norm.

Again, true or false?

 

[LarryFine]

As we all know, AC power is that which has the same heating effect as that which the given power of DC would produce.

 

[rattus]

JW,

What do "true and pure" have to do with "average"? Incandescent lamps are rated in average power as are virtually all electrical loads. There is instantaneous power and peak power as well. There is no such thing as RMS power.

 

[kingpb]

Gotta go with Rattus on this one.

According to Stevenson, Elements of Power System Analysis, 1984, P, is designated as average power, or real power. The unit for instantaneous or average power is the watt.

For all that have it, refer to pg 16, para 2.

 

[jwelectric]

What we have is a person; let’s say the average electrician, asking this question;

Ok, I know there's an easy explanation for this, hopefully someone can straighten me out....... Since P=IE, I'm thinking a 60 watt lamp on a 120 Volt circuit should draw about 0.5 Amps.

Going a step further, since E=IR, the resistance across the lamp should be roughly 240 Ohms. But, I took a new lamp out of the box and the resistance measures 16 Ohms. Is the resistance reading affected by heat or something?
I'd love to find out what I'm not understanding about this! Thanks.

Then someone gives an answer such as this;

Might also be of help to the original post to explain that true wattage can only be obtained in a DC circuit.

And this is a true statement by the way

Now we are into the engineering aspects of trying to explain that Pavg = Irms x Vrms x PF

This average electrician is not going to be able to grasp those things that the 4 year college graduate is throwing at him until he first understands the laws that govern these things.

True power or true watts can only be found in a pure resistive DC circuit and this is one of the laws of electricity that can not be changed.

When we get into the AC circuits and we no longer have pure resistive circuits then the Power Factor comes into play.

A simple answer to the original post is that the reason that P=IXE does not work on the 120 volt 60 watt light bulb is that it is not a pure resistive DC circuit.

Once again I ask is this true or false?

Edited to add;
As pointed out by King
P, is designated as average power, or real power. The unit for instantaneous or average power is the watt.

 

[rattus]

JW, this is the ultimate in nit-picking. The original question has been answered already, and the fact that a few nanohenrys of inductance, and a few picofarads of capacitance may be found in any otherwise resistive circuit has no relevance. Any error introduced by a tiny amount of inductance in the wires is miniscule when compared to the error in the instrumentation itself.

The average electrician would only be confused with your argument.

 

[kingpb]

A simple answer to the original post is that the reason that P=IXE does not work on the 120 volt 60 watt light bulb is that it is not a pure resistive DC circuit.

Once again I ask is this true or false?

Your answer is incorrect, given the question. If you want to break it down into simple terms, then false!

Correct Answer: A cold light bulb filament has less resistance than a hot one. This is especially true for tungsten filaments. The fact that tungsten, like most metals, has less resistance when it is cool and more resistance when it is hot. The rating of the bulb (i.e. 60W) is for the maximum output, which would be the watts required at the maximum temperature of the filament, for the rated voltage applied.

 

[rattus]

Furthermore:

Furthermore:

JW,

If we believe your argument, then DC measurements could not be made either because the maximum load current would never be reached due to the exponential nature of the current. An infinite number of time constants would be required for the current to reach its final value.

The electrician is really confused now.

(With due respect to all electricians who don't need to know this stuff anyway)

Then there is the inductance of the d'arsonval movement in your ammeter which must also be considered. That makes that infinite time period even longer!

 

[jwelectric]

The average electrician would only be confused with your argument.

With this I would have to beg to differ. I work with electricians and inspectors on a daily basis and with the confusion of Ohm?s Law and AC circuits as outlined in the original post.

The average electrician does not have the schooling for phasor analysis nor even something as simple as a 60 watt 120 volt light bulb that has a resistance of 16 ohms although Ohm?s law says it should have a resistance of 240 ohms.

Your answer is incorrect, given the question. If you want to break it down into simple terms, then false!

Yes I agree that when the electrical energy is changed into heat it will have an affect on the resistance of the bulb but that doesn?t change the basis law that true wattage can only be found in a pure resistive DC circuit.

The only point that I was trying to make with my original post

Might also be of help to the original post to explain that true wattage can only be obtained in a DC circuit.

was the simple formula of P=IxE and E=IxR will not get him the answer that he is seeking.

Then Rattus asked me to explain my answer and here we are debating whether or not True Power can be found in anything other than a pure resistive DC circuit.

The simple truth is that True Power can only be accomplished when the voltage and current is in phase with each other. This can not be accomplished in an Alternating Current circuit in the real world.

Now you all can engineer this any way that you like and quote all the text that you can find and every time you will come up with an Apparent Power and the NEC call this Volt-Amperes (electrician?s language).

Now I will not highjack this thread and turn it into a debate of True Power and Average Power. I have said all that I have to say about the matter.

 

[al hildenbrand]

The simple truth is that True Power can only be accomplished when the voltage and current is in phase with each other. This can not be accomplished in an Alternating Current circuit in the real world.

This is like my coffee and coffee cup. Both are mostly comprised of the subatomic void. . .there is darn little "matter" there, for my hand to grab in the morning.

But to my caffeine stimulated brain, the whole concept of subatomic void is, . . .interesting. . ., but my Newtonian experience of the dark, bitter, hot liquid is concrete and real and tactile.

 

[rattus]

Poor poco!

Poor poco!

We ought to pity the poor pocos because, according to JWelectric, they cannot correctly measure the energy they sell to us.

Pity also the thousands of poor engineers and techs who suffered though AC power courses and labs only to find that the old grey-haired profs were lying to us!