Watt's the problem??

[supergeek]

Thanks for your help and explanations

Thanks for your help and explanations

Thanks, everyone, for all the thought and effort you've put in to answer my original question. I appreciate all the explanations, even the ones that are WAY over my head. I'm perfectly happy knowing that the basic answer is that resistance increases with heat. The other theories are great, too, though and I'm glad to be presented with these various viewpoints. Thanks again. Later, I'll be in my Supergeek lab plugging in the light and reading the resistance at various intervals. It's an exciting life I lead, let me tell you!

 

[supergeek]

By the way....

By the way....

What's a Poco anyway?

 

[LarryFine]

What's a Poco anyway?

POwer COmpany.

There are already too many uses for PC.

 

[al hildenbrand]

It's an exciting life I lead, let me tell you!

Got a good laugh at that one!

You're in the right crowd.

 

[jwelectric]

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!

Now Rattus why do you want to be like this for?

Is not part of the job of an electrical engineer to figure out some type of filtering system that will improve the Power Factor of an Industrial plant so they will have a cheaper utility bill?

Why would all these capacitor banks be necessary if the power company could just figure the bill using the formula P = I x E?

In the original question this is the formula that was asked about and then why E = I x R didn?t work out to the proper numbers.
After three or four people had posted good answers I suggested the idea of also explaining that in the case of the 120 volt 60 watt light as illustrated in the original post that the wattage, resistance and amperage would not come out to match the math in Ohm?s Law as asked about which is, P = I x E and E = I x R, due to the fact that these formulas would only work in a pure resistive circuit where the current and voltage are in phase with each other.

You then asked me to explain myself and I did so using a first semester electronics text book (Delmar?s Standard Textbook of Electricity Third Edition).

Then you come back saying that my statement was False and posting a formula that was far past the basic Ohm?s Law formula that was posted by the original poster. I believe that you posted this formula, Pavg = Irms x Vrms x PF

Now if I understand what you are saying with your formula is that the Average Wattage is equal to the RMS of the Amperage times the RMS of the voltage times the Power Factor.
Why did you add all this other stuff to your formula instead of just using P = I x E?
Was this formula used to clean up the circuit a little bit?
Will this formula in any way come to the numbers of E = I x R as outlined in the original post?

This sounded like it might be just a little more in depth than what the original poster asked about. In all honesty I must say that I don?t know how much knowledge that the original poster has about this type of math, but I do know that I deal with electricians and inspectors on a daily bases that don?t understand the difference between a Delta and Wye transformer nor what is happening in there that makes the primary and secondary have different voltages. Some have never done a basic Ohm?s Law problem in their life.

To say that what I was saying about True Power is false I do believe missed the boat just a little. Go back to your first semester electronics class and rethink my statement.

I am not saying that anything you are saying is wrong and I am saying now that your formula is the correct formula to use in order to find the numbers for the bulb.
What I am saying is that using P = I x E and E = I x R and coming to the 240 ohms for the bulb, this exact number will never be found testing the bulb hot, warm or cold. I don?t think that it will even come close. I will find out tomorrow and use several testers to get an average.
Why? The simple answer is that it is not pure resistive and heat produced by the electrical energy will affect the resistance of the bulb.

A thought that I just had was about a discussion we were having a few months ago about power and energy. After a couple of days of discussion you put in the simple form that I could understand and saved my butt from failing that semester. Let me see if I remember what you said. Power is the conversion of electrical energy into some other form of energy such as heat or mechanical energy.
How did I catch on? You kept it simple instead of jumping into a lot of math as my instructor was doing. You used one paragraph that I read in less than a minute and explained something to me that he hadn?t done in four weeks of class work.
This is all I was trying to say to the original poster was a simple answer. In order for the numbers that was reached (120v, .5amps and 240 ohms) this would have to be a pure resistive circuit that has the voltage and current in exact phase with each other otherwise these numbers will never match.

I?m sorry if I said or implied anything that was offensive to you in any way. I was just looking at this as first semester entry level explanation of a simple question ask by someone that only had a couple of post by his name and listed his occupation as electrician. I took his question to be, ?Why don?t the numbers match??
I didn?t see where he was asking how to make them match so I must apologize for this mistake I must have made.

Hope you have a good night.

 

[rattus]

Sorry for the sarcasm Mike, but I am just trying to emphasize that your statement that DC is the only way to measure power accurately is incorrect. In the real world, the PF of an incandescent lamp circuit is so close to unity that we can ignore it. Now at higher and higher frequencies, stray reactances will become high enough that they must be considered.

In general, we can measure or calculate PF and compute the power quite accurately--real, reactive, and apparent.

I was introduced to PF 58 years ago while working for Utility Rate & Engineering Co. I had to take it on faith because I did not have the background to understand it. Our hard working electrician friends may have to take it on faith too. At least they should know the formula even if they rarely if ever have any need to use it.

 

[LarryFine]

In the real world, the PF of an incandescent lamp circuit is so close to unity that we can ignore it. Now at higher and higher frequencies, stray reactances will become high enough that they must be considered.

True as that may be, since we deal with 60 Hz, it would be prudent to stick to 60 Hz reactances, even when discussing theory. We raraly bring reactances into discussion for any wiring outside of loads themselves for the same reason. It might be true, but it's not relevant. (IMHO)