What is the power of an incandescent lamp at 200v 5A?

[LarryFine]

So where is the Op on this. I hate when questions are asked and the OP disappears.

Maybe he's a spy from another forum, and just wanted to see how far we'd take this. ()

 

[charlie b]

I bet most of us don't know what RMS stands for . . .

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[LarryFine]

I bet most of us don't know what RMS stands for . . .

Requires More Scrutiny.


(And, yes, I know. )

 

[realolman]

100802-1231 EST

realolman:

Yes you are wrong.

You can not just multiply Vrms and Irms and calculate load power from this. It may yield the correct result or it may not. You need more information. I discussed this in a previous post.

If you have a known pure stable resistance at the frequency of interest, then Irms^2 * R will give you the power. Same with voltage squared. But these fail if resistance is nonlinear.

For some circuits and waveforms you need to integrate the instantaneous product of voltage and current over some desired time and divide by that time period to get the average power over that period.

.

Here is a link to a wikipedia page where it seems to me that they are showing how to get the RMS values of non sinusoidal waveforms ... what would be the point of doing this?

What is the use of RMS?

 

[iwire]

So where is the Op on this. I hate when questions are asked and the OP disappears.

http://www.electriciantalk.com/f2/pf-correction-devices-again-12961/index2/#post257197

 

[LarryFine]

I was right!

 

[gar]

100802-2046 EST

realolman:

RMS measurements are very useful. In particular for an electrician, if you measure the RMS current in a wire and know the value of resistance in that wire, then you can calculate the power dissipated in the wire, which in turn is related to the temperature rise in the wire.

Consider a #12 copper wire. Its resistance is about 0.0016 ohms per foot. 20 A RMS thru this will produce about 0.64 W / ft. I have never measured a wire such as this to estimate the temperature rise.

Suppose you have a very peaked current waveform, such as from a switching power supply, then knowing the RMS current is important because peaked waveforms produce more heat than is predicted by the average current.

In the example I gave in a previous post of a rectangular pulse with a duty cycle of 1% the RMS current was 10% of the peak current. The average current is 1% of the peak value. So use of average current to calculate power would produce a gross understatement.

For a complex waveform represented by a mathematical equation one would use calculus to calculate the RMS value. For a waveform of arbitrary shape you can get a good estimate of the RMS value with a meter provided the peak to RMS ratio is not too large. This is called the crest factor. Another way is to --- determine the area under the curve that is the square of the actual current waveform, divide by the time period of the base of the area measured, then take the sq-root of this value. If you are interested I can describe how you might graphically do this. It is really the basis of integral calculus.

.

 

[realolman]

100802-2046 EST

realolman:

RMS measurements are very useful. In particular for an electrician, if you measure the RMS current in a wire and know the value of resistance in that wire, then you can calculate the power dissipated in the wire, which in turn is related to the temperature rise in the wire.

Consider a #12 copper wire. Its resistance is about 0.0016 ohms per foot. 20 A RMS thru this will produce about 0.64 W / ft. I have never measured a wire such as this to estimate the temperature rise.

Suppose you have a very peaked current waveform, such as from a switching power supply, then knowing the RMS current is important because peaked waveforms produce more heat than is predicted by the average current.

In the example I gave in a previous post of a rectangular pulse with a duty cycle of 1% the RMS current was 10% of the peak current. The average current is 1% of the peak value. So use of average current to calculate power would produce a gross understatement.

For a complex waveform represented by a mathematical equation one would use calculus to calculate the RMS value. For a waveform of arbitrary shape you can get a good estimate of the RMS value with a meter provided the peak to RMS ratio is not too large. This is called the crest factor. Another way is to --- determine the area under the curve that is the square of the actual current waveform, divide by the time period of the base of the area measured, then take the sq-root of this value. If you are interested I can describe how you might graphically do this. It is really the basis of integral calculus.

.

I am aware of all this.... At one point in my life I would have been able to do it, but not any more. I was barely able to do it then.

IMO the purpose of using the RMS value is to be able to use compatible values to make sensible calculations... like you mentioned about the heating effect in a wire.

The voltage and current are expressed in simple terms : 200 Vrms and 5 Arms,, and the object under consideration is an incandescent lamp;

I suppose there isn't enough information...

 

[mivey]

I answered ?A,? before I read any of the responses, and I got it wrong. The correct answer is ?C.?... So since we were not told the nature of the power source, that is a key piece of information that is missing. Therefore, the correct answer is ?C.?

Beaten down by the preponderance of over-thinking. :grin:

Think about it: If a client came up and said "If I have a regular bulb at 200 volts and 5 amps, how much power is that?" are you really going to go through the rest of this stuff or just give them the 1kW answer? A follow-up question might be: "Why do you ask?"

In the absence of other information, standard conditions would have to be assumed. Any additional information would really turn a simple question into a different question.

A trick question is not a trick question if there is no additional information given to address the trick. Without the information needed to solve the riddle, what would be the point? Maybe the trick was to see if someone would over-think it.

 

[dbuckley]

What is the use of RMS?

The use of RMS is that the RMS value of a repititve waveform is the equivalent to a DC value. So 5A RMS AC will have the same heating effect as 5A DC through the same load.

 

[iwire]

In the absence of other information, standard conditions would have to be assumed.

But some of us do know more about the question than what has been stated here in this thread and someone may have provided a link to that additional information. :grin:

You know what they say you can lead a horse ......

 

[mivey]

But some of us do know more about the question than what has been stated here in this thread and someone may have provided a link to that additional information. :grin:

You know what they say you can lead a horse ......

I guess I should go read it.

I was scared to go over to the dark side less my eyes be singed by the unbridled commentary. I'll wear my dark glasses. :grin:

 

[iwire]

I guess I should go read it.

I was scared to go over to the dark side less my eyes be singed by the unbridled commentary. I'll wear my dark glasses. :grin:

I usually go with these ....

 

[mivey]

I usually go with these ....

Thanks. I was able to print these out and fashion me some protective wear so I could read the link provided.

For other adventurers: you can just jump right in at post #85 (actually a little after that, but you would not want to miss the importance of the Carnot cycle in the discussion :roll::grin.

After reading the link, I am going to just have to call a technical foul on the play for dragging in bits and pieces of another thread. :grin:

 

[mivey]

Just a few questions to test the knowledge of the respondents so far (answers available on request):

1. What color is a green ground screw?______________________

2. Electricity will leak out of pipes if you do not use rain tight fittings. T F

3. To trip a circuit breaker, you must stick out your foot as it walks by. T F

4. When dealing with conduit, the O.D. must exceed the I.D. or the hole will be on the outside. T F

5. A keyless fixture cannot be unlocked. T F

6. A circuit breaker reads "20" on the handle. This means it can trip 20 times before it is worn out. T F

7. If you plug something 110 volts into a 120 volt outlet, you will have 10 volts leakage. T F

8. When hooking up a 220 volt device, you must use 220 volt wire. T F

9. The gauge number of a wire is the number of plugs the wire can handle. T F

10. A flush mount device may only be hooked up to a toilet. T F

11. An ohm is a Hindu measurement of voltage. T F

12. Electrical Inspectors are also know as: _____________________

13. Service Entrance Conductors direct from the supply house with current already in them are very dangerous. T F

14. High voltage wire is used on upper levels of tall buildings whereas low voltage wire is usually found in the basement or underground. T F

15. If you have a molded-case circuit breaker, the mold can be washed off with warm soapy water. T F

16. Three ECs enter an empty room and lock the only entrance: the door. All have on a tool-belt with a regular screwdriver, a pair of lineman's pliers, and a roll of Scotch-33. The light overhead is supplied by a sinusoidal voltage source and the light has a 0.935% power factor. Harmonics are not a factor, but the Carnot Cycle might be. The next morning, all three ECs leave the room and lock the door behind them. How many ECs are left in the room? __________

Skills Test:

S1. Guess your age: _____________________

S2. Connect the dots . .

S3. Drinking through your lips: Put liquid in a glass. Raise the glass to your mouth. Drink. Please indicate if you get any of the liquid on the back of your head. Y N

S4. Count your nose: Using your forefinger, count how many noses you have and write down your answer. ___________

 

[Electric-Light]

100801-1024 EST
Also if you operate at a sufficiently low frequency so that the bulb resistance is non-linear thru the cycle, then RMS volts * RMS current does not necessarily equal input power.

Hit the nail on the head there

 

[Dennis Alwon]

Just a few questions to test the knowledge of the respondents so far (answers available on request):

These were great...I enjoyed them tremendously. Thanks

 

[iwire]

16. Three ECs enter an empty room and lock the only entrance: the door. All have on a tool-belt with a regular screwdriver, a pair of lineman's pliers, and a roll of Scotch-33. The light overhead is supplied by a sinusoidal voltage source and the light has a 0.935% power factor. Harmonics are not a factor, but the Carnot Cycle might be. The next morning, all three ECs leave the room and lock the door behind them. How many ECs are left in the room? __________

:grin:

You might also enjoy the discussion of how a layer of paint on a cable requires the conductors be derated.


http://www.electriciantalk.com/f5/painted-into-corner-14828/index2/#post251147

 

[Electric-Light]

I said it was an incandescent lamp, not an ideal 40 ohm resistor that doesn't change resistance no matter what

Because the filament can have a hot to cold ratio well in excess of 10, it is not an ideal resistor and the current doesn't necessary change proportionately with the voltage given low enough sine wave frequency.

All I gave was that Vrms = 200v, Irms =5A, but the integration time could have been 5 hours and any thing could have happened during that period.

The only definite truth was left out, that Pavg <= 1kW

 

[mivey]

The only definite truth was left out, that Pavg <= 1kW

You forgot to bring over this piece of the truth from the other forum as well:

It's my job to complicate things.

:grin: