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

[K8MHZ]

Old Man, I am with you and probably older.

I contend that by stating RMS values, a constant load resistance is implied, and it is incorrect to apply RMS values in a non-linear problem.

Also, the way the question is stated, 200V appears across the filament which rules out any series components.

Now, it is conceivable that the frequency is high enough that the stray inducance of the filament and its connecting wires reduce the applied voltage, although that would be a special case.

In other words, it is a trick question, and the OP can come up with any scenario he wishes to make "1kw" wrong.

RMS can mean different things. In electricity it can mean the DC heating equivalent of an AC source. Mathematically it means root-mean-square and depending on the wave form it may or may not be equivalent to the DC heating value as we assume a balanced unvarying wave form and the value sparkys use is a measured value with calculations to support the measurement. Cart before horse, but it works for 99.99 percent of what we do.

So, the info has been omitted is what RMS means. Does it represent a heating equivalent or a mathematical value? If they were invariably one and the same, this thread would not exist.

As far as a trick question, perhaps. It is one that can provide some enlightenment, if you pardon the pun. It sheds some light (sorry, couldn't resist) on how our trade standards may have some exceptions, exceptions that since are very rare, may not be realized by the average sparky. But at some point we may encounter a situation like Gar has described where PAVE does not equal VRMS times IRMS and a little voice should tell us to look for non standard or imbalanced wave forms and unstable loads.

This should also teach us that an incandescent bulb is not stable and it's stability decreases as frequency decreases. Connect a watt hour meter to a bulb along with a volt meter and an ammeter. Turn the switch on and off for a half second with a 50 percent duty cycle (1HZ superimposed with 60 for half a duty cycle) and see if Pav x T = Irms x Vrms x T. Any REMF coming from the rise and collapse of the magnetic field in the coil of the bulb surely would affect the readings at 1Hz making what was once a nearly pure resistive load into an inductive and resistive load.

 

[gar]

100827-1829 EST

I will work backwards thru the last three posts.

rattus:

I agree with you that it was probably intended as sort of a trick question. On the other hand it is still a useful teaching question.

I contend that by stating RMS values, a constant load resistance is implied, and it is incorrect to apply RMS values in a non-linear problem.

There is nothing in the basic definition of an RMS measurement that implies anything relative to a load or what is measured. RMS is simply a mathematically definable function.

The question was a general question with no additional restrictions. This means you have to assume and define some restrictions to qualify your answer to 1 or 2, or answer not enough information.


Larry:

But, at 60Hz, is the filament tempreature going to fluctuate enought twice per cycle to be different than it is on DC?

There is relatively little modulation of the filament temperature of a 100 W 120 V bulb at 60 Hz and therefore answer 1 is a good answer.

Or, are you saying, because we weren't given the frequency in the OP, the answer is "Depends on the frequency." ?

This is the answer the author of the original post gave. And it is certainly true that the result depends upon frequency. But there is at least one other factor that can influence the resistance of the lamp. This is the amount external incident radiation imposed on the lamp filament.

To prove that a statement is true all possible inputs to the statement must produce a correct output of the statement. To disprove a statement all I need is one example of inputs that cause the statement to fail.

For example: design a two input binary circuit such that any input combination produces a true output. Define binary 1 as true and 0 as false. Does a two input NAND gate satisfy the design requirement? Does a two input OR or a two input AND? No to all of these. See if you can design a moderately simple circuit to accomplish the desired result. Not acceptable as an answer is simply forcing the output to 1.


realolman:

Your 1: It would be reasonable to calculate 1,000 W for the answer, and there would be little suspicion that it was a trick question if some other answers provided were like 1414. 1732, 707, and 636.

Your 2: It would be hard to calculate the RMS current from a known sine wave voltage and the static current vs voltage curve of the lamp. Considerable work would be required to develop an approximate equation that could be applied over a range of frequencies and voltages. Probably would in part be based on tables of values. Measurement would be the most practical method.

Your 3.

If the rms value of the current through the incandescent lamp were correctly calculated, why could it not be used to calculate power dissipated.... as it would be the equivelent of a dc power source

Because the resistance is not constant thru the cycle.


Your 4.

If the values given to pose the OP question are incorrect, what the heck is the point of the question?
The question is not," Will a meter correctly display the rms current of an incandescent lamp powered by a 0.1 Hz power supply?' The rms values are given... supposedly correctly.

The values given in the original question were not incorrect. They were simply two values stated in the problem, therefore, were correct for the problem by design.

The purpose of the question is to see what your answer would be given a specific RMS voltage and a specific RMS current that are associated with an incandescent lamp. The values in the question were clearly hypothetical and the author knew that if the frequency was something like 1/10 Hz that the actual power dissipation in the lamp would not be 1000 W. But the author knows that most readers of the question are unaware of the large change in resistance of a tungsten filament lamp as the voltage changes. Furthermore, the author knows that few readers have any clear idea of what an RMS measurement is and the limitations of its use. Thus, it appears the author created the question to prove his point. I am not the author, and therefore, these are my assumptions of the reason. Look at the author's answer to the question, and his various posts.

.

 

[realolman]

100827-1829 EST

To prove that a statement is true all possible inputs to the statement must produce a correct output of the statement. To disprove a statement all I need is one example of inputs that cause the statement to fail.
.

There is no reason this needs to be calculated using all possible inputs There is nothing in the question requiring the solution of ALL possible inputs... only the one for which the information was provided.

100827-1829 EST

Your 2: It would be hard to calculate the RMS current from a known sine wave voltage and the static current vs voltage curve of the lamp. Considerable work would be required to develop an approximate equation that could be applied over a range of frequencies and voltages. Probably would in part be based on tables of values. Measurement would be the most practical method.
.

Again, there is no requirement to solve this for a range of frequencies and voltages, and by virtue of the fact that the reader is given the RMS values... provided by the "author", the reader has all the facts he needs.

So, It's OK to consider the impracticality of solving the simple problem of powering an incandescent lamp by an unlimited number of frequencies and waveforms, but suddenly it's too impractical to calculate the rms values, but we must rely on the inability of a practical instrument to measure it???

100827-1829 EST

Your 3.
Because the resistance is not constant thru the cycle.
.

True... the waveform of the current from which the the rms value is calculated, measured, or otherwise discerned.will reflect that fact.

100827-1829 EST
Your 4.
The values given in the original question were not incorrect. They were simply two values stated in the problem, therefore, were correct for the problem by design.

.

Exactly

100827-1829 EST
The purpose of the question is to see what your answer would be given a specific RMS voltage and a specific RMS current that are associated with an incandescent lamp. The values in the question were clearly hypothetical and the author knew that if the frequency was something like 1/10 Hz that the actual power dissipation in the lamp would not be 1000 W. But the author knows that most readers of the question are unaware of the large change in resistance of a tungsten filament lamp as the voltage changes. Furthermore, the author knows that few readers have any clear idea of what an RMS measurement is and the limitations of its use. Thus, it appears the author created the question to prove his point. I am not the author, and therefore, these are my assumptions of the reason. Look at the author's answer to the question, and his various posts.

.

I don't think the values in the question were clearly hypothetical at all ... they were given as the rms values associated with the voltages and currents of a powered incandescent lamp, the subject of the OP, that the reader has no reason to believe were not correct... regardless of the frequency of the power source.

I think the question was "authored" by some wiseguy who thinks he's smart, but would have been much more instructive had he discussed the varying resistance of the filament due to heat in response to the slowly varying voltage of an alternating current low frequency source.

The fact that it was such an extremely simplistic question gave rise to the suspicion that the obvious answer must somehow be wrong , but the additional fact that he gave the relevant values in a form that indicated he had already taken the waveform and frequency into account, made his little question not nearly as smart as he would have liked to think it was.

What's 10+12?

No, that's not right because I was expressing them in hexadecimal... or maybe it was octal ...we better check em all...... yuk yuk yuk ...yer so dumb, an I'm so smart.... Huh, Huh, Huh

I'm done with this stupid question

 

[rattus]

Yes, but:

Yes, but:

gar, that is all well and good, but in our trade, RMS and effective are virtually synonymous. since the same formula defines each. But, apply RMS values to a non-linear load and they are no longer "effective". Worse than tricky, the question is sneaky.

 

[gpedens]

what is it

what is it

Realolman , see my answers below . Unfortunately in the real world the correct answer is always "depends" as my old buddy Sam Huertze says. But fortunately in a steady state situation it is kind of like a joke my first calculus prof told me. A good ole boy engineer and a mathmatican were told there was a sweet young thing at the end of the hall willing to do anything they wanted when they got there. The only catch was they could only move half the distance toward her every 6 seconds. The mathmatician said that was impossible and would take to infinity to get to touch her. Technically the mathmatician's answer is the correct and 100% right answer. The engineer pondered for a few seconds, pulled out his slide rule, pencil, and burger wrapper, and said no problem, in about 12 minutes I will be close enough.

1. Wouldn't the frequency, wave form, capacitance, inductance, and whatever else, already have to be known in order to get the rms values?
No. You are measuring the RMS value of a voltage which is kind of like a DC volts average or equivalent. You dont have to know the load characteristics. The RMS value of a sinusoidal 60 hz or 1 gigahertz frequency with a peak voltage of about 170 vac would be about 120 vrms. Other wave forms would be different. You dont need to know if it was capacitive, inductive, resistive, etc. The voltage is measured independently of the current and vice versa. In a way the meter doesnt need to know the frequency but it would have to be designed for all frequencys and wave forms and that just aint possible. As long as the meter is designed to respond to the frequency and wave form the resultant rms value will be ok. Thats why they give frequency ranges and accuracy specifications. At high frequencys the length of the leads and distance apart can cause false readings due to capacitive and inductive effects. The leads may "drop" a lot of the voltage such that only a small fraction of the true voltage gets to the measuring circuit. Thats why some computer cables can be no longer than about 30 feet. You put X watts in at the front and zero gets to the end. The means or way the meter uses to calculate the "true rms" will affect the results. RMS values involve integrating the function over one cycle of the wave form. Which involves a different time for each frequency. A half cycle works for symetrical wave forms.
2. How would the rms values be known without knowing all the relevent information?
You are measuring a voltage or current independently of each other with your meter or at least that is implied by stating 200 vrms and 5 arms. The wave form and frequency do matter to what the meter can respond to and the means used to derive the rms voltage or current by the meter. As long as the meter can respond properly, it will give a true rms value for the voltage or current. The voltage and amps must be measured relative to each other in time to know the power. To know this usualy involves measuring when they reach peak values. Which brings up a phase angle meter or o-scope.

3. If the frequency, wave form, and whatever else are already known, why does there need to be more information?
True power or watts depends on knowing if the load is capacitive, inductive, resistive, or a combination. In the real world there is always a little capacitance, inductance, and resistance, even in a light bulb or a straight piece of wire. Most folks think no load equals no current. And that would be moslty true in low voltage circuits. But that all changes when its 525000 volts and 100 miles of line. A few amps of no load curent at 525000 volts and 60 hertz will create a ten foot arc that sounds like the air is being ripped apart. The wire may have only a fraction of a microhenry or picofarad per foot, but over a hundred miles at a high voltage, it is awesome. At 220 volts and a few hundred feet of house wiring it is barely noticible. The higher the frequency the more the inductive and capacitive components will affect the volts and amps. True power or watts is volts times amps times a factor of the angle between them. In a sinusoidal system at one frequency----- watts = Vrms x Arms x cosine of angle between V and A. If you connect a capacitor to an ac voltage you will measure an rms voltage and an rms current but the watts will be zero. Same with an inductor, the watts will be zero. With a wave form the power depends on the direction the voltage and current are moving in time. Kind of like mules pulling a plow or a game of tug of war. The voltage and current have to be in the same direction at the same time to make power. in a capacitor or inductor they are 90 deg apart. When ever one is giving the other is taking and the net result is zero. In a resistor the current and voltage are always in the same direction and that way they can do work or power. Capacitors and inductors pull current if a voltage is applied but they do not consume power in the good ole boys world. In the mathmaticians world there will be a little power used in the resistance of the leads and internals of the capacitor or inductor.

Now back to the given question of 200 vrms and 5 arms. Given no angle between the the answer is 200 x 5 = 1000 VA or 1 KVA. This is called apparent power or Q. Given only volts and amps rms one can only know the apparent power or Q. To find watts, real power ,or P, at least the angle between the volts and current is needed. Knowing frequencys and wave forms and meter characteristics are needed to give a more acutate answer.

Now in defense of the good ole boy we are also given that the load is an incandesant lamp. Assuming most normal conditions of wanting a lamp on this earth, the frequency involved is 25, 50, or 60 cycles mostly sinusoidal wave. As such it will be almost purely resistive. Which implies that the phase angle between voltage and current is close enough to the sweet young thing above. So in that case 200 vrms x 5 arms is 1000 VA, 1 KVA or 999.999999999 W or 1.00000000 KW.

That also brings up another practical problem. We are given that it is "an" incandesent lamp. Which implies one lamp. While they make them, they are typically expensive and very short lived. Will cost you about a dollar an hour bulb cost plus electricity usage and labor to replace. Due to tungsten characteristics and filament construction, incandesent bulbs more than about 200 or 300 watts arent very cost effective. A 1000 watt halogen lamp is a better choice and is basically an incandesent lamp. But they typically arent called incandesent. Its still expensive but lasts longer and costs a little less to operate. But still nowhere near as good as HID types.

But at the same time in defense of the mathmatician don't come whining to me when your volt meter shows 200 volts rms, your amp meter shows 5 amps rms. and your watt meter shows 900 watts rms for a 1000 watt incandesant bulb. Dont be surprised if I tell you all are probably correct after asking you where its used, what frequency, what power supply, other equipment in the area, etc. Also dont be surprised if your volts and amps for the the TV speaker shows that they are the same rms for a TV commercial as for the program. As my ole buddy Sam Heurtze says the correct answer is always "depends". To be 100% correct you need to know the wave form and frequency or mathmatical equation for the volts, amps, and impedence to calculate the true power.

In the end all 3 answers are close enough to correct as long as you know all the limits. I voted for not enough info.

But then again the only IQ test I ever took said I was a smart moron or a dumb average person. Probably explains why it took me six years and repeating some courses 3 times to get a four year degree. It still amazes me that we started out with sticks, bark, rocks, and dirt and made machines that can craft to less than one thousandth of an inch. I still havent figured out accuracy of reading versus accuracy of full scale. Accuracy to me implies I have some standard to compare an unknown to. Accuracy of full scale fits that. Accuracy of reading to me implies i am using an infinite number of standards. I measured an unknown value, comparted it to itself, and stated it is correct within 0.5% plus or minus 5 digits. How do you compare something to itself and come up with any error at all doesnt make sense, much less measuring something with no standard to compare to and assigning a value.

 

[realolman]

I said I was going to quit , but you're doing the same thing... You are ignoring the obvious, missing my point, and then filling paragraphs with irrelevant stuff to try to educate us slack jawed yokels... who may be less ignorant than you assume.

My point is that wise guys who are more interested in demonstrating their intelligence than teaching, write lousy questions to try to look smart, and make people whom they could educate feel stupid and undermine their confidence.

The question has been posed with certain given information. Please stick to that information only.

It does not matter how the 5 A rms was determined. We don't care if it was arrived at using divining rods. It is stated as being 5 A rms. It is given as being correct.

The 5 a waveform had to be determined by the characteristics of the load ( the inductance ,capacitance etc.) and the voltage waveform across the incandescent lamp. ...Otherwise it has nothing to do with the question.

We don't care about anything else, because we are not posing the question. If the "author" of the question has given incorrect information, that does not make us unknowledgeable.

To use your analogy... I'd rather get close enough to the girl.

 

[iwire]

I said I was going to quit , but you're doing the same thing... You are ignoring the obvious, missing my point, and then filling paragraphs with irrelevant stuff to try to educate us slack jawed yokels... who may be less ignorant than you assume.

My point is that wise guys who are more interested in demonstrating their intelligence than teaching, write lousy questions to try to look smart, and make people whom they could educate feel stupid and undermine their confidence.

All the while you feel all your answers are the correct ones and all the rest are just silly.

At least that is how your posts come across to me.:roll:

 

[K8MHZ]

I think the crux of the biscuit is the incandescent lamp as the load.

If we assume that 1kw is the answer we are also assuming that an incandescent lamp is a purely resistive load. While it may be close, it's not a perfect load.

And that is aside from the issue of the temperature coefficient.

Varying the frequency will change the reactive impedance of the filament. Since we were not told that the frequency was stable we can't determine the reactive impedance of the filament, we don't have enough information to give a conclusive answer.

Now, considering the load as somewhat reactive, simple multiplication of volts times amps will give an answer in VA, not watts.

Thinking that through, even at 60hz there must be some reactance, even though minuscule, making the PF of a light bulb less than 100 percent and an additional needed variable.

All y'all can crab about this 'trick question', but I like problems like this and it seems that the OP has got a bunch of us to really put on our thinking caps.

You have to give him credit for that!

 

[realolman]

All the while you feel all your answers are the correct ones and all the rest are just silly.

At least that is how your posts come across to me.:roll:

OK I give up.

 

[Electric-Light]

gar, that is all well and good, but in our trade, RMS and effective are virtually synonymous. since the same formula defines each. But, apply RMS values to a non-linear load and they are no longer "effective". Worse than tricky, the question is sneaky.

Not synonymous. There is a valid reason to use RMS value for a non-linear load.

Suppose you have 75 25W CFLs with a PF of 0.6 and very non-linear waveform(this is truth).

Assume that the source is 120v, sine wave, 60 Hz
Assume each lamp consumes real power of 25W totaling to 1.875kW
Power factor is not 0.6 because of any kind of phase shift, it's all because of non-linear characteristics.

Irms is 26.04A

That information allows us to know that the wiring must be rated for 26A load and that the I^2R loss is in wiring is exactly the same as if the load was a 120v rated 3.125kW heating element.

 

[glene77is]

... in the real world the correct answer is always "depends" ...

Gpedens,

Good Point !
And kudos on the elaboration.

 

[glene77is]

I check back often enough to remember WHO started this brain-storming question, and still recognize that you are expanding the playing field enough to so that we all continue to think about the problem, which for some of the electricians has passed into obscurity.
I will check back again, maybe tomorrow schedule permitting,
to see what the next episode puts forth.

This is a good question, important topic, and mountains of posts!

 

[rattus]

Not synonymous. There is a valid reason to use RMS value for a non-linear load.

Suppose you have 75 25W CFLs with a PF of 0.6 and very non-linear waveform(this is truth).

Assume that the source is 120v, sine wave, 60 Hz
Assume each lamp consumes real power of 25W totaling to 1.875kW
Power factor is not 0.6 because of any kind of phase shift, it's all because of non-linear characteristics.

Irms is 26.04A

That information allows us to know that the wiring must be rated for 26A load and that the I^2R loss is in wiring is exactly the same as if the load was a 120v rated 3.125kW heating element.

Light,

It is true that Irms, whatever the waveshape, is properly used to determine conductor size because in effect, power is a fixed resistor is being computed. However, the VrmsIrms product in this example does not yield apparent power. The voltage and current waveshapes must be the same.

The proper way to do that is to take the average of the vi product over a period as I think you have mentioned already.

 

[Electric-Light]

Light,

It is true that Irms, whatever the waveshape, is properly used to determine conductor size because in effect, power is a fixed resistor is being computed. However, the VrmsIrms product in this example does not yield apparent power. The voltage and current waveshapes must be the same.

The proper way to do that is to take the average of the vi product over a period as I think you have mentioned already.

Yes, I'm aware. I was briefly discussing why it's useful to get RMS value even for a non linear waveform in the field.

Before PWMs, VFDs and electronics, "RMS" was not something relevant to electricians.

 

[rattus]

Afterthought:

Afterthought:

Yes, I'm aware. I was briefly discussing why it's useful to get RMS value even for a non linear waveform in the field.

Before PWMs, VFDs and electronics, "RMS" was not something relevant to electricians.

FWIW, obtaining the average vi product would yield the net power into the device. I am not sure that "apparent power" applies here. Ratio of power out to net power in would yield efficiency, not PF.

And, I would argue that effective/RMS values have been in use since day one having been applied to sinusoidal voltages and currents.

 

[gar]

100829-1302 EST

rattus and others:

Your comment

And, I would argue that effective/RMS values have been in use since day one having been applied to sinusoidal voltages and currents.

Initiated the question --- what were the first meters used to measure AC? From my folowing search results I do not have an answer. Likely possibilities are --- hot-wire, solenoid, gravity balance, electrostatic, and electrodynamometer.

http://en.wikipedia.org/wiki/Ames_Hydroelectric_Generating_Plant

From:
http://books.google.com/books?id=k3...esnum=6&ved=0CDAQ6AEwBTgK#v=onepage&q&f=false
Mentions hot-wire AC voltmeter 1880 by Major Phillip Cardew

Interesting, Weston:
http://profiles.incredible-people.com/edward-weston/
but no speccifics on AC measurement.

http://www.westonmeter.org.uk/thecompany.htm
No real details on design.

http://www.humboldt.edu/scimus/Manufac/Weston.htm
The one paragraph is important for dates. Weston meters for AC by 1891.

http://weston.ftldesign.com/1887Lab/index.htm
http://weston.ftldesign.com/Model30/index.htm
http://weston.ftldesign.com/


Interesting early references:
http://americanhistory.si.edu/lighting/scripts/s19d.htm
In particular:
Thomson wattmeter, about 1887 [318,301], from Princeton University
http://books.google.com/books?id=jF...onepage&q=theory of Thomson wattmeter&f=false
Thomson-Houston voltmeter, about 1890 [219,027], from Potomac Electric Power Co.
Stanley phase indicator [314,411], from Weston Electrical Instrument Corp.
Biddle wattmeter, about 1895 [326,921], from Trinity College
Gardiner "Electro Magnetic Meter", [319,443], from Mrs. Donald Bliss

A letter by a former Weston engineer:
http://weston.ftldesign.com/MulhernLetter/letter.htm

Certain of these instrument types were inherently RMS measuring.

In the category of interesting things found when Google searching:
http://www.archive.org/stream/chronologicalhis00natirich/chronologicalhis00natirich_djvu.txt
This is quite important and I have not yet read thru the complete thread. It is very long. Any very useful electrical equipment only evolved after about 1830.

1888 Oliver B. Shallenberger of Westinghouse (see 1887) invents
the first induction meter for measuring alternating current.

1890 Dr. Edward Weston (see 1872) produces a direct reading de-
flection type electrodynamometer.

This is an RMS reading meter on either AC or DC.

Only at:
http://wiki.answers.com/Q/How_does_a_wattmeter_work
Did I find this reference:

The theory of electrodynamometers was developed by Elihu Thomson and Thomas Edison in 1888.

Have fun studying history and put yourself back in 1800 and try to erase any knowledge of electrical theory. Then what electrical direction might you travel?

.

 

[rattus]

When?

When?

gar,

Do you have any idea when the effective/RMS formula was derived? I would guess that it preceded the development of the instruments.

 

[gar]

100830-1301 EST

rattus:

From my post in this thread numbered #158:
"I believe the trail on the concept of RMS starts in the field of statistics with Francis Galton conceiving the standard deviation function in the 1860s. This certainly predates electrical usage. See section "Statistics, standard deviation, regression and correlation" at http://en.wikipedia.org/wiki/Francis_Galton. There are many other search results that fill in some of the holes."

From an earlier post of mine in this thread:
http://en.wikipedia.org/wiki/Root_mean_square

In "Menlo Park Reminiscences", by Francis Jehl it is clear that electrical measurement was quite primitive in !879. They knew about power, energy, current, and voltage. They had voltage references, crude standard cells, a single galvanometer, Wheatstone bridge, resistance boxes, potentiometers, and thermometers. They did not have the concept of a a current shunt.

Little was done with AC before 1880. The above reference indicates that Galton worked with RMS in 1860. Weston's work in the 1880s provided a real advancement in instrumentation.

Thus, I would conclude that the mathematical concept of RMS predated its electrical usage,

.

 

[steve66]

gar,

Do you have any idea when the effective/RMS formula was derived? I would guess that it preceded the development of the instruments.

Effective and RMS values are not unique to electricity, so I would guess the equations came even long before the discovery of electricity.

They've probably been around almost as long as statistics.

 

[Electric-Light]

FWIW, obtaining the average vi product would yield the net power into the device. I am not sure that "apparent power" applies here. Ratio of power out to net power in would yield efficiency, not PF.

And, I would argue that effective/RMS values have been in use since day one having been applied to sinusoidal voltages and currents.

I'm not sure how you got yourself into efficiency thing.

When you say "average V*I product" what exactly are you referring to?