What is the power of an incandescent lamp at 200v 5A?
- Thread starter Electric-Light
- Start date
- Status
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100818-1229 EST
rattus:
Your post 158 made a good point, which I think is the basis to the original point of this thread. I think Electric-Light had in mind a low frequency of excitation to the load and thus a non-linear load. This is my conclusion from his answer in the poll, and one of his posts.
Your post 161 is certainly possible.
Using quickly available resources, meaning searching via the Internet, or books that I have, I have found it difficult to trace the evolution of certain concepts, and words.
At the time that Edison developed his electrical distributions system, 1878-1879, some if not most theoreticians believed that one should design for maximum power transfer. In other words the source resistance was equal to load resistance, 50% efficiency. Edison realized that for a practical system one needed parallel loads and a low source resistance and high efficiency in the generator. Circa this time they had galvanometers, but no voltmeters or ammeters. They had resistance boxes and bridge circuits. They apparently did not conceive of the idea at that time to calibrate the galvanometer deflection to use it for a voltmeter, nor the idea to make a low resistance shunt to measure current. Thus, very awkward means were used to get voltage and current information. However, a very useful technique of putting a thermometer in the rotor or field coil to obtain temperature rise measurements was used.
On the internet I found bits and pieces of information on the evolution of measurement from compass deflection to the Weston-D'Arsonval meter movement. But not a very clear trail.
.
rattus:
Your post 158 made a good point, which I think is the basis to the original point of this thread. I think Electric-Light had in mind a low frequency of excitation to the load and thus a non-linear load. This is my conclusion from his answer in the poll, and one of his posts.
Your post 161 is certainly possible.
Using quickly available resources, meaning searching via the Internet, or books that I have, I have found it difficult to trace the evolution of certain concepts, and words.
At the time that Edison developed his electrical distributions system, 1878-1879, some if not most theoreticians believed that one should design for maximum power transfer. In other words the source resistance was equal to load resistance, 50% efficiency. Edison realized that for a practical system one needed parallel loads and a low source resistance and high efficiency in the generator. Circa this time they had galvanometers, but no voltmeters or ammeters. They had resistance boxes and bridge circuits. They apparently did not conceive of the idea at that time to calibrate the galvanometer deflection to use it for a voltmeter, nor the idea to make a low resistance shunt to measure current. Thus, very awkward means were used to get voltage and current information. However, a very useful technique of putting a thermometer in the rotor or field coil to obtain temperature rise measurements was used.
On the internet I found bits and pieces of information on the evolution of measurement from compass deflection to the Weston-D'Arsonval meter movement. But not a very clear trail.
.
No R required.
No effective values either. Impressive equation editor though. What is it called?
The effective values come out of that.
what is it power or nothing
what is it power or nothing
Its been a few decades but best I can remember from my old VATech days power is a sumation of the integration of voltages and currents of the same wave form and frequency and angle. Throw out the angle and it is apparent power. Power typically means watts and apparent power volt-amps. The given RMS values of voltage and current have to apply to the same frequency and wave form and angle for power. In other words Prms = Vrms x Irms for the same frequency and wave form and angle. If you have different frequencys and wave forms then you have to separate them into their individual values of voltage and current at the same frequency and wave form. Prms = (vrms1 x irms1) + (vrms2 x irms2) + ( ) + ( ) +(vrmsn x Irmsn). In the question of 200 vrms and 5 arms at the same wave form and frequency with a pure capacitance or inductance the power KW would be 0. The apparent power would be 1 KVA. If the load were purely resistive (current and voltage at same angle and frequency) the power would be 1 KW. Power is measured in watts W and apparent power in volt-amps VA. The correct answer is not enough info. Even if the frequency was stated as 60 hz sinusoidal and it was 200 v 60 hz sinusoidal and 5 amp 60 hz sinusoidal the correct answer would be apparent power is 200 x 5 = 1000 VA or 1 KVA. The only choice given as a possible answer is 1 KW but that would only be true if the angle was 0 and the wave form the same.
what is it power or nothing
Its been a few decades but best I can remember from my old VATech days power is a sumation of the integration of voltages and currents of the same wave form and frequency and angle. Throw out the angle and it is apparent power. Power typically means watts and apparent power volt-amps. The given RMS values of voltage and current have to apply to the same frequency and wave form and angle for power. In other words Prms = Vrms x Irms for the same frequency and wave form and angle. If you have different frequencys and wave forms then you have to separate them into their individual values of voltage and current at the same frequency and wave form. Prms = (vrms1 x irms1) + (vrms2 x irms2) + ( ) + ( ) +(vrmsn x Irmsn). In the question of 200 vrms and 5 arms at the same wave form and frequency with a pure capacitance or inductance the power KW would be 0. The apparent power would be 1 KVA. If the load were purely resistive (current and voltage at same angle and frequency) the power would be 1 KW. Power is measured in watts W and apparent power in volt-amps VA. The correct answer is not enough info. Even if the frequency was stated as 60 hz sinusoidal and it was 200 v 60 hz sinusoidal and 5 amp 60 hz sinusoidal the correct answer would be apparent power is 200 x 5 = 1000 VA or 1 KVA. The only choice given as a possible answer is 1 KW but that would only be true if the angle was 0 and the wave form the same.
This whole thing should have died a long time ago, but I have had the same questions throughout the whole thing, that I wish you would answer:
The voltage and current values were already given in rms..
1. Wouldn't the frequency, wave form, capacitance, inductance, and whatever else, already have to be known in order to get the rms values?
2. How would the rms values be known without knowing all the relevent information?
3. If the frequency, wave form, and whatever else are already known, why does there need to be more information?
The voltage and current values were already given in rms..
1. Wouldn't the frequency, wave form, capacitance, inductance, and whatever else, already have to be known in order to get the rms values?
2. How would the rms values be known without knowing all the relevent information?
3. If the frequency, wave form, and whatever else are already known, why does there need to be more information?
Mayimbe
Senior Member
- Location
- Horsham, UK
1. YES
2. if you dont know the wave form, and the period or frecuency, it would be diifficult to know the rms values. You can always use the voltmeter or ampmeter or both.
3. Theres no need.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100826-1700 EST
realoman and Mayimbe:
1. The answer is NO. It was simply stated what the values were.
2. The values were stated. Someone concocted the values out of thin air for the problem. RMS by itself implies nothing about frequency or waveform.
3. They are not known, therefore, more information is needed.
.
realoman and Mayimbe:
1. The answer is NO. It was simply stated what the values were.
2. The values were stated. Someone concocted the values out of thin air for the problem. RMS by itself implies nothing about frequency or waveform.
3. They are not known, therefore, more information is needed.
.
Isn't that what this whole thread is about?
If you don't know anything about the frequency or waveform then you don't know the rms values... Conversely, if you know the rms values you had to know something about the frequency and the wave form... if not... you don't know the rms values.
I believe the rms values of a square, sawtooth, or whatever kiind of wave can be determined.. The shape and magnitude of the current wave would be dependent upon the load.... be it resistive, capacitive, or inductive, or any combination of the three... and the applied voltage magnitude, frequency, and wave form.
In order to determine the rms values of the current wave, all that other stuff would already be taken into account
The values were explicitly stated to be rms.... so you can use them to determine the power... or else they are not rms values.
Concocted values are not rms values.
.
Last edited:
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100826-1921 EST
realoman:
Sine waves are not a requirements for an RMS measurement, nor by stating in the problem that voltage and current are RMS does it imply that the load is a pure resistance that is invariant under all conditions.
If I have a diode for a load and a sine wave voltage source, then with appropriate instruments I can measure both the RMS voltage and current to the diode. If I have the characteristic curve of the diode, then from the voltage and this characteristic curve I can calculate the RMS current.
What I can not do is calculate the power dissipated in the diode from only the RMS voltage and the RMS current.
.
realoman:
Sine waves are not a requirements for an RMS measurement, nor by stating in the problem that voltage and current are RMS does it imply that the load is a pure resistance that is invariant under all conditions.
If I have a diode for a load and a sine wave voltage source, then with appropriate instruments I can measure both the RMS voltage and current to the diode. If I have the characteristic curve of the diode, then from the voltage and this characteristic curve I can calculate the RMS current.
What I can not do is calculate the power dissipated in the diode from only the RMS voltage and the RMS current.
.
No does not imply that the load is anything at all, BUT the load and the voltage ( in whatever weird waveform or frequency it is ) is what caused the current waveform ( in whatever weird waveform or frequency it is ) to be what it is. When you calculate the RMS values you are stating the values of the weird waveform current that will cause the same amount of heat in a resistance that a similar magnitude of DC current will cause.
The incandescent bulb is a resistance..
The imcandescent bulb in the OP is a resistance... it is not a diode.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100827-0701 EST
realoman:
The problem is that an incandescent lamp is not an invariant linear resistance under all possible frequencies and waveforms of applied voltage. For your statement to be correct you must have an invariant linear resistance.
I used a diode as an illustration of one type of non-linear resistive load for which Vrms*Irms fails to predict the dissipated power in the load because all I need is one example that fails. To prove a statement is correct you have to prove that under all possible inputs that it is correct.
Long ago in this thread the variation of the resistance of an incandescent lamp within a cycle was discussed and that at low frequencies, like 0.1 Hz or lower, a noticeable error in calculated power from the RMS voltage and current would occur.
.
realoman:
The problem is that an incandescent lamp is not an invariant linear resistance under all possible frequencies and waveforms of applied voltage. For your statement to be correct you must have an invariant linear resistance.
I used a diode as an illustration of one type of non-linear resistive load for which Vrms*Irms fails to predict the dissipated power in the load because all I need is one example that fails. To prove a statement is correct you have to prove that under all possible inputs that it is correct.
Long ago in this thread the variation of the resistance of an incandescent lamp within a cycle was discussed and that at low frequencies, like 0.1 Hz or lower, a noticeable error in calculated power from the RMS voltage and current would occur.
.
Electric-Light
Senior Member
So, there's surprising amount of lack of understanding of definition of power, appreant power and RMS values...
You are not talking nor asking about all possible frequencies, and waveforms... you already have the rms values ... where did th OP obtain those from? If not from the incandescent bulb being powered by whatever voltage source caused the amperage for which you found the rms value, then the whole question is just stupid. Same thing if they are inaccurate... and they were provided by the OP
Since you already have given the rms values, the scope of the question is limited to those values... not an infinite range of unknown frequencies.
We would not be having this discussion if the power supply to the incandescent bulb were DC... the rms values that were stated in the OP are the DC equivilent and they were given by the OP... all other possibilities or frequencies etc. are irrelevent, and not part of the question..
.
Last edited:
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
100827-1512 EST
realoman:
You are missing the point.
Suppose I power a 100 W 120 V tungsten filament incandescent lamp with a sine wave voltage of 120 V and 1/10 Hz. I measure the RMS value of the voltage and the RMS current with appropriate instruments. The product of these two measurements will not be the average power dissipated in the lamp.
The thermal time constant of this lamp is short relative to the frequency of the excitation voltage. Thru one complete cycle of the voltage there will be a large variation in the resistance of the filament, and thus a non-linear load resistance is present, and Pave = Vrms*Irms fails. However, the integral of v*i / suitable averaging time does equal the average power.
Both the voltage and current of the original post could have been measured under these low frequency conditions relative to the lamp thermal time constant. We do not know what the conditions were because not enough information was provided. The only meaningful answer to the original question is "not enough information".
.
realoman:
You are missing the point.
Suppose I power a 100 W 120 V tungsten filament incandescent lamp with a sine wave voltage of 120 V and 1/10 Hz. I measure the RMS value of the voltage and the RMS current with appropriate instruments. The product of these two measurements will not be the average power dissipated in the lamp.
The thermal time constant of this lamp is short relative to the frequency of the excitation voltage. Thru one complete cycle of the voltage there will be a large variation in the resistance of the filament, and thus a non-linear load resistance is present, and Pave = Vrms*Irms fails. However, the integral of v*i / suitable averaging time does equal the average power.
Both the voltage and current of the original post could have been measured under these low frequency conditions relative to the lamp thermal time constant. We do not know what the conditions were because not enough information was provided. The only meaningful answer to the original question is "not enough information".
.
I am not missing any point ... I have understood that from the beginning.
1. If the question of the OP were posed, using a DC power supply, could you calculate the power dissipated, or would you have to have more information?
I have little doubt that normal DMM would not display the correct value under such low frequency because of the varying resistance due to heat.... but
2. Could the rms value of the current through the incandescent lamp powered by your 0.1Hz power supply be correctly calculated, rather than measured?
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
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.
1. If the question of the OP were posed, using a DC power supply, could you calculate the power dissipated, or would you have to have more information?
I have little doubt that normal DMM would not display the correct value under such low frequency because of the varying resistance due to heat.... but
2. Could the rms value of the current through the incandescent lamp powered by your 0.1Hz power supply be correctly calculated, rather than measured?
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
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.
Last edited:
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
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.
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.
- Status