tacomafc
Senior Member
- Location
- Long Island New York
If a lamp is rated 500 watts at 115 volts. What is the watts at 120 volts?
ohms law
- Thread starter tacomafc
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tacomafc
Senior Member
- Location
- Long Island New York
Found resistance at 115 then apply that to the 120
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Aw, Gus, don't just give him the answer. :roll: Remember the saying:
"Give a man a fish, and he eats for a day. Teach a man to fish, and he sits in a boat and drinks beer all day."
"Give a man a fish, and he eats for a day. Teach a man to fish, and he sits in a boat and drinks beer all day."
Sorry, Larry...
I do recall and that's why I didn't say anything until he showed he was on the right path by figuring the resistance, but you are correct, I jumped the gun. sorry.
Tacomafc, tell us you had already derived that answer.............
I do recall and that's why I didn't say anything until he showed he was on the right path by figuring the resistance, but you are correct, I jumped the gun. sorry.
Tacomafc, tell us you had already derived that answer.............
jaylectricity
Senior Member
- Location
- Massachusetts
- Occupation
- licensed journeyman electrician
So you don't divide the watts by 115 and then multiply the result by 120?
That is correct. You need to find the fixed resistance and work it from there.
Roger
tacomafc
Senior Member
- Location
- Long Island New York
Ok here is the full story. It was actually a 3 part question on a liscense exam that I took last night.
1) What is the resistance of a 500 watt quartz lamp rated at 115 volts.
answer : R= E x E / P 115 x 115 / 500w =26.45
2) What is the amps at 120 volts?
answer: I = 500/120 = 4.16
3) what is the wattage at 120 volts?
answer: P = E x E / P 120 x 120 / 26.45 = 544 watts
thank you for your time.
1) What is the resistance of a 500 watt quartz lamp rated at 115 volts.
answer : R= E x E / P 115 x 115 / 500w =26.45
2) What is the amps at 120 volts?
answer: I = 500/120 = 4.16
3) what is the wattage at 120 volts?
answer: P = E x E / P 120 x 120 / 26.45 = 544 watts
thank you for your time.
gar
Senior Member
- Location
- Ann Arbor, Michigan
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- EE
091006-2005 EST
The problem is that the resistance at 120 V is different than at 115 V.
Following are measured resistance values of a 100 W 120 V bulb:
60.8 V --- 106.1 ohms
100.4 V --- 133.3 ohms
110.2 V --- 139.1 ohms
114.5 V --- 141.7 ohms
120.2 V --- 144.8 ohms
There will be a similar effect for any tungsten incandescent lamp designed for the same color temperature at nominal voltage.
See the curve at my web site
http://beta-a2.com/EE-photos.html
for power vs voltage for a 75 W tungsten incandescent bulb at photo
P9 --- Photo 2769A1
.
The problem is that the resistance at 120 V is different than at 115 V.
Following are measured resistance values of a 100 W 120 V bulb:
60.8 V --- 106.1 ohms
100.4 V --- 133.3 ohms
110.2 V --- 139.1 ohms
114.5 V --- 141.7 ohms
120.2 V --- 144.8 ohms
There will be a similar effect for any tungsten incandescent lamp designed for the same color temperature at nominal voltage.
See the curve at my web site
http://beta-a2.com/EE-photos.html
for power vs voltage for a 75 W tungsten incandescent bulb at photo
P9 --- Photo 2769A1
.
mivey
Senior Member
Then the answer to #2 is wrong.
tacomafc
Senior Member
- Location
- Long Island New York
2) guess I should have I = E / R 120 / 26.45 = 4.53
if this is wrong PLEASE let me know how its done. thanks
if this is wrong PLEASE let me know how its done. thanks
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
or I=P/E 544/120=4.53
tacomafc
Senior Member
- Location
- Long Island New York
oh 2 ways to skin a cat. could of done that. Maybe I did?
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
091006-2108 EST
tacomafc:
Both answers 2 and 3 are wrong for a tungsten filament lamp.
The origin of the problem is the author of the questions. This author made an invalid assumption that the resistance of the lamp is a constant with respect to voltage and it is not. Or the author needed to define it as constant.
If we suppose that the question was asked correctly for the probably intent of the author, then instead a constant resistance resistor would have been the load instead of an assumed tungsten filament bulb. Note an Edison carbon filament bulb would produce an entirely differently result than a tungsten bulb. Tungsten has a positive temperature coefficient of resistance, while carbon has a negative coefficient.
Assuming a constant resistance as the load, then your answers 1 and 3 are correct, but with question 3 rewritten as:
Using the resistance of question 1 what is the power dissipated in this resistor with 120 V applied?
Question 2 does not make a lot of sense.
What is (are) the amps at 120 volts? Amps of what?
Is it for a load resistance equal to the resistance in question 1? If so, then the answer is 120/26.45 = 4.54 A.
Or
Is the question how many amps thru a 500 W load at 120 V? This is the question you answered.
There needs to be preciseness in questions.
.
tacomafc:
Both answers 2 and 3 are wrong for a tungsten filament lamp.
The origin of the problem is the author of the questions. This author made an invalid assumption that the resistance of the lamp is a constant with respect to voltage and it is not. Or the author needed to define it as constant.
If we suppose that the question was asked correctly for the probably intent of the author, then instead a constant resistance resistor would have been the load instead of an assumed tungsten filament bulb. Note an Edison carbon filament bulb would produce an entirely differently result than a tungsten bulb. Tungsten has a positive temperature coefficient of resistance, while carbon has a negative coefficient.
Assuming a constant resistance as the load, then your answers 1 and 3 are correct, but with question 3 rewritten as:
Using the resistance of question 1 what is the power dissipated in this resistor with 120 V applied?
Question 2 does not make a lot of sense.
What is (are) the amps at 120 volts? Amps of what?
Is it for a load resistance equal to the resistance in question 1? If so, then the answer is 120/26.45 = 4.54 A.
Or
Is the question how many amps thru a 500 W load at 120 V? This is the question you answered.
There needs to be preciseness in questions.
.
Or:
Or:
Or since we know power is proprotional to the square of the voltage, you could simply say,
P = (120^2/115^2)x500 = 544
This of course assumes a constant resistance which is not quite the case in an incandescent lamp.
Or:
Or since we know power is proprotional to the square of the voltage, you could simply say,
P = (120^2/115^2)x500 = 544
This of course assumes a constant resistance which is not quite the case in an incandescent lamp.
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
Damn you and your continued empirical data messing with my theoretical knowledge!!
Seriously though, gar points out a very important reality of incandescent lights: the resistance changes by quite a lot as the applied voltage changes. This has to do with the fact that the temperature of the filament is changing; when 'cold' (room temperature) the resistance of an incandescent filament is about 1/10 that at operating temperature.
The 'rule of thumb' equations that I learned for the changes in operation when applied voltage changes are:
current scales as the square root of the change in applied voltage
power scales as the change in voltage ^ 1.5 power
light output scales as the change in voltage to the 3.5 power
life scales as the change in voltage ^ -12 power
Welch-Allyn makes various precision incandescent lamps, and gives a graph of operating characteristic changes when applied voltage is changed; see page 7 of http://www.walamp.com/lpd/files/datasheets/HPXLampCatalog.pdf
-Jon
Mayimbe
Senior Member
- Location
- Horsham, UK
I would change 500 for 544 in the question number 2.
And I wouldnt take in count what gar says about the resistance. Not even think about it.
If somebody tells you something like that in the future about the resistance of something. Ask that person this questions:
1) Did you know that ohms law is based on aproximation of a graphic where is ploted the current vs the voltage meassured of a certain material?
2) Did you know that the every meassured has an error associated?
3) Did you know that if you meassure the resistance of lamp, and I do it and the same time with a similar lamp, theres high probability that we dont get the same values??
If he says yes, he would know that you know and probably leaves the things that way. It would be an endless discusion about whos right and whos wrong.
So that said. If I were the person that have to correct this question. I will only make the remark on question 2, and check the other two. So you got 2 of 3. You got an ave of .666
broadgage
Senior Member
- Location
- London, England
As others post, one might suspect that the examiner wants you to assume that the lamp has a constant resistance, despite this not being true in the real world.
For small changes in the voltage applied to a tungsten lamp, then the current through the lamp will change by about half the % change in voltage.
In the example given we have a 115 volt 500 watt lamp.
At rated voltage the current drawn will be 4.348 amps.
At rated voltage the resistance would 26.45 ohms.
It is now proposed to run the same lamp at 120 volts, which is an increase of 5.217%
Experience has shown that the lamp current will increase by half the % in applied voltage, therefore the current should increase by 2.608%.
Therefore at 120 volts, the current would be 4.461 amps, and the wattage would be 535.3 watts.
The resistance at 120 volts would be 26.90 ohms.
However the knowledge about the change in resistance of a lamp at different voltages is not that widely known, nor is it needed by the average licensed electrician.
I therefore surmise that the examiner intended that a constant resistance be assumed.
Not a very sensible question IMHO.
A better test of the candidates knowledge and ability to perform calculations would have been a similar question involveing a water heater element, where a constant resistance is a more sensible assumption.
For small changes in the voltage applied to a tungsten lamp, then the current through the lamp will change by about half the % change in voltage.
In the example given we have a 115 volt 500 watt lamp.
At rated voltage the current drawn will be 4.348 amps.
At rated voltage the resistance would 26.45 ohms.
It is now proposed to run the same lamp at 120 volts, which is an increase of 5.217%
Experience has shown that the lamp current will increase by half the % in applied voltage, therefore the current should increase by 2.608%.
Therefore at 120 volts, the current would be 4.461 amps, and the wattage would be 535.3 watts.
The resistance at 120 volts would be 26.90 ohms.
However the knowledge about the change in resistance of a lamp at different voltages is not that widely known, nor is it needed by the average licensed electrician.
I therefore surmise that the examiner intended that a constant resistance be assumed.
Not a very sensible question IMHO.
A better test of the candidates knowledge and ability to perform calculations would have been a similar question involveing a water heater element, where a constant resistance is a more sensible assumption.
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