Another current question

Discussion in 'Question of the Day' started by derekleffew, Oct 13, 2008.

?

Which of the following currents... (see post#1 for remainder)

Poll closed Oct 20, 2008.

7 vote(s)
28.0%

7 vote(s)
28.0%

10 vote(s)
40.0%
4. d) 6.2A

1 vote(s)
4.0%

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Which of the following currents would flow in an extension cord
connecting a luminaire with a 575W 115V incandescent lamp to a
receptacle providing 120V?

a) 4.8A
b) 5.0A
c) 5.2A
d) 6.2A

No credit will be given unless work is shown. Those with prior knowledge of the answer, or more than five years in the industry, are ineligible. Edit: You may vote, but don't disclose your answer until the poll ends. Source will be cited [-]once the question has been satisfactorily answered.
[/-] when the poll closes.

Hint: You'll need more than West=Virginia.

Last edited: Oct 16, 2008
2. David AshtonActive Member

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Perhaps you should run a poll?

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Hey, great idea, David. Adding a poll. Everyone just vote, and wait a couple of days (or until responses stop coming) to post the rationale behind your answers. Again, for fairness, those who know the answer should hold off. (Otherwise everyone will just vote the same as SteveTerry.) Making the poll anonymous, just in case.

Last edited: Oct 13, 2008
4. zuixroActive Member

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Here's how I did it:

Ohm's Law
R = E^2 / P
R = 115(volts)^2 / 575(watts)
R = 13,225 / 575
R = 23 Ohms

So a 115v 575w bulb has 23 ohms of resistance
So use Ohm's law again...
I = E / R
I = 120(volts) / 23(ohms)
I = 5.2 Amps

For extra credit...
Ohm's law again....
P = E^2 / R
P = 120(volts)^2 / 23(ohms)
P = 14,400 / 23
P = 626 Watts

Last edited by a moderator: Oct 20, 2008

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C'mom people, this is pitiful. In two days, there have been 66 views but only 7 votes. The only penalty for voting wrong is that you might learn something.

6. MNicolaiWell-Known MemberFight Leukemia

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I sent my work to Derek via a PM.

7. avareActive Member

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I am disqualified. I check the thread whenever there is a new reply. The views do not indicate the appropriateness of the viewers to answer.

Andre

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[user]avare[/user], you are allowed to vote, just don't post the answer. As soon as the poll reaches 50 responses, I'll post the first correct answer, given by zuixro, who incidentally is a college freshman.

Last edited: Oct 16, 2008
9. David AshtonActive Member

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I suggested a poll as a joke, the laws of physics and maths are not governed by polls, however the results are extraordinary.

10. MNicolaiWell-Known MemberFight Leukemia

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I showed this problem to my Electronics teacher and it threw him for a loop. That said, tomorrow we're going to illustrate to everyone how it works and why it works that way, and also take current readings and such. Should be interesting.

11. waynehoskinsActive Member

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At the rate it's going, I think it might take a while.

12. TimmyP1955Active Member

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The answer is: None of the above.

13. waynehoskinsActive Member

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Actually, you're right. Even though I arrived at an answer, the real answer is we can't quite tell without measuring it, because what we must treat as a constant is actually a function, and the answer will be slightly less, though just ever so slightly, than what we can calculate.

However-- the answer that we can calculate (presuming that one value to be constant and not variable) is one of the given answers.

14. David AshtonActive Member

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Can we assume the extension cord is not a super-conductor?

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As reported earlier, [user]zuixro[/user] was the first to respond with the correct answer.

Here's more explanation:
The lamp is rated 575W at 115V. If it is operated at the rated 115V it draws 575/115 or 5A. The lamp has a resistance of 115/5 or 23 Ohms, the only constant value in the discussion. If you connect a 23 Ohm lamp to 120V you get 120/23 or 5.2A. Furthermore actual lamp power is 120 x 5.2 or 624W.

This question is one of several sample questions in the ETCP Candidate Handbook for Entertainment Electrician, and raised quite a stir on another forum a few years ago. Congratulations to the ten members who correctly chose answer "c".

16. waynehoskinsActive Member

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The only catch (that I caught after solving it) is that the filament resistance is a function of filament temperature. By burning it hotter with terminal voltage 120 instead of 115, the resistance should increase ever so slightly. That in turn will slightly decrease the current the lamp draws, and therefore it will dissipate slightly less than the calculated 626 watts.

My modified guess is that the current drawn will be more along the lines of 5.15A, being 618 watts.

17. David AshtonActive Member

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This answer is wrong and it is wrong because at the higher voltage the filament gets hotter and has a higher resistance also there is an extension cord involved which also has a resistance, so possible correct answers are a or b , depending on the cable resistance but absolutely not c
edit Wayne got there first.

18. waynehoskinsActive Member

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I had factored that out, presuming we had a perfect situation where we're measuring 120 volts at the lamp terminals. And since line voltages in my area are often nearer the 123-125 range than 117-120, it's reasonable to get 120 volts at the end of a cable plugged into a hole tied to a breaker stab-locked onto a bus bar that's got normal line voltage on it.

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Fair enough, [user]allthingstheatre[/user]. Here is Mike Wood's answer:

I agree, below are the incandescent lamp equations I've always used.
They are approximate but still useful. They all derive from the
non-linearity of resistance of the filament with temperature that you
mention:
life/LIFE = (VOLTS/volts)^13
(i.e. reduce the volts to 90% and the life increases by 393%!)

lumens/LUMENS = (VOLTS/volts)^3.4

EFFICIENCY/efficiency = (VOLTS/volts)^1.9

watts/WATTS = (volts/VOLTS)^1.6
(not 'squared' as you would get with a fixed resistance)

coltemp/COLTEMP = (volts/VOLTS)^0.42

However, after having said all that, the ETCP sample question looks like
it is testing whether or not you realise that the lamp wattage is not a
constant and changes as the voltage changes so you can't just do a
simple ratio calculation. I don't think it's necessary in most real life
situations to mess around with the incandescent lamp equations. I would
have calculated this as 'close enough' to 5.2A to choose that answer and
would only use the exact incandescent equations in lab situations. (The
above equation allowing for the resistance change would give you an
answer of 5.13A - so 5.2A is still the closest).

The bottom line is these are supposed to be highly practical exams
testing real data that electricians come across every day - not whether
an R&D geek like me can complicate the issue! <grin>

20. David AshtonActive Member

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Always read the question which states a extension cord plugged in to a 120 volt receptacle, hence my super conductor question, and given that the current had to be less than 5.2 the only options left are 5 or 4.8, assuming there was a correct answer, which is again assumed from the question.

edit again, you're answering too fast, and the green is hard to read,
However the non linearity of lamp current is one of the most important and least understood areas, just look at the intensity v currentl thread to see this.
2nd edit, you shouldn't have put the extension cord in as that must reduce current and really is a consideration.

Last edited: Oct 20, 2008