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

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

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iwire said:
You might also enjoy the discussion of how a layer of paint on a cable requires the conductors be derated.
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I started reading, but got side-tracked by the link to Mr Doubletalk videos and could not get away from that. :grin:
 
GeorgeB said:
BUT ... it it is really RMS current, the value will reflect the nonlinear resistance with (say) 1Hz power ...

I think the 1kW is correct.
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Nope. 200v RMS implies it creates a heating value equivalent to 200v DC in a constant resistor. Same with current 5A RMS means the heating value in resistance of wiring is the same as 5A DC.

Take switching power supply for example.

You have 120v, 60Hz AC sine wave RMS and 3A RMS, but not unusual to see Pavg of 200W or so.

RMS simply means, you slice a complete cycle into small pieces, say break a one whole cycle into 120 pieces. A 120v RMS sine wae ranges from 0v to 170v. When you square each piece so (v1^2+V2^2+......V120^2)
you then take the mean of the sum(divided by however many piece you sliced it up to), then you take the square root of the sum. That's the RMS value.

Vrms * Irms = VA
In this calculation, you'll get the same VA even if they're completely in phase or 90deg out of phase.

When you multiply the slices of voltage & current taken at the same time V1*I1, V2*I2... then divide by number of slices, this is the average power (average over the duration of integration period)
 
Electric-Light said:
RMS simply means, you slice a complete cycle into small pieces, say break a one whole cycle into 120 pieces. A 120v RMS sine wae ranges from 0v to 170v. When you square each piece so (v1^2+V2^2+......V120^2)
you then take the mean of the sum(divided by however many piece you sliced it up to), then you take the square root of the sum. That's the RMS value.
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Yes.
And it would give you the RMS value of current regardless of frequency or the shape of the waveform. From the original post, this is evidently known to be 5A.
Occam's Razor comes to mind.........
 
Besoeker said:
Yes.
And it would give you the RMS value of current regardless of frequency or the shape of the waveform. From the original post, this is evidently known to be 5A.
.........
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Exactly.

That is what I am not getting.... you have done some sort of mathematical manipulation that IMO should result in a value that is appropriate for use in the problem at hand... you have expressed the values in RMS... and then claim that the answer is not accurate because the values you have provided may not be appropriate for the problem at hand.???

Is it not possible to determine the RMS value for non sinusoidal wavelengths? The current waveform is going to be dependent upon the resistance, which you claim is changing because of the heat created by some unknown voltage waveform.

If you intended to pose a problem involving a circuit, would you not provide values suitable for that circuit?

If it does not compute, it is because you have provided inaccurate information.... and it's supposed to be some sort of riddle or something because it was answered using values you provided?:-?

I've seen better riddles.
 
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realolman said:
If it does not compute, it is because you have provided inaccurate information.... and it's supposed to be some sort of riddle or something because it was answered using values you provided?:-?

I've seen better riddles.
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How about:
The power to a load at steady-state is measured at 4.5 kW. After 30 minutes, how much energy is consumed?
a) 4.5 kWh
b) 2.25 kWh
c) Depends on the Carnot Cyle
d) Not enough information

Most reasonable people would answer b: 2.25 kWh. However, in this classroom, they find out they answered wrong because they forgot to allow for the fact that the load was switched off after 10 minutes. :grin:
 
realolman said:
Is it not possible to determine the RMS value for non sinusoidal wavelengths?
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You can determine the RMS value of anything.
Given +500v for 1 hour, off for one hour, and -300v for 4 hours is the same as

sqrt ((500^2 + 0^2 + (|300|^2)*4 )/6) = 318.9

With an integration time of 6 hours, the Vrms is 318.9v
 
Besoeker said:
The on period is irrelevant in calculating power - it's an instantaneous value.
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The question is asking how much energy is consumed after 30 minutes. The power is 4.5kW, but it doesn't specifically state how much of the 30 minutes the load stays on.
 
mivey said:
How about:
The power to a load at steady-state is measured at 4.5 kW. After 30 minutes, how much energy is consumed?
a) 4.5 kWh
b) 2.25 kWh
c) Depends on the Carnot Cyle
d) Not enough information

Most reasonable people would answer b: 2.25 kWh. However, in this classroom, they find out they answered wrong because they forgot to allow for the fact that the load was switched off after 10 minutes. :grin:
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Dang it.

Don't throw stuff like that out there!

I just spent most of my lunch hour reading about thermodynamics because of you.:mad::)D)
 
Electric-Light said:
The question is asking how much energy is consumed after 30 minutes. The power is 4.5kW, but it doesn't specifically state how much of the 30 minutes the load stays on.
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My answer related to the original question.
In the subsequent unrelated post about the 4.5kW for 30 minutes, it does mention steady state. One might reasonable construe that as being steady state for the 30 minutes. Otherwise, it is a meaningless question.
 
LarryFine said:
Where is that mentioned? :confused:
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While the ten minutes was arbitrary, there was a discussion on some forum a while back by somebody about energy metered over an interval period with the load being off part of the time. It should have been clear that that was within the realm of possibilities. :grin:
 
Electric-Light said:
The question is asking how much energy is consumed after 30 minutes. The power is 4.5kW, but it doesn't specifically state how much of the 30 minutes the load stays on.
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There ya go. You must have read the post from somebody on that other forum. I can't believe so many missed that thread.

Add: iwire must have read it also.
 
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