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

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

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Electric-Light said:
I'm not sure how you got yourself into efficiency thing.

When you say "average V*I product" what exactly are you referring to?
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Please note that I used lower case which denotes instantaneous values, v(t)*i(t) that is. The average of this product over a period yields the net power input which includes losses and subtracts negative power. The result is power in. Perform the same operation at the load to obtain power out. The ratio is efficiency.

Apparent power as it is commonly used is simply the product Vrms*Irms which does not subtract the negative power resulting from reactive loads.

Therefore, I question the use of the term "apparent power" in this case.
 
rattus said:
Please note that I used lower case which denotes instantaneous values, v(t)*i(t) that is. The average of this product over a period yields the net power input which includes losses and subtracts negative power. The result is power in. Perform the same operation at the load to obtain power out. The ratio is efficiency.

Apparent power as it is commonly used is simply the product Vrms*Irms which does not subtract the negative power resulting from reactive loads.

Therefore, I question the use of the term "apparent power" in this case.
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The area of curve under the product |V|*|I |is the real power. You can draw two sine waves with the two separated by 90deg, then multiply it graphically, so you get a picture of it.

The product at any given instant is zero (assuming your drawing is good).

Vrms * Irms (each computed separately, then multiplied) gives apparent power. Since the they're not multiplied together simultaneously, you'll see that the phase shift does not affect apparent power.

What you're describing as efficiency is W/VA which is power factor.
Efficiency is the comparison of two power, so if you measure the real power in to a transformer, and real power out of transformer, the efficiency is output/input and output-input will always equal dissipation where the loss occured.
 
"The area of curve under the product |V|*|I |is the real power. You can draw two sine waves with the two separated by 90deg, then multiply it graphically, so you get a picture of it. "

Voltage don't have to expressed as RMS quanities. Each can be mathamatical function.
You can multiply a 200 volt square wave at 42 cycles X sawtoothed shape current wave of 10 amps and you'll get the power.
The "X" or "Times" is a Vector cross product if you want the power. The math is not easy without some smart people and the right tools.
Sine waves and RMS make it all easy.
 
Electric-Light said:
The area of the curve under the product |V|*|I| is the real power. You can draw two sine waves with the two separated by 90deg, then multiply it graphically, so you get a picture of it.
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Not quite, the area under the v*i curve represents energy. You must divide by time to obtain average, real power.

If you integrate magnitudes, I don?t know what you get.

In your example, we see alternate lobes of positive and negative power which average out to zero as we would expect.
The product at any given instant is zero (assuming your drawing is good).
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Not so again. The vi product is zero only when either v or i is zero. The average over a period is zero in your example.
Vrms * Irms (each computed separately, then multiplied) gives apparent power. Since they're not multiplied together simultaneously, you'll see that the phase shift does not affect apparent power.
Click to expand...

Well yes, that is the idea.

What you're describing as efficiency is W/VA which is power factor.
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Oops again! W carries the units of energy while the Vrms*Irms product carries the units of power. PF is dimensionless.

What I am saying is the average value of v*i is real, not apparent, power into the device since the positive and negative reactive powers cancel in the integration. Call this value Pi, and call the power to the load, Po. Then,

Po/Pi = efficiency, not PF.

On third thought, I no longer believe you can integrate the vi product to obtain apparent power. You can however write,

Pa = Vrms*Irms

But I am not sure what the advantage is with a strongly non-linear load..
 
rattus said:
Oops again! W carries the units of energy while the Vrms*Irms product carries the units of power.
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Oops x 2. W & VA carry the same fundamental units: M?L?/T? (i.e. power, as in kg?m?/s?). Unless I misunderstood what you meant.
 
Electric-Light said:
So, there's surprising amount of lack of understanding of definition of power, appreant power and RMS values...
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It is no surprise that I lack a good understanding of many principles of electrical theory, I slept through many of my night courses.:grin:

However, it bothers me that the basis of this whole thread was not an honest attempt to learn or qualify a position.:mad:

Point blank: what is your game and why?:-?

For better or worse, I post.:)
 
I have one question, does zero equate to nothing?


Roger
 
rattus said:
Mivey, it irritates my tender psyche to hear expressions such as,

"The watts is equal to the volts times the amperes."
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Rattus, I agree, it burns my butt too, it should be "The watts are equal to the volts times the amperes" :mad: ;)

Roger
 
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