Power Consumed by Load

[teslafan01]

I always learned to divide the bottom number first. R= e^2/p. Maybe someone could shine some light on the subject?:?

I use RxV=P

so 5.51x208= 1146 watts.

so 9600w-1146w=8454watts

I am confused.

 

[JDBrown]

I always learned to divide the bottom number first. R= e^2/p. Maybe someone could shine some light on the subject?:?

I use RxV=P

so 5.51x208= 1146 watts.

so 9600w-1146w=8454watts

I am confused.

The solution shown in the image you posted is correct. It looks like what you've got incorrect is your power equation. Power is voltage squared divided by resistance. Your basic power equations are:

P=V²/R
P=I²R
P=IE

In your case, (208V)² / 5.51 Ohms = 7851 W, or 7.85 kW.

As a side note, the solution really should have written it the way I did above -- (208V)² -- instead of writing 208V². It would help prevent some confusion.

As for the part circled in red, I'm not sure where that's coming from. Whoever made that graphic wrote (208 x 208)/5.51, but the answer that's written down (0.000127358) is the reciprocal of that. In other words:

(208 x 208)/5.51 = 7851
5.51/(208 x 208) = 0.000127358

I hope this helps.

 

[teslafan01]

Thanks, I just feel like an idiot now. :ashamed1:

 

[JDBrown]

Thanks, I just feel like an idiot now. :ashamed1:

You shouldn't. You had the guts to admit you didn't understand something and ask for help. I've met plenty of people who would just assume they were right and there was a mistake in the solutions manual. They are the ones who should feel like idiots.

But I can empathize with you anyway. I think we've all done something similar. Glad to help.

 

[jumper]

You shouldn't. You had the guts to admit you didn't understand something and ask for help. I've met plenty of people who would just assume they were right and there was a mistake in the solutions manual. They are the ones who should feel like idiots.

But I can empathize with you anyway. I think we've all done something similar. Glad to help.

Nice post JD:thumbsup:

 

[dereckbc]

Thanks, I just feel like an idiot now. :ashamed1:

I agree you should not feel like an idiot. The problem today is with online courses and books. They cannot replace good ole fashion classroom instruction and homework. In a classroom room you would have been lectured for a several hours s on Ohm's Law over a period of a week and would have worked several hundred equations as homework after class. If you did not do it, you would not understand or passed the class. You cannot learn much from a book. It takes experience, passion, and money to learn.

As Eric Clapton once wrote in a song, You have been told, now it is time you learned.

 

[Besoeker]

As Eric Clapton once wrote in a song, You have been told, now it is time you learned.

There is another saying.
The beginning of understanding is the acknowledgement of ignorance.
The op made that major step and is to be congratulated for having done so.
:thumbsup:

 

[mbrooke]

The way Ive done it is converting 9.6kw with 230volts to ohms. The use the ohms value given with 208 to give the new kw rating.


Not technically perfect but it gets you there.

 

[charlie b]

Here is how I did it, before I looked at any other posts. It only takes one step.

Let?s call the power drawn at 230 volts P1, and the power drawn at 208 volts P2.
Given: P1 = 9600, then P2 will equal 9600 times (208/230) squared, = 7851.

Here?s why I know that works:

• Start with P = V²/R
• Therefore R = V²/P

More to the point, for two separate components with resistances R1 and R2,

• R1 = V1²/P1, and
• R2 = V2²/P2

But R is the same for this problem, so

• R1 = R2 = V1²/P1 = V2²/P2

From here we derive,

• P2 = P1 times V2²/ V1²
• Thus, P2 = 9600 times 208²/ 230²

QED

 

[charlie b]

The way Ive done it is converting 9.6kw with 230volts to ohms. The use the ohms value given with 208 to give the new kw rating.
Not technically perfect but it gets you there.

Actually, that is technically perfect. Indeed it is exactly what I just showed, except that I skipped the middle steps and went straight to the end.

 

[Besoeker]

Actually, that is technically perfect. Indeed it is exactly what I just showed, except that I skipped the middle steps and went straight to the end.

The principle of parsimony perhaps?

 

[RichB]

For a quick and dirty answer I have always taken the 230/240 volt Kw ratings and multiplied by 82%---and have always found a test question answer with in less than 75 watts difference--or just used as is out in the field,,

9600*.82=7872--close enough
I know it is not exact but it works

 

[charlie b]

For a quick and dirty answer I have always taken the 230/240 volt Kw ratings and multiplied by 82%. . . . I know it is not exact but it works

It works because if you take 208 and divide it by 230, and then square the result, you get 82%. So you are doing the same thing that I did.

Please note that if the item had been rated at 240 volts, and you plan to run it at 208 volts, then you should use a factor of 75% instead of 82%. Why 75%? I will leave that as your homework assignment. :happyyes:

 

[RichB]

It works because if you take 208 and divide it by 230, and then square the result, you get 82%. So you are doing the same thing that I did.

Please note that if the item had been rated at 240 volts, and you plan to run it at 208 volts, then you should use a factor of 75% instead of 82%. Why 75%? I will leave that as your homework assignment. :happyyes:

You know, I remember the 82% but not how it came about till you mentioned it--I just learned it a long time ago and stayed with it

I am familiar with and understand the math involved to get both 82 and 75%--To solve this for 240V I would follow the same derivation process you did and thus come up with 75%---One thing I am not so sure of is if there is a theoretical reason, other than what has been already mentioned, that comes into play--but I am always ready to learn!! I figure the day comes I decide I have nothing left to learn in this field is the day I quit--because I am going to seriously hurt or kill someone and that's not worth it.