Basic Questions Concerning AC Power . . . 4

[AlbertG]

Good day.

I've had a tussle with this for a bit, and just now decided to break down and float it out for community assistance.

In the scenario which is at issue, we have a "black box" load which may be essentially resistive or may be partially inductive; but has been precharacterized to consume 1000W RMS @ 100VAC (10A).

Now, we have a shortish cable feeding power to this "black box" load, which furnishes a "purely resistive" component of .1Ω to the overall mix. If anything ever stuck while studying these things, Ohm's law would seem to be directly applicable in the process of effectively elucidating the (cable) circuit element, and ultimately determining the power loss through the cable:

E = I X R
E = 10 X .1
E = 1VAC (RMS)

P = E X I
P = 1 X 10
P = 10W (RMS)

5th grade simple.

Assuming all this is so, would there be any sound reason to complicate matters with Kirchhoff power law computations to elucidate the purely resistive cable element? I've seen this hinted at; and, in my humble opinion, such a proposition seems akin to going duck hunting with a Howitzer...

Finally, this has been bugging me too: Who do we credit with the original elucidation of electric power (P = E X I)? Ohm? Watt? Joule?

Thoughts?


Thanks!

 

[waross]

Is the load 1000 Watts plus the 10 Watts loss or a total of 1000 Watts (load equals 1000 Watts minus the 10 Watts loss.)? The wrong answer may land you into the quadratic equation, or the howitzer solution.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[AlbertG]

@waross:

Thanks for stopping by!

Good point. However, that particular aspect isn't really at issue in my little thought problem (it represents a small error on the whole). Again, the hypothetical load is a simple "black box" construct; representing a fixed, precharacterized RMS load value.

What seems to be the nut contended is that if our 1000W load is indeed sufficiently inductive, one cannot take recourse to the simple application of Ohm's Law (standard DC practice) to elucidate the "purely resistive" power cable element; and, therefore, Kirchhoff would need to be consulted.

Practically speaking, this sounded like complete overkill to me; but I'm just checking with the community to see what reasoning might be at work here...

 

[davidbeach]

Scalar addition of vector quantities can sort of work when the angular differences are small but don't work as well when the angular differences are larger. It's all still Ohm's law, just using complex quantities.

Besides, is your black box a constant impedance or a constant power load? Those will respond differently.

 

[AlbertG]

@davidbeach:

Scalar addition of vector quantities can sort of work when the angular differences are small but don't work as well when the angular differences are larger. It's all still Ohm's law, just using complex quantities.

You know, that's essentially what I was thinking: "All roads lead to Ohm."

And, to (hopefully) answer your question, the load, supply, and circuit are completely stable and fixed. For the record, this whole thought problem is in the setting of standard 50/60Hz AC.

Of particular interest here is whether or not the set inductive characteristics of the fixed hypothetical "black box" load have any bearing on the purely resistive supply cable calculations. I was of the impression that, if the load and supply were completely fixed, the resistive supply cable element (also static parameterwise) would be accurately characterized with a single pass through E = I X R; irrespective of the inductive- or non-inductive nature of the downstream load itself.

Does this work?

Thanks again --

 

[marks1080]

I don't see why you would ever analyze this scenario with kirchhoffs law...

If you are saying the load is 1000W than you are implying it is purely resistive.. If it's 1000VA than it may have an inductive (or capactive) element to it... Ohm's law works just fine in either case. Practially, there will an inductive characteristic added from your cable as well.

 

[AlbertG]

@marks1080:
Thanks for the post. I thought as much; but sometimes you question your own sanity in the simplest of things
Meanwhile, I'll stick to hunting this particular game with smaller caliber tools than the great "K"...

Cheers!

 

[waross]

Ohm's law. Not just a good idea. It's the law!!

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[marks1080]

But waross; It's only illegal if you get caught...

 

[LionelHutz]

You really only need to be concerned with the cable voltage drop possibly affecting the load current. A utility source of "100VAC" is never exactly 100VAC and it could easily vary 5%. So, if the cable voltage drop causes a 1% change in the source voltage that is below the source tolerance and hardly worth worrying about.

All current inductive and reactive flows through the cable resistance and only the cable resistance (not inductance) causes a power loss so yes using I^2xR gives you the power loss in the cable.