"Char Jackson" <none@none.invalid> wrote in message
news:tfrsp5limrr2gnpcd2njbtfiplgh0o019d@4ax.com...
> On Mon, 15 Mar 2010 09:26
2 -0500, "Dave" <davidj92@wowway.com>
> wrote:
>
>>
>>
>>"Char Jackson" <none@none.invalid> wrote in message
>>news:jshrp5dnv2rc5q64sci4uin1gvntd8t289@4ax.com...
>>> On Sun, 14 Mar 2010 22:577 -0500, "Dave" <davidj92@wowway.com>
>>> wrote:
>>>
>>>>
>>>>
>>>>"Char Jackson" <none@none.invalid> wrote in message
>>>>news:dj9rp55nqq6lrchqiquhra5dup68gl6e4l@4ax.com...
>>>>> On Sun, 14 Mar 2010 22:03:16 -0500, "Dave" <davidj92@wowway.com>
>>>>> wrote:
>>>>>
>>>>>>
>>>>>>
>>>>>>"Char Jackson" <none@none.invalid> wrote in message
>>>>>>news:mboqp51ghem582m851feaf2o68o8qqjbc9@4ax.com...
>>>>>>> On Sun, 14 Mar 2010 16:28:40 -0500, "Dave" <davidj92@wowway.com>
>>>>>>> wrote:
>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>>"Peter Foldes" <okf22@hotmail.com> wrote in message
>>>>>>>>news:hnjj9t$d8b$1@speranza.aioe.org...
>>>>>>>>> Dave
>>>>>>>>>
>>>>>>>>> You do know the difference between a household 100v or a 347v
>>>>>>>>> commercial
>>>>>>>>> in the amount of their usage
>>>>>>>>>
>>>>>>>>> --
>>>>>>>>> Peter
>>>>>>>>>
>>>>>>>>
>>>>>>>>I can't give you a scientific example of the differences between
>>>>>>>>those
>>>>>>>>two.
>>>>>>>>Off the top of my head I think you might be comparing single phase
>>>>>>>>to
>>>>>>>>a
>>>>>>>>multiple, like delta or y. I can say the usage, no matter what the
>>>>>>>>supply,
>>>>>>>>is going to be dependant on the demand.
>>>>>>>>One final attempt at trying to not look stupid is that for the same
>>>>>>>>device,
>>>>>>>>if it's capable of handling the higher voltage, usually the higher
>>>>>>>>the
>>>>>>>>voltage, the lower the current so the lower the cost to use.
>>>>>>>
>>>>>>> No, the cost will be the same regardless of the voltage. Usage is
>>>>>>> measured in Watts. Watts are the product of voltage times current,
>>>>>>> so
>>>>>>> if voltage goes up by a certain factor then current comes down by
>>>>>>> the
>>>>>>> same factor. In the end, the Watts are the same, therefore the usage
>>>>>>> and the cost are the same.
>>>>>>>
>>>>>>
>>>>>>I agree with you as far a theory goes, P=I*E. But in application, if
>>>>>>you
>>>>>>have a dryer for instance. Hook it to 120V and it might use 15 Amps,
>>>>>>hook
>>>>>>it
>>>>>>to 240V and it will probably be less than half the Amperage. So, the
>>>>>>total
>>>>>>power used will be less for the 240V than for the 120V.
>>>>>
>>>>> Come on, you know that's not true.
>>>>>
>>>>
>>>>Actually, it is and IIRC it has to do with wires ability to carry power
>>>>more
>>>>efficiently at higher voltages. As I understand it, that's also one of
>>>>the
>>>>reasons utilities transmit power at kilovolts instead of 120 volts.
>>>>But, since I can't quote any statistics and am too lazy to google any
>>>>more,
>>>>I'm going to say you win. I think the difference is minor anyway so
>>>>let's
>>>>revert to theory and say you're right.
>>>>Dave
>>>
>>> You're confusing Ohm's Law with transmission theory, two totally
>>> different things. You correctly stated one form of Ohm's Law above,
>>> P=IE. That darn equals sign gets in the way of your current position.
>>>
>>>
>>
>>So, when you transmit power it acts one way, but when you apply it to an
>>appliance it acts another? I was never taught that and never found it in
>>application either. YRMV.
>
> Yes, transmission lines have known losses which need to be accounted
> for and factored into their designs, but that has nothing to do with
> the current discussion. There are no long distance transmission lines
> in a residential or even commercial property, so the losses from such
> lines are irrelevant to the current discussion. I suspect the reason
> you were never taught that is because this is far outside your area of
> expertise, which I believe is what you said in a previous post.
>
One final post and I'm done with this discussion. Now you've changed from
conductors ability to transmit, stated as MHO, to resistance or impedance, a
conductor's inability to transmit. It looks like you have trouble staying on
target and shift from one side to the other just to be right. I suspect your
errors in your postings is due to you trying to be right no matter the cost
so you don't stay on topic. Yes, I'm willing to admit my limitations, but
now realize you aren't and are limited, but won't admit it.
> Take your example of the electric dryer. If you have two dryers, one
> designed for 120v and the other designed for 240v, the 120v model will
> draw twice as many amps as the 240v model. Both dryers will use the
> same number of Watts, (because P=IE), assuming the only differences
> between the two models are the voltages they are designed to operate
> with.
>
Again, you are assuming power (P) will remain the same and current (I) will
halve when you double voltage (E or V), which I'm saying you don't know, but
are assuming.
>>I actually did not state Ohm's law, it is I=E/R. What I stated was Watt's
>>law, which is P=I*E. You do realize one is relevant to power and one is
>>relevant to resistance?
>>I would agree, except you are assuming the power is the same in both
>>instances, or more correctly, the current draw, which I am saying is not.
>
> If we're having a technical discussion, then P (power) is expressed in
> Watts, not amps. Current draw refers to amps, but power refers to
> Watts. Watts are the product of amps times voltage. (P=IE)
>
I'm sorry, but we were in fact having a technical discussion due to your
incessant postings trying to make it look like you know more than I do, or
something to that affect. I really don't care if you do, but will not let
you make me look bad when you so obviously don't know as much as you think
you do. For instance, I don't know why you'd take the time to reiterate
Watt's Law (which you erroneously called Ohm's Law) except trying to avoid
what you posted. See, you didn't recognize the difference when it was first
posted and now you're some kind of expert on Watt's Law.
>>So, in my case, you can't solve for P in both 120 and 240 by assuming I is
>>automatically going to be half if E is doubled.
>
> If you're saying that "P equals IE" is wrong, and should be restated
> as "P is approximately equal to IE", well we better get the textbooks
> updated with this new formula. In the meantime, yes, if E is doubled
> then I will be halved and P will remain unchanged. It doesn't just
> work that way in the textbooks, it works that way in the field, too.
>
Now you're trying to save face again by twisting what I wrote. It was clear
and still is clear, I did not say or suggest P=I*E is anything but exact. As
far as textbooks and formulas, when you know the difference between Watt's
Law, Ohm's Law, MHO, Impedance and etc. we can discuss again. Well, that's
probably not going to happen as you don't want to discuss, you want to
embarrass and so far I'd say it's you that needs the textbooks.
>>I hope you're not going to have us Thevonize a circuit, I hated that more
>>than anything. Plus, I'd have to dig out some reference books just to get
>>all the steps correct.
>
> Not a bad idea to dust off the books now and then.
>
Again, when you decide you're ready to read a textbook let me know, I'll
ship you some for free. You'll LOVE the one with Thevenin's Theorum.
