Making sense of watts, amps and volts -- a typo?

[W. eWatson]

Could the following be a typo. Written by someone to me on inverters (DC
to AC).


Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.

1000 watts at 120VAC is about 8.3 amps.
1000 watts at12VDC is about 83 amps. <--typo? Shouldn't it still be 8.3?

 

[[email protected]]

Could the following be a typo. Written by someone to me on inverters (DC
to AC).


Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.

This isn't strictly true for AC.

1000 watts at 120VAC is about 8.3 amps.
1000 watts at12VDC is about 83 amps. <--typo? Shouldn't it still be 8.3?

Do you have a calculator?

1000/12 = ~83 on mine.

There is also a little thing of efficiency, in here. The inverter is going to
get hot.

 

[Jamie]

Could the following be a typo. Written by someone to me on inverters (DC
to AC).


Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.

1000 watts at 120VAC is about 8.3 amps.
1000 watts at12VDC is about 83 amps. <--typo? Shouldn't it still be
8.3?

The assertion is correct. That is (83 amps) is the correct answer.

Get out your Ohms law book on Power calculations and do your
home work!



Jamie

 

[W. eWatson]

The assertion is correct. That is (83 amps) is the correct answer.

Get out your Ohms law book on Power calculations and do your
home work!



Jamie

My book are sooo old it wouldn't help. Actually I see what happened. Ah,
nuts. I read both as 12, and didn't notice the 120. So, yes, it's
correct. BTW, either Ohm's Law or Kirchoff's Law was discovered by
Cavendish. Not only was he severely anit-social, he often offered up
some of his discoveries to various scientists.

 

[W. eWatson]

This isn't strictly true for AC.


Do you have a calculator?

1000/12 = ~83 on mine.

There is also a little thing of efficiency, in here. The inverter is going to
get hot.

What I did was mis-read it, as explained by my post to Jamie.

Here's how the paragraph began to the two sentences above that followed:

Make sure cables between the batteries and the inverter are short, fat,
well-crimped and screwed down tight on both ends. They will be handling
ten times as much current as the AC side of the inverter.

He's basically telling me that I'm going to need some fat wire on one side.

 

[Jamie]

My book are sooo old it wouldn't help. Actually I see what happened. Ah,
nuts. I read both as 12, and didn't notice the 120. So, yes, it's
correct. BTW, either Ohm's Law or Kirchoff's Law was discovered by
Cavendish. Not only was he severely anit-social, he often offered up
some of his discoveries to various scientists.

Well, then he would fit right in here for the most part, as far as the
anti-social goes

Jamie

 

[[email protected]]

PF, dumbass.

 

[Fred Abse]

Could the following be a typo. Written by someone to me on inverters (DC
to AC).


Volts x Amps = Watts so as the voltage goes down, the amperage goes up to
maintain the same number of watts.

1000 watts at 120VAC is about 8.3 amps. 1000 watts at12VDC is about 83
amps. <--typo? Shouldn't it still be 8.3?

No.

Do the math:

Volts x Amps = Watts
Therefore Amps = Watts / Volts
1000/120 = 8.3 (approx)
1000/12 = 83
QED.

 

[[email protected]]

But, as any first-year engineering student knows, I * V <> W, where AC is
concerned. Go back to your 555s.

You're as dumb as DimBulb.

 

[Fred Abse]

Watts is watts...

But ain't always Volt-Amperes.

 

[Jamie]

When I see the term "VA", I know we're dealing with "REACTIVE" power.

PF (Power Factors) denotes the difference between "REACTIVE" and
"RESISTIVE (True power)" So, using the term VA is assumed power.

Having AC in the equation has nothing to do with it actually, I can
put AC into a purely non reactive load and it would simply power.. There
difference being is, you need to take measurements along the vectors to
come with a sum of power with in a time frame. Normally, with a clean
sinusoidal wave, we just assume RMS power.

if you look at this formula.

P = I+V*Cos(x), you'll notice that "I" is used as "Amperes" here.
This is a AC power formula but you don't see any distinction here with
the use of "VA" as would be in case of "REACTIVE" power.

Jamie

 

[[email protected]]

If you use the term "watts" in connection with AC, any engineer (as opposed to engineering student) will understand that you are talking about true power, which has been corrected for the phase difference or "power factor." Watts IS watts; no engineer worth the title will confuse "watts" with "volt-amps."

I * V does equal W in AC, since in AC work that HAS to be a vector calculation.

Absolute nonsense. ...as to be expected from dumbass Myers.

 

[[email protected]]

Moron, reading between the lines, anyone with as much as half a brain, would
understand that he was talking about an INVERTER, which is *NOT* resistive.

But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.

What a dumbass.

And, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.

I guess you never made it that far, though...

Like I said, you're as dumb as AlwaysWrong. Keep proving it.

Like AlwaysWrong, you insist on proving that you're *always* wrong. I have
nothing against the 555, just one-trick-ponies, like you.

AlwaysWrong.

 

[[email protected]]

What engineering school did you graduate from?

He didn't. You know, that pot and kettle diversion.

 

[[email protected]]

No, he was talking about volts and amps. You can't get to watts from there.

 

[[email protected]]

When I see the term "VA", I know we're dealing with "REACTIVE" power.

No, when you se "VA" you simply don't know "W"; not enough information given.
With an inverter you can *bet* the power factor is not unity.

PF (Power Factors) denotes the difference between "REACTIVE" and
"RESISTIVE (True power)" So, using the term VA is assumed power.

What? Did you really mean to write that nonsense?

Having AC in the equation has nothing to do with it actually, I can
put AC into a purely non reactive load and it would simply power.. There
difference being is, you need to take measurements along the vectors to
come with a sum of power with in a time frame. Normally, with a clean
sinusoidal wave, we just assume RMS power.

You've just specified the PF by stating a resistive load, so no, you're still
wrong. Without the PF explicitly stated you *cannot* get there from here.

if you look at this formula.

P = I+V*Cos(x),

Your "COS(X)" *is* the power factor, which is only true for sine waves.

you'll notice that "I" is used as "Amperes" here.
???

This is a AC power formula but you don't see any distinction here with
the use of "VA" as would be in case of "REACTIVE" power.

WTF, is the COS(X) term, if not to cover reactive power? Jamie, go back to
your ham shack.

 

[Jamie]

Oh really..

Well excuse me, I slipped with the keyboard. I hope you really don't
think I intended it to be that way ? If so, you are naive.

In the first place, it's not I+V*cos(x), it's I*V*cos(x) and, in the
second place, the cos(x) term is used to determine the actual power
dissipated.

And if you want to start punching at the bit, from what I can see with
your last assertion, It seems that It's you that has a problem with
understanding this. Maybe you should brush up on Kirchoffs laws a little
on this subject.

nuff said, And btw, there is such things as RMS power. How much in
the dark you are.

You know, I tried to actually help you but it seems obvious you have a
one way street and much of which have people going the wrong way, except
for you of course.

Jamie

 

[[email protected]]

It shows.

 

[[email protected]]

Then your comments about what first-year engineering students would
have been taught, and people not making it that far, are absurd.

I don't recall learning about reactive power first-year; that was more
like year 2 or 3, when I took the electric machinery courses.

I first encountered it in college in calc class and then in second semester
physics. ...both first year.

 

[[email protected]]

How lonely you must be.

Joining the AlwaysWrong brigade?

Isn't it a bitch when sitting there late at night scratching your
balls because you, just don't get what you though you did.

You don't have to tell us about your Saturday night. Really.

Sorry, But you failed, again.

Sorry, but you failed EE-101, as usual, Maynard.

As for Ham radio, that is just a Hobby and social hour get-together, I
actually do work in the Electrical/Electronics field and I do very well
at it. I take that opinion from bonuses, raises and royalties earned.

You *certainly* don't show it here. You're wrong about just about everything.

I've never had a problem being employed and been so for many years
performing engineering level work. I guess If "I" was so miss led as you
seem to think, our company that I work for would have been in trouble
years ago.

"Engineering level work" tells it all. Like Fields (and DimBulb), you *aren't*
an engineer, and it show. Loudly.

Oh btw, I used to work for a company that made electronic components
for the patriot missiles. Yes, the actual electronics involved in the
missiles and transport. I bet that really scares you now !

You even sound like AlwaysWrong; "I worked for". "I worked on". "I worked
with". Not "I Did".

I must say, ignorance is bliss.

I believe you! Really, I do!

Put the beer down, it's impairing your judgment.

I don't drink, dimwit. You really are slow.