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?

 

[krw@att.bizzzzzzzzzzzz]

On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com>
wrote:

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]

W. eWatson wrote:

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]

On 4/7/2012 6:14 PM, Jamie wrote:

W. eWatson wrote:

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




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]

On 4/7/2012 5:18 PM, krw@att.bizzzzzzzzzzzz wrote:

On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson"<wolftracks@invalid.com
wrote:

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.
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.

 

[John Larkin]

On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson"
<wolftracks@invalid.com> wrote:

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?

If you have an efficient switching regulator, as many DC/AC converters
are, and you connect them to a constant load, they will indeed exhibit
a negative input impedance. As you increase the DC input voltage, the
input current will drop, such as to maintain about constant input
power.

That's conservation of energy: if the converter's losses are small,
power in = power out.

Other types of devices may behave differently. Gotta know the context.

We had that problem pop up last week. We designed a laser controller
that runs from a customer's 24 volt DC supply, and it uses switching
regs to make +5 and +3.3 and such for internal use. If you connect it
to a real brute 24 volt supply, it starts right up and draws about 400
mA at 24 volts. But if we connect it to a 1-amp current-limited
supply, it bogs down the supply and won't start up. It's probably OK,
since the customer has a big supply, but we're adding better
soft-start circuits to the next rev.


--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators

 

[Jamie]

W. eWatson wrote:

On 4/7/2012 6:14 PM, Jamie wrote:

W. eWatson wrote:

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




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

 

[Tim Wescott]

On Sat, 07 Apr 2012 19:06:10 -0700, W. eWatson wrote:

On 4/7/2012 6:14 PM, Jamie wrote:
W. eWatson wrote:

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




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.

Your book must be older than anyone alive today, and possibly their
parents, as well.

The conservation of energy has been a recognized physical law for a very
long time.

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com

 

[John Fields]

On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:

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.

---
P
How does I = --- manage to not work out for AC?
E

--
JF

 

[John Fields]

On Sat, 07 Apr 2012 19:56:22 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson"
wolftracks@invalid.com> wrote:

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?

If you have an efficient switching regulator, as many DC/AC converters
are, and you connect them to a constant load, they will indeed exhibit
a negative input impedance. As you increase the DC input voltage, the
input current will drop, such as to maintain about constant input
power.

---
But, in the poster's context, that's all irrelevant since what he was
asking was that for constant power into a load, as _output_ voltage
decreases, must _output_ current not increase.

Indeed it must and - for a perfect supply with a constant voltage
input - the input current will change not one whit.

--
JF

 

[krw@att.bizzzzzzzzzzzz]

On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:

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.

---
P
How does I = --- manage to not work out for AC?
E

PF, dumbass.

 

[John Fields]

On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:

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.

---
P
How does I = --- manage to not work out for AC?
E

PF, dumbass.

---
Watts is watts...

--
JF

 

[Fred Abse]

On Sat, 07 Apr 2012 17:02:40 -0700, W. eWatson wrote:

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.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[krw@att.bizzzzzzzzzzzz]

On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:

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.

---
P
How does I = --- manage to not work out for AC?
E

PF, dumbass.

---
Watts is watts...

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]

On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:

Watts is watts...

But ain't always Volt-Amperes.

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)

 

[John Fields]

On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
<excretatauris@invalid.invalid> wrote:

On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:

Watts is watts...

But ain't always Volt-Amperes.

---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.

--
JF

 

[John Fields]

On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:

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.

---
P
How does I = --- manage to not work out for AC?
E

PF, dumbass.

---
Watts is watts...

But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.

---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was 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.

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...
---

Go back to your 555s.

---
Interesting that those of you who haven't been able to get a handle on
how to use a 555 efficaciously, for any purpose, try to use your
ignorance to make us, who use them with delight, seem inferior.
---

You're as dumb as DimBulb.

---
Perhaps, but it seems you still have a long way to go before you get
here.

--
JF

 

[Jamie]

John Fields wrote:

On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:


On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:


Watts is watts...

But ain't always Volt-Amperes.


---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.

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

 

[John Larkin]

On Sun, 08 Apr 2012 14:43:57 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:

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.

---
P
How does I = --- manage to not work out for AC?
E

PF, dumbass.

---
Watts is watts...

But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.

---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was 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.

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...

What engineering school did you graduate from?


--

John Larkin Highland Technology Inc
www.highlandtechnology.com jlarkin at highlandtechnology dot com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators

 

[Bob Myers]

On Sunday, April 8, 2012 9:57:56 AM UTC-6, k...@att.bizzzzzzzzzzzz wrote:

On Sun, 08 Apr 2012 10:18:16 -0500, John Fields


This isn't strictly true for AC.

---
P
How does I = --- manage to not work out for AC?
E

PF, dumbass.

---
Watts is watts...

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.

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.

Bob M.