I have thought about the best possible way to word this and have been stumbling around with the solution to the question. So please forgive me for the jumbled word salad that is incoming.

I was trying to figure out how many kilowatts a single 15A / 120 V circuit would consume in an hour/day/month. Assuming that I was running the circuit on the high end of consumption right before it trips.

So basically assuming that one more amp would trip the circuit how many kilowatts would it use in an hour?

Sorry again for the off the wall question, it was one of those shower thoughts that has grown into a deep question that I was trying to figure out alone but do not have enough understanding of current and general knowledge of electricity to figure out.

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-"Young" James

"To my mathematical brain, the numbers alone make thinking about aliens perfectly rational. The real challenge is to work out what aliens might actually be like." -Steven Hawking

In simplest terms, 15A X 120v = 1,800 watts. X 1 Hour = 1,800 watt/hours or 1.8kwh. x 24 hours = 43.2kwh per day.

There are caveats about power factor of the load and such things, but this will get you very close.

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"Freedom is just Chaos, with better lighting." Alan Dean Foster

Thank you for the quick reply, you do not disappoint good Sir!

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-"Young" James

"To my mathematical brain, the numbers alone make thinking about aliens perfectly rational. The real challenge is to work out what aliens might actually be like." -Steven Hawking

Thanks anyhow Chris! I was trying to figure this out, but there was a few too many conversions for me to wrap my head around it. That's what happens when you work towards a project management / org development degree. haha!

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-"Young" James

"To my mathematical brain, the numbers alone make thinking about aliens perfectly rational. The real challenge is to work out what aliens might actually be like." -Steven Hawking

Perhaps this might help in the future....the ohm's law pie chart.

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"Freedom is just Chaos, with better lighting." Alan Dean Foster

It only helps if you know that I=amps, E=volts, R=resistance, P=Power (kw).

'I' stands for current. It comes from the French word intensité , used by French scientist André-Marie Ampère to describe the 'intensity of the electric current.'

'E' stands for electric potential.

I think that key is in the middle of the pie. Great chart by the way. I need to print it out and keep it in my engineering handbook.

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I have thought about the best possible way to word this and have been stumbling around with the solution to the question. So please forgive me for the jumbled word salad that is incoming.

I was trying to figure out how many kilowatts a single 15A / 120 V circuit would consume in an hour/day/month. Assuming that I was running the circuit on the high end of consumption right before it trips.

So basically assuming that one more amp would trip the circuit how many kilowatts would it use in an hour?

Sorry again for the off the wall question, it was one of those shower thoughts that has grown into a deep question that I was trying to figure out alone but do not have enough understanding of current and general knowledge of electricity to figure out.

Hi James,

The answer to your question depends on what you define as a
"circuit." Is it a really huge incandescent lamp, a fluorescent lamp,
some kind of motor, a transformer, or a what? Do you know its PF
(power factor)?

The amount of true power, in watts, consumed varies widely
depending on the type of "circuit" you have. It can, depending on the
load PF (power factor), vary from roughly 990 watts of true power
up to the 1800 watts calculated by others. The 1800 watt figure is a
MAXIMUM and is what you would consume if the load was a
humongous incandescent lamp, or some such, which is a resistive
load.

Typical power factors are in the range below:
incandescent lamp PF = 1.0
fluorescent lamp PF = 0.95 to 0.97
synchronous motor PF = 0.80 to 1.0
squirrel cage motor PF = 0.75 to 0.92
induction motor PF = 0.55 to 0.65
plus others with a lower PF

Note: In researching the above, I found conflicting info in regard
"fluorescent lamps." I suspect the PF given above is likely for the
"old" type, the 12 to 36 inch long tubes. One source I found talks
about "CFLs," the compact fluorescent lamps that are now the
"biggie" in society since Obama "outlawed" incandescents. This
source says a CFL can have a PF as low as 0.6. A CFL does consume
fewer watts, but, if a PF of 0.6 is the new norm we use to light our
homes, instead of a PF=1 with incandescents, does the trade-off
improve anything because of the needed VA overhead which will be
explained below?

The power consumed in watts by a DC circuit is calculated simply as
watts = V(volts RMS) X I(amps RMS). In a DC circuit the PF = 1.
The power consumed in watts by an AC circuit is calculated as watts
= V(volts RMS) X I(amps RMS) X PF. As you can see, in the case of
an induction motor, if PF = 0.55, AND we used your numbers of 120V
and 15 amps, watts = 990 and not 1800.

Volts RMS is mentioned above. What is it? It is the AC voltage that
will do the same amount of "work" or generate the same amount of
"heat" as its DC equivalent.

To clarify a bit more in regard the above, in a DC circuit we speak of
watts, but in an AC circuit we use the term volt-amps, or VA. It has
two components, watts and VAR, or volt-amps reactive.

In a DC "circuit", back in the days of Edison, things were simpler.
The circuit consumed watts of power provided by a DC generator.
Fuses, breakers, and all other parts of the "grid" were calculated
using simple Ohm's Law. Tesla, a former employee of Edison,
developed AC power. DC power was "short range" in that "IR" losses
in the wiring put a limit on the distance from the power station to a
home. AC was different, it was "long range" and what the world uses
today. It had a complication though in that it had to transmit
"apparent" power and not just "real" power as with DC. Apparent
power has, as mentioned above, two components, watts and VAR. An
AC generating station has to transmit to you, over its wires, the
power in watts you use AND power that is unused, the VAR. It is
DRAWN from the power line and then RETURNED to it every cycle.
The size of fuses, breakers, and other components must be able to
handle apparent power and not just real power. It is in the best
interest of a power company if PF is as close to one as possible. This
is one reason that incandescent lamps with PF=1 were popular.

At the risk of further complicating the issue, here is a bit more. :)

In an AC system the power factor is rarely one, and that's where
the error in calculating power in watts comes in. There is a phase
difference because the load impedance, usually inductive, in what you
call a "circuit," has an impedance, Z, and it is complex. In a pure DC
system Z = R where R is a purely resistive load. In an AC system Z =
R+/- jX where X is a reactance, either inductive or capacitive, i.e., +
or - and the imaginary component is expressed by the use of "j."
Some in academia use "i," but I'm an engineer and use "j." Deal with
it. :)

It is this load reactance that introduces the VAR component and
requires the power company to generate more power than is used by
your "circuit." Its also the component that introduces calculation
error in that the PF is likely unknown. A "rule of thumb," used by
some in the industry is to say the maximum load power, i.e., real
power in watts, can't exceed 50% of the available volt-amps. This is
not "set in cement" tho, and the percentage can vary as this would
require PF to be >= 0.5

Fun stuff, huh? :)

---

River Song (aka Linda Latte on planet Earth)
"Happy I-Phone girl on the GO GO GO"

Well, of course, 100W sounds more powerful! haha But, you are 100% in the true power per stereo channel. Sometimes "white man speak with fork'ed tongue." Salesmen are like that.

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River Song (aka Linda Latte on planet Earth)
"Happy I-Phone girl on the GO GO GO"

At the museum, I rode in an electric locomotive a couple months ago. Out at the end of the line, four miles from the substation, the voltmeter showed 600... until the engineer opened the throttle. Then it dropped to about 540. Leaving the station, close to the sub, it only dropped to about 590. That illustrates just how wrong Edison was about DC.

That said, Commonwealth Edison was still providing DC to certain buildings in downtown Chicago at least into the 1980s, if not later. How do I know? My Godfather worked for Com Ed. (He also told me the stories popular among railfans about the old streetcar tracks still buried in the streets being part of Com Ed's return grid are overblown.)

---

David
Sitting on my butt while others boldly go,
Waiting for a message from a small furry creature from Alpha Centauri.

I thought Edison was totally behind DC. It was Westinghouse and Tesla that promoted AC. Didn't Edision kill a elephant to show how bad AC was? AC won and became the primary system used today but I don't think that was Edison's choice.

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Thought the real reason was you could change the voltage of AC rather easily, but not DC. No matter which you have to pump the watts through the wire. Ohms law. P = E^2/R. The R for a mile of wire is the same AC or DC, but if you can put 100KV in it AC but only 600VDC, which one puts more power at the load end? Duh. The simple transformer is what won.

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I thought Edison was totally behind DC. It was Westinghouse and Tesla that promoted AC. Didn't Edison kill a elephant to show how bad AC was? AC won and became the primary system used today but I don't think that was Edison's choice.

What you state is true enough, but then DC without superconducting transmission cables, needs a power station about every mile I've read, DC inside our PCs works fine and AC works for long distance transmissions. In any case We all owe Tesla and Edison's unknown scientific staff a great deal of gratitude, Edison's name should be replaced with Tesla's name in My opinion.

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The T1 Trust, PRR T1 Class 4-4-4-4 #5550, 1 of America's First HST's

I thought Edison was totally behind DC. It was Westinghouse and Tesla that promoted AC. Didn't Edision kill a elephant to show how bad AC was? AC won and became the primary system used today but I don't think that was Edison's choice.

Edison was #1 in promoting DC, and Westinghouse & Tesla were the AC guys. To demonstrate to people the dangers of AC Edison publically executed small animals, dogs and cats, with AC. He also invented the first electric chair.

Here was its first victim:

William Kemmler, 28, a convicted killer was next up for execution.

Here is a statement from a spectator at his execution by Edison's electric chair.

"Kemmler was strapped into the chair on August 6, 1890. The first jolt of alternating current lasted 17 seconds. Kemmler continued struggling. A second jolt lasted more than a minute, until smoke was seen rising from the body."

"The New York Times said, "an awful spectacle, far worse than hanging." The state commissioner on humane executions saw it differently. It was, he said, "the grandest success of the age."

Gee, huh?

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River Song (aka Linda Latte on planet Earth)
"Happy I-Phone girl on the GO GO GO"

War of Currents.
https://en.wikipedia.org/wiki/War_of_Currents
Both Edison and Tesla won:)

But if you read this "Remnant and existent DC systems"
https://en.wikipedia.org/wiki/War_of_Currents#Remnant_and_existent_DC_systems

DC power distribution is still common when distances are small, and especially when energy storage or conversion uses batteries or fuel cells.

Thats nonsense.

For long-distance transmission, HVDC systems may be less expensive and suffer lower electrical losses. For underwater power cables, HVDC avoids the heavy currents required to charge and discharge the cable capacitance each cycle.
https://en.wikipedia.org/wiki/High-voltage_direct_current#Advantages_of_HVDC_over_AC_transmission