Am I the only one who remembers ELI the ICE man? Voltage (e) leads
current (i) in an inductor (L) and current (i) leads voltage (e) in a
capacitor (C). Must have been the old Navy guy teaching electronics in
high school.

 

http://www.electronicstheory.com/html/e101-31.htm

 

 

Doug Hilderbrand | Systems Analyst, Information Technology | Crane
Aerospace & Electronics



From: Ken Cornetet [mailto:[email protected]] 
Sent: Thursday, October 20, 2011 1:09 PM
To: NT System Admin Issues
Subject: RE: ( OT ) Convert mA to Watts

 

Power factor is a bit difficult to explain, but here goes:

 

Ohm's law: Volts = Amps * resistance

Watts = Volts * Amps or substituting, Watts = Volts squared /
resistance.

 

 

Power factor is the ratio of "real" power divided by the apparent power.
Under what circumstances do real and apparent power differ from each
other? There are two reasons for this:

 

1.       A nonlinear load (like something with rectifiers) will present
a resistance that appears to vary with the instantaneous voltage. For a
sinusoidal alternating voltage, this leads to a non-sinusoidal current,
which creates a non-sinusoidal power. This non sinusoidal power will
deliver less average power than the equivalent sinusoidal power.

2.       A reactive load (capacitive and inductive) will have a
sinusoidal current, but it will lead or lag the voltage. This means that
any instantaneous power (voltage * current) will be less than the
average voltage times the average current.

 

This means that with a power factor, you get less "work" (power) for a
given voltage and current. In other words, a load with a power factor
less than 1 requires more "push" (current) to get the same amount of
work done as a load with a power factor of one.

 

A real world example:

 

Let's say you have a UPS running on batteries that produces 120V AC.
This feeds a reactive load that draws 60 Watts at a power factor of .5.
This means that your battery is supply 60 Watts of power (assuming 100%
UPS efficiency), BUT your UPS is delivering a current of 1 Amp (not
60/120 or .5 Amps) because of the power factor. This is why UPS are
rated in VA instead of Watts.

 

Ken Cornetet 812.482.8499

To err is human - to moo, bovine.

 

From: James Kerr [mailto:[email protected]] 
Sent: Thursday, October 20, 2011 12:58 PM
To: NT System Admin Issues
Subject: Re: ( OT ) Convert mA to Watts

 

You lost me at "factor".

On Tue, Oct 18, 2011 at 12:37 PM, Ben Scott <[email protected]>
wrote:

On Tue, Oct 18, 2011 at 12:15 PM, Stefan Jafs <[email protected]>
wrote:
> I guess the big unknown is the PF, I assumed 60% (based on Googeling),
the
> is a power supply, would it not be higher for a transformer load?

 To continue the water analogy, power factor is like a big reserve
tank right before the water tap.  It can cause your water demand to be
out-of-sync with the apparent water usage (coming out of the tap).
You run the faucet for a bit, and the tank starts to drain, but the
supply pipe isn't touched.  Then the tank starts to fill, pulling from
the supply pipe.  Then you shut the faucet off, but the tank keeps
filling.

 Or so I'm given to understand; the actual mechanism behind power
factor is magic to me.  I know a purely resistive load -- like a space
heater -- has a power factor of 1.0.  "Inductive loads" are
"reactive", whatever that means.  :)  Apparently AC motors are
"inductive".  Rectifiers -- like in an AC->DC power supply, such as in
a PC -- are also apparently "reactive".  "Power factor correction"
helps turn equipment with a lower power factor into something with a
higher power factor.

 The numbers I usually see pulled out of the air for PC power supply
units are 0.6 PF for a standard PSU, and 0.9 PF for a PFC PSU.  I have
no idea how much things vary in practice.

-- Ben


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