# Software to show electricity consumption

peterlavington@hotmail.com
07-10-2005, 12:26 AM
Anybody know of free software that will show me how much electricity
(presumably in watts) I am using whilst my PC is on?

Weird question I know but I'm interested since there are times I leave
my machine running overnight and wondering if it's costing me a lot in
power.

Thanks

David H. Lipman
07-10-2005, 12:26 AM
From: <peterlavington@hotmail.com>

| Anybody know of free software that will show me how much electricity
| (presumably in watts) I am using whilst my PC is on?
|
| Weird question I know but I'm interested since there are times I leave
| my machine running overnight and wondering if it's costing me a lot in
| power.
|
|
| Thanks

You won't find it. Computers don't monitor the current and voltage on the AC side of the
power supply.
Get a Digital Multimeter, take the current and voltage readings with the PC quiescent and
calculate it for yourself.

--
Dave
http://www.ik-cs.com/got-a-virus.htm

andy smart
07-10-2005, 12:26 AM
peterlavington@hotmail.com wrote:
> Anybody know of free software that will show me how much electricity
> (presumably in watts) I am using whilst my PC is on?
>
> Weird question I know but I'm interested since there are times I leave
> my machine running overnight and wondering if it's costing me a lot in
> power.
>
>
> Thanks
>
I'm not an electrical engineer, but I would think that the amount of
mains current consumed by your pc while it is on is governed by the

This takes the mains current and supplies it to the components. If the
PC is running then the amount of current used is steady as the
components don't change and therefore draw a consistent level of
current. So a 500w power supply left on for an hour is going to draw
500w per hour or .5 kwhour per hour (in the UK electricity tends to be
measured in kilowatt hours). Also in UK terms we tend to think in terms
of electric fires, a one bar electric fire uses 1 kilowatt per hour so
your computer with a 500w power supply would use half as much.

I'd be grateful if somebody who knows more about these things than I can
explain where I'm wrong, as I'm sure I am!

Oh yes, and don't forget your monitor/printer/etc....

Richard Urban
07-10-2005, 12:26 AM
What you want to do can not be done with software alone. You also need a
hardware interface card. They do make them for industrial use.
..

--
Regards,

Richard Urban

aka Crusty (-: Old B@stard :-)

If you knew as much as you think you know,
You would realize that you don't know what you thought you knew!

<peterlavington@hotmail.com> wrote in message
> Anybody know of free software that will show me how much electricity
> (presumably in watts) I am using whilst my PC is on?
>
> Weird question I know but I'm interested since there are times I leave
> my machine running overnight and wondering if it's costing me a lot in
> power.
>
>
> Thanks
>

David
07-10-2005, 12:26 AM
Hello, andy!
You wrote on Fri, 13 May 2005 20:36:57 +0100:

as> peterlavington@hotmail.com wrote:
>> Anybody know of free software that will show me how much electricity
>> (presumably in watts) I am using whilst my PC is on?

as> I'm not an electrical engineer, but I would think that the amount of
as> mains current consumed by your pc while it is on is governed by the
as> wattage of your power supply.

Not true!

as> This takes the mains current and supplies it to the components. If
as> the
as> PC is running then the amount of current used is steady as the
as> components don't change and therefore draw a consistent level of
as> current. So a 500w power supply left on for an hour is going to draw
as> 500w per hour or .5 kwhour per hour (in the UK electricity tends to
as> be
as> measured in kilowatt hours). Also in UK terms we tend to think in
as> terms
as> of electric fires, a one bar electric fire uses 1 kilowatt per hour
as> so
as> your computer with a 500w power supply would use half as much.

The power consumption is not constant at all. The power supply rating simply indicates how much it can supply rather than what is being used. The CPU load, disk activity, and whether the machine is in standby all affect the instantaneous power consumption. To answer the original question, I know of no system that can actually measure the power consumption. External devices are needed.

David

as> I'd be grateful if somebody who knows more about these things than I
as> can
as> explain where I'm wrong, as I'm sure I am!

See above. I am an electrical engineer.

as> Oh yes, and don't forget your monitor/printer/etc....

True

Al Dykes
07-10-2005, 12:26 AM
In article <eSwunT\$VFHA.3532@TK2MSFTNGP09.phx.gbl>,
David <someone@somewhere.com> wrote:
>Hello, andy!
>You wrote on Fri, 13 May 2005 20:36:57 +0100:
>
> as> peterlavington@hotmail.com wrote:
>>> Anybody know of free software that will show me how much electricity
>>> (presumably in watts) I am using whilst my PC is on?
>
>
>as> I'm not an electrical engineer, but I would think that the amount of
>as> mains current consumed by your pc while it is on is governed by the=20
>as> wattage of your power supply.
>
>Not true!
>
>as> This takes the mains current and supplies it to the components. If
>as> the=20
>as> PC is running then the amount of current used is steady as the=20
>as> components don't change and therefore draw a consistent level of=20
>as> current. So a 500w power supply left on for an hour is going to draw
>as> 500w per hour or .5 kwhour per hour (in the UK electricity tends to
>as> be=20
>as> measured in kilowatt hours). Also in UK terms we tend to think in
>as> terms=20
>as> of electric fires, a one bar electric fire uses 1 kilowatt per hour
>as> so=20
>as> your computer with a 500w power supply would use half as much.
>
>The power consumption is not constant at all. The power supply rating =
>simply indicates how much it can supply rather than what is being used. =
>The CPU load, disk activity, and whether the machine is in standby all =
>affect the instantaneous power consumption. To answer the original =
>question, I know of no system that can actually measure the power =
>consumption. External devices are needed.
>
>David
>
>

You need something like this:

http://www.p3international.com/products/special/P4400/P4400-CE.html

--
a d y k e s @ p a n i x . c o m

Don't blame me. I voted for Gore.

andy smart
07-10-2005, 12:26 AM
David wrote:
> Hello, andy!
> You wrote on Fri, 13 May 2005 20:36:57 +0100:
>
> as> peterlavington@hotmail.com wrote:
>
>>>Anybody know of free software that will show me how much electricity
>>>(presumably in watts) I am using whilst my PC is on?
>
>
>
> as> I'm not an electrical engineer, but I would think that the amount of
> as> mains current consumed by your pc while it is on is governed by the
> as> wattage of your power supply.
>
> Not true!
>
> as> This takes the mains current and supplies it to the components. If
> as> the
> as> PC is running then the amount of current used is steady as the
> as> components don't change and therefore draw a consistent level of
> as> current. So a 500w power supply left on for an hour is going to draw
> as> 500w per hour or .5 kwhour per hour (in the UK electricity tends to
> as> be
> as> measured in kilowatt hours). Also in UK terms we tend to think in
> as> terms
> as> of electric fires, a one bar electric fire uses 1 kilowatt per hour
> as> so
> as> your computer with a 500w power supply would use half as much.
>
> The power consumption is not constant at all. The power supply rating simply indicates how much it can supply rather than what is being used. The CPU load, disk activity, and whether the machine is in standby all affect the instantaneous power consumption. To answer the original question, I know of no system that can actually measure the power consumption. External devices are needed.
>
> David
>
>
> as> I'd be grateful if somebody who knows more about these things than I
> as> can
> as> explain where I'm wrong, as I'm sure I am!
>
> See above. I am an electrical engineer.
>
> as> Oh yes, and don't forget your monitor/printer/etc....
>
> True
>
Thanks for putting me right there David (knew somebody would if I

How much variation is there in the level of mains power taken by things
like disk activity? I can see that having it on standby would make a big
difference though.

I would have thought that the big power drain is the cooling fans which
run constantly, with the other mechanical 'bits' like spinning hard
disks next and electronic components coming last of all?

David H. Lipman
07-10-2005, 12:26 AM
From: "Richard Urban" <richardurbanREMOVETHIS@hotmail.com>

| What you want to do can not be done with software alone. You also need a
| hardware interface card. They do make them for industrial use.
| .
|
| --
| Regards,
|
| Richard Urban
|
| aka Crusty (-: Old B@stard :-)
|
| If you knew as much as you think you know,
| You would realize that you don't know what you thought you knew!

Richard:

That won't work either because you are measuring information on the DC side of the supply
and not the AC side.

Power consumption is equal to the sum of the power losses.
The two losses are..
1) sum of the DC power consumed
2) power loses in the AC power supply

So if the DC consumed is 75watts and the power supply consumes 30watts then the total power
consumption will be 105watts on the AC side.

--
Dave
http://www.ik-cs.com/got-a-virus.htm

Richard Urban
07-10-2005, 12:26 AM
The cards I used in the past monitored AC current and generated a
proportional and linear 4-20ma output for control purposes. The card also
had a computer interface to connect to a computer with monitoring software.

Now, could a computer monitor itself? I have never tried it.

--
Regards,

Richard Urban

aka Crusty (-: Old B@stard :-)

If you knew as much as you think you know,
You would realize that you don't know what you thought you knew!

"David H. Lipman" <DLipman~nospam~@Verizon.Net> wrote in message
news:OC8E3n\$VFHA.3620@TK2MSFTNGP09.phx.gbl...
> From: "Richard Urban" <richardurbanREMOVETHIS@hotmail.com>
>
> | What you want to do can not be done with software alone. You also need a
> | hardware interface card. They do make them for industrial use.
> | .
> |
> | --
> | Regards,
> |
> | Richard Urban
> |
> | aka Crusty (-: Old B@stard :-)
> |
> | If you knew as much as you think you know,
> | You would realize that you don't know what you thought you knew!
>
> Richard:
>
> That won't work either because you are measuring information on the DC
> side of the supply
> and not the AC side.
>
> Power consumption is equal to the sum of the power losses.
> The two losses are..
> 1) sum of the DC power consumed
> 2) power loses in the AC power supply
>
> So if the DC consumed is 75watts and the power supply consumes 30watts
> then the total power
> consumption will be 105watts on the AC side.
>
>
> --
> Dave
> http://www.ik-cs.com/got-a-virus.htm
>
>

T. Waters
07-10-2005, 12:26 AM
andy smart wrote:
> David wrote:
>> Hello, andy!
>> You wrote on Fri, 13 May 2005 20:36:57 +0100:
>>
>> as> peterlavington@hotmail.com wrote:
>>
>>>> Anybody know of free software that will show me how much
>>>> electricity (presumably in watts) I am using whilst my PC is on?
>>
>>
>>
>> as> I'm not an electrical engineer, but I would think that the
>> amount of
>> as> mains current consumed by your pc while it is on is governed by
>> the
>> as> wattage of your power supply.
>>
>> Not true!
>>
>> as> This takes the mains current and supplies it to the components.
>> If
>> as> the
>> as> PC is running then the amount of current used is steady as the
>> as> components don't change and therefore draw a consistent level of
>> as> current. So a 500w power supply left on for an hour is going to
>> draw
>> as> 500w per hour or .5 kwhour per hour (in the UK electricity tends
>> to
>> as> be
>> as> measured in kilowatt hours). Also in UK terms we tend to think in
>> as> terms
>> as> of electric fires, a one bar electric fire uses 1 kilowatt per
>> hour
>> as> so
>> as> your computer with a 500w power supply would use half as much.
>>
>> The power consumption is not constant at all. The power supply
>> rating simply indicates how much it can supply rather than what is
>> being used. The CPU load, disk activity, and whether the machine is
>> in standby all affect the instantaneous power consumption. To answer
>> the original question, I know of no system that can actually measure
>> the power consumption. External devices are needed.
>>
>> David
>>
>>
>> as> I'd be grateful if somebody who knows more about these things
>> than I
>> as> can
>> as> explain where I'm wrong, as I'm sure I am!
>>
>> See above. I am an electrical engineer.
>>
>> as> Oh yes, and don't forget your monitor/printer/etc....
>>
>> True
>>
> Thanks for putting me right there David (knew somebody would if I
>
> How much variation is there in the level of mains power taken by
> things like disk activity? I can see that having it on standby would
> make a big difference though.
>
> I would have thought that the big power drain is the cooling fans
> which run constantly, with the other mechanical 'bits' like spinning
> hard disks next and electronic components coming last of all?

Hi, Andy. Little fans use just a few watts each. The HDD, based on the
amount of heat it produces, is probably next, and the CPU, when active is up
there too. Frankly, you might worry more about people standing in front of
an open refrigerator! Besides, you can have your computer go into standby
overnight (better yet: Hibernate) and at that point, it is about as
energy-demanding as a nitelite.
One thing you can take away from this discussion is that any appliance you
have lists its Maximum wattage, not the actual watts consumed in normal use.

Matt Gibson
07-10-2005, 12:26 AM
> How much variation is there in the level of mains power taken by things
> like disk activity? I can see that having it on standby would make a big
> difference though.

Disk activity, not so much.

> I would have thought that the big power drain is the cooling fans which
> run constantly, with the other mechanical 'bits' like spinning hard disks
> next and electronic components coming last of all?

Fans are actually quite low, usually around 1-5W.

Matt Gibson - GSEC

David H. Lipman
07-10-2005, 12:26 AM
From: "andy smart" <anonymus@discussions.microsoft.com>

| I'm not an electrical engineer, but I would think that the amount of
| mains current consumed by your pc while it is on is governed by the
| wattage of your power supply.
|
| This takes the mains current and supplies it to the components. If the
| PC is running then the amount of current used is steady as the
| components don't change and therefore draw a consistent level of
| current. So a 500w power supply left on for an hour is going to draw
| 500w per hour or .5 kwhour per hour (in the UK electricity tends to be
| measured in kilowatt hours). Also in UK terms we tend to think in terms
| of electric fires, a one bar electric fire uses 1 kilowatt per hour so
| your computer with a 500w power supply would use half as much.
|
| I'd be grateful if somebody who knows more about these things than I can
| explain where I'm wrong, as I'm sure I am!
|
| Oh yes, and don't forget your monitor/printer/etc....

500 watts power supply for the avg. home computer is an overestimate to say the least ;-)
Most range between 125watt and 300watt power supplies. However, that's the max. rating the
supply can handle, not what is consumed.

The way to measure is with a Analogue/Digital MultiMeter. Then measure the quiescent
current and voltage on the AC side. The appraent ower consumed will be the volate
multiplied by the current.

I sacraficed an AC power chord with the following test platform...

ASUS P2B-B motherboard
200W power supply
PIII 550MHz slot 1 CPU
128MB
10GB HD
3.5" floppy drive
CDROM drive
AGP video card
3COM PCI NIC
ISA sound card

After 10 minutes of opertion, and the PC was quiescent, I measured..

121.4 VAC
..49 amps

Which yields an apparent power consumption of 59.486 watts.

--
Dave
http://www.ik-cs.com/got-a-virus.htm

David H. Lipman
07-10-2005, 12:26 AM
From: "Richard Urban" <richardurbanREMOVETHIS@hotmail.com>

| The cards I used in the past monitored AC current and generated a
| proportional and linear 4-20ma output for control purposes. The card also
| had a computer interface to connect to a computer with monitoring software.
|
| Now, could a computer monitor itself? I have never tried it.
|
| --
| Regards,
|
| Richard Urban
|
| aka Crusty (-: Old B@stard :-)
|

ASUS PC Probe monitors some of the DC voltages. However, you also need to know the current
to determine the power consumption. ASUS PC Probe is used to see if the DC voltages are "in
spec".
{ see attached }

I can't see how a ISA, EISA, MCA or PCI card can monitor the AC supply without tapping the
power supply directly. There can be NO AC voltages on the card's BUS so a cable would have
to go to the power supply. But then the card and the supply would have to be symbiotic.
Not likely with a home computer. Possible on a server platform with voltage monitors on the
motherboard and in the power supplies (note most server platforms will have two or more
redundant power supplies).

--
Dave
http://www.ik-cs.com/got-a-virus.htm

andy smart
07-10-2005, 12:26 AM
Matt Gibson wrote:
>>How much variation is there in the level of mains power taken by things
>>like disk activity? I can see that having it on standby would make a big
>>difference though.
>
>
> Disk activity, not so much.
>
>
>
>>I would have thought that the big power drain is the cooling fans which
>>run constantly, with the other mechanical 'bits' like spinning hard disks
>>next and electronic components coming last of all?
>
>
> Fans are actually quite low, usually around 1-5W.
>
> Matt Gibson - GSEC
>
>
Thanks everybody! I feel much better informed on this topic now.

Al Dykes
07-10-2005, 12:26 AM
In article <etJ4Wu\$VFHA.3636@TK2MSFTNGP14.phx.gbl>,
David H. Lipman <DLipman~nospam~@Verizon.Net> wrote:
>From: "andy smart" <anonymus@discussions.microsoft.com>
>
>| I'm not an electrical engineer, but I would think that the amount of
>| mains current consumed by your pc while it is on is governed by the
>| wattage of your power supply.
>|
>| This takes the mains current and supplies it to the components. If the
>| PC is running then the amount of current used is steady as the
>| components don't change and therefore draw a consistent level of
>| current. So a 500w power supply left on for an hour is going to draw
>| 500w per hour or .5 kwhour per hour (in the UK electricity tends to be
>| measured in kilowatt hours). Also in UK terms we tend to think in terms
>| of electric fires, a one bar electric fire uses 1 kilowatt per hour so
>| your computer with a 500w power supply would use half as much.
>|
>| I'd be grateful if somebody who knows more about these things than I can
>| explain where I'm wrong, as I'm sure I am!
>|
>| Oh yes, and don't forget your monitor/printer/etc....
>
>500 watts power supply for the avg. home computer is an overestimate to say the least ;-)
>Most range between 125watt and 300watt power supplies. However, that's the max. rating the
>supply can handle, not what is consumed.
>
>The way to measure is with a Analogue/Digital MultiMeter. Then measure the quiescent
>current and voltage on the AC side. The appraent ower consumed will be the volate
>multiplied by the current.
>
>I sacraficed an AC power chord with the following test platform...
>
>ASUS P2B-B motherboard
>200W power supply
>PIII 550MHz slot 1 CPU
>128MB
>10GB HD
>3.5" floppy drive
>CDROM drive
>AGP video card
>3COM PCI NIC
>ISA sound card
>
>After 10 minutes of opertion, and the PC was quiescent, I measured..
>
>121.4 VAC
>.49 amps
>
>Which yields an apparent power consumption of 59.486 watts.
>
>
>--
>Dave

Nope. That's 59.486 Volt-Amps unless you're measuring the current
drawn by a resistive device like a light bulb.

A computer, a motor, or just about any useful device has a capacitive
or inductive component and the current and voltage waves are out of
phase resulting in a VA fiure that's higher than the wattage used.

The Power company bills based on Watts but has to incur the costs of
generating the Volt-Amps. The obvious cost is the i-squared-r loss of
the higher current.

http://www.onewilshire.com/resource_center/colo_toolkit/ExplainingWattsandVoltAmps.htm

--
a d y k e s @ p a n i x . c o m

Don't blame me. I voted for Gore.

bumtracks
07-10-2005, 12:26 AM
>>
>>
>
>
> You need something like this:
>
>
> http://www.p3international.com/products/special/P4400/P4400-CE.html
>

I have a Kill-a-Watt hooked to my laptop power brick for the past 1514 hrs =
26.42 kwh.
17 watts is about the minimum just sitting,
20 - 22watt is pretty normal web surfing,
45 watts on boot into windows is peak.
Last time I diddled , Laptop costs me pretty darn close to \$1.12 per month

David
07-10-2005, 12:26 AM
as> Thanks for putting me right there David (knew somebody would if I

as> How much variation is there in the level of mains power taken by
as> things
as> like disk activity? I can see that having it on standby would make a
as> big
as> difference though.

as> I would have thought that the big power drain is the cooling fans
as> which
as> run constantly, with the other mechanical 'bits' like spinning hard
as> disks next and electronic components coming last of all?

Hello Andy,

Fans are very low power, less that 4-5W each. Disk drives can consume between 5 and 30W depending upon activity and the speed they are running. The CPU is actually a big variable and can vary considerably upon activity. 5W if inactive up to over 70 or 80W for a very fast CPU if 'pegged' doing a high computational task like MPEG2 encoding. Another big power hog is high end video cards. If it has a fan on it, it can probably consume 50W or so when rendering complex moving images. Most other things consume low power, but as they say, you do get nickel and dimed on them. I have rarely seen a computer (not counting the monitor) consume more than 250W at peak load.

David

David H. Lipman
07-10-2005, 12:26 AM

|
| Nope. That's 59.486 Volt-Amps unless you're measuring the current
| drawn by a resistive device like a light bulb.
|
| A computer, a motor, or just about any useful device has a capacitive
| or inductive component and the current and voltage waves are out of
| phase resulting in a VA fiure that's higher than the wattage used.
|
| The Power company bills based on Watts but has to incur the costs of
| generating the Volt-Amps. The obvious cost is the i-squared-r loss of
| the higher current.
|
| http://www.onewilshire.com/resource_center/colo_toolkit/ExplainingWattsandVoltAmps.htm
|
| --
| a d y k e s @ p a n i x . c o m
|
| Don't blame me. I voted for Gore.

Thanx Al you are correct.
I didn't want to muddy the waters by getting into phasors mathematics.

--
Dave
http://www.ik-cs.com/got-a-virus.htm

David H. Lipman
07-10-2005, 12:26 AM

|
| Nope. That's 59.486 Volt-Amps unless you're measuring the current
| drawn by a resistive device like a light bulb.
|
| A computer, a motor, or just about any useful device has a capacitive
| or inductive component and the current and voltage waves are out of
| phase resulting in a VA fiure that's higher than the wattage used.
|
| The Power company bills based on Watts but has to incur the costs of
| generating the Volt-Amps. The obvious cost is the i-squared-r loss of
| the higher current.
|
| http://www.onewilshire.com/resource_center/colo_toolkit/ExplainingWattsandVoltAmps.htm
|
| --
| a d y k e s @ p a n i x . c o m
|
| Don't blame me. I voted for Gore.

I just re-read my post and I did use the terminolgy "apparent power" so all-in-all, my post
was good enough for this level of discussion. ;-)

--
Dave
http://www.ik-cs.com/got-a-virus.htm

Not Me
07-10-2005, 12:26 AM
The obvious conclusion to this discussion is that the electricity cost
of operating most personal computers is, in most cases, insignificant
when compared to most devices [Clothes dryers, freezers, refrigerators,
air conditioners, etc]. You can ramp that figure up with usage of high
end video adapters and Raid enabled hard drive configurations, etc;
however, this does not
cover very many users past extreme game players and those who must have
the latest, newest, best of everything without regard to the need for
or cost thereof. If power cost is a consideration of running the
Computer all night just turn off the monitor plus other peripheral
devices such as the printer and exterior USB or Firewire devices.
Gene K

David H. Lipman wrote:
> From: "Richard Urban" <richardurbanREMOVETHIS@hotmail.com>
>
> | The cards I used in the past monitored AC current and generated a
> | proportional and linear 4-20ma output for control purposes. The card also
> | had a computer interface to connect to a computer with monitoring software.
> |
> | Now, could a computer monitor itself? I have never tried it.
> |
> | --
> | Regards,
> |
> | Richard Urban
> |
> | aka Crusty (-: Old B@stard :-)
> |
>
> ASUS PC Probe monitors some of the DC voltages. However, you also need to know the current
> to determine the power consumption. ASUS PC Probe is used to see if the DC voltages are "in
> spec".
> { see attached }
>
> I can't see how a ISA, EISA, MCA or PCI card can monitor the AC supply without tapping the
> power supply directly. There can be NO AC voltages on the card's BUS so a cable would have
> to go to the power supply. But then the card and the supply would have to be symbiotic.
> Not likely with a home computer. Possible on a server platform with voltage monitors on the
> motherboard and in the power supplies (note most server platforms will have two or more
> redundant power supplies).
>
>
>

Lil' Dave
07-10-2005, 12:26 AM
You need an ammeter that loops over the 120VAC power cable, and senses the
current throught the line. You will need to separate the two conductors,
but not open the insulation. There are ammeters that require opening the
power cord and separating one of the power lines including cutting the wire,
don't use this. Then multiply the shown current usage times the voltage,
then multiply that times .606 to get true power usage in watts. This is a
rough measurement as the ammeter needs calibration to get an accurate
Although PCs are low power usage items, the overall power usage has
increased due to the increase of the system bus clock frequency, and the
cpu's multiplier in the past few years. Based on the same voltage input,
which has remained the same over the years, as the frequency goes up, the
power usage goes up. This makes more heat, and is why we have such
substantial coolers on our cpus (P4/Athlon) compared to their physical size.
This is the primary culprit, followed by the system bus clock for RAM, and
other system bus items.
HDs, etc. have maintained or lowered their power usage over the years with
exception of high rpm HDs and high rpm cdreaders. This is due to the higher
motor speeds for the platters. Which, again, is a function of frequency of
power usage.
<peterlavington@hotmail.com> wrote in message
> Anybody know of free software that will show me how much electricity
> (presumably in watts) I am using whilst my PC is on?
>
> Weird question I know but I'm interested since there are times I leave
> my machine running overnight and wondering if it's costing me a lot in
> power.
>
>
> Thanks
>

Ron Sommer
07-10-2005, 12:26 AM
What is the purpose of multiplying by .606?
--
Ron Sommer

"Lil' Dave" <spamyourself@virus.net> wrote in message
news:%23Lw80PHWFHA.3840@tk2msftngp13.phx.gbl...
> You need an ammeter that loops over the 120VAC power cable, and senses the
> current throught the line. You will need to separate the two conductors,
> but not open the insulation. There are ammeters that require opening the
> power cord and separating one of the power lines including cutting the
> wire, don't use this. Then multiply the shown current usage times the
> voltage, then multiply that times .606 to get true power usage in watts.
> This is a rough measurement as the ammeter needs calibration to get an
> Although PCs are low power usage items, the overall power usage has
> increased due to the increase of the system bus clock frequency, and the
> cpu's multiplier in the past few years. Based on the same voltage input,
> which has remained the same over the years, as the frequency goes up, the
> power usage goes up. This makes more heat, and is why we have such
> substantial coolers on our cpus (P4/Athlon) compared to their physical
> size. This is the primary culprit, followed by the system bus clock for
> RAM, and other system bus items.
> HDs, etc. have maintained or lowered their power usage over the years with
> exception of high rpm HDs and high rpm cdreaders. This is due to the
> higher motor speeds for the platters. Which, again, is a function of
> frequency of power usage.
> <peterlavington@hotmail.com> wrote in message
>> Anybody know of free software that will show me how much electricity
>> (presumably in watts) I am using whilst my PC is on?
>>
>> Weird question I know but I'm interested since there are times I leave
>> my machine running overnight and wondering if it's costing me a lot in
>> power.
>>
>>
>> Thanks
>>
>
>

R. McCarty
07-10-2005, 12:27 AM
That's a good question, RMS (Root Mean Square) came to mind,
but that value is .707. Interesting thread as all those formulas and
physics fade with age and lack of use.

"Ron Sommer" <rsommer@nospam.ktis.net> wrote in message
news:%23LasdRIWFHA.2984@tk2msftngp13.phx.gbl...
> What is the purpose of multiplying by .606?
> --
> Ron Sommer
>
> "Lil' Dave" <spamyourself@virus.net> wrote in message
> news:%23Lw80PHWFHA.3840@tk2msftngp13.phx.gbl...
>> You need an ammeter that loops over the 120VAC power cable, and senses
>> the current throught the line. You will need to separate the two
>> conductors, but not open the insulation. There are ammeters that require
>> opening the power cord and separating one of the power lines including
>> cutting the wire, don't use this. Then multiply the shown current usage
>> times the voltage, then multiply that times .606 to get true power usage
>> in watts. This is a rough measurement as the ammeter needs calibration to
>> Although PCs are low power usage items, the overall power usage has
>> increased due to the increase of the system bus clock frequency, and the
>> cpu's multiplier in the past few years. Based on the same voltage input,
>> which has remained the same over the years, as the frequency goes up, the
>> power usage goes up. This makes more heat, and is why we have such
>> substantial coolers on our cpus (P4/Athlon) compared to their physical
>> size. This is the primary culprit, followed by the system bus clock for
>> RAM, and other system bus items.
>> HDs, etc. have maintained or lowered their power usage over the years
>> with exception of high rpm HDs and high rpm cdreaders. This is due to
>> the higher motor speeds for the platters. Which, again, is a function of
>> frequency of power usage.
>> <peterlavington@hotmail.com> wrote in message
>>> Anybody know of free software that will show me how much electricity
>>> (presumably in watts) I am using whilst my PC is on?
>>>
>>> Weird question I know but I'm interested since there are times I leave
>>> my machine running overnight and wondering if it's costing me a lot in
>>> power.
>>>
>>>
>>> Thanks
>>>
>>
>>
>

HeyBub
07-10-2005, 12:27 AM
Ron Sommer wrote:
> What is the purpose of multiplying by .606?

An AC voltage is measured peak-to-peak. Power is voltage times amperage.

Think of a sine curve. What's the AVERAGE voltage under the curve?

Another way to think of it is the 0.606 (but I think it's .707) is a
fudge-factor to make the answer come out right.

David Candy
07-10-2005, 12:27 AM
..6 is average and is meaningless. .707 is RMS and reflects the shape of the sine wave (and both only apply to sine waves). In AC circuits one cannot just times EMF by current. That is for DC circuits only, in AC circuits one needs to know the impedence cos capacitive or inductive circuits shift the EMF out of phase with the current.

--
----------------------------------------------------------

"HeyBub" <heybub@gmail.com> wrote in message news:ulDKmpJWFHA.3184@TK2MSFTNGP15.phx.gbl...
> Ron Sommer wrote:
>> What is the purpose of multiplying by .606?
>
> An AC voltage is measured peak-to-peak. Power is voltage times amperage.
>
> Think of a sine curve. What's the AVERAGE voltage under the curve?
>
> Another way to think of it is the 0.606 (but I think it's .707) is a
> fudge-factor to make the answer come out right.
>
>

nono91
07-10-2005, 12:27 AM
hello

"Lil' Dave" <spamyourself@virus.net> a écrit dans le message de news:
%23Lw80PHWFHA.3840@tk2msftngp13.phx.gbl...
>
> You need an ammeter that loops over the 120VAC power cable, and senses the
> current throught the line. >> (presumably in watts) I am using whilst my
> PC is on?
>>
>> Weird question I know but I'm interested since there are times I leave
>> my machine running overnight and wondering if it's costing me a lot in
>> power.
>

For information , you will find here after some figures I mesured with an
electronic wattmeter

start pc :-------------------------------------------------- 156 w max.

pc running ( with no action )----------------------------- 125 w

simple action ( writing this message ) same !about-------125 w

playing music with windows media player---------------------- 140 w

pc :PWR 450w / 2 hdd / intel celeron / no sound card ( sound module on
mother board)

/2 very small amplified speakers

regards

nono91

Don Schmidt
07-10-2005, 12:27 AM
Hey folks, we're not plotting a course to the moon here, just the
approximant cost per hour to run a computer station.

PowerChute software (comes with APC UPS) reports my station consumes 268
watts.
My power rate average is 7.36¢ per kWh

(268/1000)*.0736= 1.97¢ per hour

My station comprises of:
Computer with three 10k rpm SCSI hard drives
SoundBlaster card
CommShare phone switch
Actiontec DSL modem/router
Power Speakers
HP 960 Cse printer
HP 8200 scanner
Cruzer thumb drive
KDS power CD storage tower
Siemens 8825 Gigaset phone (needs AC power to operate)
(also have a HP CD RW, a TDK CD, DVD RW, ZIP drive and 3½" floppy drive)

Some of the items have no direct function with the computer station but all
(If we loose AC power, the USP will run the station for 30 minutes).

Budgeting costs: 2¢ per hour

--
Don
Vancouver, USA

"nono91" <nono91@labas.fr> wrote in message
news:e9uX4xJWFHA.1404@TK2MSFTNGP09.phx.gbl...
>
> hello
>
> "Lil' Dave" <spamyourself@virus.net> a écrit dans le message de news:
> %23Lw80PHWFHA.3840@tk2msftngp13.phx.gbl...
>>
>> You need an ammeter that loops over the 120VAC power cable, and senses
>> the current throught the line. >> (presumably in watts) I am using
>> whilst my PC is on?
>>>
>>> Weird question I know but I'm interested since there are times I leave
>>> my machine running overnight and wondering if it's costing me a lot in
>>> power.
>>
>
> For information , you will find here after some figures I mesured with an
> electronic wattmeter
>
>
>
> start pc :-------------------------------------------------- 156 w max.
>
> pc running ( with no action )----------------------------- 125 w
>
> simple action ( writing this message ) same !about-------125 w
>
> playing music with windows media player---------------------- 140 w
>
>
>
> pc :PWR 450w / 2 hdd / intel celeron / no sound card ( sound module on
> mother board)
>
> /2 very small amplified speakers
>
>
>
> regards
>
> nono91
>
>
>

David H. Lipman
07-10-2005, 12:27 AM
From: "Don Schmidt" <RetiredEngineer@pnb.telco>

| Hey folks, we're not plotting a course to the moon here, just the
| approximant cost per hour to run a computer station.
|
| PowerChute software (comes with APC UPS) reports my station consumes 268
| watts.
| My power rate average is 7.36¢ per kWh
|
| (268/1000)*.0736= 1.97¢ per hour
|
| My station comprises of:
| Computer with three 10k rpm SCSI hard drives
| SoundBlaster card
| CommShare phone switch
| Actiontec DSL modem/router
| Power Speakers
| HP 960 Cse printer
| HP 8200 scanner
| Cruzer thumb drive
| KDS power CD storage tower
| Siemens 8825 Gigaset phone (needs AC power to operate)
| (also have a HP CD RW, a TDK CD, DVD RW, ZIP drive and 3½" floppy drive)
|
| Some of the items have no direct function with the computer station but all
| of the above are powered by my APC Back-UPS XS 1500
| (If we loose AC power, the USP will run the station for 30 minutes).
|
| Budgeting costs: 2¢ per hour
|
| --
| Don
| Vancouver, USA
|

Additionally, the UPS is also consuming power.

--
Dave
http://www.ik-cs.com/got-a-virus.htm

Ron Sommer
07-10-2005, 12:27 AM
"HeyBub" <heybub@gmail.com> wrote in message
news:ulDKmpJWFHA.3184@TK2MSFTNGP15.phx.gbl...
> Ron Sommer wrote:
>> What is the purpose of multiplying by .606?
>
> An AC voltage is measured peak-to-peak. Power is voltage times amperage.
>
> Think of a sine curve. What's the AVERAGE voltage under the curve?
>
> Another way to think of it is the 0.606 (but I think it's .707) is a
> fudge-factor to make the answer come out right.
>

If I measure with a voltage meter, the reading will not be peak to peak.
If the reading is 110 volts, then the peak voltage would be 156 to -156.
Measuring peak to peak would be 312.
--
Ron Sommer

Don Schmidt
07-10-2005, 12:27 AM
True but the UPS wasn't part of the problem presented. Also, power rates
here in the Pacific NW are much below the national average. The cost of the
UPS I'll chock up to insurance. :o)

--
Don
Vancouver, USA

"David H. Lipman" <DLipman~nospam~@Verizon.Net> wrote in message
news:elX\$fzKWFHA.2540@tk2msftngp13.phx.gbl...
> From: "Don Schmidt" <RetiredEngineer@pnb.telco>
>
> | Hey folks, we're not plotting a course to the moon here, just the
> | approximant cost per hour to run a computer station.
> |
> | PowerChute software (comes with APC UPS) reports my station consumes 268
> | watts.
> | My power rate average is 7.36¢ per kWh
> |
> | (268/1000)*.0736= 1.97¢ per hour
> |
> | My station comprises of:
> | Computer with three 10k rpm SCSI hard drives
> | SoundBlaster card
> | CommShare phone switch
> | Actiontec DSL modem/router
> | Power Speakers
> | HP 960 Cse printer
> | HP 8200 scanner
> | Cruzer thumb drive
> | KDS power CD storage tower
> | Siemens 8825 Gigaset phone (needs AC power to operate)
> | (also have a HP CD RW, a TDK CD, DVD RW, ZIP drive and 3½" floppy drive)
> |
> | Some of the items have no direct function with the computer station but
> all
> | of the above are powered by my APC Back-UPS XS 1500
> | (If we loose AC power, the USP will run the station for 30 minutes).
> |
> | Budgeting costs: 2¢ per hour
> |
> | --
> | Don
> | Vancouver, USA
> |
>
> Additionally, the UPS is also consuming power.
>
> --
> Dave
> http://www.ik-cs.com/got-a-virus.htm
>
>

David H. Lipman
07-10-2005, 12:27 AM
From: "Don Schmidt" <RetiredEngineer@pnb.telco>

| True but the UPS wasn't part of the problem presented. Also, power rates
| here in the Pacific NW are much below the national average. The cost of the
| UPS I'll chock up to insurance. :o)
|
| --
| Don
| Vancouver, USA
|

--
Dave
http://www.ik-cs.com/got-a-virus.htm

Lil' Dave
07-10-2005, 12:28 AM
RMS, root mean square or apparent power, (.707) is an accurate
representation of power usage. Most component level technology
manufacturers use true RMS or true power (.606) factor to determine power
usage of their individual components. Reactive loads between the components
and true power usage result in apparent power usage. What you pay for at
the meter is true power. For the purposes of that, .707 is what's needed
for large power usage companies that have tariff bills. For component level
and home power usage, .606 is what's needed. The majority of home usage is
bases your bill on, a non-inductive or resistive load. Another example are
transformers and induction motors, when not energized, they're inductive.
When on, they become a resistive load. Your refigerator and AC compressor
motor are such examples and use more net energy when on. Therefore, the
resistive load is what is used for billing. We're interested, for the
purposes of this thread, in true power since that's what most of us are
paying for at the meter. Hope I made sense.

"Ron Sommer" <rsommer@nospam.ktis.net> wrote in message
news:%23LasdRIWFHA.2984@tk2msftngp13.phx.gbl...
> What is the purpose of multiplying by .606?
> --
> Ron Sommer
>
> "Lil' Dave" <spamyourself@virus.net> wrote in message
> news:%23Lw80PHWFHA.3840@tk2msftngp13.phx.gbl...
>> You need an ammeter that loops over the 120VAC power cable, and senses
>> the current throught the line. You will need to separate the two
>> conductors, but not open the insulation. There are ammeters that require
>> opening the power cord and separating one of the power lines including
>> cutting the wire, don't use this. Then multiply the shown current usage
>> times the voltage, then multiply that times .606 to get true power usage
>> in watts. This is a rough measurement as the ammeter needs calibration to
>> Although PCs are low power usage items, the overall power usage has
>> increased due to the increase of the system bus clock frequency, and the
>> cpu's multiplier in the past few years. Based on the same voltage input,
>> which has remained the same over the years, as the frequency goes up, the
>> power usage goes up. This makes more heat, and is why we have such
>> substantial coolers on our cpus (P4/Athlon) compared to their physical
>> size. This is the primary culprit, followed by the system bus clock for
>> RAM, and other system bus items.
>> HDs, etc. have maintained or lowered their power usage over the years
>> with exception of high rpm HDs and high rpm cdreaders. This is due to
>> the higher motor speeds for the platters. Which, again, is a function of
>> frequency of power usage.
>> <peterlavington@hotmail.com> wrote in message
>>> Anybody know of free software that will show me how much electricity
>>> (presumably in watts) I am using whilst my PC is on?
>>>
>>> Weird question I know but I'm interested since there are times I leave
>>> my machine running overnight and wondering if it's costing me a lot in
>>> power.
>>>
>>>
>>> Thanks
>>>
>>
>>
>

sensecam@gmail.com
07-10-2005, 12:52 AM
The "Kill a Watt" device mentioned earlier this thread looks very good.
Can it or similar be bought in UK/Europe? (UK uses approx 240V so USA
version can't be used ). I've tried the internet search tools and the
concerened about the amount of power my PCs use.

Lyn

Lil' Dave wrote:
> RMS, root mean square or apparent power, (.707) is an accurate
> representation of power usage. Most component level technology
> manufacturers use true RMS or true power (.606) factor to determine power
> usage of their individual components. Reactive loads between the components
> and true power usage result in apparent power usage. What you pay for at
> the meter is true power. For the purposes of that, .707 is what's needed
> for large power usage companies that have tariff bills. For component level
> and home power usage, .606 is what's needed. The majority of home usage is
> non-inductive load, like your lights. And that's what the power company
> bases your bill on, a non-inductive or resistive load. Another example are
> transformers and induction motors, when not energized, they're inductive.
> When on, they become a resistive load. Your refigerator and AC compressor
> motor are such examples and use more net energy when on. Therefore, the
> resistive load is what is used for billing. We're interested, for the
> purposes of this thread, in true power since that's what most of us are
> paying for at the meter. Hope I made sense.
>
> "Ron Sommer" <rsommer@nospam.ktis.net> wrote in message
> news:%23LasdRIWFHA.2984@tk2msftngp13.phx.gbl...
> > What is the purpose of multiplying by .606?
> > --
> > Ron Sommer
> >
> > "Lil' Dave" <spamyourself@virus.net> wrote in message
> > news:%23Lw80PHWFHA.3840@tk2msftngp13.phx.gbl...
> >> You need an ammeter that loops over the 120VAC power cable, and senses
> >> the current throught the line. You will need to separate the two
> >> conductors, but not open the insulation. There are ammeters that require
> >> opening the power cord and separating one of the power lines including
> >> cutting the wire, don't use this. Then multiply the shown current usage
> >> times the voltage, then multiply that times .606 to get true power usage
> >> in watts. This is a rough measurement as the ammeter needs calibration to
> >> get an accurate reading.
> >> Although PCs are low power usage items, the overall power usage has
> >> increased due to the increase of the system bus clock frequency, and the
> >> cpu's multiplier in the past few years. Based on the same voltage input,
> >> which has remained the same over the years, as the frequency goes up, the
> >> power usage goes up. This makes more heat, and is why we have such
> >> substantial coolers on our cpus (P4/Athlon) compared to their physical
> >> size. This is the primary culprit, followed by the system bus clock for
> >> RAM, and other system bus items.
> >> HDs, etc. have maintained or lowered their power usage over the years
> >> with exception of high rpm HDs and high rpm cdreaders. This is due to
> >> the higher motor speeds for the platters. Which, again, is a function of
> >> frequency of power usage.
> >> <peterlavington@hotmail.com> wrote in message
> >>> Anybody know of free software that will show me how much electricity
> >>> (presumably in watts) I am using whilst my PC is on?
> >>>
> >>> Weird question I know but I'm interested since there are times I leave
> >>> my machine running overnight and wondering if it's costing me a lot in
> >>> power.
> >>>
> >>>
> >>> Thanks
> >>>
> >>
> >>
> >

DL
07-10-2005, 12:52 AM
'Intelligent Mains Panel (IMP)'
http://www.oneclickpower.co.uk/home.htm
This what you mean?

<sensecam@gmail.com> wrote in message
>
> The "Kill a Watt" device mentioned earlier this thread looks very good.
> Can it or similar be bought in UK/Europe? (UK uses approx 240V so USA
> version can't be used ). I've tried the internet search tools and the
> usual uk electrical wholesalers but not found one yet. I'm also
> concerened about the amount of power my PCs use.
>
> Lyn
>
>
> Lil' Dave wrote:
> > RMS, root mean square or apparent power, (.707) is an accurate
> > representation of power usage. Most component level technology
> > manufacturers use true RMS or true power (.606) factor to determine
power
> > usage of their individual components. Reactive loads between the
components
> > and true power usage result in apparent power usage. What you pay for
at
> > the meter is true power. For the purposes of that, .707 is what's
needed
> > for large power usage companies that have tariff bills. For component
level
> > and home power usage, .606 is what's needed. The majority of home usage
is
> > non-inductive load, like your lights. And that's what the power company
> > bases your bill on, a non-inductive or resistive load. Another example
are
> > transformers and induction motors, when not energized, they're
inductive.
> > When on, they become a resistive load. Your refigerator and AC
compressor
> > motor are such examples and use more net energy when on. Therefore, the
> > resistive load is what is used for billing. We're interested, for the
> > purposes of this thread, in true power since that's what most of us are
> > paying for at the meter. Hope I made sense.
> >
> > "Ron Sommer" <rsommer@nospam.ktis.net> wrote in message
> > news:%23LasdRIWFHA.2984@tk2msftngp13.phx.gbl...
> > > What is the purpose of multiplying by .606?
> > > --
> > > Ron Sommer
> > >
> > > "Lil' Dave" <spamyourself@virus.net> wrote in message
> > > news:%23Lw80PHWFHA.3840@tk2msftngp13.phx.gbl...
> > >> You need an ammeter that loops over the 120VAC power cable, and
senses
> > >> the current throught the line. You will need to separate the two
> > >> conductors, but not open the insulation. There are ammeters that
require
> > >> opening the power cord and separating one of the power lines
including
> > >> cutting the wire, don't use this. Then multiply the shown current
usage
> > >> times the voltage, then multiply that times .606 to get true power
usage
> > >> in watts. This is a rough measurement as the ammeter needs
calibration to
> > >> get an accurate reading.
> > >> Although PCs are low power usage items, the overall power usage has
> > >> increased due to the increase of the system bus clock frequency, and
the
> > >> cpu's multiplier in the past few years. Based on the same voltage
input,
> > >> which has remained the same over the years, as the frequency goes up,
the
> > >> power usage goes up. This makes more heat, and is why we have such
> > >> substantial coolers on our cpus (P4/Athlon) compared to their
physical
> > >> size. This is the primary culprit, followed by the system bus clock
for
> > >> RAM, and other system bus items.
> > >> HDs, etc. have maintained or lowered their power usage over the years
> > >> with exception of high rpm HDs and high rpm cdreaders. This is due
to
> > >> the higher motor speeds for the platters. Which, again, is a
function of
> > >> frequency of power usage.
> > >> <peterlavington@hotmail.com> wrote in message
> > >>> Anybody know of free software that will show me how much electricity
> > >>> (presumably in watts) I am using whilst my PC is on?
> > >>>
> > >>> Weird question I know but I'm interested since there are times I
leave
> > >>> my machine running overnight and wondering if it's costing me a lot
in
> > >>> power.
> > >>>
> > >>> I've asked Jeeves and Google but with no joy.
> > >>>
> > >>> Thanks
> > >>>
> > >>
> > >>
> > >
>

sensecam@gmail.com
07-10-2005, 12:54 AM
Thanks!
This looks really useful.

Lyn

DL wrote:
> 'Intelligent Mains Panel (IMP)'
> http://www.oneclickpower.co.uk/home.htm
> This what you mean?
>
> <sensecam@gmail.com> wrote in message
> >
> > The "Kill a Watt" device mentioned earlier this thread looks very good.
> > Can it or similar be bought in UK/Europe? (UK uses approx 240V so USA
> > version can't be used ). I've tried the internet search tools and the
> > usual uk electrical wholesalers but not found one yet. I'm also
> > concerened about the amount of power my PCs use.
> >
> > Lyn
> >
> >
> > Lil' Dave wrote:
> > > RMS, root mean square or apparent power, (.707) is an accurate
> > > representation of power usage. Most component level technology
> > > manufacturers use true RMS or true power (.606) factor to determine
> power
> > > usage of their individual components. Reactive loads between the
> components
> > > and true power usage result in apparent power usage. What you pay for
> at
> > > the meter is true power. For the purposes of that, .707 is what's
> needed
> > > for large power usage companies that have tariff bills. For component
> level
> > > and home power usage, .606 is what's needed. The majority of home usage
> is
> > > non-inductive load, like your lights. And that's what the power company
> > > bases your bill on, a non-inductive or resistive load. Another example
> are
> > > transformers and induction motors, when not energized, they're
> inductive.
> > > When on, they become a resistive load. Your refigerator and AC
> compressor
> > > motor are such examples and use more net energy when on. Therefore, the
> > > resistive load is what is used for billing. We're interested, for the
> > > purposes of this thread, in true power since that's what most of us are
> > > paying for at the meter. Hope I made sense.
> > >
> > > "Ron Sommer" <rsommer@nospam.ktis.net> wrote in message
> > > news:%23LasdRIWFHA.2984@tk2msftngp13.phx.gbl...
> > > > What is the purpose of multiplying by .606?
> > > > --
> > > > Ron Sommer
> > > >
> > > > "Lil' Dave" <spamyourself@virus.net> wrote in message
> > > > news:%23Lw80PHWFHA.3840@tk2msftngp13.phx.gbl...
> > > >> You need an ammeter that loops over the 120VAC power cable, and
> senses
> > > >> the current throught the line. You will need to separate the two
> > > >> conductors, but not open the insulation. There are ammeters that
> require
> > > >> opening the power cord and separating one of the power lines
> including
> > > >> cutting the wire, don't use this. Then multiply the shown current
> usage
> > > >> times the voltage, then multiply that times .606 to get true power
> usage
> > > >> in watts. This is a rough measurement as the ammeter needs
> calibration to
> > > >> get an accurate reading.
> > > >> Although PCs are low power usage items, the overall power usage has
> > > >> increased due to the increase of the system bus clock frequency, and
> the
> > > >> cpu's multiplier in the past few years. Based on the same voltage
> input,
> > > >> which has remained the same over the years, as the frequency goes up,
> the
> > > >> power usage goes up. This makes more heat, and is why we have such
> > > >> substantial coolers on our cpus (P4/Athlon) compared to their
> physical
> > > >> size. This is the primary culprit, followed by the system bus clock
> for
> > > >> RAM, and other system bus items.
> > > >> HDs, etc. have maintained or lowered their power usage over the years
> > > >> with exception of high rpm HDs and high rpm cdreaders. This is due
> to
> > > >> the higher motor speeds for the platters. Which, again, is a
> function of
> > > >> frequency of power usage.
> > > >> <peterlavington@hotmail.com> wrote in message
> > > >>> Anybody know of free software that will show me how much electricity
> > > >>> (presumably in watts) I am using whilst my PC is on?
> > > >>>
> > > >>> Weird question I know but I'm interested since there are times I
> leave
> > > >>> my machine running overnight and wondering if it's costing me a lot
> in
> > > >>> power.
> > > >>>
> > > >>> I've asked Jeeves and Google but with no joy.
> > > >>>
> > > >>> Thanks
> > > >>>
> > > >>
> > > >>
> > > >
> >