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Re: Amperage pull of a computer?

Jaxon said:
I am wondering if anyone knows the approximate amperage use by a typical
home desktop computer and monitor, or if there is an easy way for me to
determine this on my own. (in U.S. with 120 V circuit).

I have other devices that I'd like to know also, such as electronic
music synthesizers and an audio mixer. I ask because I live in a very
old building with weak circuits, and I want to know how much I'm pulling
when I turn different combinations of devices on.

Thank you.

Jaxon

You might have missed the idea DarkMatter posted about
using a resistor - it is buried in the posts. It is the
best answer in the thread to give an easy way to determine
amperage use on your own.

It will let you determine the amperage used by any
of your devices with pretty good accuracy and at low
cost.
 
R

Ross Mac

Jan 1, 1970
0
You might have missed the idea DarkMatter posted about
using a resistor - it is buried in the posts. It is the
best answer in the thread to give an easy way to determine
amperage use on your own.

It will let you determine the amperage used by any
of your devices with pretty good accuracy and at low
cost.
You are assuming he has a DMM to finish the "low cost" job....Not to mention
the ability to do the conversions...
An amp clamp is a simple and easy solution for an "approximation" that he
was looking for...After all, this was not a current study on his PC but just
a power problem he has in an old building with several pieces of hardware.
 
C

Charles Perrin

Jan 1, 1970
0
After all, this was not a current study on his PC but just
a power problem he has in an old building with several
pieces of hardware.

For some reason, that made me think of "Green Acres" even though I
haven't watched re-runs of it in years. <wink>
 
R

Ross Mac

Jan 1, 1970
0
Charles Perrin said:
For some reason, that made me think of "Green Acres" even though I
haven't watched re-runs of it in years. <wink>
Brings back some old memories......
 
D

DarkMatter

Jan 1, 1970
0
It will let you determine the amperage used by any
of your devices with pretty good accuracy and at low
cost.

Extremely good accuracy. In fact. As accurate as the resistor, and
meter are. :-]
 
D

DarkMatter

Jan 1, 1970
0
You are assuming he has a DMM to finish the "low cost" job....Not to mention
the ability to do the conversions...

With a 0.1 Ohm resistor, you'd have to be a total MacTard (I owe you
several thousand of those... bwuahahahahaha) to have a problem with
the "conversion". You can do it in your head. It's called "shift
decimal, record reading".... Done! DIRECT conversion.

1V = 10A

0.1V = 1A

Any monkey can do it. Move decimal ONE place.
 
R

Ross Mac

Jan 1, 1970
0
DarkMatter said:
With a 0.1 Ohm resistor, you'd have to be a total MacTard (I owe you
several thousand of those... bwuahahahahaha) to have a problem with
the "conversion". You can do it in your head. It's called "shift
decimal, record reading".... Done! DIRECT conversion.

1V = 10A

0.1V = 1A

Any monkey can do it. Move decimal ONE place.

I see the drugs are kicking in....enjoy until you pass out!
And I am glad to see you are an expert at Ohms Law...now that's a tough
one....I will sleep better to night, SnippetTroll, knowing that you can do
simple arithmetic....
 
Ross said:
You are assuming he has a DMM to finish the "low cost" job....Not to > mention
the ability to do the conversions...
An amp clamp is a simple and easy solution for an "approximation" that he
was looking for...After all, this was not a current study on his PC but just
a power problem he has in an old building with several pieces of hardware.

No, Ross, I am assuming nothing. A DMM with an
amp clamp costs more than a DMM without an amp
clamp. A 1% resistor would cost about 5 bucks.
As to the conversion: just exactly how hard is
it for him to multiply by 10?
 
DarkMatter said:
It will let you determine the amperage used by any
of your devices with pretty good accuracy and at low
cost.

Extremely good accuracy. In fact. As accurate as the resistor, and
meter are. :-]

Yes!! It's often a pain in the ass to see the
error introduced when using an ammeter in series
with the load - and sometimes worse than that,
screwing up the circuit. I can't believe that some
respondents can't see the value of the method.
 
D

DarkMatter

Jan 1, 1970
0
It will let you determine the amperage used by any
of your devices with pretty good accuracy and at low
cost.

Extremely good accuracy. In fact. As accurate as the resistor, and
meter are. :-]

Yes!! It's often a pain in the ass to see the
error introduced when using an ammeter in series
with the load - and sometimes worse than that,
screwing up the circuit. I can't believe that some
respondents can't see the value of the method.


For some reason, when those two respond to me, they don blinders,
and not only fail to read what I post, but what others post as well.
They would rather troll me than actually participate, then they
manipulate what I posted into something I did NOT say, and claim *my*
stupidity. It don't get much worse than that. I could outline the
RossTard's omissions, and shirks from day one. He would rather sing
little wussy troll boy Usenet troll songs though.

He has his reward!

Hell, I even found (and posted) a link the next day to a national
semi site that explains the error!
 
R

Ross Mac

Jan 1, 1970
0
hardware.

No, Ross, I am assuming nothing. A DMM with an
amp clamp costs more than a DMM without an amp
clamp. A 1% resistor would cost about 5 bucks.
As to the conversion: just exactly how hard is
it for him to multiply by 10?

I was speaking of a standalone amp clamp....take your pick of available
devices...The resistor is not necesary and by the way....how do you plan to
pick the wattage.....just a guess?....Until you have an approximation of
current you cannot pick the wattage....by the time you have an
approximation...you already made your measurement......As far as
multilplying by ten....when did anyone in this thread ever bring that one
up???
 
R

Ross Mac

Jan 1, 1970
0
It will let you determine the amperage used by any
of your devices with pretty good accuracy and at low
cost.

Extremely good accuracy. In fact. As accurate as the resistor, and
meter are. :-]

Yes!! It's often a pain in the ass to see the
error introduced when using an ammeter in series
with the load - and sometimes worse than that,
screwing up the circuit. I can't believe that some
respondents can't see the value of the method.

What type of ammeter do you use for such measurements? And what type of
accuracy are you looking for? And how does that relate to the OP who wanted
an approximation?.....Just a curiosity at this point.
 
D

DarkMatter

Jan 1, 1970
0
...how do you plan to
pick the wattage.....just a guess?.

You obviously know nothing about current shunts.

For my 10 Amp power strip installed version, one would get a
resistor that allows for 15 or 20 Amp passes and still be within half
the wattage range of the resistor(s). There is nothing wrong with a
pair of resistors at twice the calculated value in parallel to
distribute the energy either. Just keep them thermally coupled.

The resistor(s) used will actually be a 0.01 or a 0.001 ohm value.
Somewhere in that range (not my original 0.1), so it take quite some
current through it before it drops enough voltage to see its rated
wattage. Sheesh, this is an easy engineering exercise to do. Try
again, RossTard!
 
Ross said:
I was speaking of a standalone amp clamp....take your pick of available
devices...

Whichever one you pick, whether it is a clamp that
you use with a DMM, or a stand-alone amp clamp that
includes a readout, will cost more than a DMM and
precision resistor to get the same accuracy. If I am
wrong, please supply me with a make/model - I'll
buy one!
The resistor is not necesary

The method I referred to requires a resistor.
Other methods do not.
and by the way ....how do you plan to
pick the wattage.....just a guess?....Until you have an approximation of
current you cannot pick the wattage....

Of course you can. You should know how.
P = I^2R. A 15W resistor will do fine.
Even if his PC draws a whopping 10 amps
(a grossly high number), the resistor would
need to dissipate only 10 watts. Likewise
for the other devices he wants to know about
such as a synthesizer and audio mixer - we're
not talking high power, here.
by the time you have an
approximation...you already made your measurement......

Not true. See above. And an approximation is NOT a
measurement.
As far as
multilplying by ten....when did anyone in this thread ever bring that one
up???

You did. You said "Not to mention the ability to do
the conversions" . The conversion from volts measured
across a .1 ohm resistor to amps requires multiplying
by ten. Did you have something else in mind when you
mentioned the ability to do the conversions?
 
Ross said:
On Tue, 27 Jan 2004 05:03:52 GMT, [email protected] Gave us:


It will let you determine the amperage used by any
of your devices with pretty good accuracy and at low
cost.

Extremely good accuracy. In fact. As accurate as the resistor, and
meter are. :-]

Yes!! It's often a pain in the ass to see the
error introduced when using an ammeter in series
with the load - and sometimes worse than that,
screwing up the circuit. I can't believe that some
respondents can't see the value of the method.

What type of ammeter do you use for such measurements?

I'll answer this question and others inline, as thoroughly
as I can, below, to try to resolve this as best I can.
That makes it long, of necessity.

That depends on what I'm doing. A couple of examples
with a project I'm currently working on:
I've got a battery pulser that sends a short pulse
of 3.2 amps into the battery. Running the pulse in
series through an amp meter, or putting a clamp on
the wire produces a pronounced affect on the waveform
due to the change in inductance. Hell - I can't even
move the short connecting wires without changing the
waveform. So I have to use a resistor in the circuit,
and measure the voltage across it with the scope to get
a current measurement. When I discharge the battery
(12V auto battery) into an auto headlamp, I am looking
for a precision discharge to exactly (as close as my
Fluke's accuracy) 11.89 volts at room temperature. I
want to be able to see the exact current drawn at every
10 mv step along the way below 12.2, again, within the
accuracy of my test equipment. I can do that with a .01
ohm 1% precision resistor in series with the headlamp,
measuring the voltage across the resistor, and get a more
accurate reading than the Fluke gives, because there is a
voltage drop across the Fluke when I put it in series with
the load to measure amps directly.

In general, I've run into this problem over and over
when I want "exact" (as close as my equipment will allow)
measurements. There is a voltage drop across the
ammeter when it is in series with the circuit being
measured. Try this yourself: set up a circuit that
draws (as measured with your ammeter in series with
the load) current that is a little less than the
maximum for the range the meter is set to. Now,
turn the meter range selector up to the next higher
range. You'll often (always?) get two different
readings. Ask yourself which one is correct? Why
are they different? (Answers below)
And what type of
accuracy are you looking for?

Try the test I mentioned above. The accuracy
needed would depend on the circuit - but I
would like readings on 2 different scales
to be within 2% of each other. That is unreasonable:
say one range is 2000 mA and the other is 200 mA.
A 1% error on 2000 mA is 20 mA. A 1% error on the
200 mA scale is 2 mA. If I am measuring 100 mA,
the meter could say 98 mA on the 200 mA scale,
and it could say 120 mA on the 2000 mA scale.
Now try the test, measuring voltage across a
precision resistor. Ask yourself why that method
yields a far more consistent current. With the
precision resistor and different voltage scales,
I get what I want. I know my measurement is within
the accuracy of my equipment. The reason is that
putting an amp meter in series with the load
changes the circuit, by adding an unknown R
(the resistance of the test leads and internal
meter shunt) - and changing the meter setting
changes the circuit again, by changing the shunt
R. With the .1 ohm precision resistor measurement,
there is no change to the circuit, other than the
known R. On voltage scales, the internal impedance
of the meter is negligible with respect to circuit,
regardless of scale. An ugly thing with DMM's is
that on the amps scales, the closer the scale is
to the current being measured, the more the meter
affects the current being drawn. The farther away
it is over the actual current being drawn, the less
it impacts the circuit - but the lower the resolution
is. Going under the range (e.g. setting the meter to
the 200 mA scale to measure 300 mA) is a non-starter:
you can't get a reading and you may blow the fuse
or cook the meter if it's not protected.
And how does that relate to the OP who wanted
an approximation?.....Just a curiosity at this point.


The op wanted 1) an approximation of the current his
PC would use OR 2) an easy way to determine it
for himself. In addition, 3) he wanted to know the
current other devices would draw. Here's the
part of his post that addresses what he wants
beyond the approximation of what his PC uses:
"or if there is an easy way for me to determine
this on my own. (in U.S. with 120 V circuit).
I have other devices that I'd like to know also, such
as electronic music synthesizers and an audio mixer. "

It is the cheapest way he can measure the current
for himself of the PC and the other devices. It also
happens to be far more accurate than an approximation.
It avoids the safety problem of opening the AC supply
line and having 120 volts AC through the meter leads
& connections. DM described putting the resistor in
the neutral side inside a power strip, so the voltage
at the meter leads would be low. As far as I know, no
one in the thread described a safe way to use an inline
ammeter. A clamp on is more expensive (I'm hoping
you'll provide a cheaper source than I have found for
them) and the use of a clamp requires opening things
up to clamp on a single lead - again a safety exposure.
It is cheaper, safer and more accurate way to provide
answers to parts 2 and 3 of what he wanted - and
in answering part 2, he gets the answer to part 1,
only it is far better than an approximation.
 
R

Ross Mac

Jan 1, 1970
0
Whichever one you pick, whether it is a clamp that
you use with a DMM, or a stand-alone amp clamp that
includes a readout, will cost more than a DMM and
precision resistor to get the same accuracy. If I am
wrong, please supply me with a make/model - I'll
buy one!


The method I referred to requires a resistor.
Other methods do not.


Of course you can. You should know how.
P = I^2R. A 15W resistor will do fine.
Even if his PC draws a whopping 10 amps
(a grossly high number), the resistor would
need to dissipate only 10 watts. Likewise
for the other devices he wants to know about
such as a synthesizer and audio mixer - we're
not talking high power, here.


Not true. See above. And an approximation is NOT a
measurement.


You did. You said "Not to mention the ability to do
the conversions" . The conversion from volts measured
across a .1 ohm resistor to amps requires multiplying
by ten. Did you have something else in mind when you
mentioned the ability to do the conversions?

The comment was not about MY ability to do the conversions. They are
rudimentary electronics. It was of the OP. Also you can use P=IE or I^2R or
whatever permutation you would like, but the OP would still need to know the
current. All the OP had, assuming he even knew the power formula was P=?E
.......so now, how do you figure the wattage for the resistor? Each device
would require a different wattage....even more guess work. An amp clamp is
cheap...and the easiest way to approximate the current.
The question was obviously not from an engineer or a technician so why would
anyone want to send him on such a wild goose chase...... While your resistor
is cheap, did you think of the other supporting hardware he might not also
have? To do this safely, you would probably want to build it up in a box
with an outlet and cord. I think by the time you spent the time and expense
with all this, you are back to a cheap amp clamp.
Enjoy the Superbowl if you are so inclined...Ross : )
 
R

Ross Mac

Jan 1, 1970
0
Ross Mac said:
PC

The comment was not about MY ability to do the conversions. They are
rudimentary electronics. It was of the OP. Also you can use P=IE or I^2R or
whatever permutation you would like, but the OP would still need to know the
current. All the OP had, assuming he even knew the power formula was P=?E
......so now, how do you figure the wattage for the resistor? Each device
would require a different wattage....even more guess work. An amp clamp is
cheap...and the easiest way to approximate the current.
The question was obviously not from an engineer or a technician so why would
anyone want to send him on such a wild goose chase...... While your resistor
is cheap, did you think of the other supporting hardware he might not also
have? To do this safely, you would probably want to build it up in a box
with an outlet and cord. I think by the time you spent the time and expense
with all this, you are back to a cheap amp clamp.
Enjoy the Superbowl if you are so inclined...Ross : )
I meant to also mention, he was not JUST measuring his PC, so a guess at 10
amps might not be a good one and if he had the current listed on the
devices, he would not need to measure it in the first place...
 
R

Ross Mac

Jan 1, 1970
0
Ross said:
DarkMatter wrote:

On Tue, 27 Jan 2004 05:03:52 GMT, [email protected] Gave us:


It will let you determine the amperage used by any
of your devices with pretty good accuracy and at low
cost.

Extremely good accuracy. In fact. As accurate as the resistor, and
meter are. :-]

Yes!! It's often a pain in the ass to see the
error introduced when using an ammeter in series
with the load - and sometimes worse than that,
screwing up the circuit. I can't believe that some
respondents can't see the value of the method.

What type of ammeter do you use for such measurements?

I'll answer this question and others inline, as thoroughly
as I can, below, to try to resolve this as best I can.
That makes it long, of necessity.

That depends on what I'm doing. A couple of examples
with a project I'm currently working on:
I've got a battery pulser that sends a short pulse
of 3.2 amps into the battery. Running the pulse in
series through an amp meter, or putting a clamp on
the wire produces a pronounced affect on the waveform
due to the change in inductance. Hell - I can't even
move the short connecting wires without changing the
waveform. So I have to use a resistor in the circuit,
and measure the voltage across it with the scope to get
a current measurement. When I discharge the battery
(12V auto battery) into an auto headlamp, I am looking
for a precision discharge to exactly (as close as my
Fluke's accuracy) 11.89 volts at room temperature. I
want to be able to see the exact current drawn at every
10 mv step along the way below 12.2, again, within the
accuracy of my test equipment. I can do that with a .01
ohm 1% precision resistor in series with the headlamp,
measuring the voltage across the resistor, and get a more
accurate reading than the Fluke gives, because there is a
voltage drop across the Fluke when I put it in series with
the load to measure amps directly.

In general, I've run into this problem over and over
when I want "exact" (as close as my equipment will allow)
measurements. There is a voltage drop across the
ammeter when it is in series with the circuit being
measured. Try this yourself: set up a circuit that
draws (as measured with your ammeter in series with
the load) current that is a little less than the
maximum for the range the meter is set to. Now,
turn the meter range selector up to the next higher
range. You'll often (always?) get two different
readings. Ask yourself which one is correct? Why
are they different? (Answers below)
And what type of
accuracy are you looking for?

Try the test I mentioned above. The accuracy
needed would depend on the circuit - but I
would like readings on 2 different scales
to be within 2% of each other. That is unreasonable:
say one range is 2000 mA and the other is 200 mA.
A 1% error on 2000 mA is 20 mA. A 1% error on the
200 mA scale is 2 mA. If I am measuring 100 mA,
the meter could say 98 mA on the 200 mA scale,
and it could say 120 mA on the 2000 mA scale.
Now try the test, measuring voltage across a
precision resistor. Ask yourself why that method
yields a far more consistent current. With the
precision resistor and different voltage scales,
I get what I want. I know my measurement is within
the accuracy of my equipment. The reason is that
putting an amp meter in series with the load
changes the circuit, by adding an unknown R
(the resistance of the test leads and internal
meter shunt) - and changing the meter setting
changes the circuit again, by changing the shunt
R. With the .1 ohm precision resistor measurement,
there is no change to the circuit, other than the
known R. On voltage scales, the internal impedance
of the meter is negligible with respect to circuit,
regardless of scale. An ugly thing with DMM's is
that on the amps scales, the closer the scale is
to the current being measured, the more the meter
affects the current being drawn. The farther away
it is over the actual current being drawn, the less
it impacts the circuit - but the lower the resolution
is. Going under the range (e.g. setting the meter to
the 200 mA scale to measure 300 mA) is a non-starter:
you can't get a reading and you may blow the fuse
or cook the meter if it's not protected.
And how does that relate to the OP who wanted
an approximation?.....Just a curiosity at this point.


The op wanted 1) an approximation of the current his
PC would use OR 2) an easy way to determine it
for himself. In addition, 3) he wanted to know the
current other devices would draw. Here's the
part of his post that addresses what he wants
beyond the approximation of what his PC uses:
"or if there is an easy way for me to determine
this on my own. (in U.S. with 120 V circuit).
I have other devices that I'd like to know also, such
as electronic music synthesizers and an audio mixer. "

It is the cheapest way he can measure the current
for himself of the PC and the other devices. It also
happens to be far more accurate than an approximation.
It avoids the safety problem of opening the AC supply
line and having 120 volts AC through the meter leads
& connections. DM described putting the resistor in
the neutral side inside a power strip, so the voltage
at the meter leads would be low. As far as I know, no
one in the thread described a safe way to use an inline
ammeter. A clamp on is more expensive (I'm hoping
you'll provide a cheaper source than I have found for
them) and the use of a clamp requires opening things
up to clamp on a single lead - again a safety exposure.
It is cheaper, safer and more accurate way to provide
answers to parts 2 and 3 of what he wanted - and
in answering part 2, he gets the answer to part 1,
only it is far better than an approximation.

Take an extension cord (lamp cord style), separate one of the power leads
and amp clamp it....No safety issue, no big expense and most likely a pretty
good approximation. How do you figure? You are going to buy a resistor with
a "guessed" wattage, then have to build some sort of box with a cord and
outlet. You make it sound like "all" you need is a resistor. Does your
resistor just plug into the wall?...You still need additional hardware. And
how do you figure the OP can handle all this since he asked a very simple
question to begin with? A 2% error, if that's what you got, certainly would
not be an issue by reading the OP. Are you even sure he can perform the
simple arithmetic to figure the current based on voltage measurement across
the resistor? Based on the OP, I would say no......
 
D

DarkMatter

Jan 1, 1970
0
The comment was not about MY ability to do the conversions. They are
rudimentary electronics. It was of the OP. Also you can use P=IE or I^2R or
whatever permutation you would like, but the OP would still need to know the
current. All the OP had, assuming he even knew the power formula was P=?E
......so now, how do you figure the wattage for the resistor?

When you read the voltage across the resistor, and convert, you get
the exact current flowing in the entire circuit.

Back to school for you!
 
D

DarkMatter

Jan 1, 1970
0
Each device
would require a different wattage....even more guess work. An amp clamp is
cheap...and the easiest way to approximate the current.

They are not cheap. Opening the line up to expose a single line is
still required as well. I have already told you this once.
The question was obviously not from an engineer or a technician so why would
anyone want to send him on such a wild goose chase...

People have been measuring current this way for well over a hundred
years. You are the wild goose, and your brain is what you are
chasing.
While your resistor
is cheap, did you think of the other supporting hardware he might not also
have?

A simple meter. Doh!
To do this safely, you would probably want to build it up in a box
with an outlet and cord.

A power strip with a line cord attached (they all have them -DOH-)

Open the line cord! D O H !!! A real toughy!
I think by the time you spent the time and expense
with all this, you are back to a cheap amp clamp.

They are not cheap. DOH!
Enjoy the Superbowl if you are so inclined...Ross : )

Grow up and admit you are clueless if you are so inclined.
 
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