I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@ h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/e...s/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 at hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@ u@e@n@t@@

Watson A.Name - 'Watt Sun' wrote in
:

I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.


Also, you must now consider the combined accuracy, of two meters, instead
of just one.

Yes it works, but, so do pliers on a hex-nut. Something about the
right tool for the job.

Congratulations. You proved Ohm's Law works. Of course it's only as accurate
as your meters, then you have to use your calculator.

Sorry, but give me a decent digital multimeter.

Your method is certainly worth remembering in a pinch.

Mark Z.


"Watson A.Name - 'Watt Sun'" wrote in message
.. .
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@ h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/e...s/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 at hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@ u@e@n@t@@

"Mark D. Zacharias" wrote in message
link.net...
Congratulations. You proved Ohm's Law works.

Actually it is the resistance formula that works. What you think is
Ohm's (V=IR) is not. See the links below and a good physics book. Ratch
http://maxwell.byu.edu/~spencerr/websumm122/node50.html
http://www.launc.tased.edu.au/online...Resistance.htm

Of course it's only as accurate
as your meters, then you have to use your calculator.

Sorry, but give me a decent digital multimeter.

Your method is certainly worth remembering in a pinch.

Mark Z.


"Watson A.Name - 'Watt Sun'" wrote in message
.. .
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@ h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/e...s/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 at hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@ u@e@n@t@@

Yes, and Ohm's Law describes the interaction of resistance, voltage, and
current. Not just resistance.
I'm not an engineer, granted, but I don't require an education on Ohm's Law.

Mark Z.

Actually it is the resistance formula that works. What you think is
Ohm's (V=IR) is not. See the links below and a good physics book.

"Ratch" wrote in message
news
"Mark D. Zacharias" wrote in message
link.net...
Congratulations. You proved Ohm's Law works.

Actually it is the resistance formula that works. What you think is
Ohm's (V=IR) is not. See the links below and a good physics book. Ratch
http://maxwell.byu.edu/~spencerr/websumm122/node50.html

http://www.launc.tased.edu.au/online...ctric/resistnc
/Resistance.htm

Of course it's only as accurate
as your meters, then you have to use your calculator.

Sorry, but give me a decent digital multimeter.

Your method is certainly worth remembering in a pinch.

Mark Z.


"Watson A.Name - 'Watt Sun'" wrote in message
.. .
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@ h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/e...s/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 at hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@ u@e@n@t@@

"Mark D. Zacharias" wrote in message
link.net...
Yes, and Ohm's Law describes the interaction of resistance, voltage, and
current. Not just resistance.
I'm not an engineer, granted, but I don't require an education on Ohm's
Law.

Mark Z.

I belive that you are missing the point. The resistance (or impedance)
formula V=IR (or V=IZ), describes the describes the interaction of
resistance (impedance), voltage, and current. While correct and true in all
cases, those formulas are NOT Ohm's law, and it is wrong to call them that.
As shown in the second link I gave, Ohm's law is a property of resistive
linearity in a material. Just as the specific gravity of a material is a
property. If it conforms to Ohm's law, it is ohmic. Otherwise it is
nonohmic. Ratch



Actually it is the resistance formula that works. What you think
is
Ohm's (V=IR) is not. See the links below and a good physics book. Ratch
http://maxwell.byu.edu/~spencerr/websumm122/node50.html


http://www.launc.tased.edu.au/online...ctric/resistnc
/Resistance.htm

"Mark D. Zacharias" wrote in message
link.net...
Yes, and Ohm's Law describes the interaction of resistance, voltage, and
current. Not just resistance.
I'm not an engineer, granted, but I don't require an education on Ohm's
Law.

Mark Z.

I belive that you are missing the point. The resistance (or impedance)
formula V=IR (or V=IZ), describes the describes the interaction of
resistance (impedance), voltage, and current. While correct and true in all
cases, those formulas are NOT Ohm's law, and it is wrong to call them that.
As shown in the second link I gave, Ohm's law is a property of resistive
linearity in a material. Just as the specific gravity of a material is a
property. If it conforms to Ohm's law, it is ohmic. Otherwise it is
nonohmic. Ratch



Actually it is the resistance formula that works. What you think
is
Ohm's (V=IR) is not. See the links below and a good physics book. Ratch
http://maxwell.byu.edu/~spencerr/websumm122/node50.html


http://www.launc.tased.edu.au/online...ctric/resistnc
/Resistance.htm

"Ratch" wrote in message
news:BZmRa.80118$Ph3.9181@sccrnsc04...
While correct and true in all
cases, those formulas are NOT Ohm's law, and it is wrong to call them
that.

Dude, go back to school, Algebra 1. V=IR I=V/R R=V/I etc.
The equation can be rearranged to any of the others by simple
multiplication and division by whichever variable.

Tim

--
In the immortal words of Ned Flanders: "No foot longs!"
Website @ http://webpages.charter.net/dawill/tmoranwms

"Ratch" wrote in
news:BZmRa.80118$Ph3.9181@sccrnsc04:


"Mark D. Zacharias" wrote in message
link.net...
Yes, and Ohm's Law describes the interaction of resistance, voltage,
and current. Not just resistance.
I'm not an engineer, granted, but I don't require an education on
Ohm's
Law.

Mark Z.

I belive that you are missing the point. The resistance (or
impedance)
formula V=IR (or V=IZ), describes the describes the interaction of
resistance (impedance), voltage, and current. While correct and true
in all cases, those formulas are NOT Ohm's law, and it is wrong to
call them that. As shown in the second link I gave, Ohm's law is a
property of resistive linearity in a material. Just as the specific
gravity of a material is a property. If it conforms to Ohm's law, it
is ohmic. Otherwise it is nonohmic. Ratch


What the heck is "nonohmic"? Is this a word you just made up? I have been
an electronis tech for 30+ years and thats a new one on me. It all doesn't
matter. Wether it is inductance, capacitive impedance, a thermistor, a
varistor, or what ever. Ohm's law still stands firm. For changing
"impedance" or fixed resistance. At any moment in time, there is a certain
resistance(impedance), a certain voltage and a certain current and ohm's
law always applies. Even in a combined circuit of capacitance and
inductance with an appplied frequency signal. At 1 instantaneous moment,
there is a vectored impedance and associated voltage and current. It is a
law of physics and there is no getting around it no matter what you call
it. A rose is a rose is a rose.
Here are the formulae and you believe what you will.

In article ,
mentioned...
"Mark D. Zacharias" wrote in message
link.net...
Congratulations. You proved Ohm's Law works.

Actually it is the resistance formula that works. What you think is
Ohm's (V=IR) is not. See the links below and a good physics book. Ratch
http://maxwell.byu.edu/~spencerr/websumm122/node50.html
http://www.launc.tased.edu.au/online...Resistance.htm

And both Mark Z and Ratch are barking up the wrong tree. If they were
to reread my post below, they would see that I used the formula,
R=V/I, which in both URLs above was shown first and named Ohm's Law.
Doh.

Of course it's only as accurate
as your meters, then you have to use your calculator.

Sorry, but give me a decent digital multimeter.

Your method is certainly worth remembering in a pinch.

Mark Z.

"Watson A.Name - 'Watt Sun'" wrote in message
.. .
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

--

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@ h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/e...s/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 at hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@ u@e@n@t@@

"Watson A.Name - 'Watt Sun'" wrote in message
.. .
In article ,
mentioned...
"Mark D. Zacharias" wrote in message
link.net...
Congratulations. You proved Ohm's Law works.

Actually it is the resistance formula that works. What you think
is
Ohm's (V=IR) is not. See the links below and a good physics book. Ratch
http://maxwell.byu.edu/~spencerr/websumm122/node50.html

http://www.launc.tased.edu.au/online...Resistance.htm

And both Mark Z and Ratch are barking up the wrong tree. If they were
to reread my post below, they would see that I used the formula,
R=V/I, which in both URLs above was shown first and named Ohm's Law.
Doh.

No, both the above URLs make a point of saying that R=V/I is not Ohm's
law, and the first refers to R=V/I as the resistance formula. In other
words, Ohm's law referring to V=IR is a misnomer. The second URL points out
that Ohm's law really and truly refers to the resistive linearity of a
material. Dah.

By the way, you did not turn your electrical energy supply into a
ohmmeter. You applied a method of using the energy supply to determine
resistance. Ratch


Of course it's only as accurate
as your meters, then you have to use your calculator.

Sorry, but give me a decent digital multimeter.

Your method is certainly worth remembering in a pinch.

Mark Z.

"Watson A.Name - 'Watt Sun'" wrote in message
.. .
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V
and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

In article tdvRa.71417$OZ2.13017@rwcrnsc54,
mentioned...

"Watson A.Name - 'Watt Sun'" wrote in message
.. .
In article ,
mentioned...
"Mark D. Zacharias" wrote in message
link.net...
Congratulations. You proved Ohm's Law works.

Actually it is the resistance formula that works. What you think
is
Ohm's (V=IR) is not. See the links below and a good physics book. Ratch
http://maxwell.byu.edu/~spencerr/websumm122/node50.html

http://www.launc.tased.edu.au/online...Resistance.htm

And both Mark Z and Ratch are barking up the wrong tree. If they were
to reread my post below, they would see that I used the formula,
R=V/I, which in both URLs above was shown first and named Ohm's Law.
Doh.

No, both the above URLs make a point of saying that R=V/I is not Ohm's
law, and the first refers to R=V/I as the resistance formula. In other
words, Ohm's law referring to V=IR is a misnomer. The second URL points out
that Ohm's law really and truly refers to the resistive linearity of a
material. Dah.

This is *still* pointless. I *never* claimed that I was doing
*anything* with _Ohm's_Law_!! Quit trying to put words in my mouth!
See my OP below.

[snip]
Ratch

Of course it's only as accurate
as your meters, then you have to use your calculator.

Sorry, but give me a decent digital multimeter.

Your method is certainly worth remembering in a pinch.

Mark Z.

"Watson A.Name - 'Watt Sun'" wrote in message
.. .
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V
and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@ h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/e...s/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 at hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@ u@e@n@t@@

"Watson A.Name - 'Watt Sun'" wrote in message
.. .
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.



Yep, that's how ohmmeters work actually. I'm currently designing a digitally
controlled PSU which includes a dot-matrix LCD that shows a lot of info
about the state of the PSU. For e.g., it multiplies V * I so the user can
see the load power in real-time. Simple, but very handy. I suppose I could
also have it display V / I to show the load resistance in real-time. I think
the PSU approach is good for measuring very small resistances (when you need
to generate a lot of current to have a voltage drop large enough to measure
accurately.

cheers,
Costas

In article ,
"Costas Vlachos" wrote:

"Watson A.Name - 'Watt Sun'" wrote in message
.. .
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.



Yep, that's how ohmmeters work actually. I'm currently designing a digitally
controlled PSU which includes a dot-matrix LCD that shows a lot of info
about the state of the PSU. For e.g., it multiplies V * I so the user can
see the load power in real-time. Simple, but very handy. I suppose I could
also have it display V / I to show the load resistance in real-time. I think
the PSU approach is good for measuring very small resistances (when you need
to generate a lot of current to have a voltage drop large enough to measure
accurately.


At a local electronics store, there was a sale of multimeters for $5. I
bought a bunch and have velcroed some to my workbench. I set them for
voltage, current or resistance and leave them there. Good cheap way of
doing some quick and dirty measurements. Surprisingly accurate too. If I
smoke one, I just toss it. After all, it is a toss away world nowadays.

Al

--
There's never enough time to do it right the first time.......

Watt Sun:
Just a little cumbersome.... isn't it??
As you indicated, there is a risk of smoking low ohm, low wattage resistors
unless you already know the value... if that is the case then why are you
measuring it??......
.....and the accuracy is compromised because you are measuring the voltage
and then measuring the current..... and you are at the mercy of the
regulation of your power supply.
2 meter operations instead of one... the inherent innaccuracy of one of the
readings is further compromised by the inaccuracy of the 2nd reading.....
give me a DMM or VOM with a dedicated OHMS function any time..
Actually, the much more used and handier version of this is the "flip-side"
where you measure the voltage across a known-value resistor in the circuit
to determine the approximate current.... most techs do this all the time
while routinely troubleshooting.
--
Best Regards,
Daniel Sofie
Electronics Supply & Repair
---------------------------------


"Watson A.Name - 'Watt Sun'" wrote in message
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

In article ,
mentioned...
Watt Sun:
Just a little cumbersome.... isn't it??

Less cumbersome than removing the DMM leads and then reconnecting
them.

As you indicated, there is a risk of smoking low ohm, low wattage resistors

Yes, as I indicated.

unless you already know the value... if that is the case then why are you
measuring it??......

If you already know the value, then there'd be no point in measuring
it. So you would be measuring the unknown value to determine it.

....and the accuracy is compromised because you are measuring the voltage
and then measuring the current.....

It still finds the value with a reasonable accuracy.

and you are at the mercy of the regulation of your power supply.

The regulation of the power supply makes no difference.

2 meter operations instead of one... the inherent innaccuracy of one of the
readings is further compromised by the inaccuracy of the 2nd reading.....

You're repeating yourself. As I said above, it still finds the value
with reasonable accuracy.

give me a DMM or VOM with a dedicated OHMS function any time..

If you reread my post, you would see that I already have the DMM. I
was using an alternate method.

If the VOM you mention above is an analog wiggle stick meter, it may
be less accurate - maybe only 3% - than using my PS method.

Actually, the much more used and handier version of this is the "flip-side"
where you measure the voltage across a known-value resistor in the circuit
to determine the approximate current.... most techs do this all the time
while routinely troubleshooting.

Right. Now you've stated something useful.

--
Best Regards,
Daniel Sofie
Electronics Supply & Repair
---------------------------------


"Watson A.Name - 'Watt Sun'" wrote in message
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@ h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/e...s/databank.htm
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Just when you thought you had all this figured out, the gov't
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@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@ u@e@n@t@@

In sci.electronics.misc Watson A.Name - 'Watt Sun' wrote:
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

This isn't especially usefull usually.
However, with low ohm resistors, it can be.

Given a constant current of an amp, the $5 meters mentioned elsewhere
can now measure with a resolution of .1mohm.


--
http://inquisitor.i.am/ | | Ian Stirling.
---------------------------+-------------------------+--------------------------
My inner child can beat up your inner child. - Alex Greenbank

In article ,
mentioned...
In sci.electronics.misc Watson A.Name - 'Watt Sun' wrote:
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

This isn't especially usefull usually.
However, with low ohm resistors, it can be.

Given a constant current of an amp, the $5 meters mentioned elsewhere
can now measure with a resolution of .1mohm.

I bought a few of those $5 DMMs from Futurlec a few months ago,
actually I think they were about $6. 9V vattery included(!)


--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@ h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/e...s/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 at hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@ u@e@n@t@@

In sci.electronics.misc Watson A.Name - 'Watt Sun' wrote:
In article ,
mentioned...
In sci.electronics.misc Watson A.Name - 'Watt Sun' wrote:
I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

This isn't especially usefull usually.
However, with low ohm resistors, it can be.

Given a constant current of an amp, the $5 meters mentioned elsewhere
can now measure with a resolution of .1mohm.

I bought a few of those $5 DMMs from Futurlec a few months ago,
actually I think they were about $6. 9V vattery included(!)

Very handy indeed.
The ones I bought were 3 pounds 99p, ($6us?) I have around 8.
Soon after buying one, I thought I'd discovered that they have an
overvoltage LED.
However, the smell of burning FR4 soon made me realise otherwise.

It's amazing how many multimeters you can use when you have them free.
I was just discharging a series string of Li-Ion batteries, to measure
capacities.
In the past I would have taken measurements every 5 minutes to ensure
none had approached 3V.

Why bother, just hook up 5 of them, and glance over every once in a while.,

--
http://inquisitor.i.am/ | | Ian Stirling.
---------------------------+-------------------------+--------------------------
Lord, grant me the serenity to accept that I cannot change, the
courage to change what I can, and the wisdom to hide the bodies
of those I had to kill because they ****ed me off. - Random

"Alan Harriman" wrote to "All" (16 Jul 03 22:18:49)
--- on the topic of " Turn Your Power Supply into an Ohmmeter - It's Free!"

Most Englishmen were also expecting biscuits and cheese on the Moon...
And so too apples bopped people on the head because the Earth sucked.

However, the German physicist George Simon Ohm found that a current
flowing through a conductor is directly proportional to the electrical
force that produces it. i.e. E=IR

So the other fellow with the ohmic and non ohmic materials discussion is
trying to pull a fast one. i.e. a troll


AH From: Alan Harriman

AH C'mon, we've been calling it Ohm's law since elementary school. Now
AH you're gonna tell us we're all wrong? If enough people call it Ohm's
AH law, then it's Ohm's law. : - )

AH Alan Harriman

.... Is reactance then illusory? No, it just appears that way...

Ian Stirling wrote:
In sci.electronics.misc Watson A.Name - 'Watt Sun' wrote:

In article ,
mentioned...

In sci.electronics.misc Watson A.Name - 'Watt Sun' wrote:

I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

This isn't especially usefull usually.
However, with low ohm resistors, it can be.

Given a constant current of an amp, the $5 meters mentioned elsewhere
can now measure with a resolution of .1mohm.

I bought a few of those $5 DMMs from Futurlec a few months ago,
actually I think they were about $6. 9V vattery included(!)


Very handy indeed.
The ones I bought were 3 pounds 99p, ($6us?) I have around 8.
Soon after buying one, I thought I'd discovered that they have an
overvoltage LED.
However, the smell of burning FR4 soon made me realise otherwise.

But, But.. Doesn't the FR in FR4 mean flame resistant? If so, how
could it burn? Char?


It's amazing how many multimeters you can use when you have them free.
I was just discharging a series string of Li-Ion batteries, to measure
capacities.
In the past I would have taken measurements every 5 minutes to ensure
none had approached 3V.

Why bother, just hook up 5 of them, and glance over every once in a while.,



--
----------------(from OED Mini-Dictionary)-----------------
PUNCTUATION - Apostrophe
Incorrect uses: (i) the apostrophe must not be used with a plural
where there is no possessive sense, as in ~tea's are served here~;
(ii) there is no such word as ~her's, our's, their's, your's~.

Confusions: it's = it is or it has (not 'belonging to it'); correct
uses are ~it's here~ (= it is here); ~it's gone~ (= it has gone);
but ~the dog wagged its tail~ (no apostrophe).
----------------(For the Apostrophe challenged)----------------
From a fully deputized officer of the Apostrophe Police!

Spammers use Weapons of Mass Distraction!

I bought some batteries, but they weren't included,
so I had to buy them again.
-- Steven Wright

FOR SALE: Nice parachute: never opened - used once.

F
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d
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f
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u
p
i
d

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n
o
u
g
h

i
n
c
l
d
u
d
e
d

t
e
x
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e
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..

In sci.electronics.misc Lizard Blizzard wrote:
Ian Stirling wrote:
In sci.electronics.misc Watson A.Name - 'Watt Sun' wrote:

In article ,
mentioned...

In sci.electronics.misc Watson A.Name - 'Watt Sun' wrote:

I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

This isn't especially usefull usually.
However, with low ohm resistors, it can be.

Given a constant current of an amp, the $5 meters mentioned elsewhere
can now measure with a resolution of .1mohm.

I bought a few of those $5 DMMs from Futurlec a few months ago,
actually I think they were about $6. 9V vattery included(!)


Very handy indeed.
The ones I bought were 3 pounds 99p, ($6us?) I have around 8.
Soon after buying one, I thought I'd discovered that they have an
overvoltage LED.
However, the smell of burning FR4 soon made me realise otherwise.

But, But.. Doesn't the FR in FR4 mean flame resistant? If so, how
could it burn? Char?

Hmm, probably, yes.
It went out when the source of 2000V (at moderate current) was turned off.

--
http://inquisitor.i.am/ | | Ian Stirling.
---------------------------+-------------------------+--------------------------
Lord, grant me the serenity to accept that I cannot change, the
courage to change what I can, and the wisdom to hide the bodies
of those I had to kill because they ****ed me off. - Random

Ian Stirling wrote:


Very handy indeed.
The ones I bought were 3 pounds 99p, ($6us?) I have around 8.
Soon after buying one, I thought I'd discovered that they have an
overvoltage LED.
However, the smell of burning FR4 soon made me realise otherwise.

It's amazing how many multimeters you can use when you have them free.
I was just discharging a series string of Li-Ion batteries, to measure
capacities.
In the past I would have taken measurements every 5 minutes to ensure
none had approached 3V.

Why bother, just hook up 5 of them, and glance over every once in a while.,


YES!!!! I had a situation where I needed to continually and
simultaneously monitor current and voltage into and out of a
dc-dc converter with two outputs, while varying the loads and
the input supply. Some asshole said he could do it with two
meters - another "sniper" who obviously did not understand
the word "simultaneously". Anyway, there's a catalog outfit
named Harbor Freight, and they had multimeters on sale for
$2.99 including the 9V battery. I bought 4 of them. As you
correctly pointed out, it's amazing how many you can use
at the same time when they are just sitting around waiting
for use!

--
http://inquisitor.i.am/ | | Ian Stirling.
---------------------------+-------------------------+--------------------------
Lord, grant me the serenity to accept that I cannot change, the
courage to change what I can, and the wisdom to hide the bodies
of those I had to kill because they ****ed me off. - Random

During a canal boat holiday, my brother enquired what current the
starter took from the battery.

I found that my cheap pocket DMM would read the millivolts across the
earth strap when we turned on all the lights of known wattage; and it
would also read the volts across the strap whilst the starter was
turning. So we were able to calculate the starter motor current as
around 200 A whilst turning steadily (but not the peaks).

I forget how we strangled the engine (diesel) to stop it starting
during this experiment. Could we have let air in somehow? Maybe there
was a stop valve?

BillJ
(Edinburgh)

On Wed, 16 Jul 2003 01:34:28 -0700, Watson A.Name - 'Watt Sun'
wrote:

I got tired of switching the leads of my DMM. Suddenly if dawned on
me that I can just set the power supply to 10.0V for exaample, and
read the current, and then divide the voltage by the current to find
the resistance. Like I put a resistance on the PS, it reads 10.0V and
the current is .018A, so 10 / .018 gives 555.6 ohms. Must be a 560
ohm resistor.

I turned my PS into an ohmmeter - FREE!

Hee-hee - Work smarter, not harder!

Of course, make sure the current stays low so the resistance doesn't
overheat. For low resistances use a volt or less.

the bottom line is that you are recreating an ohm meter except that it
has two meters.

a standard ohm meter does exactly the same thing using an internal
battery