Turn Your Power Supply into an Ohmmeter - It's Free!

  • Thread starter Watson A.Name - 'Watt Sun
  • Start date
W

Watson A.Name - 'Watt Sun

Guest
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/electronics/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' <alondra101@hotmail.com> wrote in
news:MPG.197ead80ac5f22d8989a95@news.inreach.net:

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.
Click to expand...
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'" <alondra101@hotmail.com> wrote in message
news:MPG.197ead80ac5f22d8989a95@news.inreach.net...
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/electronics/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@@
Click to expand...
 
In article <bf3lfa$n14$1@titan.btinternet.com>,
"Costas Vlachos" <c-X-vlachos@hot-X-mail.com> wrote:

"Watson A.Name - 'Watt Sun'" <alondra101@hotmail.com> wrote in message
news:MPG.197ead80ac5f22d8989a95@news.inreach.net...
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.
Click to expand...
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.......
 
"Mark D. Zacharias" <mzacharias@nospam.earthlink.net> wrote in message
news:3V9Ra.4295$Mc.389909@newsread1.prod.itd.earthlink.net...
Congratulations. You proved Ohm's Law works.
Click to expand...
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/sciences/PhysSci/done/electric/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'" <alondra101@hotmail.com> wrote in message
news:MPG.197ead80ac5f22d8989a95@news.inreach.net...
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/electronics/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@@
Click to expand...
 
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'" <alondra101@hotmail.com> 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.
Click to expand...
 
"Watson A.Name - 'Watt Sun'" <alondra101@hotmail.com> wrote in message
news:MPG.197ead80ac5f22d8989a95@news.inreach.net...
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.
Click to expand...


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 sci.electronics.misc Watson A.Name - 'Watt Sun' <alondra101@hotmail.com> 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.
Click to expand...
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/ | mailto:inquisitor@i.am | Ian Stirling.
---------------------------+-------------------------+--------------------------
My inner child can beat up your inner child. - Alex Greenbank
 
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.
Click to expand...
"Ratch" <Watchit@Comcast.net> wrote in message
news:eek:4eRa.65502$wk6.15887@rwcrnsc52.ops.asp.att.net...
"Mark D. Zacharias" <mzacharias@nospam.earthlink.net> wrote in message
news:3V9Ra.4295$Mc.389909@newsread1.prod.itd.earthlink.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/sciences/PhysSci/done/electric/resistnc
Click to expand...
/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'" <alondra101@hotmail.com> wrote in message
news:MPG.197ead80ac5f22d8989a95@news.inreach.net...
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/electronics/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@@
Click to expand...
 
"Mark D. Zacharias" <mzacharias@nospam.earthlink.net> wrote in message
news:4BmRa.5630$Mc.475640@newsread1.prod.itd.earthlink.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.
Click to expand...
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/sciences/PhysSci/done/electric/resistnc
/Resistance.htm
Click to expand...
 
X-No-Archive: Yes



"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.
Click to expand...
Dangerous and inaccurate. Inaccurate, because the resistor will warm up
and the resistance will increase.

Using a volt meter and ammeter is certainly cheap and practical for
measuring a very low resistance that the typical DMM can not.

Say you have a 25ft 12AWG extension cord and you want to know the total
resistance. The average DMM has a resolution down to 100mOhm. The
resolution is nowhere near what you need.

Connect it to ~12V DC(you don't want to use 120V, because high floating
voltage will throw off accuracy on the low voltage meter range) and
connect an ammeter in series.

Connect a load on the other end that takes about a few amps.

Measure the voltage across both ends of one conductor in mV range.

If you read 54.6mV and the ammeter reads 2.12A, you can figure out the
resistance by:


0.0546/2.12=0.02575

three sig dig=0.0258ohms=258mOhms x 2(to accomodate for return
path)=506mOhms
 
"Mark D. Zacharias" <mzacharias@nospam.earthlink.net> wrote in message
news:4BmRa.5630$Mc.475640@newsread1.prod.itd.earthlink.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.
Click to expand...
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/sciences/PhysSci/done/electric/resistnc
/Resistance.htm
Click to expand...
 
In article <o4eRa.65502$wk6.15887@rwcrnsc52.ops.asp.att.net>,
Watchit@Comcast.net mentioned...
"Mark D. Zacharias" <mzacharias@nospam.earthlink.net> wrote in message
news:3V9Ra.4295$Mc.389909@newsread1.prod.itd.earthlink.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/sciences/PhysSci/done/electric/resistnc/Resistance.htm
Click to expand...
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'" <alondra101@hotmail.com> wrote in message
news:MPG.197ead80ac5f22d8989a95@news.inreach.net...
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.

--
Click to expand...
--
@@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/electronics/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 article <vharo290904tc6@corp.supernews.com>, sofie@olypen.com
mentioned...
Watt Sun:
Just a little cumbersome.... isn't it??
Click to expand...
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
Click to expand...
Yes, as I indicated.

unless you already know the value... if that is the case then why are you
measuring it??......
Click to expand...
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.....
Click to expand...
It still finds the value with a reasonable accuracy.

and you are at the mercy of the regulation of your power supply.
Click to expand...
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.....
Click to expand...
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..
Click to expand...
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.
Click to expand...
Right. Now you've stated something useful.

--
Best Regards,
Daniel Sofie
Electronics Supply & Repair
---------------------------------
Click to expand...

"Watson A.Name - 'Watt Sun'" <alondra101@hotmail.com> 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.
Click to expand...
--
@@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/electronics/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 article <bf4ah6$a2p$3$8300dec7@news.demon.co.uk>,
root@mauve.demon.co.uk mentioned...
In sci.electronics.misc Watson A.Name - 'Watt Sun' <alondra101@hotmail.com> 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.
Click to expand...
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/electronics/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@@
 
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'
<alondra101@hotmail.com> 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.
Click to expand...
 
"Watson A.Name - 'Watt Sun'" <alondra101@hotmail.com> wrote in message
news:MPG.1980189f76ff24f6989aa3@news.inreach.net...
In article <o4eRa.65502$wk6.15887@rwcrnsc52.ops.asp.att.net>,
Watchit@Comcast.net mentioned...
"Mark D. Zacharias" <mzacharias@nospam.earthlink.net> wrote in message
news:3V9Ra.4295$Mc.389909@newsread1.prod.itd.earthlink.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/sciences/PhysSci/done/electric/resistnc/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.
Click to expand...
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'" <alondra101@hotmail.com> wrote in message
news:MPG.197ead80ac5f22d8989a95@news.inreach.net...
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.
Click to expand...
 
In article <3F163D14.16B83760@prontoREMOVETHISmail.com>,
JerC@prontoREMOVETHISmail.com mentioned...

"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.

Dangerous and inaccurate. Inaccurate, because the resistor will warm up
and the resistance will increase.

Using a volt meter and ammeter is certainly cheap and practical for
measuring a very low resistance that the typical DMM can not.

Say you have a 25ft 12AWG extension cord and you want to know the total
resistance. The average DMM has a resolution down to 100mOhm. The
resolution is nowhere near what you need.

Connect it to ~12V DC(you don't want to use 120V, because high floating
voltage will throw off accuracy on the low voltage meter range) and
connect an ammeter in series.

Connect a load on the other end that takes about a few amps.
Click to expand...
Or use a constant current PS and short the far end of the cord, and
measure across both conductors on the near end.

Measure the voltage across both ends of one conductor in mV range.

If you read 54.6mV and the ammeter reads 2.12A, you can figure out the
resistance by:

0.0546/2.12=0.02575

three sig dig=0.0258ohms=258mOhms x 2(to accomodate for return
path)=506mOhms
Click to expand...
Or you can just look up the resistance in a wire table and find that
12 AWG has 1.59 milliohms per foot. Then multiply by twice the
cord length.

As for being "dangerous and inaccurate", electronics experimenting is
frought with danger, one being 'letting the smoke out.'

As for "Inaccurate, because the resistor will warm up and the
resistance will increase", you could have the same problem if you were
measuring a resistor with a DMM, and the equipmewnt with the resistor
had been powered on before you did the measurement. The resistor
could already be hot.

--
@@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/electronics/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 article <tdvRa.71417$OZ2.13017@rwcrnsc54>, Watchit@Comcast.net
mentioned...
"Watson A.Name - 'Watt Sun'" <alondra101@hotmail.com> wrote in message
news:MPG.1980189f76ff24f6989aa3@news.inreach.net...
In article <o4eRa.65502$wk6.15887@rwcrnsc52.ops.asp.att.net>,
Watchit@Comcast.net mentioned...
"Mark D. Zacharias" <mzacharias@nospam.earthlink.net> wrote in message
news:3V9Ra.4295$Mc.389909@newsread1.prod.itd.earthlink.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/sciences/PhysSci/done/electric/resistnc/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.
Click to expand...
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'" <alondra101@hotmail.com> wrote in message
news:MPG.197ead80ac5f22d8989a95@news.inreach.net...
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.
Click to expand...
--
@@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/electronics/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:

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]
Click to expand...
It doesn't matter what your original post said, the OM found a factoid,
and wants to show it off!

-Chuck, WA3UQV
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top