Re: Simple measurement question

[SCADA]

<opm...@yahoo.com> wrote in message
news:1111439198....@f14g2000cwb.googlegroups.com...
> Hello,
>
> I have a simple ckt composed of a resistor in series with a DC voltage
> source. If I put an ohmmeter across the resistor terminal with the
> voltage source off, then I get the resistor's resistance. However, I'm
> curious how the source affects the measurment when it is turned on.
>
> I did an initial Thevenin model and found that the resistor will be
> parallel to a short, so that means that meter would be measuring 0
> ohms. However, when I tried it in the lab, a 1 VDC in series with a 500
> ohm resistor is reading a 2 Megaohm reading when I put the ohmmeter
> probe on the resistor.
>
> Any model or explanation? Thanks!
>

You don't measure resistance with power to the circuit. It will give a false
reading and can even damage the meter. Resistance can be found by ohms law.
R = E/I, Resistance = Voltage (across resistor) / current through resistor.

[opm...@yahoo.com]

[Robert Baer]

opm...@yahoo.com wrote:

Also, your thevnin model is inaccurate; all signal sources, including
bateries and power supplies, have an internal resistance.
Furthermore, you should say "parallel" and not "series".
Putting an ohmmeter across a battery is a good way to damage or burn
it out.

[bigsc...@gmail.com]

That setup will indeed give you false readings. But if you are actually
doing that, think of the test resistor (load) as a pipe. If you turn on
the power supply, currenty is already going through the load, so no
more current will be able to go through it. That means the test current
from meter will be very small or close to zero, which means that the
volt-meter in the ohmmeter will see a very large resistance due to the
small test current. This should explain the high resistance.

However, try turning off the power supply while the probe is connected
to the load. Then turn on the power supply. Your reading should still
be close to the actual resistance of the load. This is because the test
current is already "being pushed" through the load and ohmmeter's
volt-meter will be reading this value of the test current. The source's
current will not be able to push more current into the load.

Give it a try.

[Robert Baer]

bigsc...@gmail.com wrote:

Methinks you know little of electronics and have even less practical
experience.