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### Network # Simple measurement question

O

#### [email protected]

Jan 1, 1970
0
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!

S

Jan 1, 1970
0
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.

R

#### Robert Baer

Jan 1, 1970
0
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!
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.

S

#### Sir Spamalot

Jan 1, 1970
0
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 have made a small error in your analysis by considering the DC
voltage source to be a short when off. It would only be a short when
doing an AC analysis. For DC, the DC source (i.e., battery) "appears"
as a large capacitor in series with a small source resistance., which
is, to DC, an open circuit. You can verify this by adding a switch in
series and using that to "un-power" the circuit to measure the R with
your meter. R is R.

From a practical point of view, trying to measure a live circuit with
an ohm meter will more than likely fry the meter, as many others have
already pointed out. The measuring system of a meter contains it's
own source, which is not designed to source or sink current from an
external source.

Hope this helps.

B

#### [email protected]

Jan 1, 1970
0
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.

R

#### Robert Baer

Jan 1, 1970
0
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.
Methinks you know little of electronics and have even less practical
experience.

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