Why can't I measure resistance of a resistor in a live circuit?

Thread Starter

cspwcspw

Joined Nov 8, 2016
72
The superposition principle (or my not-so-good understanding of it) suggests to me that the additional flow of current through a resistor when I measure its resistance with my DVM should not impact the circuit, nor should the circuit unduly influence my measurement.

But in practice it doesn't seem to work like that.

Why cant I just measure a resistor's resistance even if there is another current flowing through it?

Thanks
 

MrSoftware

Joined Oct 29, 2013
1,860
So there are 2 cases, measuring in-circuit with the circuit powered up, and not powered up. Assuming the circuit is not powered up; rarely is a circuit made up of only a resistor, and often there are additional paths for the current to flow around the resistor. Those paths will affect the reading. In practice, you can sometimes get away with reading a resistor in-circuit when there is no power applied if you have some knowledge of the circuit and just want a ballpark reading, such as to verify the proper value is installed, etc..

Assuming the circuit is powered up; think of how a ohm meter works. It applies a very small voltage across a resistor and reads the current flow. What happens to your ohm meter if there is already a significantly large voltage, or significantly large negative voltage across the resistor? Assume your ohm meter applies 1.5V, but the resistor already has 10V across it. Your meter becomes part of the circuit, an alternative path around the resistor for the current. In the best case you get a bad reading.
 

Thread Starter

cspwcspw

Joined Nov 8, 2016
72
Very helpful, thanks! I always imagined my DVM with infinitely high impedance even when measuring resistance, but this is clearly not so. So as a fun fact, in response to your answer I hooked up two DVM's back-to-back (with leads crossed over). It is the first time my meter ever showed me that the resistance was a negative value. Clearly the one meter overpowered the other source! o_O
 

kubeek

Joined Sep 20, 2005
5,730
Superposition will work, but an ordinary meter doesn´t expect current already flowing through the resistance. If it did, it would measure voltage on the resistor, and then voltage when the meter applies its own measuring current. Provided that the original current is not changing it could then calculate the resistance. But most of the time in a working circuit the currents are not constant, so it is not a practical way to measure resistance.
 

OBW0549

Joined Mar 2, 2015
3,503
...think of how a ohm meter works. It applies a very small voltage across a resistor and reads the current flow.
That description is true for analog VOMs such as the venerable Simpson 260.

But DVMs work the other way around when measuring resistance: they pass a known and calibrated current through the resistor and measure the resulting voltage drop.
 

BobTPH

Joined Jun 5, 2013
2,494
And the problem with measuring it in an unpowered circuit, is that the injected current does not all flow through the resistor.

In a powered circuit it is worse yet because some of the injected current affects the rest of the circuit.

Bob
 

Thread Starter

cspwcspw

Joined Nov 8, 2016
72
Plain and simple.
Can you measure your body weight if someone else has a foot on the scale at the same time?
Not convinced by this analogy. I can do differential weight measurement of my backpack even if I am standing on the scale. And I can measure my walking speed, even in a moving train or a planet whizzing through space. Resistors are linear (ok, assume they're ideal), so the fact that there is already 10ma flowing through the resistor doesn't change its resistance for the next 10ma current source, or when I attach a DVM to it. So in principle, it should still be measurable.

But what other posters have said makes sense. Adding my low-resistance DVM makes a parallel resistor network, so some circuit current gets diverted to my DVM. Or vice-versa: some DVM current gets diverted to other paths in the circuit.
 

BobTPH

Joined Jun 5, 2013
2,494
Except that that requires disconnecting it from the circuit to sneak in the Ammeter. At that point, you might just as well measure it out of circuit.

Bob
 

Analog Ground

Joined Apr 24, 2019
416
All measurements are "in circuit". Just the stray impedance are very large compared to what is being measured. A resistor sitting in air has a large parallel resistance over the surface of the part, etc.
 

dendad

Joined Feb 20, 2016
3,591
Not convinced by this analogy. I can do differential weight measurement of my backpack even if I am standing on the scale. And I can measure my walking speed, even in a moving train or a planet whizzing through space. Resistors are linear (ok, assume they're ideal), so the fact that there is already 10ma flowing through the resistor doesn't change its resistance for the next 10ma current source, or when I attach a DVM to it. So in principle, it should still be measurable.
Your explaination would only work if you know the backpack weight, train speed, ....
But if you do not have a way of measuring either of the unknows, it is truly hard to find the other.
With a resistor in a powered circuit, you can measure the voltage across it easilly, but if you cannot measure the current as well, the resistance is unknown.
It come down to the need to measure 2 properties to find the third.
A meter has a known voltage, and a measured current so the resistance is calculated, or a known current and a measured voltage, so once again, the resistance is calculated. But when your circuit is powered, the voltage and/or current are no longer a known value, so neither is the resistance.
The only way you can measure the resistance in a powered circuit is if you can measure the voltage and the current through the resistor. If that is not able to be done.......
Trun it off and use a multimeter, but keep in mind you may not know if all the meter current is flowing just through the resistor under test. It is ok for a rough measurement if you have an idea of the circuit configuration.
Otherwise, the resistor will need to be isolated from the circuit to get a real measurement.
 
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