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So let's say I have a load with a resistance of 500 ohms and two ideal ohmmeters connected in parallel. And the question is, what will be the readings on those ohmmeters? This really confuses me, when one ohmmeter is measuring, then the load and the other ohmmeter have a combined resistance of Infinity? The notebook says the measured value should be 1 kiloohm but I really do not know, how they got this number. enter image description here

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    \$\begingroup\$ That would depend on how the ohmmeters perform the measurement. Is it somehow defined by the context, i.e. should we assume how your ohmmeters work or something? \$\endgroup\$
    – Justme
    Aug 28, 2023 at 19:51
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    \$\begingroup\$ What is an ideal ohmmeter? \$\endgroup\$
    – fraxinus
    Aug 28, 2023 at 20:10
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    \$\begingroup\$ Ohmmeters have a polarity (otherwise, they couldn't measure diodes). So if you connect the two ohmmeters with opposed polarity, you likely read near zero. Things could get weird if those ohmmeters are auto-ranging if you actually try connecting two together. \$\endgroup\$
    – glen_geek
    Aug 28, 2023 at 20:21
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    \$\begingroup\$ @LukaM As already pointed out, everything turns on the definition of ideal ohmmeter. But if you assume that they are identical and with perfect precision inject the exact same current and current direction (lead polarity is the same in both as shown) then the sum of the two currents would be driven through the resistor, doubling the voltage difference seen by each ohmmeter. Since each ohmmeter sees twice the expected voltage, they should report twice the actual resistance. If one of them were hooked up anti-parallel to the other (possible), then no voltage difference. Then zero Ohms results. \$\endgroup\$ Aug 29, 2023 at 1:08
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    \$\begingroup\$ Whether or not these ohmmeters are ideal or not, or what ideal even means, is irrelevant in contrast to what some people claim. The important bit is that both ohmmeters function identically. The latter will lead you to the correct answer. \$\endgroup\$
    – tobalt
    Aug 30, 2023 at 6:18

7 Answers 7

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Two simple milliohmmeters connected in parallel, measuring a 0.1Ω resistor:

enter image description here

Note how the reading is 2x the resistance measured. Those are 4-wire ohmmeters, but 2-wire ohmmeters would do the same thing.

They don't match to full precision because the sense current is not calibrated - instead, the sense amplifier gain is calibrated. So the same voltage that they both measure across the resistor is interpreted in two different ways. That's due to rather minimalistic design of those devices.

Hint to myself: v2 should use a calibrated current so that this demo will look perfect.

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    \$\begingroup\$ Beautiful experimental demonstration! \$\endgroup\$
    – Jon Custer
    Aug 29, 2023 at 15:35
  • \$\begingroup\$ You don't need identical currents, you can work out the real resistance from \$R = \frac{R_1 R_2}{R_1+R_2}\$ as I show below, so your measurements give R = 0.1002 \$\endgroup\$
    – Tesla23
    Aug 30, 2023 at 23:23
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If the two ohmmeters are the "same" model (current source), then they should measure each the "half" of the resistor which should be ... 1 kOhm (double of the 500 Ohm).

Injection of "current" into the resistor.

See this experiment: (for simplicity, I took 1 kOhm)

enter image description here

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    \$\begingroup\$ if they're "ideal" though, their current is probably infinitely small, so it might get weird \$\endgroup\$
    – BeB00
    Aug 28, 2023 at 19:48
  • \$\begingroup\$ @Antonio51 I stole the words in your answer, but corrected it. I hope you don't mind. \$\endgroup\$ Aug 28, 2023 at 20:45
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    \$\begingroup\$ You're halving the resistor "width-wise", which doubles the resistance, rather than halving it. \$\endgroup\$ Aug 29, 2023 at 12:22
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    \$\begingroup\$ @daviegravee No. Generally, nowadays ohmmeters are "current sources" ... and one measures the voltage (until a certain voltage, then it "saturates" or "overload"). Voltages read are proportional to the value of resistors. \$\endgroup\$
    – Antonio51
    Aug 29, 2023 at 14:58
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    \$\begingroup\$ I'm going with "double" too. That's what your figure shows. Both meters will read twice the voltage, but each is only aware of half of the current through the resistor. \$\endgroup\$ Aug 30, 2023 at 9:56
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Consider this circuit that shows each ohmmeter as a constant current source coupled to voltage measurement:

schematic

simulate this circuit – Schematic created using CircuitLab

It should be clear that the total current through Rx will be 2 mA, meaning that each voltmeter will measure 1 volt. Because each ohmmeter doesn't know about the other, it will compute resistance based on 1 mA through the resistor:

$$ R_{measured} = \frac{1 V}{1 mA} = 1 kΩ $$

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  • \$\begingroup\$ Good thinking... Only to note that the voltmeters should be connected in parallel to the current sources. In this way, all five devices in your schematic are connected in parallel. \$\endgroup\$ Aug 29, 2023 at 14:49
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Philosophy of ohmmeters

By their very nature, ohmmeters are devices that convert, according to Ohm's law, resistance to current (I = V/R) or voltage (V = I.R). So in these devices, resistance is an input quantity, and voltage or current are parameters. For this purpose, they need a power supply and the corresponding "meter"; so ohmmeters are sources with built-in meters. It follows that, unlike voltmeters and ammeters, which are passive (not producing energy) devices, ohmmeters are active devices that inject energy into the measuring resistance.

V-to-I ohmmeter

A V-to-I ohmmeter is a resistance-to-current converter where Iout = Vref/Rx.

schematic

simulate this circuit – Schematic created using CircuitLab

This historic first ohmmeter is implemented as a voltage source with a built-in ammeter. Since the input quantity Rx is in the denominator, at Rx = 0 the current becomes infinitely high. That is why in the CircuitLab simulation below, I have set a small internal resistance to the ammeter.

schematic

simulate this circuit

As you can see, the graph is non-linear; that is why the scale divisions of these ohmmeters are non-linear.

STEP 1

I-to-V ohmmeter

An I-to-V ohmmeter is a resistance-to-voltage converter where Vout = Iref.Rx.

schematic

simulate this circuit

It is implemented as a current source with a built-in voltmeter. At Rx = infinity (open circuit), the current becomes infinitely high. That is why in the CircuitLab simulation below, I have set some very high internal resistance to the voltmeter.

schematic

simulate this circuit

As you can see, the graph is linear. This is why modern ohmmeters are implemented this way.

STEP 2

Two ohmmeters in parallel

As we saw above, ohmmeters are actually current or voltage sources. So when an additional ohmmeter is connected in parallel to the measured resistance, it changes the current through it and the voltage across it, giving a wrong idea of its resistance value. Let's consider the possible configurations.

V-to-I ohmmeters

In the same direction: Rx = Vref/Iout = 10/0.05 = 200 Ω.

schematic

simulate this circuit

As you can see, adding a second ohmmeter (voltage source) assists the first ohmmeter by giving half the current (this was a bit hard for me to imagine, something my intuition was against). As though, the resistance Rx is virtually increased.

In the opposite direction: Rx = Vref/Iout = 10/1000 = 0.01 Ω.

schematic

simulate this circuit

This is where it gets even more interesting because adding a second ohmmeter with the same but opposite voltage creates a virtual ground at the midpoint between the two ammeters (Rx is virtually short connected). As a result, the current significantly increases.

I-to-V ohmmeters

In the same direction: Rx = Vout/Iref = 20/0.1 = 200 Ω.

schematic

simulate this circuit

Adding a second ohmmeter (current source) doubles the current through and the voltage across resistor Rx. This creates the illusion of twice the resistance.

In the opposite direction: Rx = Vref/Iout = 0/0.1 = 0 Ω.

schematic

simulate this circuit

And finally, adding a second ohmmeter with the same but opposite current creates again a virtual ground at the top midpoint between the two current sources. This creates an illusion of zero resistance (Rx is virtually short connected).

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    \$\begingroup\$ Yes, this is the good one. Covering all the cases (minus the remote possibility that the ohmmeter is really an automated bridge... 😉) \$\endgroup\$
    – Rmano
    Aug 30, 2023 at 9:11
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If the two ohmmeters are the "same" model, then they should measure each the "double" of the resistor which should be ... 1 kOhm (double of the 500 Ohm).

The current in the resistor is double, so the voltage across it double. Each meter only knows about its own current, yet sees twice the voltage, so it assumes it's a 1 kOhm resistor.

See this experiment:

enter image description here

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  • \$\begingroup\$ Rx was said to be 500 ohms. \$\endgroup\$
    – jpa
    Aug 29, 2023 at 7:13
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If this is a theoretical question, rather than practical, the answer should explain what "ideal" is assumed to mean, because there is no such thing as an "ideal" ohmeter. This is as important as the calculations that ultimately would follow.

Unlike a voltage source or current source, for which "ideal" does have an generally accepted meaning, for ohmeters there is no definition of "ideal". Here are some examples of ohmeter designs which can be considered "ideal", but which would yield vastly different final answers to the theoretical question:

Meters that inject different DC currents into the UUT

Their readings would be a R * ((I1 + I2) / I1) and R * ((I1 + I2) / I2), respectively, ie, the readings would show the same ratio as the ratio of their injected currents. @Kuba's answer shows this in practice with two meters of (slightly) different calibrations.

"Ideal" could easily mean two meters of different designs, therefore different injected currents. "Ideal" normally means that parasitic characteristics, or deviations from nominals specs can be ignored (eg, mis-calibration, current leaks, etc), and not that their nominal specifications being the same.

Meters that inject a current pulse rather than DC

If the meters read the resistance by sending a short pulse of current, and sampling the voltage drop, and they send their pulses at different times, both meters would show the "correct" reading, ie the actual resistance.

This kind of meter might be found in automated test equipment, where some "resistor" needs to be measured by more than one piece of monitoring equipment at the same time. The "resistor" might be a length of subway rail, transformer, transmission line, etc, and it is measured in-circuit.

This kind of meter can measure impulse response, and therefore measure several characteristics of the sample at the same time (resistance, capacitance, inductance).

This would allow the resistance of a sample of material to be measured by "any" number of meters at the "same" time, for reasonable values of "any" and "same".

Smart ohmeters

If the meter samples the voltage drop before and after applying a current, it can measure live, in circuit resistors by calculating the effect of its injected current. The measurement would be (V2 - V1) / I where V1 and V2 are the voltages before and after activating the current source, and I is the injected current.

In this case, the reading can be expected to be unaffected by other current sources.

Personally I am somewhat surprised to see that even 150 years of practice, engineers haven't adopted such a "smart" meter as the "basic" ohmeter. This meter would be much more capable and useful than the "classic" ohmeter.

Conclusion

The great answers above explain how to perform the numerical calculation, but a complete answer must contain the assumptions made.

I suspect most answers will assume that "ideal" meter means one that is a unit current source in parallel with a infinite-resistance voltage meter.

Some answers I've read assume the "meter" is a benchtop multimeter with typical specifications (micro/miliamp amp currents), and further assume both meters are identical (a theoretical impossibility).

If the question is meant to test your mastery of current and voltage, then you ought to demonstrate it. I can't think of a better way than this answer here. In fact, if I had to write an exam answer to this question, I would leave out the numerical calculation altogether.

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If the meters simply inject a DC current and measure the voltage across the resistance, then, assuming that the positive terminals of the two meters are connected together (so that neither meter sees a negative or zero voltage):

if the readings are \$R_1\$ and \$R_2\$, then the actual resistance is

\$R = \frac{R_1 R_2}{R_1+R_2}\$

independent of the actual currents used by the meters.

neat (but useless!)

I derived this with a few lines of algebra, but there is a very simple physical justification: if you connect \$R_1\$ across meter 1, and \$R_2\$ across meter 2, then you get the same readings, and we know these resistors have the same voltages across them, so you can connect them in parallel without changing the readings, so \$R = \frac{R_1 R_2}{R_1+R_2}\$

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