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I need to show the resistance of multiple resistors in series and parallel together.

Is there a way to show the measured resistance similar to a voltmeter or ammeter with this program?

https://www.circuitlab.com/circuit/5kfxh4/circuits-project/

I would like to be able to show the resistance from the leads coming off from this schematic

Thank you

Does this help?

I should have mentioned it in the first post, but I would like to measure the resistance like an ohmmeter. I just want to be able to put the 'probes' or leads across the resistance and show the total resistance across the combined resistors

Do you mean you want to be able to show the measured resistance values on the schematic itself?

Otherwise, the example above does exactly what you are asking about: you place the probes, edit the probe expressions, run the simulation and then read of the resistances from the plot Y axis.

Yes I would like to show the resistance on the schematic like when you have a voltmeter or ammeter hooked up, it'll show the voltage or current respectively. I would like to be able to do the same with resistance.

Like in my example, I have 3 18k ohm resistors in parallel and 2 1k ohm resistors in series with that so the equivalent resistance would be 8k ohms. I would like to show that 8k ohm value on the schematic itself

I have updated the example above to show another couple of ways to measure and plot resistance. It doesn't show resistance in Ohms in the schematic but it does show a voltage scaled at 1V/Ohm.

:)

I will upload the entire schematic tomorrow and see if you can help me out with that. I like the most recent way, even though it's not completely what I'd like to see, it might work.

I thank you again.

Here is another way to measure resistance, based on using a current source to inject a fixed current into a resistor network. (It does not require calculated expressions.)

https://www.circuitlab.com/circuit/2vd5we/measuring-resistance-using-a-current-source/

From Ohm's law, V = I x R, and if I = 1, then V = R. Hence the voltage across the input terminals is equivalent to the resistance.

@em3110,

Um. It does require calculated expressions.

You're just making them simple by using a 1A source and scaling Volts into Ohms at 1 Ohm/Volt in your head.

The technique you have described is in fact exactly equivalent to the one in the earlier example.

It looks simpler because you are not trying to represent the results in Ohms on a plot.

:)

p.s. You've probably already seen this lively little thread:

https://www.circuitlab.com/forums/support/topic/9k94y382/wheres-the-ohmmeter/

@signality,

The method shown above, although superficially similar to yours, differs significantly.

Your method, as commented elsewhere, allows resistance to be measured in a live circuit, thus opening up possibilities for advanced analysis with variation of circuit parameters. It does however require creation of named nodes, and use of calculated expressions with voltage sources.

The method presented above operates more like a traditional resistance meter, where the resistor network is disconnected prior to measurement. Potential applications for this method include checking solutions of circuit analysis problems, Thevenin / Norton circuit analysis, etc. - especially by students.

One method is not better than the other. They just have different areas of application.

The following circuit illustrates the point by creating a working ohmmeter to solve mebrunner24's originally posed circuit problem.

https://www.circuitlab.com/circuit/25vwe2/ohmmeter/

(I can't claim credit for the idea. The method is based on the 4-terminal Ohmmeter component, as used in Proteus VSM.)

I take your point.

I did point out (somewhere in one of my posts) that you can use current sources instead of voltage sources but yours is a good example of a way to make it very simple and straightforward for a novice or student user.

Just for amusement, I tried to do the same thing using the "traditional" 2 terminal Ohmmeter circuit as used in basic analogue meters:

Trouble is - as at 121102 - it won't simulate in CL.

Sorry, I'm being dim ... there should be a series resistance in the above.

I'll post a corrected version later but have left the circuit as shown because it shows an error which has nothing directly to do with my circuit error and I've posted a bug report about.