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In "Teach Yourself Electricity and Electronics, 6th Ed." [Gibilisco] (2016) a test question reads as follows:

A DC ohmmeter will give us a correct reading for a component in isolation (not connected to anything else) if that component"
(a)  contains inductance and resistance in series.
(b)  contains capacitance and resistance in parallel.
(c)  contains pure resistance.
(d)  contains pure inductance.
(e)  Any of the above.

The correct answer is claimed to be (e).

My question pertains to (b): Assuming a large capacitance inside the component, wouldn't it take some time before the correct reading could be obtained, since it would take some time for the voltage across the capacitance to creep asymptotically towards the voltage of the internal battery of the ohmmeter (during which time the ohm-reading would be too low, because of higher-than-final current through the internal ammeter of the ohmmeter)?

EDIT: When posing the question, I was unaware of the different ways of measuring resistance, e.g. keep voltage stable and measure current OR keep current stable and measure voltage, the LATTER being the preferred method in modern multi-meters. I naïvely assumed stable voltage.

With my question I was asking for confirmation that the resistance reading can take some time to settle to the actual value. This has been confirmed by the accepted answer to be the case, regardless of the type of ohmmeter employed. (With a large internal capacitance in the measured component and an ohmmeter using a small fixed current, presumably the measurement could take a LONG time to settle to the resistance value of the resistor.)

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    \$\begingroup\$ You seem to be assuming that the capacitor starts uncharged - likely a safe assumption, since the parallel R would discharge the C after awhile. I've seen ohmmeters display negative resistance when I've been too impatient to wait for discharge. And I've seen ohmmeters fall from "infinite resistance" to the proper R value, if the parallel C was charged the other way. In any case, you keep measuring until R stops changing. If you swap ohmmeter leads for a 2nd measurement, you may have to wait awhile to settle. \$\endgroup\$ – glen_geek Jun 12 at 19:34
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    \$\begingroup\$ what exactly is your question? ... are you asking for a confirmation that the resistance reading can take some time to settle to the actual value? \$\endgroup\$ – jsotola Jun 12 at 19:46
  • \$\begingroup\$ (b) false ? Can't see how - the ohmmeter specified is a DC instrument. It either applies a DC current and measures DC voltage, or it applies a DC voltage and measures DC current. \$\endgroup\$ – glen_geek Jun 12 at 19:46
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    \$\begingroup\$ @glen_geek not a safe assumption if the device has been powered recently, or even not recently if there is no discharge path eg. switching mains power supplies usually have bleeder resistors across the filter capacitors. These often fail and then the capacitors stay charged for a long time (hours/days) after the unit is disconnected. Then you try to measure the resistance and the meter gets 300V across it! \$\endgroup\$ – Bruce Abbott Jun 12 at 21:50
  • \$\begingroup\$ @jsotola Yes, I was asking for confirmation that the resistance reading can take some time to settle to the actual value. \$\endgroup\$ – Treefrog Jun 13 at 5:06
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Assuming a large capacitance inside the component, would it not be the case, that it could take some time before the correct reading could be obtained, since it would take some time for the voltage across the capacitance to creep asymptotically towards the voltage of the internal battery of the ohmmeter (during which time the ohm-reading would be too low, because of higher than final current through the internal ammeter of the ohmmeter)?

Yes, this is correct.

The time taken would depend on the test current and the capacitor value.

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    \$\begingroup\$ Yes - Test current is most common for ohmmeters. The OP suggests an ohmmeter applying a test voltage: "asymptotically towards the voltage of the internal battery". Both approaches can work, but applying a known current, and measuring the probe's voltage is the more common method. \$\endgroup\$ – glen_geek Jun 12 at 19:41
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    \$\begingroup\$ With a constant current a capacitor will charge linearly. With a resistor in parallel it will charge as the OP described. \$\endgroup\$ – Transistor Jun 12 at 19:52

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