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Jul 26, 2017 at 16:01 comment added SamGibson @tigrou - As pointed out by Todor, the markings on that 22 Ω resistor indicate a 5% tolerance, not 10% as I originally wrote. Therefore, depending on your multimeter's own accuracy, your 24 Ω reading might mean that the resistor is slightly out of tolerance (22 + 5% = 23.1 Ω). However considering that your multimeter test leads will also have a resistance, the resistor may not be outside of tolerance by much. The value of that resistor is not critical, and you've found the fault in the power supply anyway, so I wouldn't replace that resistor, even if its value is slightly high.
Jul 26, 2017 at 15:52 comment added SamGibson @TodorSimeonov - Thanks. :-) I should have checked a reference instead of relying on memory :-)
Jul 26, 2017 at 15:52 history edited SamGibson CC BY-SA 3.0
Corrected interpretation of "J" tolerance letter, as kindly pointed out by Todor Simeonov
Jul 26, 2017 at 7:17 comment added Todor Simeonov J means 5% tolerance. 10% is denoted as K!
Jul 21, 2017 at 20:35 history edited SamGibson CC BY-SA 3.0
Added warning that power should be removed before performing the measurement.
Jul 21, 2017 at 19:39 history edited SamGibson CC BY-SA 3.0
Edited to remove repetition.
Jul 21, 2017 at 19:30 vote accept tigrou
Jul 21, 2017 at 19:30 comment added tigrou I measured 24 ohms (which fit in the 10% range).
Jul 21, 2017 at 19:26 history edited SamGibson CC BY-SA 3.0
Emphasised "capacitve dropper" as type of power supply.
Jul 21, 2017 at 19:20 history answered SamGibson CC BY-SA 3.0