I would tend to think - perhaps na=EFvely - that ESR
is based on
considering a capacitor as an ideal capacitor in series with an ideal
resistor, in which case an ordinary ohmmeter could be used to measure
THat's basically it.
ESR, provided the current it runs through the
"resistance" under test
is low enough that the cap does not charge significantly during the
measurement.
And that's the problem. The ESR will quite low (you hope a few ohms at
most).
Nost digital ohmmeters work by applying a constant current to the
device-under-test and measurign the votlage drop across it. Now, to
measure small resistnaces you use a relatively high current, and that
will charge the capacitor. Surely you've tried conencitng a capaciotor
across the probes of a digital ohmmeter and watched the display increasing
at a constant rate?
You might be able to deduce soemthign fro mthe first reading, but I think
that even over the intrgration time of the DMM chip the capacitor under
teest wil lcahrge significantly.
-tony