capacitance to start with, and the ratio dv/dt to complete the product that ought to give you the

momentary charge current to the cap. But how ... ? Consulting textbooks and tutorials, you are

bound to find a phrase like: 'presents a near-short circuit', to describe the cap's behavior after

switch-on. Few if any specify the 'near'-part, however. The only thing to do then, is go look for an

R (i.e. in the cap itself, not the obvious resistances in source, wiring, etc.). A glance at one of the

many different equivalent circuit diagrams for an e-cap, reveals its series resistance ESR, which

indeed turns out to be less than one Ohm usually. And by applying this man's familiar law, we can

finally make more than an educated guess as to the value of the charge current surge. This is

because - while dv in the original formula is not easy to determine - V in Ohm's Law simply is the

full applied voltage, as there has not yet been generated a counter e.m.f. (where the 'm' stands for

'motive', not 'magnetic') directly after switch-on. So the inrush current surge equals the voltage

applied to the e-cap, divided by its ESR, or does it ? ....

I invite and await your comments and criticism as to method, reasoning and outcome of the above.

(Please try to apply as little math as possible - thank you).