I am trying to calculate the equivivalent capacitance to ground of a 3 phase LCL filter with capacitors to ground. Inductors removed from circuit as I assume they will be ignored by the meter...
Ignoring C3, it`d be simply 2 series networks in paralell with C4. Specifically:
C4 + 1/(1/C1 + 1/C5) + 1/(1/C2 + 1/C6)
This almost fits with measurements of actual circuit (3% less), small enough that could be explained by the capacitors +-10% tolerance. However what effect does C3 have in such a calculation?
Update Inductors are back in the schematic from comments suggestions, at the cost of a much uglier schematic. I think this is not necessary as the probe should only send a very low frequency current and then measure the change in voltage. And as frequency approaches zero, reactance of inductors also approaches zero, while reactance of capacitors approaches infinity. In effect making inductor values negligible. Therefore I think this simplified circuit is ok to use: