So if I used a 18,000 uf capacitor such as this one https://www.mouser.ca/ProductDetail...Z1n0r9vR22YqjN7vGN3ZSV9u2P47iI0pNPC1xmjIBdg==

Using the formula I found:

Vripple = I / ffc

Vripple = 50 / (120 x 0.018)

I get 23v. Probably good enough for my application. But I cant figure out how to calculate the ripple current in the capacitor. One source seemed to indicate you could use roughly half the load current. So if load current is 50amps what would the ripple current be? That particular cap has a ripple current rating of 27.8A. If I put 2 of them in parallel does that double my ripple current? Is it possible for the ripple current be higher then what the load is?