Sure thing, but first let’s nail the basics: the relay’s speed is limited by the inductance of the contacts and the driver’s slew rate. The energy to be delivered is \(E = \frac{1}{2}C V^2\) for a capacitor bank, so to charge a 400 V battery in two seconds you need a current of about \(I = \frac{E}{Vt}\). That means the relay must switch on with microsecond precision, and the driver must handle peak currents in the kilampere range. If we use a solid‑state relay with a fast MOSFET array, we can get to sub‑10 µs turn‑on, but you’ll need robust snubbers and a low‑inductance trace layout to keep the voltage spikes down. Also, safety first: you can’t have a relay that’s too fast without proper isolation and protection—otherwise you’ll fry the grid or worse. Let me know what part of the math you want to crunch, and we’ll draft a schematic together.