Drawing the schematic that should have been included in your question, to see if I understand what you're proposing - on the left we have ... Maine’s voltage into the DC using a full bridge rectifier which will then be fed into the transformer. If this is not the schematic you intended, please correct me.
simulate this circuit – Schematic created using CircuitLab
As the input to the transformer has a finite average DC component, the primary current will continue to rise, and will saturate any finite core in a short time. Once saturated, the core flux does not change significantly, so cannot generate any secondary voltage. The current will continue to rise until something stops it, ideally a breaker, if not then smoke.
On the right is the minimum modification needed to make the circuit work continuously. The switch will be driven on and off by some suitable circuit, with on time limited to avoid saturation. When the switch opens, the tendency of the transformer's inductance to make the current to 'want' to keep flowing means it will generate a large negative voltage across the switch. This allows the flux in the core to drop below saturation, so it's able to change and work like a transformer again. Another way to look at it is this negative voltage transient, averaged with the rectified DC input, gives you average zero DC on the input to the transformer's inductance, to keep the current finite.
In old car ignition systems, this was a mechanical switch. These days, you would use a FET or IGBT, and some additional components to prevent the 'large negative voltage' from frying it.
Using 'chopped DC' like this is using the transformer in 'flyback' mode. This needs a low permeability core for best efficiency, not the gapless iron core used in a conventional 'forward' transformer for which you have equations.
Before maths, we need to get a usable configuration.
