I had this in my head for a while so decided to post it. You take a 1:1 transformer and hook the primary and secondary in such a way that they cancel each others field and thus reduce inductance to near 0H. Then you add a switch that removes the secondary out of the circuit. What happens is that when you open the switch the inductance will suddenly rise to w/e the inductance is of the primary. The current in the circuit will almost instantly fall to zero and "charge" back up to its ohmic limit.
So I wondered. What if we placed another inductor in the circuit. This inductor will be "charged" at start. But what happens when the switch is opened? The primary of T1 wants to shoot the current down, but the now charged inductor of T2 wants to keep the current unchanged. Do these two cancel out and give us 2 charged inductors at the cost of 1?
The online falstad simulation application shows some strange results. When both inductors are rated 1H and maximum current is 1A. The current drops to 0.5A instantaneously after the switch is opened. This is odd because the initial inductive energy = 0.5J, and the final total inductive energy is 0.25J. This means that half of the energy just disappeared.
I currently don't have a good isolation transformer to test what really happens but there's really two outcomes. Either CoE holds and an above type setup should give around 0.707A after the drop or it doesn't and the value is either higher or lower. Ofcourse a real setup has to also account for coil resistance. This resistance should preferably be much lower than the circuit's resistance. Because when you open the switch you are also increasing the resistance in the circuit.
Awesome!
Nice train of thought,
This is over my head in every way; that said, It sounds like a avenue to explore
Zane