I read a paper and simulate its schematic recently,and my simulation is not the same as authors,so i want to ask something.I hope i can know the answer from theory,and not from the simulation now.
After charge \$C1\$ to a V,and flow through \$L1\$ to b V,b is higher than a.The SW1 close and SW2 opened.At this moment,what is the current direction? Theoretically shoule be "A" direction,because the current direction should be high voltage to low voltage.But my classmate told me the current in the inductor can't change suddenly,and the paper said the direction is "B".Why?i am confused now.
At the beginning,A point have aV,and B point have bV,b is higher than a.
Then,the capacitor energy will flow to the inductor to let the inductor's energy flow to the capacitor,so the A point voltage will become from a to -b,the B point will become from b to a.It is like there is a force from A point to push to B point to let the energy from B flow to A.Now if i add a vpulse at B point,and at this moment,the vpulse voltage is 0.that is,the B point voltage is 0 now,will one of these two force be disappeared to let A point won't become -b and B point won't become a ?