I was thinking about the discharging mechanism of an inductor and I have a question about it. It might sound stupid to the physics experts here, but I can't help thinking about it. Before I ask my question, I want to mention a few points I already know about an inductor.. I know an inductor resists sudden changes in current through it. I am comfortable with the differential equations and their solutions for RL circuits with different types of excitation. So, I have enough understanding of the inductor as far as my EE domain is concerned. When an inductor is charging, the energy is stored in the magnetic field and that stored energy is used to drive the current while discharging of the inductor. I understand this approach. Now my question... Consider the following circuit. The switch was closed for a long time and the inductor was carrying a current of 1A. Now, when the switch is opened at t=0, current in the inductor is maintained momentarily at 1A and voltage across the 1 ohm resistor rises to 1V. So, at t=0+ (just after t=0), inductor current is 1A and induced emf=1V. But if the current is still 1A just after the switch is closed, how is emf induced in the inductor? There has to be some change in current for emf induction. My question is, how is this voltage of 1V developed across the 1 ohm resistor just after the switch is closed? If the resistor were 10 ohm, the induced emf at t=0 would be 10V. How does the inductor "know" that resistance value? I understand the behaviour mathematically, but I don't understand what maintains the current to 1A "just after" the switch is opened. I want to know the what happens at the electronic level just after opening the switch. I am having trouble understanding "what causes what" at t=0. Thanks in advance.