# Release of energy stored in Inductor and Capacitor

I havent had much experience working with inductors and i am fairly new to them. The question is how is the energy released from an inductor.

Now if we had a capacitor circuit:

Assume switch to be always closed. Here if the source was to supply current to the resistor, now initially capacitor charges, and till then it allows the current to flow through, but as it is fully charged , it does not let any more current to flow, as this point the top plate of capacitor is +ve , and the bottom is -ve. Now if we were to remove the 15V source, and replace it by a short(thus having only capacitor and resistor in the circuit), then current will flow anti-clock wise right?

Now if were to replace the capacitor with an inductor, what would that direction of current be after current reaches its maximum value, and the magnetic energy is maximum and source is replaced by a wire(to short the 2 terminals)?

• Dude, the top plate will be +ve not -ve and the current will flow anticlockwise when the supply is replaced with a short. Please address the problems with your question because it makes this question unanswerable with your incorrect assumptions. Also, in your final paragraph, you say "switch replaced by a close circuit" - I think you mean the "supply replaced by..." Sep 7, 2013 at 20:23
• Thanks, i just realized my mistake, and have made the changes, at the end, i had by mistake replaced source by switch, but now its fixed. Sep 7, 2013 at 21:00