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: enter image description here

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)?

  • \$\begingroup\$ 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..." \$\endgroup\$
    – Andy aka
    Commented Sep 7, 2013 at 20:23
  • \$\begingroup\$ 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. \$\endgroup\$
    – Sherby
    Commented Sep 7, 2013 at 21:00

2 Answers 2


Inductors store energy in the magnetic field generated when current passes through them. When the supply is removed, the collapsing magnetic field induces a current flow in the same direction that it was traveling when it generated the magnetic field in the first place. This is why it is used as one of the storage devices in switching power supplies; the capacitor maintains the same voltage, and the inductor maintains the same current.


simulate this circuit – Schematic created using CircuitLab

(But don't try to actually build this circuit.)

  • \$\begingroup\$ I think your capacitor should be across the resistor Ignacio \$\endgroup\$
    – Andy aka
    Commented Sep 7, 2013 at 23:33
  • \$\begingroup\$ @Andyaka: So it should. Good thing I told people to not build it. \$\endgroup\$ Commented Sep 7, 2013 at 23:49
  • \$\begingroup\$ That's a useless and incorrect circuit. 1) You have a zener directly across a voltage source; what is the expected outcome ? What is the opamp supposed to do ? Even if the circuit oscillated , it's not a switching power supply; at minimum you need a diode between the junction of the FET & inductor and ground. \$\endgroup\$
    – jp314
    Commented Sep 4, 2021 at 18:47

as the circuit u have shown is dc circuit,now there are three cases: case1: initially when the dc voltage is building (yet not reached 15v) that time inductor poses opposition to the change in voltage & hence current will be in direction opposite to the direction of current of battery i.e in this case it will be anticlockwise. case2: once voltage is built to 15v then dc voltage will continue to remain 15v, so as there is no change in the current hence here inductor will behave as a short. so it's as if it's replaced by a wire. case3: now suppose u release the switch or may be you are slowly decreasing the battery voltage to zero volt then,the inductor will come into action opposing the decreasing battery voltage and hence the current direction will be clockwise as inductor tries to maintain the battery current. know that this all happens very faster within fraction of time....


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