Inductors and forward EMF

[Chai]

I'm trying to understand why the D2 led will light up after the switch is opened in this circuit. I understand that the inductor will resist the decrease in current and continue to provide current for a short period but why wouldn't it just take the easy path to ground rather than passing through D2?

[BravoV]

Cause the polarity of the inductor is reversed when the switch is opened, hence D2 will be forward biased .

[Psi]

If the switch is opened it can't get to positive at all, so why would it go to negative  ?
ie, there's no completed circuit path out ground

[bson]

Current will flow top down through the inductor and when the switch is opened it will want continue doing this.  But it has nowhere to source the current it needs, so it will form a loop through D2 until its field potential (measured in V) has dropped to where it's no longer able to light up D2.  (It's less than D2's forward voltage.)  Current escaping to ground would violate KCL.  It has to to come from somewhere.

[Richard Head]

Lenz's law.

[Chai]

Current will flow top down through the inductor and when the switch is opened it will want continue doing this.  But it has nowhere to source the current it needs, so it will form a loop through D2 until its field potential (measured in V) has dropped to where it's no longer able to light up D2.  (It's less than D2's forward voltage.)  Current escaping to ground would violate KCL.  It has to to come from somewhere.

great, thanks.