Discharging problem of a turn-off snubber circuit

[yykcw]

I am referring to the turn-on snubber on page 5 of the following website:
http://www.engr.usask.ca/classes/EE/443/notes/Snubber_Circuits.pdf
I want to ask when the switch start to turn off, isn't that there will be a reverse voltage induced in Ls? Then, the voltage across the switch has to be larger, does it really make sense? If the diode across the current source can share some voltage then it will be reverse bias and all current from the current source will flow to the switch which I think is not reasonable.
Moreover, this website saying the turn-on snubber is to delay the voltage, but my textbook tell me it is to delay the current! Why there will be a difference? I am really confuse right now.

 

[NascentOxygen]

Could you attach a jpeg? I can accommodate only jpegs.

 

[Baluncore]

The turn-on snubber is designed to reduce the current turn-on voltage. It is a partial circuit.
If you want to snub turn-off also you will need to use a different circuit.