Voltage across switch in LR circuit as switch is opened?

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nikodym
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Moved from a technical forum, so homework template missing
Handwritten problem statement with a circuit diagram: a battery, a switch, a resistor R1 and an inductor L with internal resistance RL in parallel branches, followed by parts (a) to (d) on the switch voltage.

1. I thought that when switch was closed, voltage at the right side of switch was 0 and long time after the switch is open, the voltage at the right side of the switch becomes 0.
2. After the switch is open ecause of the inductor, voltage at the right side of the switch is εL = induced emf
So voltage across the open switch = ε -εL

Is there any thing wrong with my thought1 and2?
I don't understand why voltage across the switch becomes greater if R1 is greater than RL.

Thanks for any help in advance:)
 
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  • Circuit diagram with a battery, a switch, a resistor R1 in parallel with an inductor L with internal resistance RL, and the current I labelled through the inductor branch, plus numbered problem text (a) to (d) beside it.
    20170321_152534-1.jpg
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Hello Nikodym, :welcome:

Hard to read but it looks as if the battery voltage is some kind of ##g##. If the switch is closed for a while, where does that drop to the 0 you expect in step 1 ?

What do you mean with ##\varepsilon## ? Are you aware of the relationship between current and voltage for an induction L ?