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I want to know what happens with a lightbulb immediately after a switch is turned off. We're talking about a circuit with a battery (AC), an inductor, a lightbulb and a switch. I need to know it in case the inductor and lightbulb are in serie and in parallel.

I know you can see the lightbulb as a resistor but i'm kinda lost on how to calculate this. Help would be much appreciated!

I also need to draw the current in function of the time. Will this be a wave?

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In case of parallel connection, the magnetic energy stored in inductor would be spent in the bulb after the switch is turned off. But in case of series connection when switch is turned off the circuit would be left open and the magnetic energy would not be dissipated in the resistor as no current will flow.

The current would be exponentially decreasing in case of parallel as that of discharging of capacitor and in series there would be no current. The calculation can be done just like they are done for discharging of a capacitor with "r" resistance.

Addendum(In response to comment) : In short the current in parralel connection case would behave like a R-L series circuit and the current would decay as $i(t) = I*e^{-Rt/L}$. In case os series connection no current is flowing so you do not have to worry with the equations. To derive the formula yourself you can refer to Circuit theory in Inductor.

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  • $\begingroup$ I tried the calculation but i'm not really sure how to start. I'm sorry but i'm really not that good at this... $\endgroup$
    – user2162627
    Commented Jan 12, 2014 at 13:30
  • $\begingroup$ So oscillations have nothing to do with this question? The question belongs to a chapter in my book about electrmagnetism so i don't really see how it fits in in this chapter as we've seen the discharging of a capacitor in a previous chapter. In this chapter we've seen something about elektromagnet oscillantions when there's a resistor in the circuit so i was wondering if it had something to do with that. $\endgroup$
    – user2162627
    Commented Jan 12, 2014 at 14:06
  • $\begingroup$ The question also asks to prove with formula's from elektromagnetism so i don't know how to incorporate that when i'm looking at it like a discharging capacitor. $\endgroup$
    – user2162627
    Commented Jan 12, 2014 at 14:08
  • $\begingroup$ Well you have turned the switch OFF so I think the battery does not matter anymore, sometimes questions of previous chapters are asked in current chapters, I had questions and derivations of discharging of capacitor in the chapter of EM waves, so I would not worry about the chapter. $\endgroup$
    – Rijul Gupta
    Commented Jan 12, 2014 at 14:27
  • $\begingroup$ Ok, thank you so much! You really helped me understand. $\endgroup$
    – user2162627
    Commented Jan 12, 2014 at 14:34

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