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 http://en.m.wikipedia.org/wiki/Inductor.

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.

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.

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 http://en.m.wikipedia.org/wiki/Inductor.