In the network shown, initially the switch S is open. Then just after the switch S is closed, brightness of the bulb

 

Before S is closed, magnetic flux linked with the inductor = Li

After S  is closed, ${\mathrm{L}}_{\mathrm{e}}=\frac{\mathrm{L} \mathrm{x} \mathrm{L}}{\mathrm{L}+\mathrm{L}}=\frac{\mathrm{L}}{2}$

So magnetic flux =  $\frac{\mathrm{L}}{2}\mathrm{i}\text{'}$

By Lenz's law, $\frac{\mathrm{L}}{2}\mathrm{i}\text{'}=\mathrm{Li}\Rightarrow \mathrm{i}\text{'}=2\mathrm{i}$

Since just after S is closed the current is doubled, so brightness increases