Figure shows two bulbs $B_1$ and $B_2$ resistor $R$ and inductor $L$, when the switch $S$ is turned off

When the switch S is turned off, the current in main circuit become zero instantly, but the inductor opposes the current change. Now, the inductor, resistor and bulbs will form a closed LR circuit through which current will flow the circuit and energy will dissipate which was stored in the inductor. Hence, both B₁ and B₂ die out with some delay