The circuit shown in the figure consists of a battery of emf E, resistances R1 and R2, inductance L and a two way switch ABC. First A is connected to B for a long time and then A is connected to C. The total heat produced in R2 is

When A is connected to B, inductor gains energy. After a long time,it starts acting like a short circuit, and the energy stored in the inductor becomes 12Li2= 12L(ER1)2. Thus after connecting A to C, the energy loss in the resistor  will be 12LI02= 12LER12