Initally switch `S_(1)` is closed, after a long time, `S_(1)` is opened and `S_(2)` is closed. Find the heat produced in coil after long time.

Explanation

When `S_(1)` is closed, in steady-state current in the coil `i_(0)=E//R` .
The energy stored in inductor `=(1)/(2)Li_(0)^(2)` .
After closing off `S_(2)` , after long time, heat generated in the coil i.e. in `R`
`=((R)/(R+R_(0)))((1)/(2)Li_(0)^(2))`
`=((R)/(R+R_(0)))((1)/(2)L(E^(2))/(R^(2)))=(LE^(2))/(2R(R+R_(0))`
OR After closing of `S_(2)` , current in the coil i.e. in `R`
`i=i_(0)e^-((R+R_(0))/(L))t`
Heat produced in coil i.e. in `R`
`H=int_(0)^(oo)i^(2)Rdt=i_(0)^(2)int_(0)^(00)e^(2(R+R_(0)t)/(L)dt`
`=(E^(2))/(R^(2))R(|e^(-(2(R+R_(0)))/(L)t)|_(0)^(oo))/((2R+R_(0))/(L))`
`=-(LE^(2))/(2R(R+R_(0)))|e^(-e^(oo))-e^(0)=(1)/(e^(oo))-1=(1)/(oo-1)=0-1|`
`=(LE)^(2)/(2R(R+R_(0)))`