The switch is closed for a long time and then opned at time `t=0` . Find the initial voltage across `L` after `t=o` , which end is at higher potential `a` or `b` ?

Explanation

`i_(0)=(10)/(2)=5A`
Current at any time
`i=i_(0)e^(-t//tau),i_(0)=5A`
`tau=(4)/((2+3))=0.8`
`i=5e^(-t//0.8)`
`(di)/(dt)=5e^(-t//0.8)(-(1)/(0.8))=(25)/(4)e^(-1.25t)`
`V_(L)=L(di)/(dt)=25e^(-1.25t)`
`t=0` , `V_(L)=25V`
Current is dcreasing `V_(b)gtV_(a)`