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

Explanation

Initially, current will be, `i_(0)=(12)/(1)=12A`
So, the above circuit would be now as

Current at any time is given by `i=i_(0)e^(-t//tau_(L))` and `i_(0)=12A`
`because" "tau_(L)=L//R=(2)/(1+3)=(2)/(4)=0.5`

Current, `i=12e^(-t//0.5)`
`rArr" "(di)/(dt)=12e^(-t//0.5)(-(1)/(0.5))rArr" "|(di)/(dt)|=24e^(-2t)`
So, `V_(L)=L(di)/(dt)=48e^(-2t)`
At `t=0,V_(L)=48V`
i.e., current is decreasing,
`rArr`
`rArr" "V_(Q)gtV_(P)`