How to find current through inductor when connected in series with an ideal diode?
Considering L= 3H.
When switch is closed the inductor is charging with 1A/s rate (V/L = 3V/3H = 1A/s). So at 1sec the inductor is charged with 1A.
When switch is released all inductor current (1A) starts to flow to R what couses 1V drop on it. Thats exact voltage what your diode is opening so at this time the diode current rise from zero. This voltage will be presented all rest of time so the R will take 1A from inductor current.
The inductor current rate is 2V/3H = 0.67A/s now so after 2sec it will be incresed by 1.34A.
Adding together with current at t=1s the inductor current at 2s after releasing switch is 2.34A.
The diode current is 1A less due to constant R current, so 1.34A.