Hello Kai,

here i would like to recall the circuit,


I read the description of the above inrush current limitting circuit given by you but couldnt understand fully since iam not clear on how you managed to calculate the following:

Kai said,
When the lamp is switched-on, current through it is limited to about 0.2A and up to about 2.0W will drop across the transistor, but only for a very short time.
If the current through 3R0 resistor reachs about I = 0.6V / 3Ohm = 0.2A

from reading the above note, i tried to calculate how you have said that 2W would be dissipated in the transistor (BD135) and, the 3R resistor would have drop of .6V. BUT couldnt get clear with that. also with "what would be the voltage drop in the bulb in that circuit" since i think it would not drop 12V fully on it.
So, it would be helpful for me if you could provide me with few details on those issues.

Thanking you,
Suresh.