I'm working my way through "Make: Electronics" by Charles Platt and came across the following schematic:
As someone who just recently got started with electronics, what I can't wrap my head around is how R1 relay works after Q1's voltage drop. I have built the same circuit only with different power source and relay, both 5V instead of 12V as suggested by the book and what seems to be happening is that the emitter voltage is bellow the relay trigger voltage (12V in the schematics and 5V in my case since I'm using 5V relay and source). From what I have been reading around the internet and communities like this, the emitter voltage in this configuration is supposed to be limited to the base voltage minus ~0.4V or so. Due to the 10K resistor, once the transistor is triggered, the base voltage will be under 12V, thus the emitter voltage will also be under 12V and the relay will not close. That's basically how my thoughts are at the moment and I need help spotting my mistakes.
Edit: I believe this question is being misunderstood. I'm not asking how a self-locking relay works. I'm asking how a 12V relay is being triggered when the voltage on its coil (from Q1 emitter) is bellow 12V (I'm checking it with a multimeter and theres a voltage drop across the transistor).