Door alarm

hellO!!! pls someone help me how this circuits operates?

and why it is there so much value for the resistors?

this circuits is powered by 9 volts battery..

tnx..

The resistors are high value so the idle current is very low.
When the switch is opened, Tr5 is turned ON and Tr4 is turned OFF. Tr2 and Tr1 drive the alarm B1.
At the same time C2 starts to charge via the two 3M9 resistors. After a period of time the Gate of the FET sees a voltage sufficient to turn on the FET and the alarm stops.

ahhh..okie ..tnx