This circuit demonstrates an alarm. Since on a NPN transistor the base has to be somewhat more positive by .6 volts more than the negative side to alow current to bypass the emitter. When the switches are closed the base it more negative, because the resitance at the negative side is onlt 1 k and the resitstance on the positive side it 10 k so the base it more negative because there is more resistance on the positive side. Then when the switches are open the negative side is "cut off", so there is no negative voltage going to the base, so which of course means that the base is more positive so current can pass through, lighting the LED. I have built this on a breadboard and it does the same thing. I use a 2N2222 NPN transistor and the same types of resistors.
