Electronic – How does a NPN 2N2222 transistor work

1)

In the first configuration (the collector is not connected to the power supply) your NPN essentially behaves just like a regular PN diode, and the circuit is equivalent to this one:

^{simulate this circuit – Schematic created using CircuitLab}

The reason for this behavior is that NPN transistor may be seen as two PN diodes connected back-to-back:

Leaving the Collector open and connecting the Base to the highest potential leaves you with simple forward-biased PN diode, having forward voltage drop of \$\sim 0.72V\$.

BJTs are not employed in this configuration because it is cleaner and cheaper to use a proper PN diode in such cases.

BTW, the usual way to use NPN as a switch would be to put the load (in this case the LED) between the power supply and the Collector:

^{simulate this circuit}

In the above schematic you can also see one of the approaches for utilizing the same power supply for both the Base and the Collector driving (which answers your second question). The values of the resistors should be chosen based on the parameters of the LED and the transistor. The voltage divider formed by R1 and R2 is constantly consuming power (even when the switch is open) - this is the main disadvantage of this simple scheme.

The purpose of R_B is to pull Base's voltage to ground when the switch is open - usually you don't want to leave the Base floating. The value of this resistor may be taken very big, such that its presence does not affect the voltage divider.

NOTE: as mentioned by Jippie, R2 may be completely omitted from the circuit. In this case the voltage divider will be formed by R1 and R_B when the switch is closed. When the switch is open, the absence of R2 will prevent from current to flow, therefore the OFF power consumption will be reduced. There are cases when you do need R2 though: its presence reduces the maximal voltage on the switch, and this may be desired in some cases (depends on the switch you're using).

Hope this helps.

You need to connect 0V of your RaPi to the emitter connection on your power supply because you cannot drive dc current into the base without a return path for that current.

But, before you do that make sure that the 31 V dc supply can connect to 0V on the RaPi safely i.e. is the 31 V supply isolated from ground?

It may be OK if it isn't but tread carefully because I don't want you to destroy your RaPi.