In the following circuit, the LED lights up:
I was afraid I'd misuse the transistor symbols, so I just drew a box to make sure I communicated the right thing. This is the transistor, G/D/S means Gate/Drain/Source.
My understanding is that current does not flow if there's no circuit. From this I deduce that the line going from the positive terminal of the voltage source into the Gate pin has to at some point make its way back to ground. Since the only pin coming out of the transistor connecting to ground is the Source, I deduce that the gate is internally connected to the source pin.
But from that, I would think that if I severed the connection to Drain like this, current would still flow - but my LED switches off.
Is it that the relationship between gate and drain is sort of "symmetrical"? I was always told that current flows from D to S and the G pin just controls whether that connection is open, but is it more like an AND gate? Current flows out of S as long as it's flowing into both G and D?