thanks. Got that bit about the grounds connected to the source.

Was reading art of electronics last night (rocket science to me!) and
then I saw something that you said as well:
"Always use a supression diode when switching an inductive load"

They had the following diagram:
+Vcc --+---COIL----+----C E----- GND
| | B
+---|------+ |

Ok. So I unleash the 72V through my favourite coil. When I stop the
voltage.... The magnetic field subsides. But according to you and the
book it generates somesort of spike.

So the way I figure it, is that the current must be generated in the
opposite way. So previosly the C of coil in my diagram was the + and
the L was -. Now I think the L will be + and the C -. So can I
'replace' the coil with a battery for the millisecond or two that the
field dissolves?

Anyway... that's the only way I can explain the use of the diode. When
the 72V is running through the coil it has no use. When it is switched
off, it will allow my new 'battery' with a + on the L side to short
circuit itself and that I guess takes care of the current generated...
Is this correct?

So if all this guessing of mine is correct then my question is... So
what? What will happen if the C of the BJT is more - than the E and B?