CptMichles700:
\As for my Math, it seems I forgot to account for the 100R resistors affect on the input of the Voltage regulator, though I don't know how to do that at the moment.
V = I * R
I = 260 mA = 0.26 A
R = 100 Ohm
V = 0.26 * 100 = 26V
That's your voltage drop over that resistor. Obviously at a 12V input you're never going to get 260 mA running through the resistor.
At 5 Ohm:
V = 0.26 * 5 = 1.3V
No problem there, enough voltage left for your regulator. Even if you draw quite a bit more current you're fine.
But mind the power dissipated by the resistor! It goes cubed with the current.
P = I2 * R or P = V * I
At 5 Ohm and 0.26A:
P = 0.34W (so a 1W resistor would do - it'll get warm).
Do make sure you have ventilation or your enclosure will get really warm, with the heat from that resistor and the 7805 regulator.
But if you add a bit more and go to, say, 500 mA:
Voltage drop V = 0.5 * 5 = 2.5V. No problem here, 9.5V left for the regulator, which then will be dissipating 0.5 * 4.5 = 2.25W.
Your resistor however is now dissipating 2.5 * 0.5 = 1.25W already! That's too much for a 1W rated resistor, even if well exposed to the air.