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Rowland you are trying to represent each rail with varying resistance loads via a series led, this is wrong. For example ....if you discarded the LEDs then each rail have the full 12V (Vcc) applied to each Rx. Put a LED in each rail by all means (parallel to the load) but have set resistor for each rail and try not to run all power thru each led as the leds have max Ic of ~30mA.
Lets say each led drops 2.2v, Vf = 2.2v. Then 12-2.2 = 9.8V (across Rx). If we set the led current, Iled @2mA this is R=V/I = 9.8/.002 = 4.9K ohms Rx required for each led (around about, resistors and LEDs in parallel to the load).
The important thing is to remember not to run all cct power thru each LED here, run the LEDs in parallel with a dropping Rx.
Oh and consider you voltage drop across your supply cable. If you draw lets say 8Amps max thru this cable then power loss (V drop) acorss the cable can be significant, P=I^2R. Industry std is a 3% V drop across supply cables.
Lest say you use 3m of a supply cable with .00252 ohms per m (a good quality cable, Jaycar 8 gauge). Thus power loss @ 8A 20degrees C is P = 8^2 x 3 x .00252 = .484v drop across the cable.
This is .484/12 x 100 = 4% V drop across supply cable! You electronics will only get ~11.5V @8A.
Ok all this is worst case scenario with I draw but consider that cable resistance I quoted is a good quality cable with low resistance per m (8 gauge). Thus I'd be looking at around at minimum an 8 gauge cable of no longer than 3m...preferably maybe even a lower gauge and maybe a power supply of 13.8V DC!.
I have thrown a cat amongst the pigeons for you and given you lots to think about but I'm happy to supply further info as I have recently built my own power distribution box mounted on my scope dovetail. Ok mine has a USB hub and +5V and +8V PSU in it but the principle is the same.
Oh one last thing..I wouldn't run dew heater control power (if you are using PWM dew controller) on the same cct as the camera etc power. It can get a little electrically noisy!
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