Your circuit is not quite correct. If V1 is present and you close the switch V1 and the battery will be shorted together, and you don't want that. Their voltages will never be exactly the same, and that will cause a large short-circuit current. More about this in a minute.
When V1 is present the LED will be on through the battery, regardless of the position of SW1.
If V1 is gone the relay will be off, and so will the LED.
How to solve these issues? Use a switch-over relay to switch between V1 and battery and place it at the node next to the LED's anode.
A minor thing, though it will not change the circuit's functionality: you're interrupting the ground with your relay. Don't. Swap the relay contacts and the LED, so that you interrupt the positive voltage instead. Just a good habit. Has been fixed anyway in the previous paragraph.
And don't forget to place a series resistor with the LED. For a 20mA indicator LED a 1k\$\Omega\$/1W will do.
And clean it up a bit. The ground line at the bottom, going to the contact of SW3 is superfluous, and draw a single ground line for both power sources, so that it's obvious that they're connected.
edit (re your revised circuit)
We're almost there. The switch is in series with the relay coil, so closing it will create a path from coil through R1 and LED to ground. The - of the coil should go directly to ground, and the switch should go between V1 and the unused relay contact.
But schematic-wise it looks already much better, don't you think? It may even pass the olin test ;-).
The printer draws large amounts (20+ Amps) of current at 12V and these supply exactly that - nothing more, nothing less!
The power supply will certainly deliver less current if the load demands less. If the 3D printer draws 1A, the supply will supply 1A. Your statement that the power supply delivers a certain power level - no more, no less - is incorrect.
I'm looking for a more stable power supply for my 3d printer.
You haven't mentioned stability in your posting. You did say that at high loads, the output voltage sags. The stiffness of a power supply is related to how well the voltage regulates at high load. The stability of a power supply is related to how well the voltage regulates when the output is subjected to a rapidly-changing dynamic load.
There could be two things going on causing the voltage sag:
1) The voltage sensing point is close to the power supply; it is regulating the voltage at that point and the loss you see at the load is due to resistive losses between the sense point and your measurement point
2) The power supply is entering a protection state and is limiting the voltage to limit the output power and keep the power supply thermally safe.
I just have a few concerns about the safety of using one of these devices.
A 'safe' power supply, in industry parlance, means it has been evaluated by a regulatory authority and found to comply with certain national / international safety standards for the application in which it was intended to be used. A single abnormal should not cause a safety hazard (shock / fire / shrapnel). The unit should bear one or more well-recognized safety marks (UL, CSA, TUV, etc.)
My main concern is that I noticed there is nowhere to simply plug a mains cable as an input to the power supply.
As Olin pointed out, this is a unit meant to be permanently installed into some other piece of equipment, not something that a user would be expected to swap-in or swap-out often.
However, isn't there a fair likelihood that someone may just pick it up by putting their fingers and short the live and neutral together? The exposed screw contacts look awfully prone to accidental contact with not just fingers, but nearby metallic objects. A built-in fuse won't exactly help here would it?
Notice in the photo that there's a clear insulating shield over the terminal block. That shield is part of the inherent safety of the unit and should prevent against mains shock from casual contact with the unit. If someone wants to hot-screw a powered mains cord onto this UUT, well, they deserve what they get. Not trying to be facetious, but these sorts of power supplies are meant to be installed by 'qualified' personnel who have some basic knowledge.
It's quite common for power supplies like this (meant for use inside other equipment) to get fed from a feed that has a fuse or breaker in it; there's the possibility that the internal wiring may make contact with the equipment itself. That doesn't mean there isn't a fuse in the power supply though (there should be!)
The other concern I have is whether it will be extremely dangerous if I accidentally reverse the live and neutral wires?
Yes, but only if there's a fault. The power supply will 'work' with reversed L and N. However, the power supply internal fuse is in series with the terminal marked L (line). Blowing the fuse means the neutral is now floating, which is a big no-no (netural must never be interrupted) and a big risk that your chassis can become a shock hazard.
At the moment, I am using a typical 320W PSU for a desktop computer to handle the power workload.
PC power supplies have minimum load requirements on various rails and are rarely a good choice for industrial applications like a 3D printer. Spend the money and get a single-output 12V supply that can deliver the power plus has remote sensing capabilities so that you get the best regulation possible.
Best Answer
A 60W incandescant light bulb requires 500 mA @ 120 V to light it up to full brightness. You can easily pass 10 mA through it — enough to operate a control module like Lightwave — without causing it to light up at all.
This is why such modules are often specified to work only with incandescant (resistive) loads. CFLs and LED bulbs will often glow or flicker on the current that they pass.