Electrical – The right gauge wire for a 12v parallel wiring to LEDS and heat sink

Cutting the Alu in half and sticking it to the back will do almost nothing to help. What you need to do is increase the surface area of the Alu.

If you're stuck for space, then you have a couple of options. If you have space in the vertical direction, you might try just bending the alu so it's not as wide and fits into the shape of the available space. However, you do still need to make sure the heat can get to the outside world. There needs to be some kind of air flow, preferably over the alu.

If there is a fairly constant trickle of water to the plants, then you can use this to extract a lot of heat from the LEDs. Instead of using a sheet of Alu, use some box section. Seal up the ends and attach a push-fit pneumatic connector at each end. Pump the water supply through this, and you should be good to go. Even a tiny flow of water should be enough, but it probably needs to flow fairly constantly, not just once a day.

Most of these strips are laid out in a repeating pattern of three LEDs and their current-limiting resistors along a flexible circuit. There will be scissors marks every 10 cm or so.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. Cut along the dotted line.

If it's this type of strip then the LEDs are effectively wired in series parallel along the length of the strip. You can connect the strips end to end but it should be clear from the schematic that you're actually connecting the individual strings of LEDs in parallel.

If you daisy-chain the strips then be aware that the first strip has to carry the current for all the downstream LEDs and the copper may overheat.

Alternatively wire each strip individually back to the PSU.

Figure 2. All you need to know.

• You need a 12 V supply.
• The current rating needs to be at least that required by the LEDs.

\$ 10~m \times 1.5~A/m = 15~A \$.

\$ P = VI = 12 \times 15 = 180 W \$.

Be careful daisy-chaining the strips. The start of the first one will have to carry the full 15 A, falling to zero at the end of the second. It would be safer to feed 12 V in both ends. That way each end would carry 7.5 A, falling to zero in the centre.