I am planning on buying this LED.
This is for my my project of doing a video light that I can carry to the field. Since these are 10 W, 12-14 V, my math (P/V = I) (10/14) equals to 710 mA (assuming 10 W is achieved at 14 V).
My plan is to buy three of them and wire them up in parallel. That means I will have a load of 0.71 mA × 3 = 2.13 A for the project.
If I computed this correctly (7 Ah/2.13 A), I will have around 3.2 hours of use before I run out of juice. The 7 Ah in the equation because I'm planning on using a 12 V, 7 Ah lead acid battery. It's a deep-cycle battery. The reason I am going to use this is because it is cheap and available, and though heavy, it's tolerable.
These computations are accurate if I'm going to use it at full capacity, but I can't drive the LEDs at full capacity because my supply is only 12 V. Also, I'm going also to use a step-down motor with a potentiometer to dim it when needed.
All-in all, I'm hoping if anyone can advise me on anything I have missed? Are my computations correct?