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Hi Fam! I'm looking into getting some cute lil led lights to string up around the edges of a 10x10 canopy or two. Not intended to throw much light, but just for decoration and atmosphere. Any product/brand/website recommendations? There's so many options on Amazon lol
Very roughly: assuming the specs are correct (you can measure that), and assuming that stringing together the two units means they are connected in parallel, so that both run at 4.5 V, would mean that the two units (3 W each) at 4.5 V draw 6 W / 4.5 V = 1.33 A.
So for 24 hours you would need a 4.5 V battery with a capacity of 1.33 A · 24 h = 32 Ah.
You will never get all the quoted capacity from a battery, and there are always losses, so you should add capacity to compensate; I would just add 50%, or even double it.
The battery must provide energy to a load, of power P, for a time T. The nominal voltage of the battery is Vcc. Based on the definition of direct electric current I, the electric charge supplied is: Q=I∙T. Given that 1∙Coulomb= 1A∙s= 1A∙hr/3600 so 1Ah=3600Coulomb. The electric charge supplied to the user in the time T is Q=P∙T/Vcc. For example: T= 12h, the power dissipated by the load is equal to 100W, then the energy supplied by the battery in the time T is equal to: E=P∙T=1.2kWh=4.32MJ and the charge supplied in this time is equal to Q= E/Vcc=100Ah. Without taking into account the fact that the battery discharges in this time and therefore the actual voltage of the battery is lower than the nominal voltage as time goes by. Therefore a 200Ah battery should be used to ensure that the load receives the required energy for the requested time.
In the case of LEDs, if each LED dissipates a power of P=20mW with a total supply LED string of Vcc=12V and the number of LEDs is N=100, it can be deduced that the total dissipated power is PT=N∙P=2W, so if the time when the LEDs are on is 24hr, you will need to use a battery with a charge of at least P∙T/Vcc=2W∙24hr/12V= 4Ah or 4000mAh, but it would be advisable to double this value to ensure that the load receives the required energy for the requested time.