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I have assembled a 300 units of 3650 LED lights which draws a total power of roughly 6 Amps (130Watts) and I want to power it using a 12V 18Ah Lead Acid Battery.

My question is, how long or how many hours the assembled light will last before the battery gets fully discharged?

Is there a certain formula that I can use to get the total hours consumed?

Many thanks in advance for your help.

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    \$\begingroup\$ 130 watts at 6 amps requires 21.67 volts, not 12v. So you have some "how is this going to work at all" engineering to do, or mistakes to correct, before you can make this calculation. If you involve a boost converter, you have to consider its efficiency. \$\endgroup\$ – Chris Stratton Aug 15 '17 at 11:48
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    \$\begingroup\$ You should include the minimum voltage needed by your LED array in your question. If you give some more details, you get better answers. \$\endgroup\$ – Uwe Aug 15 '17 at 17:19
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    \$\begingroup\$ As Mr. Stratton says, the load seems not to be a 12v one. You will need a voltage multiplier. Based only on power consumption, if the load is 130 W and battery is 12v*18Ah=216Wh, the battery will fail before 216Wh/130W=1.6 hours \$\endgroup\$ – pasaba por aqui Oct 31 '17 at 16:06
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Battery capacities are measured in Amp-hours. You know how much current your circuit draws, so this is pretty simple: 18Ah/6A = 3 hours.

That assumes a constant current. Without a regulator, the voltage will drop as the battery discharges, so the current will drop too.

The capacity will also change depending the how much current is drawn, and the temperature. Here's an example from a Panasonic battery datasheet: Panasonic SLA excerpt

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    \$\begingroup\$ While the general idea is correct an SLA battery is normally rated based on something like a 10 or 20 hour discharge, so you normally get a reasonable bit less than the rated capacity if discharged over three hours. \$\endgroup\$ – PeterJ Aug 15 '17 at 11:57
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    \$\begingroup\$ Also, it's a very bad idea (in the long run) to discharge regular lead-acids to more than 1/2 of full charge, so 1 to 1.5 hours is a better goal. If you do more than this, you'll find that the batteries don't last nearly as long as you might expect before they start getting weak and need to be replaced. \$\endgroup\$ – WhatRoughBeast Aug 15 '17 at 12:39
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    \$\begingroup\$ calculation based only in current is valid if voltage of battery is the same than voltage of circuit. This is not the case of the original question, note it is said load of 6A, 130 W. \$\endgroup\$ – pasaba por aqui Oct 31 '17 at 16:01

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