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Emergency lighting solution

Continuing from the previous post above. Firstly I know how to calculate current in a DC situation.

However, since I am using an inverter that produces 220V at its output I would like some assistance in calculating the current consumption of a 30W LED and NOT a 50W LED. The specs given from the box for the 30W LED is 30W LED website:

  1. Supply voltage: 200-240AC

  2. Power: 30W

  3. CRI: RA>70.PF>0.9

Is it possible to determine the current consumption of the 30W LED from the specs given above?

Is the current consumption of 151mA correct, assuming a supply voltage of 220V AC? Using the following site of rapid tables

If it is correct then using a 18A-h is an overkill and since I will be running the light for only 2 hours(duration of the black out) a 3.5A-h will therefore suffice. Is this assumption correct?

The smallest SLA battery is 3.5A-h battery in my region.

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  • \$\begingroup\$ links to any available spec useful. \$\endgroup\$
    – Russell McMahon
    Feb 2 '20 at 21:44
  • \$\begingroup\$ @RussellMcMahon the site only shows the power output. The specs given is what is shown on the box. There is nothing else. I have included the site though. \$\endgroup\$
    – JoeyB
    Feb 2 '20 at 21:50
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You need to calculate using energy. For this problem Wh (watt-hour) is the most convenient.

Batteries are normally rated in A-h (Amp-Hour), not A/h (amps/hour) as you have specified

Output Energy = 30W * 2 hours = 60 Wh

Input energy = 3.5 A-h * 12 V = 42 Wh

So, you need more than a 3.5 A-h battery. And this doesn't even account for inefficiencies or the fact that many batteries shouldn't be discharged all the way if you want a long life.

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  • \$\begingroup\$ From your experience whats is the normal percentage or watt-hour given to account for inefficiencies that needs to be added to the output energy? \$\endgroup\$
    – JoeyB
    Feb 2 '20 at 22:01
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    \$\begingroup\$ Look up or calculate the efficiency of your inverter. I would guess about 75%. Also consider that the efficiency is rated at the maximum output, if the output is lightly loaded, it could be much less. In your other question, someone already said to reserve 50% in your battery. So a ballpark figure is 60 Wh / (75% * 50%) = 160 Wh. For a 12V battery that would be 13.3 A-h. \$\endgroup\$
    – Mattman944
    Feb 2 '20 at 22:18
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    \$\begingroup\$ Rather than guess, measure the input current and multiply by the voltage. \$\endgroup\$
    – Mattman944
    Feb 2 '20 at 23:28
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    \$\begingroup\$ Efficiency is not related to safety. But, I would read the user's manual for the Inverter carefully before I used it. \$\endgroup\$
    – Mattman944
    Feb 2 '20 at 23:43
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    \$\begingroup\$ 30 W Out / 57.26 W In = 52% \$\endgroup\$
    – Mattman944
    Feb 5 '20 at 16:10

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