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I want to run a coolbox in my campervan that has a 12v leisure battery.

The given wattage is 35w

It would be plugged into an inverter that provides 230v and has roughly 85% efficiency

How should I calculate the amp usage so I can judge the amount of time I can run the fridge from my battery?

a) (35W/230V)*/0.85 giving 0.18 Amps

b) (35W/12V)*/0.85 givig 3.4 Amps

c) ???

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    \$\begingroup\$ The one that's actually the current going through the battery. \$\endgroup\$ – Hearth May 19 '18 at 18:23
  • \$\begingroup\$ You cannot plug a 12v coolbox into the 220v inverter. Perhaps you should do a diagram of what you mean. Also a coolerbox and a fridge or is it one and the same? \$\endgroup\$ – Solar Mike May 19 '18 at 18:24
  • \$\begingroup\$ 35W is most likely a peltier, and not a compressor cooler. Peltiers run on low voltage. Check if your cooler has a power supply inside... if you're lucky it uses a 12V peltier and a 230V to 12V power supply which would be redundant... \$\endgroup\$ – peufeu May 19 '18 at 18:28
  • \$\begingroup\$ @SolarMike Thanks I have a 12v battery but the cool box would normally run off the mains, hence the inverter. \$\endgroup\$ – aaaaargZombies May 19 '18 at 18:56
  • \$\begingroup\$ Could someone please just explain the maths and I'll worry about those other things you mentioned myself! \$\endgroup\$ – aaaaargZombies May 19 '18 at 18:59
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You have to consider the voltage and the power on the battery side, to calculate the current:

$$P_\mathrm{bat} = V_\mathrm{bat} \cdot I_\mathrm{bat} \Rightarrow I_\mathrm{bat} = \frac{P_\mathrm{bat}}{V_\mathrm{bat}}$$

To get \$P_\mathrm{bat}\$ you need to know the output power of the inverter (which is equal to your load) and the efficiency:

$$P_\mathrm{bat} = \frac{P_\mathrm{load}}{\eta_\mathrm{inv}}$$

Putting everything together leads to your suggestion "b":

$$ I_\mathrm{bat} = \frac{P_\mathrm{load}}{\eta_\mathrm{inv} \cdot V_\mathrm{bat}} = 3.43\,\mathrm{A}$$

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