Formula
Current is taken from the battery, after the inverter. Usable capacity is the rated amp-hours times the depth of discharge. The energy limit is that capacity divided by the current, which is the same as the Peukert expression with the exponent set to 1. The lead-acid line raises the exponent.
I = watts ÷ (volts × inverter efficiency)
t = H × ((C × DoD) / (I × H))^k
H is the rating period, 20 hours unless you change it. A lithium pack with k = 1 does not get a second reduction. Some LiFePO4 data sheets sit near 1.05. At 1.05 the change is small, and this page’s lithium preset keeps k at 1 so you can see the energy limit on its own. Flooded lead-acid is started at 50% discharge and k = 1.3. AGM at 50% and k = 1.1. Those are manufacturer-practice figures, not a standard. If the watts came from a voltage and a current, that product is the Ohm’s law calculator.
Worked examples
100 Ah, 12 V, 120 W, 85% inverter, 50% discharge, k = 1.3, 20-hour rating. Battery current is 11.76 A. The energy limit is 4.25 hours. Peukert brings it to 2.67 hours.
The same battery with k = 1. Runtime stays at 4.25 hours. That is the lithium preset’s exponent, though the lithium preset also raises the discharge limit to 80%, which lengthens the run.
Reference
The two hours are always both on the readout. Use the Peukert line for flooded and AGM. Use the energy line when k is 1.
Questions
How long does a 100 Ah, 12 V battery run a 120 W load?
At 85% inverter efficiency the battery supplies 11.8 A. Using half the capacity, the energy limit is 4.25 hours. With a lead-acid Peukert exponent of 1.3 that falls to about 2.67 hours. A lithium exponent of 1 stays at 4.25 hours.
What is Peukert’s exponent?
A lead-acid battery delivers fewer of its rated amp-hours as the current rises. The usual form against the 20-hour rating is t = 20 × ((usable Ah) / (amps × 20))^k. k near 1.3 is a flooded starting point. k = 1 means the effect is ignored, which is the right default for LiFePO4.
Does inverter efficiency change the current?
Yes. The watts are on the AC side. The battery current is watts ÷ (volts × efficiency). An 85% inverter makes the battery work harder than the load’s wattage suggests.
Sources
- Peukert’s equation in the C20 form t = H ((C · DoD) / (I · H))^k. H defaults to 20 hours.
- Inverter efficiency is applied to the battery current, not as an after-the-fact fudge on the hours.
- Flooded k ≈ 1.3 and 50% DoD, AGM k ≈ 1.1 and 50%, LiFePO4 k ≈ 1 and 80% are typical practice. A k near 1.05 is sometimes quoted for LiFePO4. This page’s lithium preset uses 1.