Key takeaways
- Runtime comes from usable energy ÷ load, not nameplate amp-hours.
- Usable energy = Ah × volts × depth of discharge, then trimmed by inverter efficiency.
- A 200 Ah / 12 V battery gives about 1,080 Wh usable at 50% DoD and 90% efficiency.
- Heavy loads drain lead-acid faster than the math suggests; lithium holds up better.
Capacity is not the same as runtime
A battery's amp-hour rating tells you how much charge it holds, but it doesn't tell you how long it will run your gear. To get runtime you first convert the rating into usable energy, then divide by the power your loads draw. Three things stand between the label and reality: voltage, depth of discharge, and conversion losses.
Start with energy. Capacity in amp-hours multiplied by the battery voltage gives watt-hours — the real currency of runtime. A 200 Ah battery at 12 V holds 2,400 Wh on paper. But you can't safely use all of it.
Depth of discharge limits what you can use
Pulling a battery all the way to empty shortens its life, so you only draw down a fraction of the total. That fraction is the depth of discharge (DoD). Lead-acid batteries are typically held to about 50% DoD; LiFePO₄ tolerates 80–100%. At 50% DoD, our 2,400 Wh battery offers only 1,200 Wh before you should stop.
Inverter efficiency takes its cut
Most loads run on AC through an inverter, which is roughly 90% efficient — about a tenth of your stored energy is lost as heat in the conversion. Apply that to the 1,200 Wh and you're left with roughly 1,080 Wh of genuinely usable AC energy. Runtime is then simply that usable energy divided by your continuous load in watts.
Plug your own numbers into the battery backup runtime calculator to skip the arithmetic, or size the bank itself with the battery bank sizing calculator.
Runtime table: a 200 Ah / 12 V battery
Using 1,080 Wh usable (200 Ah × 12 V × 0.5 DoD × 0.9 efficiency), here's how long that one battery lasts across common loads:
| Load | Usable energy | Approx. runtime |
|---|---|---|
| 100 W | 1,080 Wh | ≈ 10.8 h |
| 300 W | 1,080 Wh | ≈ 3.6 h |
| 600 W | 1,080 Wh | ≈ 1.8 h |
| 1000 W | 1,080 Wh | ≈ 1.1 h |
Worked example: running a 300 W load
Take the same 200 Ah, 12 V battery. Usable energy is 200 Ah × 12 V × 0.5 × 0.9 = 1,080 Wh. Divide by a steady 300 W load: 1,080 ÷ 300 = 3.6 hours, or about 3 h 36 min. To estimate your own load first, itemize appliances in the off-grid load calculator and feed the total back into the runtime math.
Why real-world runtime falls short
The table assumes the rated capacity is fully available — but on lead-acid batteries it isn't under heavy loads. The Peukert effect means high discharge currents reduce the effective capacity, so a 600 W or 1,000 W draw empties the bank faster than the simple division predicts. Lithium (LiFePO₄) chemistry holds its rated capacity far better under high loads, which is one reason it dominates serious off-grid builds. At the other extreme, an inverter's standby or idle draw quietly consumes energy even when nothing is plugged in, shortening runtime on light loads. For honest planning, build in margin rather than treating the calculated figure as a hard floor.
Frequently asked questions
How long will my battery last?
It depends on usable energy and load. A 200 Ah / 12 V battery holds ~2,400 Wh, but at 50% DoD and 90% inverter efficiency you get about 1,080 Wh usable — roughly 3.6 hours at a 300 W load.
How do I calculate battery runtime?
Runtime (h) = Ah × V × depth of discharge × inverter efficiency ÷ load (W). Convert to usable watt-hours first, then divide by your continuous load.
Why do high loads drain a battery faster?
On lead-acid the Peukert effect cuts effective capacity at heavy current, so big loads empty the bank faster than the math predicts. Lithium holds capacity better, and inverter idle draw eats into light-load runtime.