Worked problem

UPS Runtime: Energy, Efficiency and Usable Capacity

Estimate the runtime of a 2 kW load on a 48 V, 100 Ah battery bank from usable capacity and inverter efficiency, and see the limits of the model.

Electrical & Power · Last reviewed October 17, 2026

Given

A constant 2.0 kW load runs from a 48 V (nominal), 100 Ah battery bank through an inverter. For this exercise assume 80% of the nominal capacity is usable before the battery must stop discharging, and an inverter efficiency of 92%.

Find

The estimated runtime, and what the answer depends on.

Solution

Nominal stored energy = 48 V × 100 Ah = 4,800 Wh.
Usable energy = 0.80 × 4,800 = 3,840 Wh.
Power taken from the battery = 2,000 ÷ 0.92 = 2,174 W, because the inverter has losses.
Runtime = 3,840 ÷ 2,174 = 1.77 h, about 106 minutes.
Discharge current at nominal voltage = 2,174 ÷ 48 = 45.3 A, which is 0.45C for a 100 Ah bank. Compare it with the discharge rating of the battery.

Answer

about 1.77 h (roughly 106 minutes)  |  discharge about 45 A

Usable fraction	Runtime
50%	66 min
70%	93 min
80% (this problem)	106 min
90%	119 min

Where the simple energy model fails

Capacity depends on discharge rate. A 100 Ah rating is normally quoted at a slow rate such as 10 or 20 hours. At 0.45C a lead-acid battery delivers noticeably less than 100 Ah.
The battery voltage falls as it discharges. For a constant-power load the current rises as the voltage falls, which shortens the runtime further.
Low temperature, battery age and the end-of-discharge voltage setting all reduce usable capacity. A battery at the end of its life has about 80% of its original capacity.
Inverter efficiency changes with load. Fans, charger and controls draw power too.

Use the energy estimate to compare options, and the manufacturer discharge table (watts per cell to a stated end voltage) for a runtime you will rely on.

Related tools and labs

Calculators, labs and simulators are part of the All-Access Pass.

Questions

Why not use the full 4,800 Wh?

Discharging a battery to empty shortens its life and may not be possible, because the inverter or battery management stops at a minimum voltage. The usable fraction is a planning assumption.

How do I get a longer runtime?

Add battery capacity, reduce the load, or use batteries rated for a higher discharge rate. Doubling the energy roughly doubles the runtime in this model.

Is the usable fraction a code requirement?

No. Here it is an exercise input. Battery makers give the real limits for their product.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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