What do kWh, Wh, Ah, and A actually mean?

kWh (kilowatt-hours) and Wh (watt-hours) measure energy. 1 kWh = 1,000 Wh.

A (amps) is electrical current.

Ah (amp-hours) is current over time (how many amps for how many hours).

To connect them: Wh = Ah × Volts, and Ah = Wh ÷ Volts.

Why don’t real-world results match the calculator exactly?

Variations in inverter efficiency at different loads, temperature effects (cold reduces capacity), battery age, voltage sag under heavy load, Peukert effect on lead-acid at high currents, and BMS cut-offs can all shorten runtime or increase the Ah you need.

What do “Efficiency” and “Usable %” change?

Efficiency covers losses in your system (inverter, wiring, charging).

Usable % is the depth of discharge you plan to use.
Capacity with both applied: Ah_required = Ah_ideal ÷ Efficiency ÷ Usable
Example: 2.4 kWh at 12 V = 200 Ah ideal. With 90% efficiency and 80% usable:
200 ÷ 0.90 ÷ 0.80 ≈ 278 Ah recommended.

Which system voltage should I pick (12 V, 24 V, 48 V)?

Match the voltage your equipment supports. Higher voltages reduce current for the same power, which helps with cable size and voltage drop, but all components must be compatible. For small systems 12 V is common; for larger loads 24 V or 48 V is typical.

How do I estimate runtime from a battery I already have?

If your load is in watts: Hours ≈ (Ah × Volts × Usable × Efficiency) ÷ Watts

If your load is in amps: Hours ≈ (Ah × Usable × Efficiency) ÷ Amps
Example: 100 Ah at 12 V, 80% usable, 90% efficient, 60 W load →
Deliverable Wh = 100 × 12 × 0.80 × 0.90 = 864 Wh → Runtime ≈ 864 ÷ 60 = 14.4 hours.

How do I convert kWh to amp-hours?

Use: Ah = (kWh × 1,000) ÷ Volts.
Example: 2.4 kWh at 12 V → 2,400 Wh ÷ 12 V = 200 Ah (ideal).