CourseWiki/Engineering Fundamentals/Electrical Systems

Battery Capacity and Packs

Convert Ah to Wh, use C-rate and current to estimate performance, and predict the effect of series and parallel cells.

Ampere-hours measure charge; watt-hours measure energy; Practice

Ampere-hours measure charge; watt-hours measure energy

A 2 Ah cell and a 2 Ah battery can store very different amounts of energy if their voltages differ. Ah tells you charge. To estimate energy, you also need voltage.

Energy (Wh) ≈ capacity (Ah) × nominal voltage (V)
Energy (J) = energy (Wh) × 3600

For an ideal 2.4 Ah, 3.6 V cell: 2.4 × 3.6 = 8.64 Wh, or 8.64 × 3600 = 31,104 J. Convert mAh to Ah by dividing by 1000 before multiplying by voltage.

Why the battery estimate uses an approximate sign

Wh = Ah × V is exact for constant voltage. A real cell's voltage changes during discharge, so nominal voltage gives an energy estimate. Usable charge also changes with load, temperature, age and the cutoff voltage. These calculations assume the stated capacity is available.

See Work and Energy for energy units and Current, Voltage, and Resistance for voltage and charge.

Practise battery energy

C-rate sets a current scale; load current sets ideal runtime; Practice

C-rate sets a current scale; load current sets ideal runtime

A discharge rate of 1C means a current numerically equal to the capacity in Ah divided by one hour. The rate has units of h⁻¹. Here C is a rate notation, not the coulomb symbol.

Rated maximum current (A) ≈ maximum discharge C-rate (h⁻¹) × capacity (Ah)
Ideal runtime (h) = capacity (Ah) ÷ load current (A)

A 2.4 Ah cell rated for continuous 3C discharge has a rated current estimate of 3 × 2.4 = 7.2 A. At a constant 1.2 A load, ideal runtime is 2.4 ÷ 1.2 = 2 h. At 3C, the ideal duration is one-third of an hour, or 20 minutes.

A C-rate may describe a test rate, a charge rate, a continuous limit or a short pulse limit. Only a specified maximum discharge rating belongs in the maximum-current equation. Actual pack limits may be lower because of wiring, temperature or protection hardware. A constant-power load also changes current as voltage changes.

Practise ideal runtime

Series adds voltage; parallel adds charge capacity

For matched cells, connect cells in series to increase voltage. Put identical series strings in parallel to increase Ah at the same voltage. 3s2p means three cells per series string and two strings in parallel: six cells altogether.

Two identical 3.6 V 2.4 Ah cells in series give 7.2 V and 2.4 Ah; the same two in parallel give 3.6 V and 4.8 Ah. Each arrangement stores an ideal 17.28 Wh.

Pack V = series count × cell V
Pack Ah = parallel count × cell Ah
Pack Wh ≈ pack V × pack Ah

With 3.6 V, 2.4 Ah cells, a 3s2p pack is 10.8 V, 4.8 Ah and 51.84 Wh. At 3C it has an ideal rated-current estimate of 14.4 A. A 1.2 A pack load gives 4 h ideal runtime.

This is an electrical model, not a pack assembly procedure. It assumes matched cells, balanced series operation and equal current sharing between parallel strings. Real packs need appropriate cell matching, interconnections and protection. See Series and Parallel Circuits.

Explore a pack by changing one input at a time; Interactive; Practice

Explore a pack by changing one input at a time

First double the series count: voltage and energy double, but Ah and runtime at the same pack current do not. Reset it, then double the parallel count: Ah, energy, rated current and ideal runtime double. The load current below is measured at the pack terminals.

Predict first: same current or same ideal power?

Before checking the results, predict whether ideal runtime will stay the same or double in each comparison, and explain what must happen to current. Start with 3.6 V, 2.4 Ah cells in a 3s2p pack at 1.2 A, then try two different meanings of “the same load”:

Same pack current: change only the series count from 3 to 6 and leave the load-current input at 1.2 A.
Same ideal nominal pack power: keep 6s2p, calculate I = P ÷ V = 12.96 W ÷ 21.6 V = 0.6 A, then enter 0.6 A in the existing load-current input.

Paper example: 3.6 V, 2.4 Ah cells; 3s2p; continuous discharge rating 3C; load 1.2 A.

V = 3 × 3.6 = 10.8 V; Ah = 2 × 2.4 = 4.8 Ah; Wh = 10.8 × 4.8 = 51.84 Wh; Imax = 3 × 4.8 = 14.4 A; t = 4.8 ÷ 1.2 = 4 h.

With no JavaScript, use these equations to explore other packs. This example remains in the print view.

Check the fixed-current and fixed-power comparison

Same cells and parallel count, with two different load bases
Case	Nominal voltage	Capacity	Nominal energy	Pack current	Nominal load power	Ideal runtime
3s2p starting case	10.8 V	4.8 Ah	51.84 Wh	1.2 A	12.96 W	4 h
6s2p, same current	21.6 V	4.8 Ah	103.68 Wh	1.2 A	25.92 W	4 h
6s2p, same ideal power	21.6 V	4.8 Ah	103.68 Wh	0.6 A	12.96 W	8 h

Check: at the same current, both packs still have 4.8 Ah, so t = Ah ÷ I stays at 4 h while the 6s2p pack supplies twice the nominal power. At the same ideal power, twice the nominal voltage needs half the current, so the ideal runtime becomes 8 h.

The 6s2p case uses twelve cells instead of six and stores twice the nominal energy. The longer fixed-power runtime is not a free benefit.

This comparison assumes matched cells, constant nominal voltage, the stated Ah is available and the entered load is ideal. It does not predict changing discharge voltage, load or converter losses, reserve capacity, aging, cutoff or protection behavior. The model reports when the requested current exceeds the entered rating; a result below that rating is not a safety certification.

Practise series and parallel packs

Appears in

DET 403 DET Energy Generation and Storage