Learning objectives
By the end of this lesson you will be able to:
- calculate current from nominal capacity and C-rate;
- calculate C-rate from a measured or datasheet current;
- calculate the ideal duration associated with a C-rate;
- use charge as the area under a current–time curve to explain why the ideal duration is 1 / C-rate;
- separate exact C-rate arithmetic from real-cell voltage-cutoff behaviour.
Before you start: Part 01 · Cell anatomy. Arithmetic with units is all the maths you need.
01Nominal capacity is charge, not energy
Nominal capacity, written , is usually reported in ampere-hours (Ah). It describes electric charge delivered under stated test conditions. It is not an energy rating, and on its own it says nothing about the cell’s voltage.
So a 20 Ah rating means that 20 amperes sustained for one hour would transfer 20 Ah of charge — in the ideal arithmetic sense. Whether a particular cell actually delivers that charge before reaching its voltage limit is a separate question, answered in section 05.
02C-rate is a normalised current
C-rate divides the current by the nominal capacity, so the same number means the same relative load on a 3 Ah phone cell and a 300 Ah stationary cell:
The units show why C-rate is tied to time. Amperes divided by ampere-hours leaves reciprocal hours:
Rearranging gives the two ideal relationships used everywhere in battery engineering:
The second expression contains no capacity, because the current has already been normalised by it. It is an ideal charge-transfer duration, not a prediction of when a real cell reaches its cutoff voltage.
| Label | C-rate | Current | Ideal duration |
|---|---|---|---|
| C/10 | 0.1 h⁻¹ | 2 A | 10 h |
| C/5 | 0.2 h⁻¹ | 4 A | 5 h |
| C/2 | 0.5 h⁻¹ | 10 A | 2 h |
| 1C | 1 h⁻¹ | 20 A | 1 h |
| 2C | 2 h⁻¹ | 40 A | 30 min |
| 5C | 5 h⁻¹ | 100 A | 12 min |
| 10C | 10 h⁻¹ | 200 A | 6 min |
03Three patterns in the rate table
1C is the reference point. For the 20 Ah example, 1C is 20 A and an ideal duration of one hour.
Doubling the C-rate doubles the current and halves the ideal duration. Moving from 1C to 2C changes 20 A to 40 A, and one hour to thirty minutes.
At the same C-rate, the ideal duration is the same for every capacity. A 3 Ah cell and a 300 Ah cell both have an ideal duration of 30 minutes at 2C — but they need 6 A and 600 A respectively.
That normalisation is the whole point of C-rate. A current on its own does not tell you how demanding a load is until you know the capacity of the cell carrying it.
04Why duration is reciprocal: charge is area
Electric charge is the time integral of current. For a constant current it reduces to a product:
On a current-versus-time graph, transferred charge is therefore the area under the curve. Each rectangle below holds exactly 20 Ah:
05Ideal duration is not a runtime prediction
The equal-area picture assumes the full nominal capacity is available at the chosen current. A real cell is normally stopped when its terminal voltage reaches a specified limit, and the charge it can deliver before that limit depends on the cell and on the test.
| Effect | Why the measured duration can change |
|---|---|
| Ohmic voltage drop | A larger current produces a larger instantaneous terminal-voltage drop, so the lower cutoff can be reached sooner. |
| Polarisation and transport | Concentration gradients and reaction overpotentials generally grow with current, changing how much charge is accessible before cutoff. |
| Temperature | Temperature changes resistance, kinetics, transport and usable capacity. |
| Chemistry and design | Electrode thickness, particle size, electrolyte and other design choices set the rate capability. |
| State of health | Aging changes both capacity and resistance — see Part 04. |
| Rating and test definition | Nominal capacity is tied to a specified current, temperature, rest conditions and voltage limits. |
At high current a cell often reaches its lower voltage limit before delivering its nominal charge. But the ideal duration is not a universal upper or lower bound either: usable capacity can be below, near or occasionally above the nameplate value, depending on the rating procedure and conditions.
This lesson does not show a rate-versus-capacity curve. Predicting cutoff time needs measured discharge data or a validated voltage model with cell-specific parameters — which is where Chapter 2 begins.
06Worked example: reading a datasheet
A datasheet lists a 3.2 Ah cell with a maximum continuous discharge current of 15 A. What C-rate is that, and what is its ideal duration?
That 12.8 minutes is a reference calculation, not a promise that the cell stays above its lower voltage limit for 12.8 minutes at 15 A. The real duration comes from the datasheet discharge curve, measured data, or a validated model at the stated temperature, voltage limits and cell condition.
07Calculate it in Python
Every number on this page comes from the same three tested functions in the open-source battery_core package:
from battery_core import (
c_rate_from_current,
current_from_c_rate,
ideal_duration_hours,
)
current_from_c_rate(nominal_capacity_ah=20.0, c_rate=10.0) # 200.0 A
ideal_duration_hours(10.0) * 60.0 # 6.0 min
c_rate_from_current(current_a=15.0, nominal_capacity_ah=3.2) # 4.6875
The two conversion functions accept keyword arguments only. Capacity is the first parameter of one and the second of the other, so a transposed positional call would return a plausible but wrong number with no error. Requiring names makes that mistake impossible.
Want to change the code and plot it yourself? Run the Part 02 notebook in your browser (no installation; the first launch can take a few minutes).
08Assumptions and limits
- Current is a positive magnitude; charge and discharge sign conventions belong to a later module.
- Nominal capacity must be strictly positive. Current and C-rate may be zero — a rest step is physical, and its C-rate is exactly 0.
- The ideal duration still needs a strictly positive C-rate, because is undefined.
- No voltage, energy, heat, aging, kinetics or rate-dependent capacity is calculated.
09Check yourself
Work each one out before opening the answer.
A 50 Ah cell is discharged at 25 A. What is its C-rate and ideal duration?
25 / 50 = 0.5C, so the ideal duration is 1 / 0.5 = 2 hours.
Which has the shorter ideal duration at 2C: a 5 Ah cell or a 100 Ah cell?
Neither. Both have an ideal duration of 30 minutes. Their currents differ: 10 A for the 5 Ah cell and 200 A for the 100 Ah cell.
A 20 Ah cell is discharged at 40 A for 25 minutes before reaching its cutoff voltage. How much charge was delivered, and how does that compare with the nominal capacity?
40 × (25 / 60) = 16.7 Ah, about 3.3 Ah below the nominal rating under those test conditions. On its own this does not tell you how much charge would be usable at a different current, temperature or cutoff voltage.
Why is “10C always lasts exactly 6 minutes” not a valid real-cell prediction?
Six minutes follows from the ideal reciprocal relationship. Real cutoff time also depends on voltage limits, resistance, polarisation, temperature, chemistry, design, state of health and the capacity-rating procedure.
10Practice quiz
Three questions on this lesson, marked in your browser. The numeric drills generate fresh numbers every time, so you can repeat them until the method is automatic.
11References and further reading
These sources give broader battery-modelling context. The explanation, examples, code and tests in Battery Core are written independently.
- T. B. Reddy (ed.), Linden’s Handbook of Batteries, 4th ed., McGraw-Hill, 2011.
- G. L. Plett, Battery Management Systems, Volume I: Battery Modeling, Artech House, 2015.