Math Bridge: Why can a 5 Ah cell fail a 220 mA pulse?

← Back to M2M Case Studies
Math BridgeEmerging ParadigmsStruggle-friendly runway

Why can a 5 Ah cell fail a 220 mA pulse?

Connect nameplate energy, decade self-discharge, internal resistance, pulse sag, and current capability for the chapter’s meter battery.

Battery Bex, the guideBattery Bex guides
The one targetShow why a correct mAh ledger can still fail at transmit time.
The chapter case5 Ah, 3.6 V, ten years, 220 mA pulse, 5–30 Ω internal resistance.
What it buys youA clear reason for a pulse capacitor and load testing.

A field team faces an unresolved physical question: Why can a 5 Ah cell fail a 220 mA pulse? They must answer it before changing internal resistance on the real device. Predict the direction first.

See the relationship before changing it

The figure reads from left to right. The blue card is internal resistance. The middle card applies this page's relationship. The green card is nameplate energy. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.

Internal resistance changes nameplate energy An input card leads through the page relationship to the nameplate energy result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The decade mAh ledger says how much charge exists. Internal resistance separately limits how quickly that charge can reach the modem.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for internal resistance is 5.

  2. 2

    Name the relationship. E=18.0 Wh; Qremain=93.22%; Vsag=IRint Rint=5 ohm: Vsag=1.10 V; Vmodel=2.50 V; Imax=0.720 A

  3. 3

    Substitute the chapter fixture. Set internal resistance to 5. The page ledger gives nameplate energy as 18.00 Wh.

  4. 4

    Read the result. Keep Wh beside the value. Use it only inside the technical boundary on this page.

Predict, then change internal resistance

Try Predict the direction of nameplate energy. Move one control, calculate, then check your prediction.

5
Chapter baseline
Nameplate energy

Observe The decade mAh ledger says how much charge exists. Internal resistance separately limits how quickly that charge can reach the modem. Reset the control to 5 and compare nameplate energy.

Explain Only internal resistance moves here. The other chapter fixtures remain fixed.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only internal resistance moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only internal resistance. It replays the chapter fixture; it is not a field prediction.

1. Start with the physical question

Show why a correct mAh ledger can still fail at transmit time. A clear reason for a pulse capacitor and load testing.

Battery Bex: Keep the units and the model boundary visible from the first line.

2. Name every algebra move

1

Turn charge into energyE=QV.

2

Apply annual storage lossQremain=Q0(1−r)^t.

3

Find pulse sagVsag=IRint.

4

Subtract from open circuitVmodel=Voc−Vsag.

5

Find the simple current ceilingImax=Voc/Rint and compare with 0.220 A.

3. Reproduce the chapter case

E=18.0 Wh; Qremain=93.22%; Vsag=IRint
Rint=5 Ω: Vsag=1.10 V; Vmodel=2.50 V; Imax=0.720 A

The arithmetic reproduces the chapter case while keeping its assumptions explicit.

4. Try the controlling input

TryMove the control and watch every displayed result come from the shown formula.

Internal resistance
Nameplate energy
Charge remaining
Pulse sag
Resistor-model terminal
Ohmic current ceiling
Pulse shortfall
Direct pulse possible

ObserveAt 5 Ω, the pulse sag is 1.10 V, the resistor-model terminal is 2.50 V, and the simple ohmic current ceiling is 0.720 A.

ExplainThe decade mAh ledger says how much charge exists. Internal resistance separately limits how quickly that charge can reach the modem.

Technical boundaries.

This compact engine isolates one relationship; it is not a deployment certificate.

Cell model
Real polarization and voltage recovery are not one fixed resistor
Threshold
The modem’s minimum voltage and capacitor ESR are not included
Chemistry
Resistance varies with cell design, age, storage, temperature, and pulse history

Measure the real system and reopen the decision when its inputs change.

5. Add the pulse source

Size and test the hybrid capacitor, its ESR, recharge interval, regulator dropout, and cold/aged-cell behavior under the real modem waveform.

6. Keep the battery evidence

Record chemistry, lot, capacity, voltage, storage time, temperature, pulse waveform, resistance or impedance, capacitor, cutoff, recovery, result, owner, and retest trigger.

7. Check yourself

Why is nameplate energy 18.0 Wh?
Answer: 5 Ah multiplied by 3.6 V equals 18.0 Wh.
Why does 30 Ω signal failure?
Answer: 0.220 A×30 Ω=6.60 V, more sag than a 3.6 V cell can provide; the simple model has crossed its physical supply limit.
Is a negative resistor-model terminal a real battery voltage?
Answer: No. It is a failure flag: the demanded pulse cannot be supplied by the cell alone.
Honesty boundary.

The worked values are traceable chapter examples or explicitly labelled teaching assumptions.

5 Ah and 3.6 V
Explicit chapter cell case
0.7% per year
Catalog-typical storage assumption
5–30 Ω
Illustrative fresh-to-passivated range

Correct, not complete: field evidence still decides acceptance.

Return to M2M Case Studies with the maths unpacked →