Math Bridge: Cell Sag and Meter Burden

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Math BridgeEnergy & PowerStruggle-friendly runway

Which voltage drop belongs to the meter, and which belongs to the cell?

Treat the cell resistance and meter shunt as two series resistors carrying the same pulse current.

Battery Bruno, the energy and power guideBattery Bruno guides
The one targetSeparate instrument error from a source that cannot support the pulse.
The chapter caseA 200 mA pulse from a 3.0 V coin cell through a 1 Ω shunt.
What it buys youA pulse measurement that does not blame the meter for the cell's own sag.

A field team faces an unresolved physical question: Which voltage drop belongs to the meter, and which belongs to the cell? They must answer it before changing cell 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 cell resistance. The middle card applies this page's relationship. The green card is meter drop. 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.

Cell resistance changes meter drop An input card leads through the page relationship to the meter drop result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. Series drops add, but their ownership stays separate. Removing the instrument can improve accuracy without making an unsuitable coin cell able to source 200 mA.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for cell resistance is 10.

  2. 2

    Name the relationship. Vshunt=0.200x1=0.200 V Vcell=0.200x10=2.00 V Vtotal=2.20 V Vterm=3.00-2.20=0.80 V

  3. 3

    Substitute the chapter fixture. Set cell resistance to 10. The page ledger gives meter drop as 0.20 V.

  4. 4

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

Predict, then change cell resistance

Try Predict the direction of meter drop. Move one control, calculate, then check your prediction.

10
Chapter baseline
Meter drop

Observe Series drops add, but their ownership stays separate. Removing the instrument can improve accuracy without making an unsuitable coin cell able to source 200 mA. Reset the control to 10 and compare meter drop.

Explain Only cell 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 cell resistance moves. Field effects named in the page's technical boundary stay fixed.

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

1. Start with the physical story

A real battery is not an ideal voltage source. We can model it as an open-circuit voltage in series with an internal resistance. A shunt meter adds another series resistance. During a current pulse, both resistors carry the same current, so both voltage drops use V=IR and then add.

Battery Bruno: Keep the units beside every number. They show whether a result is charge, energy, voltage, time, or a ratio.

2. Name every algebra move

1

Name the pulseUse I=0.200 A for the chapter's radio event.

2

Find meter burdenMultiply I by the 1 Ω shunt.

3

Find native cell sagMultiply I by the cell's internal resistance.

4

Add the dropsThe total drop is I(Rint+Rshunt).

5

Subtract from open circuitVterm=Voc−total drop.

6

Remove only the instrumentWith an ideal ammeter, subtract cell sag but not shunt burden.

3. Reproduce the chapter case

Vshunt=0.200×1=0.200 V
Vcell=0.200×10=2.00 V
Vtotal=2.20 V
Vterm=3.00−2.20=0.80 V

Removing the meter shunt raises the terminal voltage from 0.80 V to 1.00 V. That 0.20 V recovery is real, but the cell's own 2.00 V drop remains and still makes the pulse unusable.

4. Try one real input

TryMove the control and predict which outputs should change before reading them.

Cell resistance
Meter drop
Cell drop
Total drop
Terminal with meter
Terminal without meter
Meter share of drop

ObserveAs cell resistance rises, native sag grows and terminal voltage falls. The meter's 0.20 V burden stays fixed because pulse current and shunt value stay fixed.

ExplainSeries drops add, but their ownership stays separate. Removing the instrument can improve accuracy without making an unsuitable coin cell able to source 200 mA.

Technical boundaries.

This is a small formula ledger, not a complete source qualification.

Internal resistance
Ten ohms is a catalog-typical example, not a fixed CR2032 value.
Linear model
Real cells polarise and current-limit; resistance changes with time and temperature.
Pulse shape
A single resistance does not model recovery between bursts.

Correct, not complete: this model diagnoses dominant sag; it does not qualify a battery without measured pulse tests.

5. Use the result in the design

Measure open-circuit voltage, pulse current, terminal voltage, and shunt burden. Repeat at temperature and state of charge. If possible, compare with a low-burden probe.

6. Record the evidence state

Keep cell chemistry, age, temperature, open-circuit voltage, pulse waveform, shunt value, meter range, and minimum terminal voltage.

7. Check yourself

Why do the two drops add?
Answer: The same pulse current flows through both series resistances.
What voltage returns if the shunt is removed?
Answer: Only I×Rshunt, which is 0.20 V in this case.
Does a perfect ammeter rescue the example cell?
Answer: No. The cell's own 2.00 V sag still leaves only 1.00 V.
Honesty boundary.

The arithmetic uses the chapter's named or clearly labelled catalog-typical inputs.

Internal resistance
Ten ohms is a catalog-typical example, not a fixed CR2032 value.
Linear model
Real cells polarise and current-limit; resistance changes with time and temperature.
Pulse shape
A single resistance does not model recovery between bursts.

Correct, not complete: this model diagnoses dominant sag; it does not qualify a battery without measured pulse tests.

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