Estimate a source model from measurements

Infer a relation, then try to disprove it

A single loaded terminal voltage does not uniquely identify a source’s emf and internal resistance. For example, a 5 V terminal reading at 25 mA could fit emf 6 V with r = 40 Ω, or emf 7 V with r = 80 Ω. Both satisfy V = emf − I r at that one point. Additional conditions help distinguish models.

This lesson uses original synthetic readings, not physical battery observations. The source is assumed approximately steady during the readings. The browser holds emf at 6 V; it can illustrate the candidate below, but is not a general fitting tool and does not collect experimental data.

Fit using two different currents

Use only these two invented observations initially:

Observation	Current	Terminal voltage
A	20.0 mA	5.20 V
B	40.0 mA	4.40 V

For V = emf − I r, the slope of voltage against current is −r. Thus r = (V A − V B)/(I B − I A). Convert current to amperes: r = (5.20 − 4.40)/(0.040 − 0.020) = 40 Ω. Then emf = V A + I A r = 5.20 + 0.020 × 40 = 6.00 V. Substitution into B gives the same emf by construction; that is a consistency check on the fit, not independent validation.

Two identical current settings would give a zero denominator and would not determine the slope. If current rises while voltage also rises, a positive-resistance steady discharge model does not fit that trend. Check polarity, source control behavior, timing and model assumptions before interpreting a negative fitted r as a physical component.

Compare the fitted model with the browser

Set internal resistance to 40 Ω. At R = 200 Ω the model predicts 25 mA and 5 V; at R = 400 Ω it predicts approximately 13.636 mA and 5.455 V. Capture both and verify their positions on the source line. These computed values illustrate the inferred relation; they do not create two more independent measurements.

Observations A and B imply external loads V/I = 260 Ω and 110 Ω, respectively, if each load is purely ohmic. Those exact loads are outside the slider presets but within the same physical relation. Do not force them onto the 200 Ω or 100 Ω presets and call the mismatch an error.

Check a reading withheld from the fit

Before opening the answer, predict terminal voltage at 30.0 mA using the inferred parameters. Then compare with a separate synthetic check reading of 4.79 V. State the sign of observed minus predicted voltage.

Check the source prediction

The model predicts 6.00 − (0.0300)(40) = 4.80 V. The observed-minus-predicted residual is 4.79 − 4.80 = −0.01 V. This is a small discrepancy under the stipulated voltage bounds below; it is not an exact match. If you use this reading to revise the parameters, it becomes fitting evidence and you need another reading for a fresh check.

One successful check does not establish a linear relation at every current or after prolonged discharge. Collect several separated currents, repeat an earlier condition, and record time and temperature to distinguish curvature from drift. No such measurements have been performed by this website.

Attach bounds to the inferred resistance

For this exercise only, assume each voltage reading has a bound of ±0.02 V, and each current reading has a bound of ±0.1 mA. These are stipulated worst-case bounds, not confidence intervals or actual meter specifications. Let endpoint errors vary independently inside them.

The voltage difference then lies between 0.76 and 0.84 V; current difference lies between 0.0198 and 0.0202 A. The positive resistance slope therefore lies in 0.76/0.0202 to 0.84/0.0198, approximately 37.62–42.42 Ω. Quoting exactly 40.000 Ω would hide this uncertainty.

The fitted emf and resistance are correlated because they use the same readings. The resistance interval alone is not a complete uncertainty interval for every predicted voltage; propagate the allowed endpoint readings jointly if you need one. The −0.01 V check residual is smaller than the assumed ±0.02 V check-voltage bound alone, but that observation does not prove the full model correct. Shared calibration error, changing source state and load or meter effects require separate consideration.

Transfer to an unfamiliar source

Another synthetic source reads 8.6 V at 10 mA and 7.8 V at 30 mA. Infer its emf and internal resistance, then predict its voltage at 20 mA without changing the browser’s fixed-emf model to pretend it supports a different source.

Check the source-estimation transfer

r = (8.6 − 7.8)/(0.030 − 0.010) = 40 Ω. Emf = 8.6 + 0.010 × 40 = 9.0 V. At 20 mA the candidate predicts 9.0 − 0.020 × 40 = 8.2 V. This is a hand calculation for a different source; the browser remains a 6 V model.

Record fitted readings, current conversions, inferred parameters, the unused-reading residual, error assumptions and one proposed test that could reject the model. A fitted line identifies a behavior over a tested range; it does not identify battery chemistry, remaining capacity or every cause of voltage sag.