Infer internal resistance and audit source power

Establish the model and evidence

A source’s emf is energy supplied per unit charge, measured in volts despite the historical word “force.” A simple nonideal source model puts an ideal emf in series with internal resistance r. During discharge into an external load, terminal voltage is Vterminal=emf−Ir. It equals emf only at zero current in this model, not at every operating point.

Plot terminal voltage against current for several loads. The vertical intercept gives emf and the negative slope gives internal resistance. A changing load varies the operating point; it does not by itself mean the source’s internal resistance changed. Real battery chemistry, temperature and history can make the constant-r model fail, so the synthetic straight line is a model testbed rather than a universal battery characterization.

Current is emf/(R+r). The source converts energy at rate emf×I, while the external load receives I²R and the internal resistance dissipates I²r. All three terms belong in the power ledger. Reducing R increases current, but terminal voltage falls and a larger fraction of converted energy becomes internal heat.

Maximum external power and maximum efficiency are different goals. For the constant-r model, external power peaks when R=r, where only half the source power reaches the load. Larger R improves the fraction delivered but lowers current. This is an inference within a bounded model, not advice to stress-test a battery; no short circuit, heating trial or real source experiment is authorized by this lesson.

Worked evidence

With emf=6 V, r=2 Ω and R=8 Ω, I=0.60 A and terminal voltage=4.8 V. Source power is 3.6 W, load power 2.88 W and internal loss 0.72 W. The delivered fraction is 80%.

Predict and investigate

If external load resistance decreases, can current rise while terminal voltage falls?

At r=2 Ω, compare loads 2, 8 and 20 Ω. Fit the terminal line from two points and verify the full power ledger for each. Separate the maximum-power observation from the efficiency trend.

Transfer to changed conditions

A 9 V emf with r=3 Ω powers a 6 Ω load. Find current, terminal voltage, internal loss and external power.

Compare the worked transfer

I=1 A, terminal voltage=6 V, internal loss=3 W and external power=6 W. The ideal emf supplies 9 W, closing the ledger.

Explain the boundary

What new evidence would challenge the assumption that one constant internal resistance explains all operating points?

Sources and authorship

Physical relationship reference. Current AP Physics 2 framework, checked 30 October 2026. Original cases and computed/synthetic evidence, not actual laboratory measurements. These activities are virtual only: do not reproduce wiring, short-circuit, battery charging, heating or charged-capacitor experiments. Any real practical work requires qualified supervision and an independently approved safety procedure.