Primer: Rosemary Ainslie's "over-unity" heater circuit

Companion files: arbiter.md (final verdicts), physicist.md, mathematician.md, contrarian.md, example.py (reproduces every number in Section 5).

Source gap (repeated from claims.md and arbiter.md). The October 2002 Quantum article (5 pages), the Pretoria News clipping (1 page) and the patent bundle "Harnessing a back EMF" (43 pages; priority application ZA 98/0541 filed 22 January 1998; bundle dated 19 January 1999; WO 99/38247 filed 22 January 1999 [read: patent p.1 and p.38 OCR; WO date from Kelly's digest]) were OCR'd and read in full, and the patent's Examples 1-3 (OCR pp.19-32) were read as page images. The remaining gaps: the folder does not contain the raw data (the oscilloscope waveforms and the spreadsheet behind the 1.13 W figure), the company reports (Sasol, BP and Power Engineers, and the ABB / Cape Hope Metrology Laboratory calibration records), or any independent replication. So everything below about the experiment rests on the authors' own published tables plus arithmetic, and everything about the "independent tests" rests on the author's own report.

Source tags: [read] = from the documents in the folder; [web] = internet source with URL; [memory] = recalled, with confidence.

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1. The idea in one paragraph

Rosemary Ainslie, a South African researcher, built a small battery-powered heater: a coil-shaped resistor that is switched on and off very quickly by a transistor. Each time the switch opens, the coil's magnetic field collapses and pushes a burst of current back around the circuit (the "back EMF" or "inductive kick"). She and her co-authors measured the heat coming out of the resistor (about 17.5 watts, by comparing its temperature with a DC-heated control) and the power going in from the battery (about 1.13 watts, from an oscilloscope reading across a small resistor). That ratio, about 15 to 17, would mean the resistor gave off far more energy than the battery supplied. She explained this with her own "magnetic field model", in which electric current is a flow of tiny magnetic particles and the usual energy limits do not apply. The reviews found that the heat measurement is believable but the battery measurement almost certainly is not: ordinary circuit theory says this circuit cannot produce that much heat without the battery supplying about the same amount, and the input was measured in a way that is very easy to get wrong. The decisive check (count the charge leaving the battery) was never done.

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2. What you need to know first

Ohm's law and Joule heating. A resistor R carrying current I has voltage V = IR across it and turns electrical energy into heat at the rate P = I²R = V²/R (watts = joules per second). For a current that changes in time, the average heating is R times the average of I², not R times (average I)². Read: Halliday, Resnick and Walker, Fundamentals of Physics, the chapters on current, resistance and circuits; or Horowitz and Hill, The Art of Electronics, chapter 1.

Inductance and back EMF. A coil with inductance L (unit: henry, H) resists changes in its current: the voltage across it is v = L dI/dt. If you try to stop the current suddenly, dI/dt is huge and so is the voltage; this is the "back EMF" (EMF = electromotive force, an old word for a voltage produced by something other than a static charge). A coil carrying current I stores energy E = ½LI² in its magnetic field. That energy has to be put in by whatever drove the current, and it comes back out when the current falls. Read: Griffiths, Introduction to Electrodynamics, chapter 7 (Faraday's law, inductance, energy in magnetic fields).

RL circuits (first-order ODEs). A resistor R and inductor L in series driven by a battery V obey V = IR + L dI/dt. The solution is an exponential approach to V/R with time constant τ = L/R. If you remove the battery and give the current a closed loop, it decays as e^(−t/τ). This is the only differential equation in this primer. Any introductory physics text covers it (Halliday/Resnick "RL circuits").

Switching circuits and the flyback (freewheel) diode. In switching power supplies, a transistor turns a supply on and off thousands to millions of times per second. The fraction of each cycle that the switch is on is the duty cycle D (0 to 1). A diode placed across an inductive load lets the coil's current keep circulating when the switch opens, instead of producing a destructive voltage spike. This is called a flyback or freewheel diode and is in every relay driver and buck converter. A MOSFET (metal-oxide-semiconductor field-effect transistor) is the usual switch; it also contains a built-in "body diode" that conducts backward when its drain is pulled below its source. Read: Horowitz and Hill, The Art of Electronics; Erickson and Maksimović, Fundamentals of Power Electronics (the buck converter and "inductor volt-second balance").

Conservation of energy (first law) and the second law. The first law of thermodynamics says energy is neither created nor destroyed: in a circuit, the heat produced must equal the energy delivered by the sources (minus any energy still stored). The second law is a different statement, about entropy (heat cannot flow spontaneously from cold to hot; no engine converts heat entirely into work). For electromagnetic systems, the first law takes the specific form of Poynting's theorem (Griffiths, chapter 8). Read: Schroeder, An Introduction to Thermal Physics, early chapters.

Measuring power with an oscilloscope and a shunt. To measure current with an oscilloscope you put a small known resistor (a "shunt") in the current path and measure the voltage across it: I = V_shunt / R_shunt. At high frequencies the shunt's own small inductance, the probe's ground lead and the scope's zero offset can all add voltages that are not I × R. Read: any oscilloscope application note on current measurement; Horowitz and Hill on grounding and probes.

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3. The textbook view

The circuit [read: 2009 reprint, Figure 1 and Table 1]

• BAT1: two 12 V, 20 ampere-hour lead-acid batteries in series, V = 24 V.
• RL: the load, a 10 Ω wire-wound resistor on a hollow ceramic former, 48 turns, chosen for its inductance L = 8.64 μH (microhenry = 10⁻⁶ H).
• Q1: an IRFPG50 N-channel MOSFET (the switch), between the load and the battery's negative side.
• Rshunt: four 1 Ω, 2 W carbon resistors in parallel = 0.25 Ω, in series with the battery's negative terminal.
• D1: a 1N4007 rectifier diode across the load (the flyback diode).
• A 555 timer chip drives the MOSFET gate; set to 2.4 kHz at 3.7% duty. The 2002 Quantum article also discloses a separate 12 V battery powering the gate drive; the 2009 reprint omits it [read: page image Quantum p.3, p.5; checked by mathematician and contrarian].

Symbols used below: V = 24 V battery voltage; R = 10 Ω load resistance; R_s = 0.25 Ω shunt; L = 8.64 μH; f = switching frequency (Hz); D = duty cycle; t_on = D/f, t_off = (1−D)/f; I = current in the load.

What the circuit does, cycle by cycle

Switch on. Current flows battery(+) → load → MOSFET → shunt → battery(−). Kirchhoff's voltage law:

V = I (R + R_s) + L dI/dt,

so the current rises toward I_∞ = V/(R + R_s) = 24/10.25 = 2.34 A with time constant τ₁ = L/(R + R_s) = 0.84 μs.

Switch off. The battery and shunt are disconnected. The coil keeps the current flowing round the small loop load → D1 → load:

0 = I R + L dI/dt (ideal diode, no voltage drop),

so the current decays with τ₂ = L/R = 0.86 μs.

Two facts follow directly, and they are the heart of the textbook view:

1. In the ideal circuit the battery current is never negative. When the switch is off, the coil's returning current circulates through D1 and the load; it does not pass through the battery or the shunt. Current can be pushed back into the battery only through the MOSFET's body diode, which requires the drain to ring below the source (parasitic capacitance and stray inductance); that is a small, second-order effect.
2. Energy balance. Over one steady cycle the coil starts and ends with the same current, so its stored energy ½LI² returns to the same value. Therefore (energy from battery) = (heat in load) + (heat in shunt) + (heat in MOSFET and diode). The back-EMF energy is real, but it is energy the battery put into the coil during the on phase. It is recycled, not created.

A useful shortcut: the buck-converter rule

When the switching period is short compared with τ (so the current barely changes during a cycle), the load sees an average voltage D × V, and

P_load ≈ (D V)² / R, P_battery ≈ V × (D × I_avg) = (D V)²/R (the same number).

This is the "volt-second balance" rule of every buck converter (the circuit that turns 12 V into 5 V in a phone charger). When the period is long compared with τ (e.g. 2.4 kHz, period 417 μs ≫ 0.86 μs), the coil hardly matters and P ≈ D V²/(R + R_s) × R/(R + R_s).

How the heat was measured [read: 2009 reprint, "Measurement of energy dissipated at the load"]

A platinum temperature probe was fixed to the resistor; the room was draught-free. The resistor reached 52 °C above ambient. Then the same resistor was put on a DC supply and the voltage adjusted until it again reached 52 °C above ambient: 13.32 V, so P = V²/R = 13.32²/10 = 17.74 W. This "caloric control" method is sound in principle: it avoids every high-frequency measurement problem by comparing like with like. Its weaknesses are that it was "crude" by the authors' own word (they allow 10%), and that the table converts other temperatures to watts by assuming a straight-line law (0.341 W per kelvin) over a range where convective cooling is not exactly linear [read; mathematician, contrarian].

How the battery power was measured [read: 2009 reprint, "Measurement of energy delivered by the battery"]

The scope sampled the voltage across the 0.25 Ω shunt (time base 1 μs/div; the Quantum text says a 1.2 μs sample range), computed I = V_shunt/R_s for each sample, multiplied by a fixed V = 24 V, treated positive samples as power out of the battery and negative samples as power returned, and averaged:

P_battery = (1/x) Σ V × I_n, n = 1 … x.

Result: 1.13 W. The textbook problem with this is that an average over a window shorter than (or not an integer number of) switching cycles is not a cycle average, and the answer is a small difference of positive and negative samples, so any zero offset or inductive pickup in the shunt signal changes it a lot (Section 5, step 5).

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4. Where this researcher departs

The exact point of departure is the second row. The experiment itself is ordinary; the departure is the claim that the coil's returning current is a net source. Note also the logical slip in the fourth row: the experimental claim violates the first law (energy conservation), but the argument offered is about the second law. Even if the second-law argument were right, it would not license creating energy.

Fairness point: the design is not foolish. Ainslie published parts lists with supplier codes, used a calorimetric method precisely to avoid inductive phase errors, disclosed the gate battery in 2002, and openly invited replication and disproof [read: Quantum p.2-3; contrarian]. That is how a testable claim should be presented.

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5. Worked example: what should this circuit do, and do the published numbers add up?

All numbers here are reproduced by example.py.

Step 1. Check the paper's own energy total

Table 2 [read: 2009 reprint] lists, for each time, the load "Control Watts" and a "Joules" figure. Divide each Joules entry by elapsed seconds:

• 29 min: 30,333 J / 1,740 s = 17.43 W, which is exactly that row's watts.
• 997 min: 1,222,429 J / 59,820 s = 20.44 W, exactly the last row's watts.

So "Joules" is (watts at that moment) × (total time so far), not an integral. The paper's stated average is 17.53 W, and

E_load = 17.53 W × 59,820 s = 1.049 MJ (megajoules), not 1.22 MJ.

(A trapezoid integral of the watts column gives 1.024 MJ, close.) Battery energy at the reported 1.13 W: 1.13 × 59,820 = 67.6 kJ. Ratios:

• Paper's figures: 1.222 MJ / 67.6 kJ = 18.1.
• Corrected: 1.049 MJ / 67.6 kJ = 15.5.
• Adding the 12 V gate battery (12.45 V × 0.039 A = 0.486 W, 29.0 kJ): 1.049 MJ / 96.6 kJ = 10.85 (the arbiter rounds to 10.8).

So the published "greater than 16" does not follow from the authors' own watts column. This does not settle anything physical, but it is a real arithmetic error, and it is why the arbiter labels the published energy figure "Internally inconsistent".

Step 2. Solve the switched RL circuit exactly

From Section 3, with a = e^(−t_on/τ₁) and b = e^(−t_off/τ₂), the requirement that the current at the end of a cycle equals the current at the start gives (solved with SymPy in example.py):

I_peak = I_∞ (1 − a) / (1 − a b), I_valley = b × I_peak.

Then integrate I (for battery power, on phase only) and I² (for load heat, both phases) over one cycle. Results:

(The arbiter's table gives 18.0 W of load heat at 500 kHz; my 18.3 W uses an ideal diode loop without the shunt. The conclusion is the same.) In every row battery power = load heat + shunt heat, to machine precision; the script asserts this, and an independent time-stepping simulation agrees.

Why the 1.3% rows are so small: the on-time is 0.013/170,000 Hz = 76 ns, much shorter than τ = 0.84 μs, so the current only climbs to about V t_on / L = 24 × 76×10⁻⁹ / 8.64×10⁻⁶ ≈ 0.21 A before the switch opens.

Reading. No stated operating point gives "1.13 W in, 17.5 W out". The stated 2.4 kHz setting predicts about 2 W in and 2 W of heat. A 55% duty, which is what Kelly reports, predicts about 18 W of heat, matching the measured heat, but also about 18 W from the battery.

Step 3. What duty gives the measured 17.74 W?

Buck-converter rule: (D V)²/R = 17.74 W → D V = √(17.74 × 10) = 13.32 V → D = 0.555.

Notice that D V equals 13.32 V, the very DC voltage the authors used in their calibration. That is no coincidence: in the smooth-current limit, a switched supply at duty D heats the resistor exactly like a DC supply of D × V. The full model at 500 kHz needs D = 0.54 for 17.74 W, and then the battery supplies 18.0 W. So the measured heat is exactly what ordinary physics predicts for the circuit running at about 55% effective duty, the duty Kelly describes.

Step 4. The back-EMF energy bound

The most energy the coil can ever hold is at the largest possible current, I = V/(R + R_s) = 2.34 A:

E = ½ L I² = ½ × 8.64×10⁻⁶ × 2.34² = 23.7 μJ.

Even if every bit of it were released every cycle at 200 kHz: 23.7 μJ × 200,000 /s = 4.7 W, and this energy was taken from the battery first. Back EMF cannot be the source of 16 extra watts.

Step 5. How fragile is the input measurement?

• Mean battery current at 1.13 W: 1.13/24 = 47 mA, so the mean shunt voltage is 0.047 × 0.25 = 11.8 mV, the quantity everything rests on.
• Sensitivity: P_battery = V × V_shunt / R_s, so each 1 mV of error in the average shunt reading shifts the answer by 24 × 0.001 / 0.25 = 0.096 W.
• To make a true 17.7 W read as 1.13 W, the average shunt reading would have to be wrong by (17.7 − 1.13)/96 ≈ 0.17 V. That is large for a scope's DC offset, but the shunt is four carbon resistors with leads, at hundreds of kHz, with spikes; a stray inductance of 20 nH and a 10 ns current edge gives L dI/dt ≈ 4.8 V, eight times the full resistive signal (mathematician). Over a whole number of cycles that inductive voltage averages to zero, but over a 1.2 μs or 10 μs window of an irregular waveform it need not. Which error actually occurred cannot be determined without the raw waveforms; this step shows only that the method could not resolve the answer, not what went wrong.
• Battery check: if 17.53 W really came from the battery, it would draw 0.73 A for 16.6 h = 12.1 ampere-hours, about 60% of the 20 Ah pair. At 1.13 W it would be 0.78 Ah. The difference is easily measured with a coulomb counter or by discharging the batteries to a cutoff afterwards. The authors left out the capacity comparison, citing "too many vagaries" [read]. Battery terminal voltage (24.3-25.1 V in the table) does not distinguish these cases: lead-acid voltage under light load changes only slowly with state of charge [memory: high].

Step 6. A check that supports the authors

The gate battery was reported to supply 0.039 A; how it was measured is not stated (the Quantum text says only that the battery's voltage fall from 12.45 V to 12.04 V "was consistent with" that current) [read: Quantum journal p.8]. A MOSFET gate needs charge Q_g each cycle; for the IRFPG50, Q_g is up to about 190 nC [web: https://www.vishay.com/docs/91254/91254.pdf, via the arbiter's search summary]. At 200 kHz: 190×10⁻⁹ × 200,000 = 0.038 A. So the reported gate current is consistent with the 2002 article's 143-200 kHz oscillation. The parasitic oscillation was probably real; it is the energy accounting that fails.

Ainslie's earlier bench tests (patent Examples 1-3)

Before the 2002 heater, Ainslie's patent application (priority application ZA 98/0541 filed 22 January 1998; bundle dated 19 January 1999; WO 99/38247 filed 22 January 1999) reported three bench tests [read: patent pp.19-32, page images; summarised from arbiter.md "QE revision" and the deep read in deep/]. They are worth working through because each one shows a different way a power comparison can go wrong. Two terms: COP (coefficient of performance) here means "power out divided by power in", so COP > 1 is a claimed gain. A sense resistor is the same thing as a shunt (Section 2). These numbers are checked by hand below; example.py does not reproduce them.

Example 1: a 68 Ω control against a 22 Ω test resistor [read: patent pp.19-22]. - Control: a 68 Ω resistor on a 12.28 V battery reached 95 °C. Battery power: 0.196 A × 12.28 V = 2.406 W. - Test: a 22 Ω resistor in parallel with an inductor, switched by a MOSFET at about 5 kHz (0.2 ms period), no diode. The duty was adjusted until the resistor reached 93 °C. Mean battery current 0.1718 A, so battery power = 0.1718 A × 12.28 V = 2.1097 W. - The patent then adds a 5% error allowance to the test side only: 2.1097 × 1.05 = 2.215 W, and COP = 2.406 / 2.215 = 1.086. (Without the allowance, 2.406 / 2.1097 = 1.14.) - What is wrong: the method treats equal temperature as equal power, but it compares two different resistors (68 Ω and 22 Ω) at a single temperature reading, with no calorimetry and no stated room temperature. A resistor's temperature depends on power and on how well it sheds heat (its size, winding, mounting and probe contact), so equal temperature means equal power only for the same part in the same mounting. Also, the 8.6% "gain" is no larger than the error of reading a 5-19 mV sloped waveform by eye on a scope. (There is also an arithmetic slip, "0.043 V" for 0.0043 V, that does not change the final current.) - Arbiter verdict: Never properly tested (a single-point temperature match; the 8.6% sits inside reading error). Net gain: Refuted by calculation (by energy conservation the resistor's heat cannot exceed what the battery supplies minus the inductor, MOSFET and wiring losses).

Example 2: a 100 Ω control against a 47 Ω test resistor [read: patent pp.22-29]. - Control: 100 Ω on 6.12 V (two 6 V batteries in parallel): 0.0612 A, 6.12 × 0.0612 = 0.374 W, resistor at 30 °C (room 22 °C). - Test: a 47 Ω resistor in series with an inductor, diode and MOSFET at 50% duty, also reaching 30 °C. Three 0.05 Ω sense resistors: SR1 on the load, SR2 on battery 1, SR3 on battery 2. - Battery 2's shunt (SR3) shows peaks of about ±5.5 mV, i.e. 0.0055 / 0.05 = ±0.11 A, the largest current signal in the circuit. The author "discounted" it as too small. - Counting battery 1 only: 0.025 A × 6.12 V = 0.153 W. Against the control's 0.374 W, that is 0.374 / 0.153 = 2.45 (the patent also states it as 0.0612 A / 0.025 A, calls it "100 percent greater than unity", and elsewhere gives about 10% gain). - What is wrong: dropping battery 2 drops part of the input; the three shunt currents as reported do not obey Kirchhoff's current law, and the discarded SR3 average decides whether the ratio is above or below 1. The load "rms" is taken as peak/2, which is the mean of a triangle wave, not its rms (peak/√3). The patent also argues that battery voltage × load average current is the load's power; for a resistor the power is I_rms² R, so that step is wrong. And again, a 47 Ω part and a 100 Ω part at the same temperature need not be dissipating the same power. - Arbiter verdict: Internally inconsistent (rms slip, incompatible efficiency figures, Kirchhoff violation); "V × I_avg is the load power": Refuted by calculation. Net gain: Refuted by calculation.

Example 3: duty-cycle table, 90% down to 50% [read: patent pp.30-32, Table 1 on page image p.32]. - Circuit: a 12 V battery, a 0.05 Ω sense resistor S1, a 6.3 Ω 50 W load resistor R1, and a 350 VA coupled inductor whose second winding returns current through a diode. S1 and R1 are in series, so they carry the same current, which reverses during the off phase (current flows back into the battery). - Method: battery power = V × I_avg (net average current through S1); load power = V_rms × I_rms across R1. The printed ratios (load / battery) are 1.176, 1.148, 1.033, 1.082 and 0.976 at 90, 80, 70, 60 and 50% duty: COP 0.976 at 50% rising to 1.176 at 90%. - The mechanism, as arbiter.md gives it: this is not an average-versus-rms artefact. Net average current × constant battery voltage is the correct power for the source, and V_rms × I_rms is the correct power for a resistor (voltage and current in phase). The defect is in the measurements. V_rms / I_rms across R1 must equal R1 = 6.3 Ω, but the table gives 4.89, 5.18, 5.24, 4.68 and 2.96 Ω at 90, 80, 70, 60 and 50% duty. Using the known R1 instead (P = V_rms² / 6.3 Ω), the ratios become 0.91, 0.94, 0.86, 0.80 and 0.46: all below 1, as energy conservation requires. In addition, at 50% duty the net battery current is only 1.2 mV across S1 (0.024 A) while its rms is 15.9 mV (I_rms / I_avg = 13), so a 1 mV scope offset would change the battery power by 100%. - Worked check (90% row): V_rms = 10.02 V, I_rms = 2.05 A → 10.02 / 2.05 = 4.89 Ω, not 6.3 Ω. With R1 = 6.3 Ω: 10.02² / 6.3 = 15.94 W, against battery power 1.390 A × 12.57 V = 17.46 W, ratio 0.91. - Arbiter verdict: Internally inconsistent (the table contradicts its own resistor); with the correct R1 the data show no gain. Net gain: Refuted by calculation.

Taken together. The three examples use different supplies, circuits, instruments and efficiency definitions (8.6%; 10% or 145%; 0.976-1.176), so they do not corroborate one another [read; arbiter]. As with the heater, an ampere-hour count on the battery would settle each one.

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6. Evidence table

Verdicts are the arbiter's final verdicts (see arbiter.md).

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7. What a learner should take away

What is solid. - Back EMF is real, and it can be large: a current of 2.34 A falling to zero in 0.1 μs through 8.64 μH gives v = L dI/dt ≈ 200 V. Flyback diodes and body diodes really do carry returning current. - The thermal comparison method (match the temperature with a DC supply) is a good way to measure heat in a messy high-frequency circuit. The 17.7 W of heat is probably roughly right. - The circuit almost certainly did oscillate parasitically at a much higher frequency than the 555 was set to; the gate-current check supports it. Parasitic oscillation in MOSFETs with a large gate resistor is ordinary engineering [memory: high].

What is wrong, and why it is instructive. - The battery-side number. Energy conservation, applied to the stated parts, says the battery must supply what the resistor dissipates (Section 5, steps 2-4). The lesson: when two measurements disagree with a conservation law, first ask which measurement is weakest. Here it is a 12 mV average extracted by subtracting large positive and negative spikes over a few microseconds, and that is the measurement to distrust. - The energy total. A column labelled "Joules" was not an integral of power. Always check that a table's units and operations match its labels. - The theory. An argument about the second law (entropy) cannot rescue a violation of the first law (energy). And "a proton is three electrons" fails on charge before any other test. - The protocol. A self-oscillating circuit tuned by hand "until the effect appears" has no defined operating point, which makes a careful replication impossible to compare against.

What is genuinely open. - What the circuit was actually doing during the 16.5 h run (frequency, duty, gate waveform), and why the resistor heated by another 11 K in the last 37 minutes while battery voltage stayed steady [read: Table 2; physicist]. - What the ABB and BP tests found, especially BP's battery-duration test, and what Stellenbosch actually declined. - Whether any of the 2009 hobbyist reproductions included a coulomb count or calorimeter. - These are open as facts about this history, not as open physics: no result here challenges energy conservation until a test with full energy accounting shows a gain.

A test you could run yourself. Build the circuit (all parts are cheap). Power it from a large capacitor bank instead of batteries and measure the energy taken as ½C(V₁² − V₂²), including the gate supply; measure the resistor's heat by DC matching as the authors did. Or simply put an inexpensive coulomb counter (ampere-hour meter) between battery and circuit. If the heat clearly exceeds the energy drawn, you would have a first-class anomaly. All three reviewers and the arbiter proposed this test.

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8. Check yourself

1. In the ideal circuit (perfect switch, ideal diode, no stray capacitance), can the current through the shunt ever be negative? Why or why not?
2. A coil with L = 8.64 μH carries 2.34 A. How much energy does it store, and what is the most power it could deliver if completely discharged 200,000 times per second? Where did that energy come from?
3. The paper reports 17.53 W average over 997 minutes and 1.22 MJ total. Which is right, and how can you tell what the "Joules" column actually computes?
4. Using the buck-converter rule, what duty cycle on a 24 V supply heats a 10 Ω resistor at 17.74 W, and how much power does the battery then supply?
5. Ainslie's model says a proton is made of three electrons. Give one conservation-law check that rules this out in one line.
6. Patent Example 3 forms a COP from battery power V × I_avg and load power V_rms × I_rms. What, if anything, is wrong with that pair of formulas, and what actually went wrong in the table? Separately, Examples 1 and 2 match a test resistor to a control resistor of a different value at the same temperature. Why does equal temperature not show equal power?

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9. Further reading

Mainstream. - D. J. Griffiths, Introduction to Electrodynamics: chapter 7 (Faraday's law, inductance, magnetic energy) and chapter 8 (conservation laws, Poynting's theorem). - P. Horowitz and W. Hill, The Art of Electronics: inductors, diodes, MOSFET switching and grounding/probing. - R. W. Erickson and D. Maksimović, Fundamentals of Power Electronics: the buck converter, volt-second balance, and why switched inductive circuits obey energy balance. - D. V. Schroeder, An Introduction to Thermal Physics: first and second laws. - D. J. Griffiths, Introduction to Elementary Particles: quark structure of the proton (the evidence against "proton = three electrons"). - Any oscilloscope maker's application note on current measurement with shunts and probe grounding (for the measurement pitfalls in Section 5, step 5).

The researcher's own works (in the project folder). - R. Ainslie and D. Martin, "Counter Electromotive Force Enables Overunity Results in Electric Systems", 2 February 2009 (electric_heater_experiment.pdf), the reprint of the experiment with Table 2. - Quantum magazine, October 2002 (quantum_october_2002.pdf), the original article with the 12 V gate battery and the 143-200 kHz description. - R. Ainslie, magnetic field model (magnetic_field_model.pdf), the "zipon" theory and the proton mass sum. - Patent application "Harnessing a back EMF" (harnessing-a-back-emf-id00000000130311.pdf; priority application ZA 98/0541 filed 22 January 1998; bundle dated 19 January 1999; WO 99/38247 filed 22 January 1999). The body was read (OCR, with Examples 1-3 checked on the page images) and its bench tests are summarised in Section 5, "Ainslie's earlier bench tests (patent Examples 1-3)". Its claimed regime is 50-99% duty (preferably 75-95%) at 40 Hz or more [read, per arbiter.md], the opposite of the 3.7% / 1.3% duty of the heater circuit. - P. Kelly's chapter on the circuit (Ainslie_heater_circuit.pdf), a third-party description with the 500 kHz / 55% tuning notes and the environmental-energy interpretation. - Pretoria News, 14 November 2002 (pretoria_news_2002.pdf), the Stellenbosch refusal and the company list.

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Answers

1. No. When the switch is on, current flows from the battery's positive terminal through the shunt in the forward direction. When it is off, the coil's current circulates through D1 and the load and never reaches the shunt. Negative shunt readings require the MOSFET body diode to conduct during ringing, or else they are measurement artefacts. That is why a measurement method that subtracts large negative samples deserves suspicion.
2. E = ½ × 8.64×10⁻⁶ × 2.34² = 23.7 μJ; × 200,000 /s = 4.7 W at most. That energy came from the battery during the on phase, so it is already counted in the battery's output and adds nothing.
3. 17.53 W × 59,820 s = 1.049 MJ is right. Dividing each "Joules" entry by its elapsed seconds returns that row's own watts (e.g. 1,222,429 / 59,820 = 20.44 W, the last reading), so the column is instantaneous watts × elapsed time, not an integral.
4. D = √(17.74 × 10)/24 = 0.555, and the battery then supplies (D V)²/R = 17.74 W (a little more in practice because of the shunt). Same heat, same input.
5. Charge: three electrons carry −3e; a proton carries +e. (Spin and deep-inelastic scattering also rule it out, but charge alone is enough.)
6. The formulas themselves are a legitimate pair, and arbiter.md corrects an earlier claim that this was an average-versus-rms mix. With a steady battery voltage, V × (net average current) is the battery's power; for a resistor, whose voltage and current are in phase, V_rms × I_rms is its power. The problem is the measurements. V_rms / I_rms across the load must equal its resistance, 6.3 Ω, but the table gives 4.89 Ω at 90% duty, falling to 2.96 Ω at 50%. Using the known resistance (P = V_rms² / 6.3 Ω), the load-to-battery ratios become 0.91 down to 0.46, all below 1. At 50% duty the net battery current is only 1.2 mV on the shunt against 15.9 mV rms, so a 1 mV offset changes the battery power by 100%. On the second question: a resistor's temperature rise equals its power times its thermal resistance (kelvin per watt), which depends on its size, winding, mounting, probe contact and the room. Two different resistors can sit at the same temperature while dissipating different powers. In Example 2 the 47 Ω part reached the same 30 °C as the 100 Ω control while carrying only about 0.17-0.23 W of electrical power against 0.374 W, which points to a different thermal resistance or to heat from nearby parts (inductor, MOSFET, diode), not to extra energy. A fair control is the same resistor in the same mounting, on DC, as in the 2009 heater test.