Power, and the part that does no work

The energy a unity power factor doubles

Reactive power was computed three ways on this site and the agreement was called a verification. Two of the three are one theorem written twice and cannot disagree about anything; only the third is independent, and what it computes is a difference. Correcting a 20 Ω, 50 mH load to a power factor of 1.000000 takes its reactive power from 1,285 var to nothing and takes the energy stored in the installation from 2,044.9 mJ to 4,089.7 mJ — exactly twice, at every load and every frequency.

Assumes: The current that does no work · The capacitor that was right once · The inductor that is an amplifier

The current that does no work computed one load’s reactive power three separate ways and said that a quantity computed three ways by routes sharing no arithmetic is very unlikely to be wrong. That is true. It is also not what those three routes establish, and the difference matters, because two of them agree for a reason that has nothing to do with the circuit being right.

The first sums S=VIS = V\cdot\overline{I} over the sources. The second sums the same product over the elements. The third is Q=2ω(WLWC)Q = 2\omega(W_L - W_C) — twice the frequency times the difference of the energies stored in the inductances and the capacitances — and it names no impedance and no source at all.

The first two cannot disagree. Not for this circuit: for any circuit whatever, correctly assembled or not, made of any elements at all. That leaves one independent route, and the interesting thing about it is that it computes a difference of two quantities that the other two never see separately.

Two things follow, and they point in opposite directions. The demonstration is weaker than it looked, because a three-way agreement in which two of the routes are one theorem is a two-way agreement wearing a third hat. And the third route is stronger than it looked, because a route that measures something the others cannot see is not a check at all — it is an instrument, and it has been pointed at nothing.

What follows points it at three things: a load being corrected, a circuit at resonance, and an inductance made of an amplifier. The first two say what the discarded sum is. The third is where the route stops agreeing with the other two, and is not wrong to.

Correcting a 20 Ω, 50 mH load to unity, and what it storescomputed by solving, not by drawing. A capacitor across a 230 V, 50 Hz supply is swept from nothing to 300 µF against a load drawing 1636 W and 1285 var. The reactive power falls through zero at 77.31 µF and keeps going; the energy stored in the installation rises from 2044.9 mJ to 4089.7 mJ at that point — exactly twice, because unity power factor means the two stores are equal — and goes on rising afterwards. Only the cable current has a least value, 7.113 A against 9.044 A.-20240100200300correction capacitance across the supply (microfarads)as a multiple of the same quantity with no capacitorenergy storedreactive powercable currentunity at 77.31 µFtwice the stored energystored energy, no capacitor2044.9 mJ…at unity power factor4089.7 mJ…which is2.000000×reactive power, no capacitor1285 var…at unity power factor0.0e+0 varcapacitor that does it77.31 µFcable current, before9.044 A…after7.113 Asolved, then checked — 61 capacitors, one loadthe difference is nulled, the sum is doubled
Fig. 1 A capacitor across the supply, swept from nothing to 300 µF against a load drawing 1,636 W and 1,285 var. The reactive power falls through zero at 77.31 µF; the energy stored in the installation rises from 2,044.9 mJ to 4,089.7 mJ there and goes on rising afterwards, because the two curves are not two views of one quantity. Only the cable current has a least value — 7.113 A against 9.044 A. The slider is the load inductance and the doubling holds at every value of it.

The reproduction, which comes before the departure

An instrument that has never been made to agree with something already known is an instrument whose disagreements mean nothing. So the case to start from is the one where all three routes must give the same answer, and where that answer is known independently: a resistor and an inductor in series across a stiff supply, which is the load this whole field is quoted at.

A 20 Ω, 50 mH load on 230 V at 50 Hz. computed by solving, not by drawing. The load draws 1636 W and 1285 var, an apparent power of 2080 VA at a power factor of 0.786. The reactive side is confirmed by a route that touches no impedance: 2ω times the energy stored in the inductor gives 1285 var. The cable carries 9.04 A and only 7.11 A of it does anything.
Fig. 2 Twenty ohms and fifty millihenries on 230 V at 50 Hz, drawn as the triangle its three powers make. The load takes 1,636 W and 1,285 var, an apparent power of 2,080 VA at a power factor of 0.7864 and an angle of 38.15°. The energy route agrees at 1,285 var without naming an impedance anywhere. The cable carries 9.04 A and 7.11 A of it does anything.

Here the three agree to a part in 10⁹, which is the tolerance the figure asserts rather than the tolerance the arithmetic achieves. That agreement is the calibration, and it is worth being precise about what each half of it is evidence for. The first two agreeing says the netlist was assembled correctly. The third agreeing with them says that the inductor in the netlist behaves like an inductor — that the energy in it really is 12Li2\tfrac12 L|i|^2, and that this really is what the terminals report as reactive. Nothing else in the collection makes that second statement, and every number below leans on it.

Two routes that are one theorem

Tellegen’s theorem says that for any network satisfying the two circuit laws, the complex powers absorbed by the elements sum to zero. It uses the topology and nothing else: not what the elements are, not whether the stamps are right, not whether the answer means anything. Given a set of branch voltages consistent with a loop law and a set of branch currents consistent with a node law — of the same graph, not even of the same circuit — the sum is zero.

So the first route and the second route are one statement partitioned two ways. Summing over the sources and summing over the elements are the two halves of a sum that is identically zero, and their agreement is a restatement of the two circuit laws that what a network answers already imposes on every solve. It is worth having: it catches an assembly error, a sign reversed in a stamp, a current unknown attached to the wrong branch. It is a check on the arithmetic. It is not evidence about the physics, because it holds just as exactly for a circuit whose every element value is nonsense.

The distinction is easiest to see in what would have to go wrong for each to fail. To break the first two, the solved voltages and currents would have to violate a circuit law — which is what the solver imposes, so the failure would be in the linear algebra rather than in the model. To break the third, an element would have to store an amount of energy inconsistent with the voltage across it or the current through it, which is a statement about the element and not about the matrix. The two failures live in different places, and only one of them is about the subject.

Where the power is in a 20 Ω, 50 mH load. computed by solving, not by drawing. Each bar is the complex power absorbed by one element, positive to the right. The resistor takes 1635.9 W and no reactive power at all; the inductor takes 1284.8 var and no real power, and the capacitor row is empty because none is fitted. The supply's own entry is -1635.9 W and -1284.8 var, and it is negative in both, which is what delivering looks like. Every column sums to 1.1e-13 VA, which is Tellegen's theorem and is why the two power routes to the reactive half cannot disagree. The stored energy is 2044.9 mJ and appears in no bar.
Fig. 3 The same solve, element by element, with the sign kept. The resistor takes 1,635.9 W and no reactive power; the inductor takes 1,284.8 var and no real power; the supply’s entry is −1,635.9 W and −1,284.8 var, which is what delivering looks like, and the capacitor row is empty because there is no capacitor yet. Every column sums to 1.1 × 10⁻¹³ VA of a 2,080 VA apparent power. The stored energy — 2,044.9 mJ — appears in no bar at all.

The third route is a different kind of statement. It uses the constitutive laws of the reactances: WL=12Li2W_L = \tfrac12 L|i|^2 for an inductor and WC=12Cv2W_C = \tfrac12 C|v|^2 for a capacitor, with the root-mean-square phasors this field uses throughout. Those are claims about what an inductor is. A network of resistors that has been mis-stamped as inductors passes Tellegen and fails this.

That is the whole reason the third route is worth running, and the reason the site’s habit of demanding two routes is not proof against a shared assumption — the same trap the assumption that is a geometry found in a winding, where three independent computations of an alternating-current resistance turned out to be one sentence written three times.

The difference, and the sum

Q=2ω(WLWC)Q = 2\omega(W_L - W_C) is a difference. Nothing in either power route reads WL+WCW_L + W_C, and nothing in the collection had until now.

A correction capacitor drives the difference to zero. It cannot touch the sum, and the arithmetic of why is short enough to state. The capacitor sits across the supply, so the load branch still sees 230 V and its current is unchanged: WLW_L is fixed at 2,044.9 mJ for the whole sweep. The capacitor’s own store is 12CV2\tfrac12 C|V|^2, which is 26.45 mJ for every microfarad added, exactly linear, and it only ever grows. At the point where WC=WLW_C = W_L the difference is zero and the power factor is one — and the sum is therefore exactly twice what it was, not approximately.

That identity does not depend on the load. It falls out of the definition of the correction: the capacitance chosen is Q/(ωV2)Q/(\omega|V|^2), so its stored energy is 12Q/ω\tfrac12 Q/\omega, and QQ was 2ωWL2\omega W_L. Twenty millihenries gives 1,204 mJ becoming 2,407; a hundred and twenty gives 1,743 becoming 3,486. The generator asserts the ratio at 2.000000000 on every frame of the slider rather than checking it once.

The capacitance that does it is not monotonic in the load, which is worth a line because it is the kind of thing a formula hides. Twenty millihenries wants 45.51 µF, fifty wants 77.31, eighty wants 77.55, and a hundred and twenty wants 65.89 — the requirement peaks near the inductance whose reactance equals the resistance and falls away on both sides, because past that point the load’s own reactance throttles the current that produces the reactive power in the first place. A designer sizing a bank from a nameplate sees a single number and not a maximum.

The cable current is the one quantity of the three with a genuine optimum. It falls from 9.044 A to 7.113 A at the correction point and rises again afterwards, reaching 17.593 A at 300 µF — nearly twice the uncorrected figure, in copper sized for the uncorrected figure. That is what the correction buys, and it is bought against the sum rather than for nothing: less current in the cable, more energy in the installation.

So the corrected installation is a resonant circuit at the supply frequency, which is exactly what the reactance cancelled, and the resonance it buys is about in the series position and what the capacitor that was right once prices in the shunt one. Cancelling a reactance and building a resonator are the same act described twice, and only one of the two descriptions makes it obvious that something is now storing twice as much energy and exchanging it internally rather than with the supply.

The supply, taking it back

Past the correction point the difference changes sign rather than getting smaller, and the element-by-element reading keeps the sign rather than taking a magnitude, so the condition is visible in the solve instead of having to be inferred from it.

Where the power is, with 150 µF across the supply. computed by solving, not by drawing. Each bar is the complex power absorbed by one element, positive to the right. The resistor takes 1635.9 W and no reactive power at all; the inductor takes 1284.8 var and no real power, and the capacitor -2492.9 var. The supply's own entry is -1635.9 W and 1208.0 var, and it is POSITIVE in the reactive column: the supply is absorbing reactive power from an installation that has been over-corrected. Every column sums to 1.1e-13 VA, which is Tellegen's theorem and is why the two power routes to the reactive half cannot disagree. The stored energy is 6012.4 mJ and appears in no bar.
Fig. 4 The same load with 150 µF fitted rather than 77.31. The capacitor now carries −2,492.9 var against the inductor’s +1,284.8, and the supply’s own bar has gone positive: it absorbs 1,208.0 var. The power factor is 0.8044 leading, which is as bad as the 0.7864 lagging the installation started from, and the stored energy has reached 6,012.4 mJ. The columns still sum to 1.1 × 10⁻¹³ VA, because Tellegen does not care which way the energy is going.

A meter that reads a magnitude reports 0.8044 with the 150 µF fitted and 0.7864 with none, and cannot say that the first of those is leading and the second lagging. That is the fault an installation carries for years, and it is now three separate readings rather than one: the power factor is no worse, the cable current is worse, and the stored energy is worse still — 2.94 times the uncorrected figure at 150 µF, rising to 4.88 at 300.

The third of those has a consequence the first two do not. A supply absorbing reactive power is a supply whose own impedance sits across a capacitance large enough to have resonated with something, and the energy available to that resonance is the sum rather than the difference. The difference reads 1,208 var and the sum reads 6,012.4 mJ, and it is the second that decides how much there is to ring with when a contactor opens. Nothing in a power-factor calculation contains that number.

What the discarded sum turns out to be

The sum is not an abstraction. Divide it by the power dissipated and multiply by the frequency and it is the quality factor.

A 20 Ω series resonance, and the number the power meter cannot reach. computed by solving, not by drawing at 85 frequencies. 20 Ω, 50 mH and 202.6 µF resonate at 50 Hz, and there the reactive power is -7.5e-5 var — the two stores exchange with each other rather than with the supply. The quantity both power routes discard, ω(W_L + W_C) per watt dissipated, does not go to zero with it: it reaches a floor of 0.785398 at 50.0000 Hz, found by golden section, and that floor is ωL/R = 0.785398. The model behind every number here is one sinusoid at one frequency.
Fig. 5 Twenty ohms, fifty millihenries and 202.6 µF in series, swept from 10 Hz to 250 Hz. The falling and rising curve is the reactive power divided by the real power, which reaches zero at resonance — the two stores exchange with each other and the supply sees a resistor. The other curve is ω(WL+WC)\omega(W_L + W_C) per watt dissipated, and it does not go to zero with it: it bottoms out at 0.785398, located by golden section at 50.0000 Hz, and that floor is ωL/R = 0.785398. The slider is the series resistance, and at every value the two agree to nine figures.

That is the standard energy definition of QQ arriving from the power field rather than from the frequency one, and it is worth noticing which of the three routes it came out of. A power meter on that circuit at resonance reads 2,645 W, 0 var, power factor 1.000000, and has no access whatever to the 6,612.5 mJ the inductor and the capacitor are passing back and forth between them. Whether that matters is the subject of resonance, and the bandwidth it sets exactly and of the Q the components allow; the point here is only that the number is in the solve and that two of the three routes to reactive power throw it away by construction.

The minimum is at resonance and not at either end, and the reason is a race between the two stores: below resonance the capacitor holds most of it and above it the inductor does, and the crossover is where the total per watt is least. At 20 Ω the floor is 0.785398, at 5 Ω 3.141593 and at 2 Ω 7.853982 — π/4, π and 5π/2, since ω₀L is 5π for the parts drawn. The generator finds it by golden section and then checks the located frequency against the one the components predict, rather than evaluating at 50 Hz and asserting that the answer is a minimum.

Where the third route stops agreeing

A gyrator is an inductance made of resistors, a capacitor and two amplifiers. The inductor that is an amplifier measures it as an inductance and says in passing that nothing in it stores energy in a magnetic field. That sentence has a number, and the third route is what reads it.

A one-henry inductor with nothing magnetic in it, good for 3.6 decades. computed by solving, not by drawing. The impedance at the input of an Antoniou impedance converter, read as an inductance: a current source drives the node and the voltage is solved for, and the imaginary part divided by ω is what is plotted. Five components — four resistors of 10 kΩ and a 10 nF capacitor — behave as 1000 mH, which as a wound coil would be several henries of wire. It is that inductance to within one per cent from 1.00 Hz to 3.65 kHz, 3.56 decades, and the upper edge belongs to the amplifiers rather than to the arrangement: with ideal ones in the same netlist the inductance is exact everywhere drawn. Nothing in it stores energy in a magnetic field — the current lags because an amplifier is holding a capacitor's voltage somewhere else in the loop.
Fig. 6 The arrangement from the rung below, as an impedance. Four 10 kΩ resistors and a 10 nF capacitor behave as 1,000 mH at the input node, to within one per cent from 1.00 Hz to 3.65 kHz. Every terminal measurement says inductor. The upper edge belongs to the amplifiers rather than to the arrangement: with ideal ones in the same netlist the inductance is exact everywhere drawn.

Drive that node and read the reactive power at it, then read the reactive power the stored energy accounts for. With ideal amplifiers the two are the same size and opposite in sign, to nine figures, at every frequency in the band.

The magnitudes matching is the surprising half. It is not obvious that a capacitance held at the right voltage by an amplifier should store exactly what an inductor of the designed value carrying the terminal current would need — the arrangement was derived to produce an impedance, and an impedance is a ratio of two phasors that says nothing about energy. It comes out at 1.000000000 anyway, which says the equivalence is stronger than the derivation claimed: the synthetic inductor is not merely an impedance imitation but an energy-accurate one, in the wrong element and with the wrong sign.

A 1000 mH inductance whose energy is all in a capacitor. computed by solving, not by drawing at 140 frequencies. Four 10 kΩ resistors, a 10 nF capacitor and two amplifiers present an inductance at one node. With ideal amplifiers the energy stored in that capacitor is 1.000000000 times the energy a real inductor of the same value carrying the same terminal current would store — the same to nine figures, everywhere drawn — and the reactive power computed from it is -4.973e-4 var against 4.973e-4 var at the terminals: equal, and of the opposite sign. With amplifiers of 1 MHz gain–bandwidth the ratio is 7.31e-4 J against 2.24e-7 J, 3272 times over, because the compensation inside them stores far more than the inductance does.
Fig. 7 The energy the arrangement actually stores, divided by the energy a real 1,000 mH inductor carrying the same terminal current would need. With ideal amplifiers it is 1.000000000 across the whole sweep — every joule the inductance requires is present, and it is in the capacitor. The reactive power computed from that store is −4.973 × 10⁻⁴ var against +4.973 × 10⁻⁴ var at the terminals: the same number with the sign reversed, because the energy is in the wrong kind of element. The slider is the amplifiers’ gain–bandwidth.

Both facts are exact rather than approximate, and neither is available to a power meter. An instrument at the terminals reports an inductance absorbing reactive power. The energy route reports a capacitance of exactly the right size delivering it. Both are right; they are answering different questions, and in a passive one-port the two questions have the same answer because there is nothing in the circuit that can move energy between the two kinds of store without holding it.

An amplifier can. That is what the two controlled sources in the arrangement do, and it is why the third route is the only one of the three that carries information: Tellegen locks the first two together whatever is in the network, while the energy route asks each element what it is and gets a different answer here. The same asymmetry underlies the reading that does not care which way round — reciprocity is a property of the elements, and the elements that break it are exactly these.

The mechanism is worth stating as a theorem rather than as an observation, because it says exactly which circuits are exempt. For a resistor, an inductor or a capacitor, the reactive power absorbed is 2ω2\omega times that element’s own store with a fixed sign — zero, plus, minus. Sum those over a passive one-port and Tellegen turns the terminal reactive power into 2ω(WLWC)2\omega(W_L - W_C) with no further assumption: the two routes coincide because every element in the sum obeys the same relation between what it absorbs and what it holds. A controlled source obeys no such relation. What it absorbs is set by a gain and a voltage somewhere else, and it holds nothing, so it enters Tellegen’s sum and not the energy sum. Every circuit on this site containing a nullor, a gyrator or a controlled source is therefore a circuit where the two quantities are free to differ — and where the difference between them is a measurement rather than an error.

What is inside the amplifiers

Ideal amplifiers are a limit. Real ones have a dominant pole, and a dominant pole is a capacitor.

How little of a synthetic inductor's energy is the inductance. computed by solving, not by drawing at 169 gain–bandwidths, 1000 Hz throughout. The energy an Antoniou arrangement actually stores is mostly in the compensation that gives its amplifiers their dominant pole, and a slower amplifier has more of it. At 1 MHz parts the inductance accounts for 2.218e-2 per cent of 5.552e-5 J. The share rises in proportion to the gain–bandwidth — an exponent of 1.0022 fitted over four decades, with the worst point 8.19 per cent off it — and it never reaches one, so no choice of parts makes the arrangement store only what the inductance needs.
Fig. 8 The share of the arrangement’s stored energy that the inductance it imitates accounts for, against the gain–bandwidth of the parts, read at 1 kHz. With 1 MHz amplifiers it is 0.02218 per cent: the inductance needs 1.231 × 10⁻⁸ J and the parts hold 5.552 × 10⁻⁵ J, four and a half thousand times over. The share rises in proportion to the gain–bandwidth — an exponent of 1.0022 fitted over four decades, worst point 8.19 per cent off the fit — and never reaches one at any value drawn.

The compensation that makes an amplifier stable is the largest energy store in the circuit by three or four orders of magnitude, and it is a store nobody put on the schematic. The ideal amplifier, and where it stops being one prices that pole in gain and phase; this is the same pole priced in joules, and the price is much larger than the thing the circuit was built to make. A slower amplifier costs more of it, in exact proportion, which is the opposite of the direction the bandwidth argument pushes and is one more reason one inductor, and ten components is a trade rather than a substitution.

None of this makes the gyrator a bad inductor. It makes the phrase stores energy in a magnetic field a measurement rather than a description, and gives it a number.

What it does not say

It does not say the first two routes are worthless. They catch the errors they were written to catch, they cost nothing, and a solve that fails them is broken beyond interpretation. What they cannot do is corroborate the third, and quoting three agreeing routes as three independent confirmations overstates the evidence by exactly one.

It does not say reactive power is the wrong quantity. It is the right one for the thing it decides — the current in the cable, the rating of the transformer, the bill — and none of those cares how much energy is stored anywhere. The sum matters for different questions: how long a circuit takes to settle, how sharply it selects, how much a capacitor bank rings when something upstream switches, which is the pair that is worse than either arriving at its own frequency.

And the sign reversal at the gyrator is not a failure of the energy route. It is the honest report of a circuit where the terminal behaviour and the stored energy have genuinely come apart, and it can only happen where a controlled source is present. Nor is the ideal-amplifier case a convenience: a nullor is the element that constrains its input to zero volts and zero amps and lets its output take whatever the rest of the network needs, so a netlist built with nullors is the arrangement with the amplifiers’ own limits removed and nothing else changed. It is the calibration for the second half of the argument in the same way the passive load was for the first, and the exact 1.000000000 it returns is what entitles the 3,272 to be quoted as a property of the parts. In a network of resistors, inductors and capacitors the three routes agree, always, and their agreement is then a check on all three.

Every number above assumes one sinusoid at one frequency. A current that is not a sinusoid has no single phasor, WLW_L and WCW_C are not single numbers either, and the whole apparatus stops applying — a limit every model has an edge states in general and which this field’s own rectifier essays measure.

The number worth carrying

Two thousand and forty-four millijoules becoming four thousand and eighty-nine, exactly, at the point where the reactive power reads zero.

The habit with it is shorter, and it is two habits rather than one.

When a quantity is defined as a difference, find out what the sum does before treating the difference as a description of the circuit. A null in a difference is a statement about a balance and says nothing whatever about the size of the things being balanced, which is why unity power factor and twice the stored energy are the same operating point.

And when several routes to one number agree, count how many of them could have disagreed. Two routes locked together by a theorem about topology are one route with two names, however different their arithmetic looks — the same reading a winding’s three alternating-resistance calculations turned out to deserve. Here the count was one, and the one that could disagree was also the only one that ever did.

Part 2 on reactive power

One argument about Reactive power, and one of 2 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

The objects named here

The third axis, after the field and the idea: the things themselves, and every essay that touches each one.

Complex powerEnergy-storageGyratorNullorPower factor correctionQuality factorReactive powerSeries resonanceSynthetic inductorVerification