Circuit Response
every model has an edge, and the figure carries it
Power, and the part that does no work

The capacitor that was right once

Cancelling a load's reactive power needs one division and no iteration, and the answer is exact. It is exact for the load it was computed from, at the frequency it was computed at, and the figure here is about what it does everywhere else — which includes making the installation worse than it was before anything was fitted.
16 min read 7 figures Every model has an edgeThe constants decideRefused, not extrapolated

A capacitor across a supply carries reactive power of −ωC|V|². A load carries some reactive power Q. Setting the sum to zero gives C = Q/(ω|V|²), and that is the whole of power factor correction: one division, no search, and an answer that is exact rather than approximately right.

Checking it takes one solve. For the twenty-ohm, fifty-millihenry load of the current that does no work, which draws 1,636 W and 1,285 var on a 230 V supply at 50 Hz, the capacitance is 77.311 µF. Adding it and solving again gives a power factor of 1.00000000, a reactive power of 0.0 var, and a real power unchanged to the last digit the arithmetic carries. Nothing about that is an approximation.

One capacitor of 77.3 µF, against every load it was not sized forcomputed by solving, not by drawing. Sized from the 20 Ω load, the capacitor takes the power factor to 1.000000 there and leaves 0.0e+0 var of 1636 VA. At 178 Ω the same installation sits at 0.23 leading. The correction is exact at one point on this axis and nowhere else on it.00.250.500.75110100load resistance (ohms)power factorno capacitorsized here: pf 1.000000leading above 21 Ωthe 0.95 an installation is usually required to holdsolved, then checked — 61 loads, one capacitorexact at 20 Ω, 0.23 elsewhere
Fig. 1 The same 77.3 µF capacitor measured against sixty-one loads, only one of which it was sized for. On the design load the power factor is one; on the lightest load on the axis it is 0.23 leading, which is worse than the 0.79 lagging the installation started with. The dashed line is the uncorrected power factor, and the region where the corrected curve sits below it is the region where the fix is doing harm. The slider moves the load the capacitor was sized for.

What “exact” is exact about

The claim the calculation supports is narrow, and every word in it is doing work: this capacitance cancels this load’s reactive power at this frequency.

Change any of the three and the cancellation fails, and it fails in a way that has no restoring tendency. A capacitor does not know what it is cancelling. Its reactive power depends on the voltage across it and the frequency, both of which are held nearly constant by the supply, so it delivers the same 1,285 var whatever the load is doing. When the motor is unloaded and drawing a third of the reactive power, the capacitor is still delivering all of it, and the excess flows back into the supply.

That is the whole of overcorrection, and it is visible in the figure as the region where the two curves cross. The corrected installation is not gently less good outside its design point; it is on the other side of unity, with reactive power of the opposite sign, and it gets worse in exactly the same way an uncorrected one does — as the inverse square of the power factor in every conductor.

A 20 Ω, 50 mH load on 230 V at 50 Hzcomputed 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.0500500100010001500150020002000real power (watts)reactive power (volt-amperes reactive)1636 W1285 var2080 VAreal power1636 Wreactive power1285 varapparent power2080 VApower factor0.7864angle38.15°line current9.04 A…doing work7.11 Athe load230 V20 Ω50 mHsolved, then checked — three routes to Qonly 0.79 of the current works
Fig. 2 The load before correction, as a triangle. The vertical side is what the capacitor is sized to cancel. What the sizing calculation cannot see is that this triangle is drawn for one operating condition: a motor at a quarter of its rated load has a much smaller real power and a barely smaller reactive one, so the triangle changes shape rather than scaling.

Why the shape changes rather than scaling

The reason overcorrection is a practical problem rather than a curiosity is in how an inductive load actually varies.

A motor’s magnetising current is set by the voltage and the machine’s inductance, and it is very nearly the same whether the motor is doing work or spinning free. The current that does work is the one that varies with the load. So as the mechanical load falls, the real power falls a long way and the reactive power hardly falls at all — the triangle grows thinner rather than shorter, and the power factor of an unloaded motor is famously bad for exactly this reason.

Modelled here as a rising resistance in series with a fixed inductance, that behaviour comes out directly: at twenty ohms the load draws 1,636 W and 1,285 var, at eighty ohms it draws 637 W and about a fifth less reactive power. The capacitor is unchanged. The mismatch between what it delivers and what is wanted is therefore largest exactly when the installation is lightest — overnight, at weekends, whenever the plant is idling — and that is when the supply sees a leading power factor and a rising voltage.

One capacitor of 9.7 µF, against every load it was not sized forcomputed by solving, not by drawing. Sized from the 70 Ω load, the capacitor takes the power factor to 1.000000 there and leaves 0.0e+0 var of 719 VA. At 624 Ω the same installation sits at 0.47 leading. The correction is exact at one point on this axis and nowhere else on it.00.250.500.751100load resistance (ohms)power factorno capacitorsized here: pf 1.000000leading above 73 Ωthe 0.95 an installation is usually required to holdsolved, then checked — 61 loads, one capacitorexact at 70 Ω, 0.47 elsewhere
Fig. 3 The same figure with the capacitor sized for a much lighter load. It is a smaller capacitance, so it overcorrects less at the light end and undercorrects at the heavy end. There is no setting that is right everywhere, which is why real installations switch capacitance in and out in steps and why the control that does the switching is the expensive part.

The resonance nobody asked for

There is a second consequence, and it is the one that does damage rather than merely wasting current.

A correction capacitor is not alone across the supply. The supply itself has inductance — the transformer’s leakage inductance and the cable’s, which is small but not zero — and a capacitance across an inductance is a resonant circuit. It has a frequency, and nothing in the sizing calculation looked at it.

A resonant circuit of Q = 8, and its measured bandwidthThe half-power points are 1.50 kHz and 1.69 kHz, a bandwidth of 198.9 Hz. The components predict f₀/Q = 198.9 Hz. They differ by 0.000%.00.200.400.600.8011001k10kfrequency (hertz)fraction of the source across the resistorhalf the power198.9 Hz measuredresonance 1.59 kHzsolved, then checked — half-power points by bisectionf₀/Q predicts 198.9 Hz — exactly
Fig. 4 A series resonance, and the width of the band around it. The correction calculation is performed at 50 Hz and the resonance formed by the correction capacitor and the supply’s own inductance sits several hundred hertz higher, which is empty as far as the calculation is concerned. It is not empty in practice: it is where the harmonics are.

For plausible numbers — a 77 µF capacitor and a few hundred microhenries of supply inductance — the resonance lands in the region of a few hundred hertz, which is to say somewhere between the fifth and the eleventh harmonic of a 50 Hz supply. Those harmonics are not hypothetical. Every rectifier on the installation produces them, and the figure in the next section shows what a rectifier’s current actually looks like.

At resonance, a circuit’s currents can greatly exceed the driving current — that is what the quality factor measures, and resonance and its bandwidth makes the point that the half-power width is exactly f₀/Q. A modest harmonic current at a frequency the installation happens to resonate at becomes a large circulating current between the capacitor and the transformer, and the capacitor is the part that fails.

This is the field’s clearest instance of the site’s rule. The correction calculation is exact and it is exact about a quantity — the reactive power at 50 Hz — that is not the quantity that decides whether the installation survives.

Impedance of a series RLC of Q = 4, measured by driving itOne ampere is forced into the terminals at each frequency and the resulting voltage is the impedance. The minimum is 7.91 Ω at 5.03 kHz.1101001k10k1001k10k100k1Mfrequency (hertz)impedance magnitude (ohms)reactances cancel at 5.03 kHz7.91 Ωsolved, then checked — one ampere in, 201 frequenciesnot a component value: what the pair does
Fig. 5 The impedance a supply presents, swept. Below the resonance it is inductive and above it capacitive, and at the resonance itself it collapses — which is the same statement as “a small harmonic voltage drives a large harmonic current”. The correction calculation is a single number read off the extreme left of an axis like this one, and everything that happens further along it is outside the calculation’s field of view.

The arithmetic is worth doing once explicitly, because the frequency it lands on is not a matter of opinion. A distribution transformer’s leakage inductance seen from its secondary is of the order of a hundred microhenries for a small installation. With 77 µF across it, ω₀ = 1/√(LC) gives about 11.4 krad/s, which is 1.8 kHz — the thirty-sixth harmonic, comfortably above anything a rectifier produces in quantity. Scale the installation up, though: ten times the capacitance for ten times the load, on a transformer with a tenth the leakage inductance, and the frequency stays put. Scale it the other way — a large bank on a weak supply, which is the case that actually gets built — and it falls as the square root of the product. A bank of 800 µF on 400 µH resonates at 281 Hz, which is the fifth and sixth harmonic of a 50 Hz supply and precisely where rectifier current lives.

The engineered answer to that is a detuning reactor: a small inductance deliberately put in series with each capacitor to move the resonance below the lowest harmonic present, usually to about 189 Hz for a 50 Hz supply so that the fifth harmonic at 250 Hz sees an inductive rather than a capacitive circuit. The reactor is typically around seven per cent of the capacitor’s reactance and it costs real money and real losses. That is the price of the fact this essay is about: the sizing calculation is exact, and the component it sizes cannot be installed on its own.

Why the exactness is the trap

There is a general pattern here worth separating from the electrical detail, because it recurs across this collection and it is the reason this field’s second essay is about a fix rather than about a problem.

An approximate answer carries its own warning. A curve fitted to five points invites the question of what happens at the sixth; a series truncated at three terms invites the question of the fourth. The error is visible in the form of the answer, so the reader knows to ask.

An exact answer carries no such warning, and the exactness is doing something quite specific: it is telling the truth about the model. C = Q/(ω|V|²) is not an approximation to the capacitance that cancels the reactive power of a fixed load at a fixed frequency; it is that capacitance, and the solved network agrees to a part in 10⁹. Every digit of that agreement is a statement about the model and not one word of it is a statement about the installation.

The same shape appears elsewhere on this site and is worth collecting. Resonance and its bandwidth finds that the half-power width of a series resonance is exactly f₀/Q at every quality factor tested — and that the band is not centred on the resonance, which the usual picture draws as though it were. The exact relation and the false one sit side by side, and the exact one gives no clue that its neighbour is wrong. The quarter-wave transformer in the lines field reflects 5×10⁻¹⁷ of the incident wave at the frequency it was cut for, and something quite different a few per cent either side.

An exact model with an unstated domain is more dangerous than an approximate one with a stated error bar, and this field supplies the cleanest example of it in the collection.

The current that made the harmonics

A load conducting for 60° either side of each peakcomputed by solving, not by drawing over 1024 samples of one cycle. The current's fundamental is exactly in phase with the voltage, so the displacement factor — the cos φ a phasor calculation returns — is 1.000000. The true power factor is 0.7803, and the difference is the distortion factor 0.7803: the same 100 W drawn as 0.558 A rather than the 0.435 A a sinusoid would need.supply voltagecurrent drawnone cycle of the mainsharmonics of the current, against the fundamental11.0002—30.6654—50.1986—70.1448—90.222displacement (cos φ): 1.000000distortion: 0.7803true power factor: 0.7803solved, then checked — Parseval on the currentcos φ says 1.000, the meter says 0.780
Fig. 6 A load drawing current only near the peaks of the supply, which is what a rectifier with a reservoir capacitor does. The fundamental of that current is exactly in phase with the voltage, so the displacement factor — the cos φ a phasor calculation returns — is 1.000000. The true power factor, real power over apparent power, is 0.780. The bars are the harmonic content that accounts for the difference, and they are what the resonance above amplifies.

The measured displacement factor for that waveform is 1.000000 and the true power factor is 0.7803, and the two are related exactly: the true power factor is the displacement factor times the distortion factor, and the identity holds to the arithmetic on every waveform tested.

What that means for correction is unusually stark. A capacitor cancels reactive power, which is a displacement effect. This load has no displacement to cancel — its cos φ is already one — so a correction capacitor does nothing for its power factor at all, while still forming a resonance with the supply and still amplifying the harmonics that made the problem in the first place.

An installation that measures cos φ, finds it acceptable, and concludes that no correction is needed has reached the right conclusion by an argument that would have been just as confident if it had been wrong. An installation that measures cos φ, finds it poor because of a motor, fits a capacitor, and has rectifiers on the same board has fitted a resonant circuit to a harmonic source.

What a correction calculation cannot be asked

It is worth being precise about what the sizing routine on this site refuses to do, because the refusal is the honest part.

Given a real power, a reactive power, a voltage and a frequency, it returns a capacitance. It does not take a load model, because it does not need one — and that is exactly why it cannot warn about anything. Everything that goes wrong above goes wrong because the four numbers it was given describe one operating point of a system that has many, and the routine has no way of knowing that.

A 10 kΩ + 10 kΩ divider, solved with its loadThe unloaded answer is 6.00 V. It is 1% low at a load of 495 kΩ and 4.00 V at a load equal to the divider's own resistance. The ratio does not predict any of this; the magnitude does.02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ
Fig. 7 The argument in its simplest form, from the first field. A divider’s ratio is a property of the divider and of what is across it, so two dividers of identical ratio give 5.970 V and 1.000 V into the same load. A correction capacitor is the same mistake at a larger scale: a component chosen from a calculation that did not include everything it would be connected to.

The general shape is the one the divider and its load sets out. A component’s effect is a property of the component and of its surroundings, and a calculation that takes only the component’s own parameters gives an answer that is right about the component and silent about the installation.

It is worth naming what would have to be supplied for the routine to be able to warn about any of this, because the list is short and none of it is exotic.

The range of the load, not one point of it. Two solves rather than one — heaviest and lightest — would put both ends of the figure above in front of whoever is choosing the capacitance, and would turn “0.23 leading at light load” from a discovery into an input.

The supply’s inductance. One number, available from the transformer’s nameplate as a percentage impedance, and enough to place the resonance to within a factor of two. Without it the resonant frequency is not merely unknown; it is not even a quantity the calculation has a symbol for.

The shape of the current. Not a phasor but a waveform, or at minimum the amplitude of the harmonics present. That is what decides whether the resonance matters, and it is the one input a 50 Hz measurement cannot produce at all.

None of those three is difficult to obtain and none of them is in the formula, and that gap is the essay. A calculation is not made careless by being simple; it is made careless by returning a number of the same apparent authority whether or not the conditions it assumed are present.

What holds and what does not

Three statements, in decreasing order of how far they can be trusted.

The capacitance is exact. C = Q/(ω|V|²) takes the reactive power to zero and the solved network confirms it to a part in 10⁹, with the real power unchanged to the last digit. There is no approximation in it.

It is exact at one point. The same capacitor across a load of three or four times the resistance gives a power factor below 0.5 leading, and the figure sweeps sixty-one loads to show the shape of the failure rather than asserting that one exists. The measured worst case on the default axis is 0.23.

It cannot see the resonance or the harmonics. Both of those need information the sizing calculation was never given: the supply’s inductance, and the shape of the current rather than its phasor. Neither is available from the four numbers that produce the capacitance, and neither shows up in any check performed on the result.

The unusual thing about this corner of the subject is how clean the wrong answer looks. There is no residual to inspect, no iteration that fails to converge, no tolerance to widen. The number is exact, the meter agrees with it on the day of commissioning, and everything that goes wrong afterwards goes wrong somewhere the calculation never looked.

Where this goes next

0500500100010001500150020002000real power (watts)reactive power (volt-amperes reactive)1636 W1285 var2080 VAreal power1636 Wreactive power1285 varapparent power2080 VApower factor0.7864angle38.15°line current9.04 A…doing work7.11 Athe load230 V20 Ω50 mHsolved, then checked — three routes to Qonly 0.79 of the current works The current that does no work A solved network has been reporting its own real power on every page of this collection, as the second of the two checks each answer passes before it is drawn. What that check discards is the imaginary half — the power that flows out to a reactance and back again, does nothing, and is still carried by the cable, still heats the transformer, and is still on the bill. 00.200.400.600.8011001k10kfrequency (hertz)fraction of the source across the resistorhalf the power198.9 Hz measuredresonance 1.59 kHzsolved, then checked — half-power points by bisectionf₀/Q predicts 198.9 Hz — exactly Resonance, and the bandwidth it sets exactly The half-power bandwidth of a resonant circuit is f₀/Q — not approximately, but to every digit the arithmetic has, which is rare enough to be worth checking. What is not exact, and is drawn as though it were, is the idea that the band sits centred on the resonance. At a quality factor of one its middle is twelve per cent above. 02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ The divider, and the thing it does not know about A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.