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

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.
16 min read 7 figures Two routes to a numberThe constants decideEvery model has an edge

Every solve on this site returns twice. Once as node voltages and branch currents, and once as a pair of numbers that must agree: the power dissipated in the resistors, computed as |i|²R and never touching a node voltage, against the power delivered by the sources, computed as Re(v·conj(i)) and never touching a resistance. That second pair is the check described in what a network answers — the one that constrains the voltages against the currents, which current law alone leaves open.

It has been running on every figure in this collection since the first one, and its output has been thrown away every time. The number it computes is the real power.

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. 1 A twenty-ohm, fifty-millihenry load on a 230 V supply at 50 Hz, drawn as the right-angled triangle its three powers make. The horizontal side is what the load consumes; the vertical side is what flows out to the inductor and back again each cycle; the hypotenuse is what the cable and the transformer are sized for. The slider is the inductance, and the third route to the vertical side — 2ω times the energy stored in the inductor — is computed on every frame and required to agree.

What the imaginary half is

A solve at s = jω returns complex node voltages and complex branch currents. The product S = V·conj(I) is therefore complex too, and both of its parts mean something.

The real part is the average power over a cycle: energy that goes into the circuit and does not come back, because a resistor converts it to heat. The imaginary part is energy that goes in during one part of the cycle and comes out during another, because a reactance stores rather than dissipates. Averaged over a whole cycle it is exactly zero, and that is why it can be dismissed. It should not be, because averaged over a cycle is not the same as not there: the current that carries it is in the wire the whole time.

For the load in the figure the numbers are 1,636 W of real power and 1,285 var of reactive. The apparent power — the product of the root-mean-square voltage and the root-mean-square current, which is what the two meters on a switchboard read — is 2,080 VA, and it is the hypotenuse of those two because S is a complex number and |S| is its magnitude.

The ratio of the first to the third is the power factor: 0.786. What it says, in the most concrete form available, is that the cable carries 9.044 A and 7.113 A of it does something.

Phasors have been carrying this since the second essay

Nothing new is being introduced here. Three voltages that close draws a series circuit’s voltages as arrows in the complex plane and shows that they sum to the source not because their magnitudes add — they do not, and the sum of the magnitudes exceeds the source by a wide margin — but because their directions differ. The angle between the current and the voltage in that picture is the whole of the power factor.

Three element voltages closing on one source, at 1.59 kHzSolved at 1.59 kHz, which is 1.00× the frequency at which the two reactances cancel. The three phasors add head to tail to the 1 V source exactly; their magnitudes sum to 5.26 V, which is not the same statement.realimaginaryacross Racross Lacross Cthe source, 1 Vmagnitudes|v_R| = 1.000 V|v_L| = 2.128 V|v_C| = 2.128 Vsum 5.255 Vvector sum 1.000 Vsolved, then checked — one solve at 1.59 kHzsteady state only: 3 cycles to settle
Fig. 2 The same construction the frequency field opens with. The three element voltages of a series circuit are arrows in the complex plane, and the reason their magnitudes do not add to the source’s is that two of them are at right angles to the third. That right angle is the reactive power’s right angle in the triangle above; the two pictures are the same fact drawn on two different axes.

The angle in the triangle is 38.15°, and cos 38.15° = 0.786. A reader who has the phasor picture already has this one; what is added is a convention about which product to take and a name for each part of the answer.

That convention is worth stating explicitly, because it is the first place on this site where the absolute size of a quantity matters. Everywhere else the answers are ratios — a Bode magnitude is the same number whichever scaling produced it — and a factor of two hiding in a definition would cancel out. Here it would not. Phasors on this page are root-mean-square, so a 230 V source means 230 V RMS and S = V·conj(I) with no factor of a half. The alternative convention, peak phasors and S = ½V·conj(I), is equally correct and puts a half in front of every product in the file, which is one more place for it to go missing.

Three routes, and the third touches no impedance

The reactive power is computed three ways in the figure above and they are required to agree.

The first sums V·conj(I) over the sources. The second sums the same product over the elements — |i|²R for a resistor, jωL|i|² for an inductor, −j|v|²ωC for a capacitor — and never looks at a source. Those two agreeing is a real check, and it is a stronger one on the imaginary half than on the real: real power balance is very nearly guaranteed by current law alone, while reactive balance requires the energy stored in every reactance to be consistent with the voltages across it.

The third route is the interesting one. Reactive power is also 2ω(W_L − W_C), where W_L is the energy stored in the inductor and W_C the energy stored in the capacitor. That expression contains no impedance at all — only stored energy and a frequency — and it is what makes “reactive” a physical statement rather than the name of an imaginary part. It agrees with the other two to a part in 10⁹, which is the tolerance the figure asserts.

A quantity computed three ways by routes that share no arithmetic is a quantity that is very unlikely to be wrong, and that is the whole reason for going to the trouble.

There is one detail in those three that is easy to pass over and is load-bearing. The sign of the reactive power is not a convention chosen for convenience; it falls out of the conjugate. Taking S = V·conj(I) puts a lagging current — one that peaks after the voltage, which is what an inductor produces — into the positive imaginary direction, and a leading one into the negative. The opposite convention exists in the machines literature, and the entire content of the words leading and lagging is which of the two is in use. It matters here because the correction capacitor computed later takes Q to zero by having the opposite sign, and a routine that silently took a magnitude would size it correctly and be unable to report the one condition worth reporting: an installation that has overshot and is now leading.

The element-by-element route is also where a subtle assembly error would show. A sign reversed in one stamp gives a smooth, plausible response and passes every visual inspection. It does not pass this: the sources and the elements would disagree about where the energy went, and the disagreement is returned as a number rather than hidden behind a magnitude.

A 60 Ω, 50 mH load on 230 V at 50 Hzcomputed by solving, not by drawing. The load draws 825 W and 216 var, an apparent power of 853 VA at a power factor of 0.967. The reactive side is confirmed by a route that touches no impedance: 2ω times the energy stored in the inductor gives 216 var. The cable carries 3.71 A and only 3.59 A of it does anything.025025050050075075010001000real power (watts)reactive power (volt-amperes reactive)825 W216 var853 VAreal power825 Wreactive power216 varapparent power853 VApower factor0.9674angle14.67°line current3.71 A…doing work3.59 Athe load230 V60 Ω50 mHsolved, then checked — three routes to Qonly 0.97 of the current works
Fig. 3 The same solve with a sixty-ohm load rather than twenty. The real power falls because the resistance rose; the reactive power falls too, but not in the same proportion, because the reactance did not change. The result is a load with a better power factor drawing much less useful power — a reminder that the power factor is a ratio and says nothing on its own about how much work is being done.

Why a number with no work in it is expensive

A reactive current does no work and it is not free, for a reason that has nothing to do with economics and everything to do with |i|²R.

The cable between the supply and the load has resistance. The current in that cable is the whole current, 9.044 A, not the 7.113 A that does something, and the loss in the cable is proportional to the square of it. At a power factor of 0.786 the cable dissipates 1/0.786² = 1.62 times as much as it would at unity for the same delivered work — sixty-two per cent more heat, for nothing.

The same argument sizes every piece of equipment between the generator and the load. A transformer rated at 2,080 VA is fully loaded by this circuit while delivering 1,636 W. Its core saturates on volts and its windings heat on amps, and neither of them knows or cares about the angle between them. That is why apparent power has its own unit: a volt-ampere is not a watt, and equipment is rated in the first.

The size of the effect is worth holding on to, because the power factor is a number between zero and one and small-looking departures from one are not small. A power factor of 0.9 — which sounds close enough to unity to ignore — costs 1/0.9² = 1.23, so twenty-three per cent more heat in every conductor carrying it. At 0.7 the figure is a factor of two. The loss goes as the inverse square of the power factor, so the penalty accelerates exactly where an installation is least able to absorb it, and a supply designed with a comfortable margin at 0.95 has none at all at 0.7.

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. 4 The impedance the supply sees, swept over frequency. The power factor is nothing more than the cosine of this impedance’s angle, so the two figures carry the same information — and this one makes visible what the triangle cannot, which is that the angle is a function of frequency. A load with a power factor of 0.786 has it at one frequency.

The cost of the fix, and the fact that it has one

The reactive power can be cancelled, and this is the part of the subject where the arithmetic is unusually clean. A capacitor across the supply carries reactive power of the opposite sign, −ωC|V|², so the capacitance that takes Q to zero is Q/(ω|V|²) — one division, no iteration. For the load above that is 77.3 µF, and adding it takes the power factor to 1.000000 and leaves 0.0 var of a 1,636 VA apparent power.

The real power does not change by a part in 10⁹, which is the point: the load still does exactly what it did, and the cable now carries 7.113 A instead of 9.044 A.

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. 5 The same capacitor measured against every load it was not sized for. It is exact at the twenty ohms it was computed from and the power factor there is 1.000000; a few times that resistance and the installation sits at 0.23 leading, which is worse than the 0.79 it started at. The correction is a model with a very sharp edge, and the edge is a load rather than a frequency.

That figure is the argument of the capacitor that was right once and is left here as a warning rather than developed. What matters at this point is only that the fix exists and is exact where it was computed.

The other thing that is not free

There is a second trade in this area, and it is usually taught a chapter away from the first: the load that draws the most power from a source is not the load that uses it most efficiently, and the two cannot both be had.

The load that takes the most power, and the load that wastes the leastcomputed by solving, not by drawing at 71 load values. The power into the load peaks at a ratio of 1.000000, where the efficiency is exactly 0.500000 — the source dissipates as much as the load receives. Ninety per cent efficiency needs a ratio of 9.7 and delivers 34% of the available power.00.250.500.751100m110load resistance ÷ source resistancefraction of the maximumpower into the loadefficiencymatched: most power, half of it wasted90% efficient at 9.7×solved, then checked — the maximum searched, not quotedthe matched load wastes exactly half
Fig. 6 Power into the load and efficiency out of the source, against the ratio of the two resistances. The maximum is found by golden-section search on the solved network rather than by quoting R_L = R_S, which is the claim being checked; it comes out at a ratio of 1.00000001 and an efficiency of exactly 0.50000000. Every load larger than that gives up power to gain efficiency.

The peak is at a ratio of one, and the efficiency there is exactly one half: the source burns as much as the load receives. Nine times the source resistance gives ninety per cent efficiency and thirty-six per cent of the available power. A tenth of it gives nine per cent efficiency and thirty-three per cent of the power — the same power, wasted eleven times over.

Which of those is wanted is a question about the application. A radio receiver’s front end wants the first, because what it is short of is signal and it has power to spare. A power station wants the second, because it is selling the thing that gets wasted. The figure’s only claim is that the choice is real and that the arithmetic is not a matter of opinion: the maximum was searched for, not asserted, and the efficiency at it came out as a half to eight decimal places.

Where the phasor picture gives out

Everything above assumes one frequency. That assumption is invisible because it is built into the word phasor, and it is worth making visible, because for a large fraction of what is plugged into a wall socket it is false.

A phasor represents a sinusoid. If the current is not a sinusoid it has no single phasor, the angle between voltage and current is not defined, and cos φ is a calculation performed on a quantity that does not exist. What does still exist is the ratio of real power to apparent power — that only needs the two root-mean-square values and an average, none of which cares about shape — and the two numbers part company completely.

For a load conducting only near the peaks of the supply, which is what a rectifier with a reservoir capacitor does, the displacement factor is exactly 1.000000 and the true power factor is 0.780. A phasor calculation on that load reports a perfect result. The meter reports a cable carrying forty-three per cent more current than the real power requires. Both are right about what they measure, and only one of them is measuring the thing that decides the cable.

That is three phases and the wire that carries nothing and the field’s third essay between them; it is named here so that the triangle above is not read as a general description of power. It is a description of power at one frequency, in a circuit whose current is a sinusoid, and those conditions are stated rather than assumed.

Where four of this site's models stop being trueIn order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is.101001k10k100k1M10M100M1G10Gfrequency (hertz)the ideal operational amplifier1.42 kHz — a gain of 100 from a 1 MHz part is 1% low herea 10 V output at full amplitude7.96 kHz — above this the output cannot move fast enoughthe ideal 100 nF capacitor4.69 MHz — 1.2 nH of lead makes it 10% wrong hereKirchhoff's laws on 10.0 cm3.97 MHz — the board is one degree long hereeach bar is where the model may be used; the rule at its end is the numbersolved, then checked — each boundary from its own modeland one that is not a frequency: 7.3 mV
Fig. 7 The boundaries this collection has measured, on one frequency axis. The power triangle’s own boundary is not on it, because it is not a frequency: it is the condition that the current be a sinusoid, and there is no frequency at which that starts to fail. It either holds or it does not, and for most modern loads it does not.

What was actually added

Very little machinery, which is the pleasant part. The solver already returned complex voltages and complex currents; the verifier already summed one of the two products. Real, reactive and apparent power are read out of quantities that were being computed anyway and discarded.

What is new is a convention — root-mean-square phasors, so that S = V·conj(I) — and the observation that a check written to catch assembly errors was, incidentally, computing the quantity a utility bills for.

The site’s rule applies to this field like any other, and here the edge is not a frequency but a shape. Every number on this page is exact for a sinusoidal current at 50 Hz and means nothing for a current that is not one. That is a strong condition and it is stated on every figure, because the alternative — a power factor quoted with no indication of what it assumes — is how an installation ends up with a correction capacitor that makes things worse.

Where this goes next

realimaginaryacross Racross Lacross Cthe source, 1 Vmagnitudes|v_R| = 1.000 V|v_L| = 2.128 V|v_C| = 2.128 Vsum 5.255 Vvector sum 1.000 Vsolved, then checked — one solve at 1.59 kHzsteady state only: 3 cycles to settle Three voltages that close on one, and the steady state they assume Kirchhoff's voltage law drawn as a polygon in the complex plane. The three element voltages of a series circuit add head to tail to the source exactly — while their magnitudes add to five times it. And the whole picture is a statement about a settled circuit, which takes a computable number of cycles to arrive. 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 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. a bridge, which no series-parallel reduction reachesnode a7.5566 Vnode b4.7993 Vcurrent law, rebuilt from the element laws2.71e-16 of the largest branch currentpower delivered against power dissipated4.33e-16 apart · 48.07 mWsolved, then checked — 6 elementsa linear network has no edge: this one is exact What a network answers, and how the answer is checked A circuit has exactly one answer and a matrix finds it. The part that matters is not that the answer exists but that it can be checked twice, by routes that share no arithmetic — and that a circuit with no answer is refused by name rather than returned as a large plausible number.