Circuit Response
every model has an edge, and the figure carries it
Concept

Real power — where it appears

The part of a load's power that is dissipated rather than returned, being the average of the instantaneous voltage times the instantaneous current. Summed over a network it must equal what the sources deliver, and requiring that is the only check here that constrains the voltages against the currents.

Named by 15 essays across 3 fields — each of them below, with the objects they name alongside it.

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.

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.

power · Reactive power What a charge through 1 kΩ costs, against how long it is given. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. A step loses 1.00000 of ½CV² — 12.5 µJ here — and that number is the same through 10 Ω and through 100 kΩ, which is what the faint curves show. What the resistance decides is where the fall starts: the loss goes as 2τ/T once the ramp is long against τ = 1 ms, with a fitted exponent of -0.994 and no floor beneath it. The circles are marched; the line through them is a closed form that never sees a netlist.

The half that never arrives

Charging a capacitor from a step loses exactly as much energy as it stores, and the resistance it is lost in does not appear in the answer — the same 12.5 microjoules through ten ohms and through a hundred kilohms, to nine figures. Drive the same network with a ramp instead and the loss falls as two time constants over the ramp, with no floor beneath it at all.

transients · Switching energy The load that takes the most power, and the load that wastes the least. computed by solving, not by drawing at 71 load values. The power into the load peaks at a ratio of 1.000000, which is the magnitude of the source impedance to six figures, and the efficiency there is 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.

The load that takes the most

A load equal to the source resistance takes more power than any other, and it does so at exactly fifty per cent efficiency — the source burns as much as the load receives. Ninety per cent efficiency needs a load nine times the source and delivers 36% of what was available, and a load half the source resistance delivers exactly as much as one twice it.

power · Power transfer Superposition holds for the solution and not for its square. computed by solving, not by drawing, at 81 settings of the second source. Two ten-volt sources reach one hundred-ohm load through a hundred ohms each. Every node voltage and every branch current is the sum of the two single-source solves to 3.8e-16 of itself — superposition, exactly. The power is not: the difference is 2·Re(I₁·conj(I₂))·R to 1.0e-15, and at 0° between the sources and equal size it is 444.4 mW against the 222.2 mW that adding gives. Adding the two powers is within one per cent of the truth only when one source is below 0.00505 of the other. The slider is the angle between them: at 90° the two curves coincide to the last bit, and at 180° the true power falls to zero while the sum does not.

Two solves that add, and the one that does not

Every node voltage and every branch current in a linear network is the sum of the per-source solves, here to the last bit of a double at eighty-one settings. The power is not, and the gap is not a correction: two equal sources in antiphase put nothing at all into a load while adding their powers gives 222 milliwatts, and the sum is within one per cent of the truth only when one source is two hundred times the other.

networks · Superposition Two meters, one current, and neither of them measuring the heat. computed by solving, not by drawing, at 35 conduction angles. The first curve is an average-responding meter: it rectifies, averages and multiplies by 1.1107, which is exactly right for a sinusoid — -7.8e-5% here — and exactly 11.07% high on a square wave, because the error is the ratio of two form factors and contains neither the amplitude nor the frequency. On a rectifier drawing its 100 W in sixty degrees of conduction it is -35.90% low. The second curve is a true-RMS meter that reaches 9 harmonics, which has no shape assumption in it and a bandwidth instead: it returns the root-sum-square of the lines it can see, and is one per cent low below every angle here of conduction. The crest factor at sixty degrees is 1.733, which is inside every instrument's rating — neither meter is failing because the peak is large. One is failing because the shape is not a sinusoid and the other because the spectrum is wider than it is.

What a meter multiplies by

An average-responding meter rectifies, averages and multiplies by 1.1107, which makes it exactly right for a sinusoid and wrong for everything else by the ratio of two form factors — 11.07 per cent high on a square wave and 35.9 per cent low on a rectifier drawing its current in sixty degrees. It is also exactly right at one other waveform, a 145.90 degree conduction angle, which is nobody's sinusoid. Beside it a true-RMS meter that reaches nine harmonics is two per cent low on a square wave and never within one per cent of anything narrower.

power · Rms and average A staircase costs one Nth, computed rather than quoted. computed by solving, not by drawing, marched, with the energies rebuilt from the element laws. The charge is broken into N equal risers, each held for 16 time constants so that it completes. The measured losses are 1.00000, 0.500001, 0.250000, 0.125000, 0.0625000, 0.0312500 of ½CV² — which is 1.000004, 1.000002, 1.000001, 1.000001, 1.000000, 1.000000 times 1/N, so the law is exact to four parts in a million at the worst rather than approximately true. The fitted exponent is -1.00000 and the energy account closes to 2.17e-6 at the worst.

The half a switch keeps

Charging a capacitor from a step loses half the delivered energy whatever the resistance, and a ramp takes the loss down as 2τ/T with no floor. A staircase of N settled risers costs one Nth of the step, exact to four parts in a million, and the law ends at a dwell of 5.272 time constants. A switch is the other half of the same product and buys nothing at all: with the supply held at five volts and the channel conductance ramped over a thousand time constants, the loss is 1.00000000 of ½CV².

transients · Switching energy One rms value, and a peak that moves 41 per cent with a phase. A 10 V fundamental and a 3.33 V 3th harmonic, summed, with the harmonic's relative phase swept. The two are orthogonal over a period, so the root-mean-square value is the quadrature sum 7.45356 V at every phase — to 6.2e-15, which is the one case superposition allows and is a theorem rather than an approximation. The peak is not a sum of anything: it runs from 9.428 V to 13.33 V, a factor of 1.414, against a bound of 13.33 V that it reaches exactly at 180° — where this harmonic's own crest lands on the fundamental's, which for n ≡ 3 (mod 4) is that end of the sweep and not the other. At the far end it flattens the crest to 9.428 V, and there the peak has no closed form at all. So one signal has one heating and a headroom requirement that varies by 41 per cent, decided by a phase that no magnitude spectrum carries.

The peak that only has a bound

Power does not superpose and the gap has a closed form — two times the real part of one current times the conjugate of the other, exact at every setting. A peak has nothing of the kind. A ten-volt fundamental with a third harmonic a third its size has one root-mean-square value, 7.454 volts at every relative phase because the two are orthogonal, and a peak running from 9.428 to 13.333 volts across the same sweep — reaching the sum of the two peaks exactly at 180 degrees, where the harmonic's own crest lands on the fundamental's, and having no closed form at all at the other end.

networks · Superposition Adding powers is wrong by the number of sources, when they are coherent. computed by solving, not by drawing, at five source counts. 8 ten-volt sources reach one hundred-ohm load through a hundred ohms each. With every source in phase the load takes 790.1 mW against the 98.77 mW that adding their separate powers gives — exactly 8 times, at every count, to a part in 10¹², because all 28 cross terms are positive and equal. With seeded random phases it takes 0.9865 times the sum over 64 runs, against a standard error of 0.125 — the cross terms average to nothing, which is the reason noise powers add and a root-sum-square is the right arithmetic for uncorrelated interference. The factor between the two limits is the source count itself, so it grows without bound: adding powers is not wrong by a bounded amount, it is wrong by how many things are being added.

The cross terms that outnumber the sources

With two sources the power in a load has one cross term, and adding the separate powers is wrong by a factor of two. With thirty-two there are four hundred and ninety-six of them, and the factor is thirty-two: 940 milliwatts delivered against the 29.4 that adding gives. With the same thirty-two sources at seeded random phases it is 26.9 milliwatts — the sum itself, inside the standard error of sixty-four runs, because every cross term averages to nothing. One identity, two limits, and the two are a phased array's gain and a noise sum's.

networks · Superposition With each source 0.8 of the one before, 32 coherent sources cost 8.99 times the sum of their powers, and never more than 9.00. computed by solving, not by drawing, at five counts and five amplitude ratios. Sources of 10 V, 10q V, 10q² V … each reach one 100 Ω load through 100 Ω; with all in phase the load's power over the sum of each source's power alone is (Σ|a|)²/Σa² to ten figures. Equal sources (q = 1) give the count itself, 2, 4, 8, 16, 32. Falling amplitudes give less, and a limit: for q = 0.8, 1.98, 3.77, 6.41, 8.51, 8.99, tending to (1 + q)/(1 − q) = 9; for q = 0.5, 3.000 against 3. The coherent penalty counts sources by their participation, not their number.

How many sources the cross terms count

N equal coherent sources on one load deliver N times the sum of their separate powers. Sources of unequal size deliver (Σ|a|)²/Σa² times it — solved on the network to ten figures — which is the number of equal sources that would carry the same penalty: a participation ratio. Amplitudes falling by 0.8 a source cap it at (1 + q)/(1 − q) = 9 however many there are; one large source and thirty-one at a tenth of its amplitude give 12.8, not 32, and for small sources the excess is 2(N − 1)ε — their amplitudes, not their powers, so thirty-one sources carrying a hundredth of the large one's power each turn 1.31 times its power into 16.8 times it when they line up. Random amplitudes take a fixed fraction off the count, (mean)²/(mean square): 0.75 for uniform ones. And with random phases none of it matters: the load takes the sum.

networks · Superposition Two 230 V sources joined by 0.05 + j0.2 Ω: the most that can be received is 194 kW, at 76.0°; with 50% of the reactance cancelled it is 262 kW, at 63.4°. computed by solving, not by drawing, at 50 Hz: two 230 V sources, the sending one leading the receiving one by an angle δ, joined by a line of 0.05 Ω and j0.2 Ω with a series capacitor cancelling a fraction of the reactance; the real power absorbed at the receiving end against δ. Uncompensated (dashed) the maximum is 194.37 kW at 75.96°, the line's impedance angle. With 50% cancelled (solid) it is 261.55 kW at 63.43°. Both maxima match V² cos β (1 − cos β)/R, found by a search on the solve to a part in a million. Past its maximum the received power falls as the angle grows: a larger angle drives more current, and past the impedance angle the extra current's loss outgrows what it carries.

Sixty degrees, and the limit compensation stops raising

Series capacitors are put in transmission lines to raise how much power can pass between two sources before the angle between them runs out. The lossless textbook limit, V²/X(1 − k), rises without bound as the reactance is cancelled. With the line's resistance kept, the limit is V² cos β (1 − cos β)/R for an impedance angle β, and it peaks where β is exactly sixty degrees: 56.7 per cent cancellation and 264.5 kW on a 230 V tie of 0.05 + j0.2 Ω, against 194.4 uncompensated — then falls, to nothing at full cancellation. A line of R/X 0.1 can nearly triple its limit; one above 1/√3 cannot raise it at all. And at the best compensation the line delivers exactly half of what it is sent.

power · Series compensation With 0.05 Ω of machine reactance at each end the sixty-degree peak needs 107% of the line cancelled — out of reach — so the limit rises all the way to full cancellation, 261.6 kW against 145.3. computed by solving, not by drawing, at 50 Hz, searched over the angle between the two EMFs on the solve (dots) and from V²·cos β·(1 − cos β)/R with the totals in it (line): the 230 V tie of 0.05 + j0.2 Ω with each source a machine whose EMF is held behind 0.05 Ω of its own reactance, against the fraction of the LINE's reactance cancelled. The capacitor cannot reach the machines' reactance, so the total impedance angle reaches sixty degrees only at k* = (X + 2Xₘ − √3·R)/X = 106.7%, beyond full cancellation; the limit rises from 145.34 to 261.55 kW and never turns. Dashed: the bare tie, peaking at 56.7%. The slider is the machines' reactance.

The reactance the capacitor cannot reach

A series capacitor raises a tie's transfer limit until the loop's impedance angle is sixty degrees, and on a bare 0.05 + j0.2 Ω line that is 56.7% cancellation. Put a machine at each end, its EMF held behind a reactance of its own, and the capacitor still cancels only the line: the sixty degrees now needs the line's reactance plus twice a machine's, less √3 times the resistance, as a fraction of the line's reactance, which passes 100% as soon as each machine's reactance exceeds √3/2 of the line's resistance — 0.043 Ω here. Beyond that the limit rises at every step to full cancellation, and the lossless estimate of the gain, wildly high on a bare line, becomes nearly right: 1.43 against 1.50 at 0.2 Ω. The higher limit is reached with the receiving terminal lower — 180 V falling to 148 at 0.1 Ω — and the capacitor's own resistance is negligible for any power capacitor: a Q of 222 keeps the peak within 1%.

power · Series compensation With the receiving end at 0.9 of the sending voltage the limit peaks at 62.6% cancellation, where the impedance angle is 56.3° — and the peak is still Vₛ²/4R = 264.5 kW. computed by solving, not by drawing, at 50 Hz, searched over the angle between the sources on the solve (dots) and from (Vᵣ·cos β/R)·(Vₛ − Vᵣ·cos β) (line): the 0.05 + j0.2 Ω tie with the sending source at 230 V and the receiving one at 207 V, against the fraction of the reactance cancelled; dashed, equal ends. The limit is a parabola in cos β with its top at cos β = Vₛ/(2Vᵣ) = 0.5556, an impedance angle of 56.25°, reached at k = 62.58%; its height there is Vₛ²/4R = 264.50 kW, the same as for equal ends. Uncompensated, this tie carries 180.53 kW at most. The slider is the receiving voltage as a fraction of the sending.

The peak a sagging receiver keeps

The sixty degrees at which series compensation stops raising a tie's transfer limit is a statement about two ends at equal voltage. With the receiving end lower, the limit is a parabola in the cosine of the impedance angle whose top is at cos β = Vₛ/2Vᵣ: 56.3° and 62.6% cancellation for a receiver at 0.9 of the sender, 33.6° and 83.4% at 0.6, full cancellation at one half. But the height of that top is Vₛ²/4R whatever the receiving voltage — 264.5 kW for the 0.05 + j0.2 Ω tie from a receiver at 1.1 of the sender down to one at 0.5 — and at it the receiving end takes exactly half of what is sent, to four decimal places, for every receiving voltage tried. The best compensated tie is the sending source driving a matched load, and the receiving source's voltage only decides how much capacitor that takes.

power · Series compensation A capacitor of 67% of the load's reactive power, moved along a feeder whose load is spread evenly over 10 km. computed by solving, not by drawing at each of 30 positions: one phase of an 11 kV feeder of 0.3 + j0.35 Ω/km, carrying 2 MW at 0.8 lagging spread over 30 equal loads. Uncorrected it loses 16.8 kW a phase; with every load corrected to unity at its own terminals, 11.1 kW — so 5.66 kW is the reactive current's share. This capacitor does best at 67% of the way out, where the loss is 11.8 kW and 87.6% of the reactive share is gone. At the substation end it removes nothing; at the far end it overshoots.

The capacitor two-thirds of the way out

A feeder's reactive current costs loss in every section it passes through, and on a feeder whose load is spread along its length that current is largest at the substation and falls to nothing at the far end. Correcting it at the substation removes no loss at all. One phase of an 11 kV feeder carrying 2 MW at 0.8 lagging over ten kilometres loses 16.8 kW, of which 5.66 kW is the reactive current's; the best single capacitor is about two-thirds of the load's reactive power placed about two-thirds of the way out, and it removes 87.6% of that share — against the eight-ninths the textbook rule gives for an idealised line. Two capacitors at 2/5 and 4/5 remove 95.6%. The same capacitor left in at light load costs more than it saves below half load, and the rule for a continuous load is a statement about many taps.

power · Reactive power A capacitor charged from 5 V through 1 mH, with and without a diode, at a circuit Q of 10. computed by solving, not by drawing, from the netlist's poles. Without a diode the capacitor rings about 5 V and settles there, and the resistance takes 1.00000 of ½CV² — exactly what a resistor alone would take. With an ideal diode in series the current stops at its first zero, 99.5 µs in, and the capacitor is left at 9.272 V, which is V(1 + d) with d = 0.8545 the step's overshoot. That charge cost 0.2699 of ½CV².

The swing a diode keeps

Charging a capacitor through a resistor loses half the energy whatever the resistance, and putting an inductor in the path changes nothing: twenty-four settled charges from 10 µH to 10 mH and Q from 0.3 to 100 lose ½CV² to 1.4 parts in a hundred million. A diode that stops the current at its first zero changes everything. The capacitor keeps the overshoot, V(1 + d), the loss falls to ½CV²(1 − d²), and a charge of duration T costs π²/8 of what a ramp of the same duration costs.

transients · Switching energy A 1 µF capacitor at 5 V joined through 100 Ω to an empty one of the same size. computed by solving, not by drawing, from the netlist's poles. Charge is conserved and both settle at 2.500 V. Energy is not: the pair holds 50.0% of the 12.50 µJ the first one started with, and the resistor has taken 50.00% — C2/(C1 + C2), with no resistance in it.

The energy two capacitors cannot share

Join a charged capacitor to an empty one and the charge is shared exactly while the energy is not: the pair loses C₂/(C₁ + C₂) of what the first one held, through 1 Ω or 10 kΩ, with an inductor ringing for hundreds of cycles or without one. An inductor and a diode turn the sharing into an exchange — equal capacitors swap 96.9% of the energy at a loop Q of 50. And a charge pump is the sharing loss billed once a cycle: a doubler's efficiency is its output voltage over twice its input, to four parts in ten million, at every load and every clock.

transients · Switching energy

Named alongside it

The objects these essays reach for when they reach for this one.

Design tradeoffModel rangeCharging exponentialEnergy-storageMaximum power transferPhasorSuperpositionVoltage regulationClosed formQuadratureReactanceReactive power

All concepts