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

Voltage regulation — where it appears

The rise in a supply's or a feeder's voltage between full load and no load, as a fraction of the loaded value. It rises with the reactance of the feeder as well as its resistance, and a series capacitor that cancels part of that reactance buys a resonance in exchange.

Named by 11 essays across one field — each of them below, with the objects they name alongside it.

50 Ω + j100 Ω of line, and the load angle past which the far end rises. computed by solving, not by drawing at 71 load resistances and four load angles. A source of 50 Ω + j100 Ω feeding loads of the same resistance and different power factor: at unity power factor the voltage across the load climbs towards the source's and stops there, reaching 0.9675 of it at the largest load drawn. A lagging load leaves less. A leading one leaves more, and past a computable angle it leaves more than the source has: the condition is 2Rₗ(Rₛ + Xₛ·tanφ) + |Zₛ|² < 0, which for the largest load here is 28.35° of lead — bisected on the solve at 28.35° — tending to atan(Rₛ/Xₛ) = 26.57° as the load grows. So the edge is a property of the line and the load angle together, and "voltage regulation" quoted as a percentage carries neither.

The far end that rises

Voltage regulation is quoted as a percentage: how far the voltage at the end of a line falls when the load is applied. The percentage carries neither of the two things that decide it. Past a computable angle of leading load the voltage at the far end goes above the source's — 28.35 degrees for a line of fifty ohms and a hundred of reactance — and with no reactance in the line there is no such angle at all, because the rise is a partial resonance and needs both halves.

power · Voltage regulation 25% compensation: 4.26% regulation, and a resonance at 25.0 Hz. computed by solving, not by drawing. The current a 230 V feeder draws, swept from two hertz to a hundred and fifty, with 25% of the line's reactance cancelled by a series capacitor. The regulation falls from 5.17% to 4.26%, which is what the capacitor was fitted for. What comes with it is a series resonance at 9.59 Hz with the load connected and 24.99 Hz with the far end shorted — the latter being exactly f₀√k = 25.00 Hz, a frequency the compensation fraction chooses on its own. The line's inductance does not appear in it and neither does the voltage. It is always below the fundamental, which is where the machines are.

The reactance cancelled, and the resonance it buys

Putting a capacitor in series with a feeder cancels part of its reactance and the far end falls less: five per cent of regulation becomes four and a quarter at a quarter compensation and 2.7 per cent at seventy per cent. What comes with it is a series resonance that was not there before, at the line frequency times the root of the fraction cancelled — so a quarter compensation resonates at exactly half the line frequency and a ninth at exactly a third. The line's own inductance is not in that answer and neither is the voltage.

power · Series compensation 50 Ω + j100 Ω of line: every power below the nose at two voltages, and a leading load's nose at 1.055 of the source. computed by solving, not by drawing: a load of fixed angle swept in size from open circuit to short circuit on a line of 50 Ω + j100 Ω, with the power it takes against the voltage it is left with, for four load angles. Each curve rises to a most power and turns back while the voltage keeps falling, so every smaller power is delivered at two voltages and every larger one at none. A 30° lagging load reaches 0.4222 of a matched resistive line's power with 0.5221 of the source voltage left; a unity power factor load reaches 0.6180 of a matched resistive line's power with 0.5878 of the source voltage left; a 30° leading load reaches 0.8240 of a matched resistive line's power with 0.7293 of the source voltage left; a 60° leading load reaches 0.9960 of a matched resistive line's power with 1.0553 of the source voltage left. Each nose is found by golden-section search on the solved network, agrees with V²cos φ/(2|Z|(1 + cos(θ − φ))), and falls where the load impedance's magnitude equals the line's.

The load that has two voltages or none

A load that takes a fixed power takes more current as its voltage falls, and on a line with impedance in it every power below a limit is delivered at two voltages and every power above it at none. The limit sits at the load the maximum-power theorem describes, half the source voltage on a resistive line. On fifty ohms and a hundred of reactance a load leading by sixty degrees reaches that limit with its far end at 1.055 of the source, and at nine tenths of it reads 1.172 — so a far end that reads high is not a far end with margin. On a direct-current bus the lower of the two voltages is not a state at all: one per cent below it the bus runs down to nothing in 3.48 milliseconds.

power · Voltage regulation With its own capacitance at B|Z| = 0.2, a line's nose for a unity-power-factor load moves from 0.618 to 0.658 of a matched line's power, at 0.635 of the source. computed by solving, not by drawing: a unity-power-factor load swept in size from open circuit to short circuit on a line of 50 Ω + j100 Ω carrying its own shunt capacitance as a π, half at each end, with the total susceptance stated as B|Z|. At B|Z| = 0 the nose is 0.6180 of a matched resistive line's power with 0.5878 of the source voltage left, and the unloaded far end reads 1.0000. At B|Z| = 0.1 the nose is 0.6376 of a matched resistive line's power with 0.6108 of the source voltage left, and the unloaded far end reads 1.0465. At B|Z| = 0.2 the nose is 0.6582 of a matched resistive line's power with 0.6353 of the source voltage left, and the unloaded far end reads 1.0969. At B|Z| = 0.4 the nose is 0.7025 of a matched resistive line's power with 0.6895 of the source voltage left, and the unloaded far end reads 1.2107. Each nose is found by golden-section search on the solved network and agrees with the nose of the Thevenin equivalent V/(1 + jBZ/2) behind Z/(1 + jBZ/2).

The headroom that is the line's own charge

A line of 50 + j100 ohms delivers at most 0.6180 of a matched resistive line's power to a unity-power-factor load, with 0.5878 of the source voltage left. Give the line its own shunt capacitance — a π, half at each end, B|Z| = 0.4 in all — and the nose moves to 0.7025 at 0.6895: 13.7 per cent more power and 17.3 per cent more voltage. It is headroom, but not the headroom the far end advertises, which with no load rises 21.1 per cent. The capacitance turns the source and line into a Thevenin equivalent with more voltage behind more impedance, and it moves a voltage threshold's meaning in opposite directions depending on whether it is referred to the source or to the unloaded far end.

power · Voltage regulation A regulator holding 10 V on a load that asks for 101% of the nose power collapses it in 256 s; at 110%, in 79 s. Marched with a fourth-order rule. A 10 V source behind 1 Ω feeds a load resistance through an ideal ratio n, and a regulator raises n at 0.05 per volt-second of error to hold the load at 10 V. The load's resistance is chosen so that at 10 V it takes the stated fraction of the most the line can deliver, 25 W. At 90% the regulator settles at n = 1.5195, below the nose ratio √(Rₗ/R) = 2.1082. At 99% the regulator settles at n = 1.8182, below the nose ratio √(Rₗ/R) = 2.0101. At 101% the voltage climbs to 9.950 V — the most the line allows that load, V₀√(Rₗ/R)/2 — at 101.7 s, and falls below half the setpoint at 256.3 s while the regulator keeps raising the ratio. At 110% the voltage climbs to 9.535 V — the most the line allows that load, V₀√(Rₗ/R)/2 — at 23.5 s, and falls below half the setpoint at 78.6 s while the regulator keeps raising the ratio. The regulator's gain, the slope of the load's voltage against the ratio, is positive below √(Rₗ/R) and negative above it.

The regulator that pushes past the nose

A regulator that raises a ratio whenever its load's voltage is low is a stabiliser only while raising the ratio raises the voltage, and on a line that stops being true at exactly the nose: the load's voltage, n·V₀ times the load resistance over n²R plus that resistance, peaks at a turns ratio of the square root of the load over the line and falls beyond it. Ask the load for 90 per cent of the nose power and the regulator settles at n = 1.519 — unless it starts above n = 2.925, where the same setpoint is met on the wrong side of the peak, and then it collapses the voltage. Ask for 101 per cent and the voltage climbs to 9.950 volts, the most the line allows, and is below half its setpoint 256 seconds later. Near the nose the collapse takes a time that grows as the inverse square root of the excess: 2,521 seconds at a hundredth of a per cent.

power · Voltage regulation Cancelling 50% of the line's reactance: a 0.3 lagging load's regulation goes from 7.78% to 4.18%, a unity one's from 2.17% to 1.96%, a 0.7 leading one's from −3.78% to −1.26%. computed by solving, not by drawing, at 50 Hz: a 230 V feeder of 0.05 + j0.2 Ω carrying 20 kVA, its regulation against the load's power factor from 0.7 leading to 0.3 lagging, with no compensation (dashed) and with 50% of the line's reactance cancelled by a series capacitor (solid). The solve agrees with the voltage divider in closed form to 9e-10 of a per cent. At 0.3 lagging the regulation falls from 7.781% to 4.176%; at 0.9 lagging, from 5.168% to 3.381%; at unity, only from 2.171% to 1.960%; and at 0.7 leading the far end, above the source at −3.778%, comes down to −1.255%. The capacitor cancels what the load's reactive current does in the line's reactance, in either direction, and pulls every load towards the regulation the line's resistance alone would give.

The drop a series capacitor can reach

A series capacitor cancels part of a line's reactance, so it can only cancel the part of the drop that the reactance makes — and that part is set by the load, not the line. On a 230 V feeder of 0.05 + j0.2 Ω carrying 20 kVA, half compensation takes a 0.3 lagging load from 7.78% regulation to 4.18% and a unity load only from 2.17% to 1.96%, and it brings a 0.71 leading load's raised far end back down. Cancel all of it and a leading load and its lagging twin sit at exactly the same voltage. Cancel more and the far end rises with its load; at exactly 100% the resonance is the fundamental and a far-end fault draws 4,600 A. Where the capacitor goes, the load cannot tell.

power · Series compensation 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 Asked for 105% of the nose power, a tap changer that blocks on a wrong-signed step stops at 9.759 V and holds it; one that does not keeps raising, and the voltage keeps falling. Marched. The 10 V source behind 1 Ω of the regulator essays, its load drawing 105% of the line's 25 W nose power at the 10 V setpoint, regulated by a tap changer that raises the ratio by 1.25% every 5 s while the load is more than 20 mV low. With no blocking rule (dashed) it climbs past the nose ratio √(Rₗ/R) = 1.9518 and keeps going, to n = 3.250 and 8.61 V by 900 s, still falling; the integrating regulator of the earlier essay (dotted) collapses the same way, faster. With the rule (solid) the tap at n = 1.9625 is the first to lower the voltage, the regulator stops at 385 s, and the load holds 9.759 V — within a tap of the most this line can give it, 9.759 V. The setpoint is out of reach either way; the rule converts a collapse into a stall.

The rule that reads the sign of the gain

A regulator that raises a ratio whenever its load is low collapses the voltage once the load asks for more than the line's nose power, because past the nose raising the ratio lowers the voltage. The standard protection is a rule: after each tap, compare the voltage with the one before, and if it fell, stop. On a 10 V line asked for 105% of its 25 W, a 1.25% tap changer with the rule stops at n = 1.9625, the first tap past the nose, and holds 9.759 V — the most the line can give — where without it the ratio climbs to 3.25 and the voltage is 8.61 V and still falling. The rule is exact without noise and fragile with it, because it reads a difference that vanishes at the nose: with 3 mV of measurement noise it wrongly stops a raise that could have reached the setpoint 0.3% of the time at 98% of the nose and 15% at 99.5%. Trusting it to 1% costs margin — 97.9% of the nose at 3 mV, 79.9% at 10 mV — or coarser taps.

power · Voltage regulation

Named alongside it

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

Maximum power transferDesign tradeoffReactive powerPhasorPower factorReactanceLine impedanceModel rangeReal powerOperating pointResonanceSeries resonance

All concepts