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

Line impedance — where it appears

The impedance of the conductors between a source and its load, which turns every current the load draws into a voltage the load never receives. It decides how far the far end sags or rises under load, and it is the divider through which an unbalance in the current is read as an unbalance in the voltage.

Named by 5 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 The two sequences a neutral current says nothing about. computed by solving, not by drawing at 61 imbalances. Three 20 Ω loads on a 230 V, 50 Hz star supply, one of them raised by a fraction of itself, with the neutral in place. The zero-sequence current is the one the neutral carries three times and is the only one this collection has read; the negative sequence is a balanced set of three phasors rotating the other way. At 30.0 per cent imbalance it is 0.8846 A against 10.6154 A of positive sequence, 8.333 per cent, against 8.333 per cent from x/(3 + 2x). Two per cent arrives at 6.250 per cent imbalance, bisected on the network.

The half the neutral does not carry

A star load unbalanced in one phase produces two things, not one. The neutral carries three times the zero-sequence current, which is the half this collection has read; the other half is a negative-sequence set of exactly the same size, rotating backwards, that the neutral never sees. With 0.5 Ω of line in front of 20 Ω loads, losing a phase entirely puts 50.00 per cent negative sequence in the current and 0.8265 per cent in the voltage a switchboard meter reads.

power · Three-phase 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

Named alongside it

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

Maximum power transferVoltage regulationModel rangeOperating pointPower factorReactive powerPhasorApparent powerDesign tradeoffImbalanceLine lossMeasurement condition

All concepts