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

Series compensation — the series

5 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · power
  2. 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.

    part 2 · power
  3. 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.

    part 3 · power
  4. 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%.

    part 4 · power
  5. 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.

    part 5 · power

All series