Concept

Series resonance — where it appears

The frequency at which an inductance and a capacitance in one loop cancel, leaving a minimum of impedance set by the loop's resistance. Its depth is set by the loss and its width by the quality factor, so a null and a bandwidth are two readings of one number.

Named by 4 essays across 2 fields — each of them below, with the objects they name alongside it.

One capacitor of 77.3 µF, against every load it was not sized for. computed 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.

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.

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

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

A follower with 1000 pF on it looks like -1182 Ω of negative resistance. computed by solving, not by drawing. The impedance looking into the base of an emitter follower carrying 5.0 mA, with 1000 pF on its emitter. The real part is negative from 1.25 MHz upward and reaches -1182 Ω at 3.40 MHz: the load's reactance multiplied by a complex current gain, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it — the total loop reactance passes through zero at a frequency the inductance chooses, and the loop resistance there goes negative above 74.9 nH with 10 Ω of source, which is a few centimetres of wire. A hundred ohms of source raises that to 913 nH: the repair is a resistor in the base, and it works by making the source worse.

The input that pushes back

An emitter follower with a capacitor on its emitter has a negative resistance looking into its base — 1182 ohms of it at 3.4 megahertz for a nanofarad, with nothing added to the model. A negative resistance is not an oscillator until a reactance cancels, and the base lead supplies it: with ten ohms of source the loop goes unstable above 74.9 nanohenries, which is seven centimetres of wire. The repair is the opposite of the instinct — a hundred ohms of source raises the threshold to 913 nanohenries, so the fix for a follower that oscillates is to make the thing driving it worse.

semiconductors · Emitter follower

Correcting a 20 Ω, 50 mH load to unity, and what it stores. computed by solving, not by drawing. A capacitor across a 230 V, 50 Hz supply is swept from nothing to 300 µF against a load drawing 1636 W and 1285 var. The reactive power falls through zero at 77.31 µF and keeps going; the energy stored in the installation rises from 2044.9 mJ to 4089.7 mJ at that point — exactly twice, because unity power factor means the two stores are equal — and goes on rising afterwards. Only the cable current has a least value, 7.113 A against 9.044 A.

The energy a unity power factor doubles

Reactive power was computed three ways on this site and the agreement was called a verification. Two of the three are one theorem written twice and cannot disagree about anything; only the third is independent, and what it computes is a difference. Correcting a 20 Ω, 50 mH load to a power factor of 1.000000 takes its reactive power from 1,285 var to nothing and takes the energy stored in the installation from 2,044.9 mJ to 4,089.7 mJ — exactly twice, at every load and every frequency.

power · Reactive power

Named alongside it

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

Power factor correction Reactive power Complex power Design tradeoff Dynamic resistance Emitter follower Energy-storage Gyrator Harmonic amplification Input capacitance Lead inductance Model range

All concepts