Concept

Short-circuit current — where it appears

The current a source delivers into a short across its terminals, which divided into the open-circuit voltage gives the resistance behind them. The pair of measurements it belongs to is exact at the terminals and wrong inside, since an equivalent reproduces every reading and not the dissipation.

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

A 9 V source with 500 mΩ inside it. The ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A.

The source that is not a source

An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object.

networks · Source model

The equivalent of six elements, and the heat it does not account for. computed by solving, not by drawing at 73 loads. The network reduces to 8.56032 V behind 599.768 Ω, obtained twice and agreeing to twelve figures. Into every load on the axis the two put the same voltage to 2.2e-16 of it and deliver the same power to 7.8e-16 — there is no load that can tell them apart. Inside, at a load of 5.62 kΩ, the real network is dissipating 49.168 mW and the equivalent claims 1.1349 mW, a factor of 43.3. With the port open the equivalent says nothing at all is being burned and the network is burning 48.03 mW. The slider moves the resistor bridging the two sources, which joins two ideal sources and so cannot change anything the port can reach: the equivalent stays 8.56032 V behind 599.768 Ω to the last bit at every setting, while the heat inside moves by two and a half times.

Exact outside and wrong within

Six elements reduce to one source and one resistor that no load can distinguish from them: the same voltage into every load across six decades, to the last bit of a double. The reduction is wrong about the heat by a factor of forty-three, and with nothing connected it says the network is dissipating nothing while it burns 48 milliwatts.

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

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

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

Named alongside it

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

Internal resistance Reactance Reactive power Series resonance Thevenin equivalent Voltage regulation Design tradeoff Ideal voltage source Maximum power transfer Model range Output impedance Power factor

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