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

Power, and the part that does no work

The solver has been computing real power on every solve since the first commit, to check itself, and then discarding it. This field reads it out — and then the imaginary half beside it, which sizes the cable, heats the transformer and is billed for. A correction capacitor is exact at the load it was computed for and at no other; a power factor is cos φ only while the current is a sinusoid, and a rectifier's is not.
0500500100010001500150020002000real power (watts)reactive power (volt-amperes reactive)1636 W1285 var2080 VAreal power1636 Wreactive power1285 varapparent power2080 VApower factor0.7864angle38.15°line current9.04 A…doing work7.11 Athe load230 V20 Ω50 mHsolved, then checked — three routes to Qonly 0.79 of the current works

The current that does no work

A solved network has been reporting its own real power on every page of this collection, as the second of the two checks each answer passes before it is drawn. What that check discards is the imaginary half — the power that flows out to a reactance and back again, does nothing, and is still carried by the cable, still heats the transformer, and is still on the bill.

 00.250.500.75110100load resistance (ohms)power factorno capacitorsized here: pf 1.000000leading above 21 Ωthe 0.95 an installation is usually required to holdsolved, then checked — 61 loads, one capacitorexact at 20 Ω, 0.23 elsewhere

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.

 the three line currents, and their sum2.65 A in the neutral00.2000.40000.2000.4000.6000.8001imbalance in one phaseneutral current ÷ line currenta tenth of a line current11.1%solved, then checked — neutral by two routesa tenth of a line at 11.1% imbalance

Three phases, and the wire that carries nothing

Three sources a third of a cycle apart, feeding three equal loads, return a current of 5×10⁻¹⁵ amperes down the wire between the star points. That is zero, and the whole of three-phase distribution rests on it. What is worth measuring is how fast it stops being zero, and the answer is that an eleven per cent imbalance in one phase puts a tenth of a line current down a conductor often sized on the assumption that it carries none.

All essays