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

Kirchhoffs current law — where it appears

The requirement that currents meeting at a node sum to zero, which rebuilt from the element laws is the first check on any solved network. Rebuilding the currents from the element laws and summing them is a route that shares no arithmetic with the solve, which is what makes it a check.

Named by 5 essays across one field — each of them below, with the objects they name alongside it.

A network solved, and checked: a bridge, which no series-parallel reduction reaches. Node potentials from modified nodal analysis. The branch currents are then recomputed from each element's own law and summed at every node; the residual is 2.7e-16 of the largest current in the circuit, which is floating-point rounding and nothing else.

What a network answers, and how the answer is checked

A circuit has exactly one answer and a matrix finds it. The part that matters is not that the answer exists but that it can be checked twice, by routes that share no arithmetic — and that a circuit with no answer is refused by name rather than returned as a large plausible number.

networks · Nodal analysis A network the solver will answer, and should not be asked, into 0.01 Ω. computed by solving, not by drawing at 61 spreads. A hundredth of an ohm either side of a wire whose resistance is made smaller and smaller, into 0.01 Ω — an element written where the right answer is no element at all. The solution stays exact: 0.499999998422607 at the last spread before the refusal, against 0.500000000000000. What grows is the current-law residual, as the 0.96 power of the spread. The solve is refused at a spread of 2.5e+8, where the smallest pivot falls under 1e-12; the residual tolerance of 1e-7 would have been reached at about 3.3e+9. Two guards written for unrelated reasons, arriving within a decade of one another.

The answer that is perfect and absurd

A network with a wire written into it as a small resistance returns the right node voltage to fifteen figures, passes both verifications with a residual of two parts in ten to the sixteenth, and reports two hundred thousand amperes. The solver refuses it one decade further on, and by then it has been answering for eight decades.

networks · Singular network The circuit does not care which node is called zero, and the matrix does. computed by solving, not by drawing. The condition number of the nodal matrix for a 12-section chain, against which of its nodes was taken as the reference. The network, its elements and its physics are identical in every case — only a label has moved — and every branch voltage and branch current comes back the same to 1.1e-13. The condition number runs from 5.25e+4 at "n6" to 1.72e+5 at "n12", a factor of 3.28, which is 0.52 decimal digits of the arithmetic's own margin.

The node that is not in the circuit

Nodal analysis needs a node to call zero and no circuit contains one. Moving it changes every node voltage by the same amount and no branch voltage or branch current at all — to a part in ten to the fourteenth on a well-behaved chain. What it does change is the matrix: the condition number of a twelve-section chain moves by a factor of 3.3 with the reference, and on a star whose resistances span nine decades by 8.0. Measured against the same matrix solved in twice the precision, the worst reference costs four decimal digits of the answer, and it is the reference the condition number named before the error was looked at.

networks · Nodal analysis 4 branches read with 10 Ω ammeters: one at a time every reading is low, all at once they add up and the small branches are high. computed by solving, not by drawing. A 1 A source into 4 parallel branches of 100 Ω, 1 kΩ, 10 kΩ, 100 kΩ, each read with an ammeter of 10 Ω. With one ammeter moved from branch to branch, each reading is low — by −0.989%, −0.902%, −0.099%, −0.010% — exactly r over the resistance that meter looks back into plus r, and the four readings add to 990.275 mA rather than 1000.000. With an ammeter in every branch at once the readings add to 1000.000 mA and are wrong by −0.892%, +7.939%, +8.910%, +9.008%: the meter in the largest branch pushes current out of it, and the small branches take it.

The readings that add up and are wrong

Four branches of 100 Ω, 1 kΩ, 10 kΩ and 100 kΩ share a 1 A source. Read them one at a time with a single 10 Ω ammeter moved from branch to branch, and every reading is within one per cent of the truth, but the four add to 990.275 mA. Put a 10 Ω ammeter in every branch at once and the four readings add to 1000.000 mA exactly, satisfying Kirchhoff's current law to the last digit. Yet three of them are 7.9 to 9.0 per cent high. The check a reader would run on the readings is passed by the wrong set and failed by the right one.

networks · Divider What a simulator's small conductance reports: at 10⁻¹² S a resistive divider reads 5 V, 1 µA into a capacitor reads 1 MV with a residual at machine precision, and the slope says which of those belongs to the circuit. computed by solving, not by drawing on the nodal matrix at direct current, with a conductance g from each node to ground. A divider of two 1 kΩ on 10 V (flat, 5 V at every g); 1 µA into a capacitor, which the solver refuses without g and which reads I/g with it (slope −1, 1 MV at 10⁻¹² S, 100 mV at 10⁻⁵ S); a biased stage behind a coupling capacitor, 100 mV (flat until g is comparable with its 1 MΩ); and two capacitors in series on 10 V, whose node reads 0 mV at every g, since a conductance to ground is the only thing at that node and it has nothing to divide against. Every one of these passes Kirchhoff's law with g included to machine precision. The residual has no way to tell these answers apart.

The conductance added so it would invert

A solver that refuses a node with no direct-current path can be made to answer by adding a tiny conductance from every node to ground, which is what a circuit simulator does. Kirchhoff's law then holds to machine precision, and the numbers are of three kinds. A real divider's node is exact at any conductance. 1 µA into a capacitor reads a megavolt at 10⁻¹² S, ten times that at 10⁻¹³, and is purely the reciprocal of what was added. A floating 9 V source reads ±4.5 V, a centre tap nobody built, and follows the ratio of two conductances that do not exist. Scaling the conductance finds the first kind and misses the third. One test current per node finds them all, because a node held only by what was added has a resistance to ground near one over it.

networks · Singular network

Named alongside it

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

Modified nodal analysisSingular matrixVerificationNumerical errorModel refusalComplex frequencyConditioningCurrent dividerLoadingMeasurement errorModel rangeNorton equivalent

All concepts