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

Singular network — the series

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

    part 1 · networks
  2. 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.

    part 2 · networks

All series