A network solved, and checked: a bridge, which no series-parallel reduction reaches
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At those defaults the edge it states is a linear network has no edge: this one is exact — the right-hand slot
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It belongs to Networks, and how a solve is checked, which is to say a change to it is a
change to lib/figures/networks.js.
It carries no slider, so the frame above is the whole of what it draws.
Called by 17 essays
which is the blast radius of changing it
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.
The divider, and the thing it does not know about
A two-resistor divider's output is set by the ratio of its resistances — with nothing connected. Connect anything at all and what decides the answer is the quantity the ratio was built to discard: the magnitude. Two dividers of identical ratio give six volts and one volt into the same load.
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.
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.
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.
Two solves that add, and the one that does not
Every node voltage and every branch current in a linear network is the sum of the per-source solves, here to the last bit of a double at eighty-one settings. The power is not, and the gap is not a correction: two equal sources in antiphase put nothing at all into a load while adding their powers gives 222 milliwatts, and the sum is within one per cent of the truth only when one source is two hundred times the other.
The reading that does not care which way round it is
Inject a current at one end of a network and read the voltage at the other; then swap the two. A passive network gives back the same number — not a similar one, but the same to five parts in ten to the fourteenth over four decades of frequency, and for a coupled pair of windings the same double. Put a transconductance of eight tenths of a femtosiemens anywhere in it and the two readings part company.
The matrix that is ill, and the answer that is not
Every other essay here treats the solve as exact, and it is not. A bridge walked towards balance loses one digit per decade of imbalance and has none left at a part in 10¹⁵ — on a matrix whose condition number never moves and whose smallest pivot stays four orders above the threshold this solver refuses at. A feedback amplifier does the opposite: the solver declines to answer at a gain of 10⁹, one decade after it returned an answer that was exact to the last bit.
The bridge that is linear near one point
The expression for a Wheatstone bridge's output is V delta over four, and the solve says it is V delta over two times two plus delta — low by half the fractional change, exactly, at every change tested. So the tangent is worth one per cent only up to a two per cent change and a tenth of a per cent only up to two tenths. Driving the bridge from a current source instead halves the departure at every point and doubles both edges, which is a change of one component and no change at all to the four resistors being measured.
The derivative of a root
An earlier essay turned one transposed solve into the derivative of a response with respect to every element, and found a doubly terminated ladder filter stationary at its ripple peaks to a part in ten to the eighth. A pole is a different object — a value of s at which the matrix loses rank — and its derivative comes from two null vectors and a division. Pointed at the same two realisations, the ladder filter's advantage is a factor of 2.17, not eight orders of magnitude: what is stationary is the magnitude at one frequency, and it says nothing about where the poles are.
The three tolerances that do nothing
An earlier essay computed every second derivative of a ladder filter's magnitude at a ripple peak, found them all negative, and built six hundred ladder filters to argue that no combination of tolerances could raise the response. The whole matrix says so outright — and says something six hundred samples could not have found, because a sample of a five-dimensional box never lands on a three-dimensional subspace: two of the five eigenvalues are of order one and the other three are nine decades down.
The reading that does care which way round
An earlier essay measured a network's departure from reciprocity and left its size as a constant — 62.8 per siemens, for that network at that frequency. It is not a constant and it is not the network's: it is 2πfL for the single inductor between the controlled source's control node and its output node, 62.832 ohms at ten kilohertz, and the ratio of the two readings is 1 − gm·Z to 8.8 parts in 10¹⁴ over ninety-nine readings. So 4.5455 millisiemens across a 220 ohm branch makes a ladder that transmits a hard zero forwards at every frequency at once, and 206.13 ohms back at ten kilohertz.
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.
The source that must not be zeroed
Superposition says to zero all the sources but one, and the care usually taken over that instruction is that zeroing is not deleting — a voltage source with no voltage is a short circuit and it stays in the netlist. A dependent source needs the opposite care: it must not be zeroed at all, because it is an element whose law involves another node rather than a source the theorem is about. Zeroed, it is a short circuit, and the answer that comes back is 4.0625 volts at every gain from one to three hundred — while the true answer moves from 3.82 volts to 0.206.
The reference a sweep should keep
At direct current the best node to call zero in a resistor chain is its middle. Put a capacitor to ground at each node and solve at s = jω, and the best reference moves: the middle of an eight-section RC ladder below 2.37 kHz, ground above it, where every capacitor returns. So a solver that chooses its reference once for a whole sweep is choosing wrongly somewhere, as the earlier essay suspected — but the two wrong choices are not alike. Kept for the whole sweep, ground is never worse than 1.27 times the best condition number, a tenth of a digit. The middle node, best at direct current and kept, is 143 times worse at a megahertz, 2.2 digits. A fifth-order LC filter says the same: ground within 2.18 times throughout.
The readings that come back the same
Reciprocity is usually stated as 'swap the source and the meter and the reading is unchanged', and that is true of some readings and false of others. On one passive ladder network the transimpedance, the transadmittance, and the reverse voltage gain against the forward current gain each come back the same to 10⁻¹³, while the voltage gain forwards and backwards differ by 13.1 times at 82.5 kHz — exactly z₂₂/z₁₁. A 600-to-50-ohm pad has voltage gains of 0.0426 and 0.511 and one power gain both ways. A controlled source breaks all three reciprocal pairs by the same factor.
The resistance colder than the room
In a passive network the noise at a port is 4kT times the real part of its impedance, and reciprocity is why: the transfer from each resistor to the port equals the transfer from the port to that resistor, so one solve from the output gives every resistor's share and the shares add to Nyquist's expression, here to 4.4 × 10⁻¹⁶. With gain in the network both fail. A 50-ohm termination made by feedback around an amplifier of gain A presents 50.000 Ω and has the noise of a 50-ohm resistor at 290/(1 + A) kelvin — 2.90 K at a gain of 99 — but only where the amplifier still has its gain.