Superposition holds for the solution and not for its square
Drawn above at its default parameters, which is almost never how an essay calls it. A placement states the numbers that essay is arguing about, so the figure a reader meets is about that argument rather than about the generator — 98% of the placements on this site pass one, and the phase that raised that number from 12% found eight captions describing a figure the page was not showing.
At those defaults the edge it states is powers add only at 90°, or beyond 198:1 — the right-hand slot
of the caption strip, which on this site is never used for anything else, and which is read
back out of the drawing above rather than out of the code that wrote it.
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 takes a slider on the angle between the two sources (degrees) with 7 settings,
and every one of them has passed the same assertions as the frame above — a figure whose
circuit stops doing what its caption says at any setting stops the build.
Called by 5 essays
which is the blast radius of changing it
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 peak that only has a bound
Power does not superpose and the gap has a closed form — two times the real part of one current times the conjugate of the other, exact at every setting. A peak has nothing of the kind. A ten-volt fundamental with a third harmonic a third its size has one root-mean-square value, 7.454 volts at every relative phase because the two are orthogonal, and a peak running from 9.428 to 13.333 volts across the same sweep — reaching the sum of the two peaks exactly at 180 degrees, where the harmonic's own crest lands on the fundamental's, and having no closed form at all at the other end.
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 cross terms that outnumber the sources
With two sources the power in a load has one cross term, and adding the separate powers is wrong by a factor of two. With thirty-two there are four hundred and ninety-six of them, and the factor is thirty-two: 940 milliwatts delivered against the 29.4 that adding gives. With the same thirty-two sources at seeded random phases it is 26.9 milliwatts — the sum itself, inside the standard error of sixty-four runs, because every cross term averages to nothing. One identity, two limits, and the two are a phased array's gain and a noise sum's.
How many sources the cross terms count
N equal coherent sources on one load deliver N times the sum of their separate powers. Sources of unequal size deliver (Σ|a|)²/Σa² times it — solved on the network to ten figures — which is the number of equal sources that would carry the same penalty: a participation ratio. Amplitudes falling by 0.8 a source cap it at (1 + q)/(1 − q) = 9 however many there are; one large source and thirty-one at a tenth of its amplitude give 12.8, not 32, and for small sources the excess is 2(N − 1)ε — their amplitudes, not their powers, so thirty-one sources carrying a hundredth of the large one's power each turn 1.31 times its power into 16.8 times it when they line up. Random amplitudes take a fixed fraction off the count, (mean)²/(mean square): 0.75 for uniform ones. And with random phases none of it matters: the load takes the sum.