Circuit Response
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Concept

Norton equivalent — where it appears

A current source in parallel with a single impedance that reproduces what a linear network delivers at a pair of terminals. It is the Thévenin equivalent redrawn, with the source current equal to the short-circuit current, and it is the natural form for loads that add in parallel.

Named by 3 essays across one field — each of them below, with the objects they name alongside it.

The current divider, and the resistance that is not in the branch. computed by solving, not by drawing. A current source into two parallel branches, the metered one 10.0 kΩ and the other 1.00 kΩ. The metered branch takes 0.090909 of the current, which is the OTHER branch's resistance over the sum; writing the subscripts the way a voltage divider writes them gives 0.90909, a different number at every ratio but one. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with the roles exchanged: one per cent of error at 111.1 Ω, which is that resistance over ninety-nine to 1.7e-12%. And the headline of the loaded divider holds in the dual too — two current dividers of identical ratio read 0.04762 and 0.09090 into one hundred-ohm meter — while a perfect ammeter reads them identically.

The branch the other resistance decides

A voltage divider's output is set by the resistance the output is taken across; a current divider's is set by the resistance the current does not go through. The dual of a voltmeter's finite resistance is an ammeter's non-zero one, and the threshold has the same shape with one word changed: one per cent at a meter resistance of R/99 where R is what the meter looks back into — and removing an ideal current source means OPENING it, so that R is the two branches in series, 11 kΩ here, not the 909 Ω of their parallel combination — a factor of twelve in the same construction on the same network.

networks · Divider 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 The bootstrap reduced both ways: Thévenin's resistance diverges at A = 2, Norton's conductance crosses zero there. computed by solving, not by drawing at 121 gains, R₁ = R₂ = 10 kΩ. The Norton current is Vs/R₁, 0.500 mA, at every gain to a part in a billion (dashed, at 1), because shorting the node also shorts the bootstrap's input. The Norton conductance times R₁ is a straight line, 2 at A = 0 through zero at A = 2 to −1 at A = 3 (solid). The Thévenin resistance over R₁ is its reciprocal (faint, clipped at ±3): 0.5, 1 at A = 1, infinite at A = 2, −1 at A = 3. One network, two coordinate systems, and only one of them is singular exactly where the network is interesting.

The reduction that stays finite

The bootstrap's Thévenin pair diverges at A = 2: its resistance and its open-circuit voltage both pass through infinity and the solve that finds them refuses. Its Norton pair does not. The short-circuit current is Vs/R₁, 0.5 mA, at every gain to a part in a billion, and the conductance is a straight line through zero. A bench that drives a current and reads a volt loses the network for every gain from 1.992 to 2.008; a bench that drives a volt and reads a current sees it to a nanosiemens throughout. A 0.1% error in the gain moves the conductance by a constant 0.1 µS and the resistance by an amount that diverges at the pole — and four hundred bootstraps built from 0.1% parts to sit on it have conductances within ±0.3 µS of zero and resistances that are a median 10.6 MΩ, 196 of them negative.

networks · Equivalent circuit

Named alongside it

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

Current dividerLoadingModel rangeThevenin equivalentComponent toleranceDualityKirchhoffs current lawMeasurement conditionMeasurement errorNegative resistanceOutput impedanceSingular matrix

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