Networks, and how a solve is checked

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.

Assumes: The divider, and the thing it does not know about · The source that is not a source

The branch the other resistance decides priced one ammeter in one branch of a current divider with two branches. The price had the same shape as a voltmeter’s on a voltage divider, with the roles exchanged: an ammeter of resistance rr reads low by r/(R+r)r/(R + r), where RR is the resistance it looks back into with the source removed. Removing an ideal current source means opening it, so for two branches RR was the pair in series, and the one-per-cent threshold fell at that resistance over ninety-nine.

That result is Norton’s theorem, and Norton’s theorem does not care how many branches there are. So the price of one ammeter in one of many branches is already known. What is not known is what happens to the set of readings, and the set is what gets checked. Anyone who has metered the branches of a divider adds the readings up and compares the total with the source, because Kirchhoff’s current law says they must agree. This page measures that check, and it turns out to reward the wrong way of taking the readings.

One branch of two, for reference

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.
Fig. 1 The two-branch divider the earlier essay measured: a current source into a 10 kΩ branch, which is metered, and a 1 kΩ branch, which is not. The metered branch takes 0.090909 of the current. The meter looks back into the two branches in series, and its reading is one per cent low at 111.1 Ω, which is that resistance over ninety-nine.

The figure is the starting point: the meter’s error depends on the meter’s resistance against one number, the resistance it sees looking out of its own terminals. Nothing about the error depends on which branch carries more current. With two branches there is only one way to meter them, which is one at a time, and only one reading at a time to worry about.

Four branches and two ways to read them

The network here is a 1 A source feeding four branches in parallel, of 100 Ω, 1 kΩ, 10 kΩ and 100 kΩ. A decade apart, they carry 900.1, 90.01, 9.001 and 0.9001 mA, and those four currents add to 1 A. There are two ordinary ways to measure them.

The first is to own one ammeter and move it: break a branch, put the meter in, read, restore the branch, move on. Each reading is taken with three branches undisturbed and one carrying the meter. The second is to put an ammeter, or a current shunt, in every branch and read them all at once. Every branch then carries its meter at the moment every reading is taken.

Both use the same meters. Both take four readings. They do not give the same four numbers.

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.
Fig. 2 A 1 A source into branches of 100 Ω, 1 kΩ, 10 kΩ and 100 kΩ, each read with a 10 Ω ammeter. One meter moved from branch to branch reads every branch low, by 0.989, 0.902, 0.099 and 0.010 per cent, and the four readings add to 990.275 mA. A meter left in every branch reads the 100 Ω branch 0.892 per cent low and the other three 7.939, 8.910 and 9.008 per cent high, and the four add to 1000.000 mA.

One meter, moved

With a single meter moved around, each reading is exactly what Norton’s theorem says. The meter in the 100 Ω branch looks back into 100 Ω in series with the other three branches in parallel, about 901 Ω, and reads low by 10 Ω over roughly 1011: 0.989 per cent. In the 100 kΩ branch it looks back into 100 kΩ plus the other three, and 10 Ω in a hundred thousand is 0.010 per cent. Every reading is low, every one is within a per cent, and the error in each is exactly r/(Rth+r)r/(R_{\text{th}} + r) with RthR_{\text{th}} taken for that branch — to ten decimal places, on the solved network, for all four.

Add them up and the total is 990.275 mA. The source is delivering 1 A, so a reader checking the readings against Kirchhoff’s law finds a milliampere in a hundred missing. It is not missing. It is the sum of four under-readings, each honest, each taken while a different branch was disturbed, and no single state of the circuit ever had all four of those currents flowing at once.

A meter in each

With a meter in every branch the picture inverts. The four readings now describe one state of the circuit — the state with four meters in it — so they must satisfy Kirchhoff’s law in that state, and they do: 1000.000 mA, to the last digit carried.

The individual readings are another matter. The 100 Ω branch reads 0.892 per cent low, close to what one meter alone would cost it. The other three read high, and by far more than any meter could cost its own branch: 7.939, 8.910 and 9.008 per cent. The 100 kΩ branch, whose meter costs it one part in ten thousand, is reading nine per cent too much current.

The mechanism is the big branch. Putting 10 Ω in series with 100 Ω raises that branch’s resistance by a tenth, and it carries nine tenths of the source. The source is a current source, so the current the big branch sheds has to go somewhere, and the only place is up: the voltage across the parallel combination rises until the four branches together take the whole ampere again. The three small branches, whose own meters barely change them, see that voltage rise and carry it straight into their currents. Each small branch’s error is, to a close approximation, the node voltage’s rise, which the big branch’s meter decided.

So each small reading in the in-place set is an accurate measurement of the current flowing in that branch at that moment. It is the moment that is wrong. The meters have redistributed the current, the set of readings describes the redistributed circuit perfectly, and Kirchhoff’s law is a law about whatever circuit is there.

Two factors in every reading

The in-place error has a closed form, and it separates cleanly into a part every branch shares and a part that is each branch’s own. With a current source II feeding branches of conductance Gk=1/RkG_k = 1/R_k, the voltage across them is V0=I/GkV_0 = I/\sum G_k. With a meter of resistance rr in every branch, each branch’s conductance becomes Gk=1/(Rk+r)G'_k = 1/(R_k + r) and the voltage becomes V=I/GkV = I/\sum G'_k. Each reading is VGkV G'_k against a truth of V0GkV_0 G_k, so

readingktruek=GkGkRkRk+r.\frac{\text{reading}_k}{\text{true}_k} = \frac{\sum G_k}{\sum G'_k} \cdot \frac{R_k}{R_k + r}.

The first factor is the node voltage’s rise, and every branch has it: for these four branches and 10 Ω meters it is about 1.090, nine per cent. The second is the branch’s own meter, and it is what a moved meter would cost that branch if the others had no meters and the source were a voltage source. For the 100 Ω branch the second factor is 100/110 and nearly cancels the first, which is why that reading is only 0.892 per cent low. For the 100 kΩ branch the second factor is one part in ten thousand from unity, and the first is left standing.

The moved-meter reading has a different structure entirely. It has only its own Norton factor, Rth/(Rth+r)R_{\text{th}}/(R_{\text{th}} + r), and that factor is always below one — which is why every moved reading is low and none is high. The two sets of readings differ not by a correction but by a shape: one is a common factor times a local one, the other a local factor alone.

Smaller and larger meters

The size of both effects follows the meters’ resistance, and the shape of the comparison does not change with it.

4 branches read with 1 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 1 Ω. With one ammeter moved from branch to branch, each reading is low — by −0.100%, −0.091%, −0.010%, −0.001% — exactly r over the resistance that meter looks back into plus r, and the four readings add to 999.019 mA rather than 1000.000. With an ammeter in every branch at once the readings add to 1000.000 mA and are wrong by −0.091%, +0.808%, +0.898%, +0.907%: the meter in the largest branch pushes current out of it, and the small branches take it.
Fig. 3 The same four branches read with 1 Ω ammeters. One meter moved reads every branch low, by 0.100, 0.091, 0.010 and 0.001 per cent, and the readings add to 999.019 mA. A meter in each reads the 100 Ω branch 0.091 per cent low and the others 0.808, 0.898 and 0.907 per cent high, adding to 1000.000 mA.

With one-ohm meters every error is a tenth of what it was, to three figures. A moved meter is never more than a tenth of a per cent out, and the in-place readings of the small branches are still nine times worse than that, at 0.9 per cent. The ratio between the two methods’ worst errors is set by the branches, not by the meters: the big branch’s meter always sheds about r/R1r/R_1 of nine tenths of the source into branches that carry one tenth, which multiplies its effect by roughly nine.

A shunt of a few milliohms in each leg is the usual answer to that, and it works for the same reason the one-ohm meter does. But a shunt of a few milliohms in a branch of a few milliohms — a paralleled set of switching devices, or of cells, each sharing a load — is the 100 Ω branch and its 10 Ω meter again at a different scale. The ratio of the shunt to the branch it sits in is the whole of the matter, and a shunt that is small against a wire is not small against another wire.

The sum against the meters

Moved from branch to branch the readings never add up; left in place they add up exactly, unless the source has a branch of its own. computed by solving, not by drawing. The sum of four branch readings, 100 Ω to 100 kΩ in decades, against the ammeters' resistance from 0.1 to 1000 Ω. One meter moved from branch to branch: 0.99028 A at 10 Ω and 0.91064 at 100. A meter in every branch with an ideal source: 1 A at every resistance. With a source of 5 kΩ output resistance, which is a branch nobody meters, the in-place sum is 0.98075 A at 10 Ω, and the unmetered branch's share grows from 17.683 mA as the meters are made worse.
Fig. 4 The sum of the four readings against the ammeters’ resistance from 0.1 Ω to 1 kΩ. One meter moved: 0.99028 A at 10 Ω and 0.91064 at 100. A meter in every branch with an ideal source: exactly 1 A at every resistance. With a source whose own output resistance is 5 kΩ, the in-place sum is 0.98075 A at 10 Ω, and the unmetered branch across the source, which takes 17.683 mA with perfect meters, takes more as the meters get worse.

The moved-meter sum falls away smoothly as the meter gets worse: a per cent short at 10 Ω, nine per cent at 100. The in-place sum with an ideal source does not move from one ampere at any meter resistance, because a set of readings taken in one state always closes. So the sum is not a test of the meters at all when they are all in place. It passes when the meters are fine and passes when they are appalling.

The one thing that breaks the in-place sum is a branch nobody meters, and a real current source always has one: its own output resistance. A source with 5 kΩ across it sends 17.683 mA through that resistance even with perfect meters, and when the meters go in and the branches get harder to drive, the voltage rises and the source’s own resistance takes a larger share still. The in-place sum then comes up short, by 19.2 milliamperes at 10 Ω meters — and a reader who sees that shortfall has found the source’s resistance, not an error in any meter.

The source that is not a source makes the same point from the other side: a real source is an ideal one with an element attached, and that element carries current whatever the instrument does. Here the element is a fifth branch in parallel with the four, and it is the only branch whose current the check can see.

Where the in-place error comes from

The in-place readings’ error came from the difference between the branches — one branch much larger than the others, whose meter changed it much more. That suggests a test: make the branches equal.

What the ammeters in place get wrong is made of the branches' spread, and equal branches read exactly. computed by solving, not by drawing. The worst reading error among four branches of 100 Ω × spread^k, each read with a 10 Ω ammeter, one at a time and all at once, against the spread from 1 to 100. With equal branches the in-place readings are exact — every meter changes its branch alike and the shares do not move — while one meter alone reads 6.977% low. The two are equal at a spread of 1.74, and at a spread of ten the in-place error is 9.008% against 0.989% for one meter moved.
Fig. 5 The worst reading error among four branches of 100 Ω, 100 Ω × ss, × s2s^2 and × s3s^3, each read with a 10 Ω ammeter, against the spread s from 1 to 100. With equal branches the in-place readings are exact, while one meter alone reads 6.977 per cent low. The two methods’ worst errors are equal at a spread of 1.74; at a spread of ten the in-place error is 9.008 per cent against 0.989 per cent for one meter moved.

With four equal branches and a meter in each, every branch’s resistance rises by the same ten per cent and the shares do not move at all: the in-place readings are exact. One meter alone in one of four equal branches is the worst case for the moved meter, since the branch it sits in has only a third of its own resistance in parallel behind it, and it reads 6.977 per cent low.

So the two methods trade places with the spread. Below a spread of about 1.74 the meters in place do better; above it, the moved meter does, and by a factor that keeps growing with the spread: nine at a spread of ten, where the in-place error has levelled off near nine per cent and the moved meter’s is still falling. Nothing about the meters decides which method to use. The branches do.

Equal branches are not a curiosity. Paralleled parts that are meant to share a current equally — a set of identical resistors, or matched transistors each with its own emitter resistor — are close to this case, and metering all of them at once changes their sharing by almost nothing. It is when the paralleled parts differ that the meters redistribute, and it is precisely when they differ that someone wants to measure how they share.

The source that has a branch of its own

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 −1.145%, −0.903%, −0.099%, −0.010% — exactly r over the resistance that meter looks back into plus r, and the four readings add to 971.385 mA rather than 982.317. With an ammeter in every branch at once the readings add to 980.752 mA and are wrong by −1.050%, +7.767%, +8.736%, +8.834%: the meter in the largest branch pushes current out of it, and the small branches take it.
Fig. 6 The four branches read with 10 Ω ammeters from a 1 A source with 5 kΩ of its own across it. With perfect meters the branches carry 982.317 mA between them. One meter moved reads them low by 1.145, 0.903, 0.099 and 0.010 per cent, adding to 971.385 mA; a meter in each reads them −1.050, +7.767, +8.736 and +8.834 per cent, adding to 980.752 mA.

With the source’s own resistance included, the moved meter’s errors grow slightly — the meter in the big branch now looks back into the other branches and the source’s resistance, a smaller total, and reads 1.145 per cent low instead of 0.989. The in-place errors shrink slightly for the opposite reason: the source is no longer ideal, so the voltage rise the big branch’s meter causes is partly absorbed by the source’s own resistance, and the small branches see less of it.

Neither method’s readings now add to the 1 A the source is nominally delivering, and neither adds to the 982.317 mA the branches actually carry without meters. The in-place set comes closer to both, and is wrong in the same pattern as before.

What the check can and cannot say

The lesson is narrower than “Kirchhoff’s law does not apply”, which is false: the law applies to every state of every circuit on this page, to the last digit. What fails is the habit of using it as a test of measurements. A set of readings that closes is a set taken from one state of the circuit, and nothing more: it says the readings are mutually consistent, not that they describe the circuit without the instruments in it. A set that does not close was taken from several states, and its shortfall is the sum of the instruments’ disturbances, each honest.

The divider, and the thing it does not know about made the voltage divider’s version of this point with one load. The stage that is wrong is the far one found that in a chain of dividers the error sits away from where anyone probes. Here the error sits in branches whose own meters are nearly perfect, put there by a meter somewhere else. In all three the lesson is that an instrument’s error lives wherever the network carries it, and the reading taken at the instrument is the last place to look for it.

The ammeter that is a resistor and the ammeter that is not in the circuit take the practical routes out. A shunt small against its branch shrinks both methods’ errors in proportion. A clamp or a field sensor that never enters the branch removes the redistribution entirely, because it changes no branch’s resistance, and so it is the one instrument whose in-place readings are the unmetered currents. Two terminals measure the leads as well supplies the reason a four-wire shunt is still a series resistance: its sense connections remove the leads from the voltage it reads and do nothing about the resistance its current has to pass through.

Getting the unmetered currents back

Because the circuit is linear, the in-place readings are not merely wrong: they are wrong by a known amount, and the right answer can be computed from them. With every meter in place, the voltage across the branches and each meter’s own drop are both measurable. The drop gives each meter’s current, which is the reading; the branch voltage divided by that current, less the meter’s resistance, gives each branch’s own resistance. From the resistances and the source current, the unmetered voltage V0=I/GkV_0 = I/\sum G_k and every unmetered branch current follow.

So the in-place method with one extra voltage reading is a complete measurement, and the moved-meter method is not: its four readings come from four different states and no one voltage describes them. That reverses the conclusion the sums seemed to point to. The set of readings that closes is the set from which the truth can be rebuilt, because it describes one state; the set that does not close is closer to the truth reading by reading and can only be corrected branch by branch, each with its own Norton resistance, which needs the other branches to be known first.

The practical choice is therefore not between accurate and inaccurate readings but between two kinds of arithmetic afterwards. Readings taken all at once need one global correction that uses every branch; readings taken one at a time need a local correction per branch that uses the rest of the network. Neither can be skipped once the meters are not small against the branches they sit in.

How the numbers were obtained

Each state of the circuit is solved by nodal analysis: a 1 A current source into four branches, each branch a resistor into either a zero-volt source (a perfect ammeter, whose current is read directly) or a resistor of the meter’s value, and optionally a resistor across the source. Each branch current is read from its resistor’s voltage. The moved-meter readings take four solves with one meter each, the in-place readings one solve with four. Every moved-meter reading is checked against Norton’s prediction r/(Rth+r)r/(R_{\text{th}} + r), with RthR_{\text{th}} the branch plus every other branch and the source’s resistance in parallel, to ten significant figures. The spread sweep takes twenty-five spreads spaced logarithmically from 1 to 100, and the point where the two methods’ worst errors are equal is found by interpolating between the two spreads that bracket it.

What it leaves out

It uses resistive branches and resistive meters at direct current. A branch with a reactance and a meter with an inductance — a shunt is both — make the redistribution a function of frequency, and a branch current with harmonics in it would be redistributed differently at each one.

It uses identical meters. Real shunts in a set differ by their own tolerance, and a meter in each branch then adds a tolerance of its own to each branch’s resistance, which is a second source of redistribution that equal branches do not cancel.

And the source is a current source throughout. A voltage source driving the same branches in parallel has no redistribution at all — each branch sees the source’s voltage whatever the others do — so the whole of this page’s in-place error belongs to the source’s being stiff in current rather than in voltage, and a source that is stiff in neither sits somewhere between.

Still open: shunts that differ, a voltage source with a resistance, and currents that are not steady

Shunts with a tolerance of their own. With a meter in every branch, a one-per-cent spread among the shunts changes each branch’s resistance by a different amount, and for equal branches that is the only redistribution left. Solving a set of equal branches with shunts drawn from a seeded tolerance would say how closely matched the shunts must be before the in-place method’s advantage for equal branches is gone.

A voltage source with its own resistance. Between the ideal voltage source, where there is no redistribution, and the ideal current source, where there is the most, lies every real supply. The in-place error of the small branches should scale with the ratio of the source’s resistance to the branches’ parallel resistance, and measuring that ratio’s effect would say which supplies make multi-branch current measurement safe.

A load that switches. Paralleled switching devices share a current that changes within a cycle, and their shunts have inductance. Whether the in-place redistribution at the switching edge is larger or smaller than at direct current — the inductances, like the resistances, may be unequal in a different ratio — is the measurement a designer checking current sharing on a bench would actually be making.

Part 5 on divider

One argument about Divider, and one of 5 essays on it so far, each part numbered by how much of the idea it assumes. What sits either side of it:

The objects named here

The third axis, after the field and the idea: the things themselves, and every essay that touches each one.

Current dividerKirchhoffs current lawLoadingMeasurement errorModel rangeNorton equivalent