The branch the other resistance decides

Two resistors and a source. Draw the source as a voltage and the resistors in series and the output is a voltage divider; draw it as a current and the resistors in parallel and the output is a current divider. The two pictures are duals, the algebra is the same algebra with the words exchanged, and the second one is the one that gets written down backwards.

$$\frac{V_{out}}{V_{in}} = \frac{R_2}{R_1+R_2}, \qquad \frac{I_1}{I} = \frac{R_2}{R_1+R_2}$$

Those look identical and the subscripts have moved. In the first, $R_2$ is the resistor the output is taken across. In the second, $R_2$ is the resistor the measured current does not flow through. The expression that a reader carries over unchanged from the voltage case gives $R_1/(R_1+R_2)$ — which is a different number at every ratio except one.

The branch with the larger resistance takes the smaller share #

The sanity check is worth doing once, because it is the fastest way to remember which subscript goes where.

Ten kilohms in parallel with one kilohm, a milliamp in. Most of the current goes through the kilohm — obviously, since it is the easier path — so the metered ten-kilohm branch takes the smaller share. Solved, it takes 0.090909 of the total, which is $1/(1+10)$ and is the other branch’s resistance over the sum.

The expression carried over from the voltage divider gives 0.90909, which is a factor of ten out and in the direction that looks plausible on a schematic. The two agree only when the resistances are equal, which is exactly the case a reader is most likely to check against and the one case that cannot distinguish them.

That is the whole of the elementary content and it is not what this essay is for. What is worth having is the dual of that essay’s actual subject, which is what happens when the instrument is real.

The dual of a voltmeter is an ammeter, and the dual of an open is a short #

The divider, and the thing it does not know about measured what a finite load does to a voltage divider, and the current the instrument draws put that in the form an instrument specification uses: a voltmeter costs one per cent when its own resistance falls to ninety-nine times the resistance it looks back into.

The dual statement exists and the word that changes is not the one expected.

A voltmeter’s defect is that it draws current, so its ideal is an open circuit and its failure is a finite resistance in parallel. An ammeter’s defect is that it drops voltage, so its ideal is a short circuit and its failure is a non-zero resistance in series. That much is the ordinary duality and everybody has it.

The part that goes wrong is the resistance to compare against. For the voltmeter it is the Thévenin resistance — what the meter looks back into with the source killed, and killing a voltage source means shorting it, so the answer is the two resistors in parallel. For the ammeter it is what the meter looks back into with the source killed, and killing a current source means opening it, so the answer is the two branches in series.

Ten kilohms and one kilohm: the voltmeter would be compared against 909 Ω and the ammeter is compared against 11 kΩ. A factor of twelve, in a quantity that is supposed to be the same construction applied to the same network.

The figure measures it rather than claiming it — a probe current into the ammeter’s terminals with the current source removed, solved — and gets 11.000 kΩ against $R_1 + R_2$ exactly, and not the 909.09 Ω of the parallel combination.

Move the slider a little either way and the two expressions separate immediately, which is why the equal case is the one setting a reader should never check an expression against.

The threshold, bisected #

With the resistance to compare against settled, the threshold follows and has exactly the voltage case’s shape.

An ammeter of resistance $r$ in the metered branch makes the reading

$$\frac{I_1}{I} = \frac{R_2}{R_1 + r + R_2}$$

so the fractional error is $r/(R_1+R_2+r)$, and one per cent of it is at $r = (R_1+R_2)/99$. Bisected on the solved network at ten kilohms and one kilohm, that is 111.1 Ω against $(11\,\mathrm{k}\Omega)/99 = 111.1$ Ω.

Put the two statements side by side and they are one sentence:

The right-hand column is where the duality is most easily lost. A voltmeter wants to be a hundred times larger than something and an ammeter wants to be a hundred times smaller than something, and the two somethings are different resistances computed by different kill rules on the same network.

At the other end of the slider the two expressions are as far apart as they get.

The headline, in the dual #

That essay’s sharpest statement is that two dividers of identical ratio give different voltages into one load, because the ratio discards the magnitude. The dual holds and the figure measures it.

Two current dividers of identical ratio — ten-to-one, both of them — one built from a hundred ohms and ten, the other from a megohm and a hundred kilohms. Into a perfect ammeter both read 0.090909, identically, which the figure claims as a refusal so that what follows is known to belong to the instrument.

Into a hundred-ohm ammeter, the first reads 0.0090090 and the second reads 0.0099009 of the total. A nine per cent difference between two circuits with the same ratio, from the quantity the ratio throws away.

The direction is the reverse of the voltage case and it is worth noticing. A voltage divider is spoiled by being high-impedance relative to its load; a current divider is spoiled by being low-impedance relative to its meter. So the same instinct — “use large resistors, they draw less current” — helps one and ruins the other, which is the practical cost of carrying a rule instead of the construction it came from.

Why the two thresholds are not reciprocals of each other #

A reader who has both rules might reasonably expect them to be symmetric: a voltmeter needs to be ninety-nine times something and an ammeter a ninety-ninth of something, and if the somethings were the same number the two instruments would face the same problem seen from two sides.

They are not the same number, and the gap between them is a property of the network rather than of the instruments.

On ten kilohms in parallel with one, the voltmeter’s reference is $R_1 \parallel R_2 = 909.1$ Ω and the ammeter’s is $R_1 + R_2 = 11$ kΩ. Their ratio is $(R_1+R_2)^2/R_1R_2$, which is $2 + R_1/R_2 + R_2/R_1$ — four when the resistances are equal and growing without limit as they separate. At ten to one it is 12.1; at a hundred to one it is 102.

So the more unequal the two branches, the further apart the two instruments’ requirements become, and they are least different in the case where duality is easiest to believe. A designer checking the two rules against each other on a symmetric network finds a factor of four and might take it for a constant; on a ten-to-one divider it is twelve, and on a divider with a hundred-to-one ratio it is a hundred and two.

The reason is that a parallel combination is dominated by the smaller resistance and a series combination by the larger, so the two references diverge as the resistances do. It is the same observation that makes the resistor that is not made of the resistors worth its own essay: a network’s equivalent resistance is not one of its resistors and is not close to their average, and which combination applies depends on a question that has to be asked rather than assumed.

Where the current divider is actually met #

Three places, and in each the dual reading is the useful one.

A shunt with a meter on it. The ammeter that is a resistor balances a shunt’s insertion error against an amplifier’s offset. The insertion error is this: the shunt is one branch of a current divider whose other branch is the rest of the circuit, and the burden voltage is the price of the divider being imperfect. That essay’s insertion term, burden over supply, is the expression above with the load resistance written as $V/I$.

A mirror. A current mirror’s output is a current divider between the mirror’s own output resistance and whatever it drives — which is why an output resistance matters at all and why the copy, and its two errors has to solve it rather than quote the ratio of the two transistors.

A parallel connection of anything. Two capacitors sharing a ripple current, two conductors sharing a return, two devices sharing a load: each of them divides a current in the inverse ratio of the impedances, and the branch with the lower impedance carries the higher current. The essay where the current comes back is the distributed version of the same statement, with a continuum of branches instead of two.

In all three the quantity that decides the answer is the other branch, and in all three the mistake that is actually made is not writing the ratio upside down — it is failing to notice that there is a divider at all.

What an ammeter’s resistance costs, in the units an instrument uses #

The threshold above is a resistance and an instrument’s specification is a burden voltage, so the translation is worth making once.

An ammeter dropping $r$ at a reading of $I_1$ has a burden of $I_1 r$, and the error is $r/(R_1+R_2+r)$ — so the burden voltage at the one-per-cent threshold is $I_1(R_1+R_2)/99$, which for a branch current of 90.9 µA through 11 kΩ is 10.1 mV. A common bench meter’s 200 µA range has a burden of about 100 mV at full scale, so it is at the one-per-cent threshold at roughly a tenth of full scale and worse below that.

That is a specification worth reading off a meter’s front panel and almost nobody does. The number quoted is the burden at full scale; the error is the burden divided by the voltage available to drive the circuit, and on a low-voltage circuit the two can be comparable. Measuring a 30 µA supply current in a battery-powered circuit running from 3 V with a meter whose burden is 100 mV is a 3.3 per cent insertion error before anything else goes wrong, and the reading that comes back is low — the meter has reduced the current it was asked to measure.

The repair is the same one the shunt essays use: choose the range for the burden rather than for the resolution, or use a meter whose burden is a servo rather than a resistance. The arithmetic is the same arithmetic and the only thing that has changed is which side of the duality it is written on.

Where the duality stops #

That the two expressions are hard to tell apart in principle. They are not; the sanity check at the top takes ten seconds. What the figure prices is the second-order version of the same confusion, where the kill rule for the source is carried over instead of the ratio, and that one is genuinely easy to get wrong because it produces a plausible number.

That a real current source is an open circuit when killed. An ideal one is. A real one has a finite output resistance, which appears in parallel with the two branches and changes the resistance the meter looks back into — from $R_1+R_2$ towards $R_1 + (R_2 \parallel R_{src})$. The figure uses an ideal source because the claim is about the construction, and the source that is not a source is where the finite one is measured.

That an ammeter’s resistance is its whole defect. It is the direct-current one. A real ammeter has inductance too, which is the subject of a different field, and the series inductance behaves in this divider exactly as the series resistance does with $\omega L$ in place of $r$.

That the duality extends to everything. It extends to linear two-element networks and it stops where reciprocity does. Duality is a statement about planar networks with the roles of voltage and current exchanged, and a network containing an element that is not its own dual — a transistor, a gyrator — has a dual that is a different circuit rather than the same one relabelled.

The construction executed: a probe, a removed source, and a bisection #

The share is solved rather than substituted, and checked against $R_2/(R_1+R_2)$ — the other branch’s resistance over the sum — to a part in $10^{10}$.

The resistance the meter looks back into is measured, by a probe current with the source removed, and required to be $R_1+R_2$ and explicitly not the parallel combination, which is the step the dual gets got wrong at.

The one-per-cent threshold is bisected on the solved reading and checked against that resistance over ninety-nine, to five parts in a thousand.

The identical-ratio pair is checked to differ by more than five per cent with a real meter across four decades of magnitude.

And it is refused with a perfect one: with a zero-ohm ammeter the two magnitudes read identically to twelve digits, which is what says the spread belongs to the instrument and not to the network.

One construction, two kill rules #

The reason this is worth a whole essay is that duality is normally taught as a table and used as a substitution, and a substitution is exactly what fails here.

The table says: voltage ↔ current, series ↔ parallel, resistance ↔ conductance, Thévenin ↔ Norton, open ↔ short. Every line of it is correct. Applying it to “the resistance the meter looks back into with the source killed” requires exchanging two things at once — the kind of source and therefore the kill rule, and the topology and therefore whether the result is a series or a parallel combination — and a reader who exchanges one of them gets a number that is dimensionally right and factually wrong.

The repair is not a better mnemonic. It is to do the construction rather than the substitution: put a probe on the instrument’s own terminals, remove the source in whatever way that source is removed, and solve. That is what the figure does, it takes one line, and it cannot go wrong in the way the table can.

Which is the general form of the habit here, applied to a piece of elementary theory: a construction can be executed and a rule can only be remembered, and the elementary results are the ones where nobody bothers to execute.

Still open: the source with a finite resistance, the divider with three branches, and the dual of the chain #

The current source that is not ideal. Its output resistance appears across the two branches, so the resistance the meter looks back into moves from $R_1+R_2$ towards something smaller, and the one-per-cent threshold moves with it. Solved against the source resistance it would say at what point a real current source’s own imperfection is the dominant error rather than the meter’s — and it would connect this essay to the source that is not a source directly.

Three branches instead of two. With $n$ parallel branches the share of one of them is its conductance over the total conductance, and the resistance the meter looks back into is its own branch in series with the parallel combination of all the others. That is a different expression from the two-branch one and it does not reduce to it by inspection, so a three-branch case would say whether the threshold rule survives the generalisation.

And the dual of the chain. The essay before this one cascades three voltage dividers and finds the error at the far end from the output. The dual is three current dividers in cascade — a chain of shunt branches driven by a current source, read at the last one — and whether the error is at the near end or the far one is a question the duality ought to answer and nobody here has asked.