Circuit Response
every model has an edge, and the figure carries it
Measurement, which is a circuit on a circuit

Two terminals measure the leads as well

Fifty milliohms in each lead makes a two-wire measurement one per cent high at ten ohms, ten per cent high at one ohm, and a hundred per cent high at a tenth. Not approximately — the reading is the resistance plus the leads, and below about ten ohms most of what is being reported is the wire between the instrument and the thing.
16 min read 7 figures Two routes to a numberEvery model has an edgeThe constants decide

An ohmmeter forces a current through something and measures the voltage that appears. If it forces the current down the same pair of wires it measures across, it measures the wires too.

That sentence is the whole of the subject, and everything interesting about it is in how large the consequence is at what resistance.

Measuring with 50 mΩ of lead in each wirecomputed by solving, not by drawing at 61 resistances, twice each. The two-wire arrangement measures the leads too, so its error is 2×50 mΩ over whatever is being measured: one per cent at 10 Ω, and 10000% at 1 mΩ. The four-wire arrangement senses on a separate pair that carries almost no current, and its error stays under 1.0e-2% across the whole range.-6-4-201m10m100m1101001kresistance being measured (ohms)log₁₀ of the reading's errortwo wiresfour wiresone per centtwo wires are 1% out at 10 Ωsolved, then checked — two arrangements, not two formulastwo wires fail below 10 Ω
Fig. 1 The same resistance measured two ways, at sixty-one values spanning six decades, with fifty milliohms in each lead. The two-wire error is 2×50 mΩ over whatever is being measured: one per cent at ten ohms, and a thousand per cent at a milliohm. The four-wire error stays under 10⁻³ of a per cent across the whole range. The slider is the lead resistance.

Two netlists, not two formulas

Both arrangements here are built as networks and solved, and the reason is that the four-wire case has a subtlety that a formula would hide.

The two-wire netlist is a current source, a lead, the resistance under test, a second lead, and a voltmeter across the point where the current enters and leaves. The voltmeter is a ten-megohm resistor, because a voltmeter is a resistor.

The four-wire netlist forces the current down the same pair, and adds a second pair tapping the resistance’s own terminals, with the voltmeter across those. The sense leads have the same fifty milliohms as the force leads — they are the same kind of wire — and the point is that it does not matter, because almost no current flows in them.

“Almost no” is the subtlety. The sense pair is not carrying zero current; it is carrying whatever ten megohms takes at the voltage being measured, and the sense leads’ resistance appears in series with that. The four-wire error is therefore not zero either. It is the ratio of the sense lead resistance to the voltmeter’s input resistance, which for fifty milliohms and ten megohms is 5×10⁻⁹ — and the measured four-wire error across the whole range is between 10⁻⁶ and 10⁻³ per cent, which is that, plus the divider the voltmeter forms with the resistance under test.

A formula for the four-wire case would have said “zero”. The netlist says what it actually is, and what it actually is tells a reader which instrument specification decides the answer.

The numbers, and where the crossing is

Two-wire, with fifty milliohms per lead: 0.099% at a hundred ohms, 1.000% at ten, 10.000% at one, 100.000% at a tenth, 1,000% at a hundredth.

The relation is exact and trivially so — the reading is R + 2R_lead, so the error is 2R_lead/R — and the one-per-cent crossing is at 2R_lead/0.01, which is 10 Ω. The figure asserts that: at exactly ten ohms the solved two-wire network gives 10.09999 Ω, which is 1.000% high to a part in a thousand of the claim.

The residual is worth a word, because it is not rounding. The solved network also contains the voltmeter’s ten megohms in parallel with everything, and a milliohm of return resistance, so the solved answer is not exactly R + 2R_lead. It differs from it by about a part in ten thousand, which is the size of those two effects, and asserting the simple relation to a part in a thousand is therefore asserting the dominant term while leaving room for the ones that are genuinely present.

Measuring with 500 mΩ of lead in each wirecomputed by solving, not by drawing at 61 resistances, twice each. The two-wire arrangement measures the leads too, so its error is 2×500 mΩ over whatever is being measured: one per cent at 100 Ω, and 100000% at 1 mΩ. The four-wire arrangement senses on a separate pair that carries almost no current, and its error stays under 1.0e-2% across the whole range.-6-4-201m10m100m1101001kresistance being measured (ohms)log₁₀ of the reading's errortwo wiresfour wiresone per centtwo wires are 1% out at 100 Ωsolved, then checked — two arrangements, not two formulastwo wires fail below 100 Ω
Fig. 2 The same measurement with half an ohm in each lead — a long, thin pair of test leads, or a poor connection. The one-per-cent crossing moves up by a factor of ten to a hundred ohms, so a two-terminal reading of a hundred-ohm resistor is now suspect. The four-wire curve has not moved, because the sense leads’ resistance still does not matter.

Why fifty milliohms is a realistic number

The lead resistance is the parameter the whole page turns on, so it is worth saying where fifty milliohms comes from rather than treating it as an arbitrary choice.

A metre of ordinary test lead is a few tens of milliohms: copper’s resistivity is 1.7×10⁻⁸ ohm metres, so a metre of wire a millimetre in diameter is about 22 mΩ. Two of those, plus the resistance of the two banana plugs at each end and the two crocodile clips, comfortably reaches fifty milliohms per lead and often exceeds it.

The connections are the larger and less predictable half. A clean plug-and-socket pair is a few milliohms; a tarnished one is tens; a crocodile clip on an oxidised surface can be hundreds and can change while the measurement is being taken. That variability is the second reason the two-wire arrangement is unsatisfactory at low resistance: not only is the error large, it is not repeatable, so it cannot be subtracted.

Subtracting it is what a two-wire instrument’s “relative” or “null” function does — short the leads together, record the reading, and subtract it from everything afterwards. That works, and it works exactly as well as the assumption that the lead resistance has not changed since the null was taken. Flexing a lead changes it. Warming it changes it: copper’s temperature coefficient is about 0.4% per kelvin, so a lead that warms ten degrees in a rack changes by four per cent of fifty milliohms, which is 2 mΩ — an error the null cannot remove and which is a fifth of a per cent at one ohm.

What the four-wire arrangement is actually doing

It is worth being precise, because the usual summary — “it eliminates lead resistance” — is right about the outcome and unhelpful about the mechanism.

What it does is separate the two functions. The force pair’s job is to establish a known current, and its resistance affects only how much voltage the source has to produce, which the source can handle. The sense pair’s job is to report a voltage, and its resistance affects only the current the voltmeter draws, which is nearly zero. Neither lead resistance appears in the answer, because neither is in a path where it would.

Two consequences follow that the summary does not give.

Where the sense pair connects is the whole measurement. The four-wire arrangement measures the resistance between the two points where the sense leads touch. Connect them a centimetre away along the same conductor and that centimetre is included. This is why a four-terminal shunt resistor is a manufactured component with four terminals on it rather than a resistor with two extra wires soldered on — the sense connections have to be at defined points, and the definition of what is being measured is where they are.

The force current has to be large enough to measure and small enough not to heat. The voltage across a milliohm at one milliampere is a microvolt, which is at the edge of what an instrument can resolve against its own noise; raising the current to an ampere gives a millivolt, which is comfortable, and dissipates a milliwatt in the component, which may not be. That trade is the noise field’s argument arriving in a measurement, and it is where the lower end of this figure’s axis actually stops in practice.

A 10 kΩ + 10 kΩ divider, solved with its loadThe unloaded answer is 6.00 V. It is 1% low at a load of 495 kΩ and 4.00 V at a load equal to the divider's own resistance. The ratio does not predict any of this; the magnitude does.02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ
Fig. 3 The general form, from the networks field: a ratio that is a property of the network and of what is across it. Every essay in this field is that statement about a different instrument, and the four-wire arrangement is the one case where the fix is structural rather than numerical — the loading is not reduced, it is moved to a place where it does not enter the answer.

The same problem in the source

There is a mirror image of this argument that this collection has already made, and putting the two side by side is worth doing.

The source that is not a source is about a voltage source with resistance in series with it: what it delivers depends on what is drawn from it, and the number on the label is what it produces into nothing at all. That is the supply side of exactly this problem. A four-terminal measurement is what a sense connection does about it, and a regulated supply with remote sensing is the same technique applied in the other direction — the regulator’s feedback is taken at the load rather than at its own terminals, so the drop in the supply leads is inside the loop and corrected rather than outside it and delivered.

The two are the same circuit with the roles exchanged, and neither is available without a second pair of wires. That is the honest cost of the technique: four conductors instead of two, and a component or a connector designed for it.

A 9 V source with 500 mΩ inside itThe ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A.024681010m100m110current drawn from the source (amperes)terminal voltage, solvedthe ideal source: 9 V at any current1% low at 180 mAthe model9 Vrloadsolved, then checked — the load swept over four decadesthe ideal source is 1% wrong above 180 mA
Fig. 4 A source with its own resistance, from the networks field. Remote sensing is the four-terminal arrangement applied to this circuit rather than to a measurement — the feedback is taken where the voltage matters instead of where it is generated — and it needs the same extra pair of wires for the same reason.

Where the two-wire measurement is fine

The figure is drawn to make two-wire measurement look bad, and it is worth being fair about where it is not.

Above about a kilohm with ordinary leads, the two-wire error is under a hundredth of a per cent and is smaller than the instrument’s own accuracy. Almost every resistance anybody measures is in that range, which is why two-wire measurement is the default and why most people never think about this at all.

The trouble starts at the low end, and the low end is where the interesting components are: current shunts, contact resistances, cable resistances, winding resistances, the resistance of a connector, the resistance of a solder joint. Those are milliohms to ohms, and every one of them is in the region where a two-terminal reading is mostly leads.

The crossing point is the useful number and it scales exactly: with 5 mΩ leads it is 1 Ω, with 50 mΩ it is 10 Ω, with 0.5 Ω it is 100 Ω, with 2 Ω it is 400 Ω. Multiply the lead resistance by two hundred and that is where a two-wire reading is one per cent high.

That factor of two hundred is worth carrying around, because it converts a quantity nobody measures into a rule anybody can apply. Estimate the lead resistance — a few tens of milliohms for ordinary leads, a few hundred for a clip on a corroded surface — multiply by two hundred, and any resistance below the answer is being measured mostly through the leads. Everything above it is fine to a per cent, and the transition is a smooth inverse rather than a threshold, so a factor of ten below the crossing is ten per cent and a factor of ten above is a tenth of a per cent.

There is no ambiguity anywhere in that, which is unusual for a rule of thumb. It follows from R + 2R_lead being the exact reading of the two-wire arrangement, and the only estimate in it is the lead resistance itself.

Two probes on a 2.0 kΩ sourcecomputed by solving, not by drawing twice per frequency: the node alone, and the node with the probe's elements across it. The one-to-one probe's 115.0 pF makes the reading one per cent wrong at 6.79 kHz. The ten-to-one probe puts 12.8 pF in series with the cable, so its tip sees 11.5 pF and the same error arrives at 69.2 kHz — 10 times further up, bought with a factor of ten in signal — the two edges stand in the ratio of the tip capacitances, 10.00. At direct current neither probe is capacitive at all and the ten-to-one still reads 0.02% low, because 10 MΩ across 2.0 kΩ is a divider.-6-4-201101001k10k100k1M10M100Mfrequency (hertz)log₁₀ of the error in the reading1× probe, 115.0 pF at the tip10× probe, 11.5 pFone per cent1× is 1% out at 6.79 kHz10× at 69.2 kHzsolved, then checked — the node with and without the probe1% wrong at 6.79 kHz with a 1× probe
Fig. 5 The other instrument in this field, and the same statement in the frequency domain. A probe’s loading is a capacitance across a node; an ohmmeter’s is a resistance in series with a current. Both are the instrument appearing in the answer, and in both cases the size of the effect is computable from the instrument’s specification and the circuit’s own impedance.
Impedance of a series RLC of Q = 4, measured by driving itOne ampere is forced into the terminals at each frequency and the resulting voltage is the impedance. The minimum is 7.91 Ω at 5.03 kHz.1101001k10k1001k10k100k1Mfrequency (hertz)impedance magnitude (ohms)reactances cancel at 5.03 kHz7.91 Ωsolved, then checked — one ampere in, 201 frequenciesnot a component value: what the pair does
Fig. 6 An impedance swept over frequency. Everything on this page is at direct current, and the reason is that lead resistance is the dominant effect there. Move the measurement up in frequency and the leads’ inductance takes over — a metre of lead is about a microhenry, which is a milliohm at 160 Hz and an ohm at 160 kHz — so a four-terminal measurement of a milliohm at any appreciable frequency has a second problem this page does not model.

What the field adds up to

Three instruments, three mechanisms, one argument.

A probe puts capacitance across a node and its reading is one per cent wrong above a frequency set by that capacitance and the source’s resistance. A compensated divider has two ratios and they agree only inside a four-and-a-half per cent window of one component’s value. An ohmmeter’s leads are in series with the thing being measured and dominate below a resistance set by their own.

None of the three is an inaccuracy in an instrument. Every one of them is the instrument doing exactly what it is built to do, in a circuit that now includes it — and in every case the size of the effect is computable in advance from two numbers that are both on the specification sheet.

The three fixes are worth comparing, because they are structurally different and the differences are instructive. The probe’s fix is attenuation: divide before the capacitance, and pay a factor of ten in signal for a factor of ten in bandwidth. The compensated divider’s fix is cancellation: make a pole and a zero coincide so that the frequency dependence disappears, and pay for it with an adjustment that has to be maintained. The four-terminal arrangement’s fix is separation: move the offending resistance into a path where it does not enter the answer, and pay for it with two more wires.

Only the third of those is free of a residual. An attenuator still loads; a cancellation still drifts; a separated path is simply not in the answer, and its resistance can be anything at all. That is why the four-terminal arrangement is the one of the three that is described as eliminating its error rather than reducing it, and the description is very nearly accurate — the residual it leaves is the voltmeter’s own input resistance, six orders below the effect it removed.

A network solved, and checked: a bridge, which no series-parallel reduction reachesNode potentials from modified nodal analysis. The branch currents are then recomputed from each element's own law and summed at every node; the residual is 2.7e-16 of the largest current in the circuit, which is floating-point rounding and nothing else.a bridge, which no series-parallel reduction reachesnode a7.5566 Vnode b4.7993 Vcurrent law, rebuilt from the element laws2.71e-16 of the largest branch currentpower delivered against power dissipated4.33e-16 apart · 48.07 mWsolved, then checked — 6 elementsa linear network has no edge: this one is exact
Fig. 7 A network solved and checked. Every reading on every page of this field is one of these: a netlist that contains the circuit and the instrument, solved together, with current law rebuilt from the element relations and the energy counted twice. There is no separate theory of measurement here — only the observation that the instrument is an element.

One thing this page does not model

The two-wire error described here is entirely resistive, and there is a second error at low resistance that is not.

Every junction between two dissimilar metals is a thermocouple, and a measurement circuit contains several: copper to solder, solder to the resistive element, copper to the connector’s plating. Each produces a few microvolts per kelvin of temperature difference, and at low resistance a few microvolts is a large signal — the voltage across a milliohm at a milliampere is one microvolt, so a one-kelvin gradient across a single junction can be several times the quantity being measured.

The standard answer is to reverse the current and average the two readings, since the thermal voltage does not reverse and the resistive one does. That is a genuinely different technique from the four-wire arrangement, it solves a genuinely different problem, and this page does not model it at all: the netlists here contain no sources other than the forcing current and no temperature anywhere.

Naming it is the honest thing to do, because a reader who takes this page as a complete account of low-resistance measurement would have the four-wire technique and none of the thermal one, and at milliohms the second matters as much as the first. The site’s rule asks every model to carry the condition under which it stops applying, and the condition here is a statement about physics that is not in the model rather than a frequency or an amplitude that is.

Where this goes next

-6-4-201101001k10k100k1M10M100Mfrequency (hertz)log₁₀ of the error in the reading1× probe, 115.0 pF at the tip10× probe, 11.5 pFone per cent1× is 1% out at 6.79 kHz10× at 69.2 kHzsolved, then checked — the node with and without the probe1% wrong at 6.79 kHz with a 1× probe The probe is part of the circuit A one-to-one oscilloscope probe on a two-kilohm source gives a reading that is one per cent wrong at 6.8 kHz. Not because the instrument is inaccurate — it is reading correctly — but because the hundred and fifteen picofarads at the end of the cable are across the node, and above that frequency the trace on the screen is a picture of a circuit that only exists while the probe is attached. 024681010m100m110current drawn from the source (amperes)terminal voltage, solvedthe ideal source: 9 V at any current1% low at 180 mAthe model9 Vrloadsolved, then checked — the load swept over four decadesthe ideal source is 1% wrong above 180 mA The source that is not a source An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object. 02461001k10k100k1M10Mload resistance across the output (ohms)output voltage, solved with the load in place6.0 V with nothing connected1% low at 495 kΩthe circuit12 VR₁R₂R_Lsolved, then checked — the load swept over six decadesthe ratio is 1% wrong below 495 kΩ 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.