Circuit Response
every model has an edge, and the figure carries it
Networks, and how a solve is checked

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.
15 min read 5 figures The constants decideEvery model has an edgeTwo routes to a number

Two resistors in series across a supply, with the output taken from the join. It is the first circuit anybody meets, its behaviour is a single fraction, and the fraction is memorable enough that most people never write it down again. It is also, in exactly the form it is usually taught, a model with a range — and the range is not stated anywhere in the fraction.

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. 1 A divider of two equal resistances, solved with a load across its output at every value from a thousandth of its own resistance to three thousand times it. The flat line is the answer the ratio gives. The curve is what the network actually does. Move the slider: the ratio never changes, and the point at which the answer collapses moves by four decades.

What the fraction says, and what it leaves out

With nothing connected to the output, the current through the two resistors is the same, so the potential at the join divides in proportion to the resistances. Two equal resistances give half the supply. That is exact, it needs no approximation, and it is true for any pair of values with the same ratio: a hundred ohms and a hundred ohms, ten kilohms and ten kilohms, a megohm and a megohm all give half.

The moment anything is connected across the output, that stops being true, and the reason is immediate: the two resistors no longer carry the same current. Some of the current arriving through the upper resistor now leaves through whatever has been attached, and only the remainder goes on through the lower one. Less current through the lower resistor means less potential across it, which means a lower output. The divider has not changed; the network has.

The interesting part is what decides how much lower, because it is not the ratio. Attaching a load across the output puts it in parallel with the lower resistor, and the answer depends on how the load’s resistance compares with that lower resistance — an absolute comparison between two quantities in ohms, not a ratio. The two equal resistances that gave half the supply are equal whether they are a hundred ohms or a megohm, and a hundred-kilohm load leaves the first almost untouched and destroys the second.

The figure above makes that concrete, and the numbers are worth having in the prose. With a twelve-volt supply and a hundred-kilohm load across the output:

  • a divider of two 1 kΩ resistors gives 5.970 V — 0.5% below its unloaded six volts
  • a divider of two 10 kΩ resistors gives 5.714 V — 4.8% low
  • a divider of two 1 MΩ resistors gives 1.000 V

The third one is not a degraded version of the first. It is a sixth of the answer, from a circuit with the same ratio, into the same load. The specification that the divider is “a two-to-one divider” has told a reader nothing at all about what will come out of it.

The number the model is missing

Every model on this site is drawn together with the point at which it stops being true, and the divider’s is a resistance rather than a frequency. It can be computed exactly by solving the loaded network and asking where the answer falls one per cent below the unloaded value.

For a divider of two equal resistances R, the answer is a load of about 49.5 R. Not a suspicious number and not a rule of thumb: it falls out of requiring the parallel combination of R with the load to be 0.9802 R, and it says something a reader can carry away. A divider is accurate to one per cent only when what it drives is fifty times its own resistance. Below that, the ratio has stopped being a prediction and become an upper bound.

That is the number in the caption strip of the figure above, and it moves with the slider exactly as it should: a divider of two 100 Ω resistors is good to one per cent into 4.95 kΩ, and a divider of two megohms is good to one per cent only into 49.5 MΩ, which is not a load anybody has.

There is a second, blunter number for the case where a reader wants a feel rather than a specification. A load equal to the divider’s own resistance takes a third of the answer away. Two equal resistances loaded by a third of the same value give four volts from a twelve-volt supply rather than six, whatever the value is. That figure is worth remembering because it is exactly the situation that arises when one divider is asked to drive another.

Why the error is one-sided

There is a structural fact hiding in the curve that is worth drawing out, because it is the reason this failure mode is dangerous in a way that a symmetric error would not be.

Adding a load can only ever reduce the output. Whatever is connected across the lower resistor draws current that would otherwise have flowed through it, and there is no arrangement of passive components that can make the join sit higher than the unloaded fraction says. The error therefore has a sign, always the same sign, and it accumulates rather than averaging out.

That matters in a chain. Three dividers in a row, each perfectly adequate on its own, do not have independent errors that partially cancel; every one of them is low, and the product of three one-sided errors is worse than any of them. The ladder later in this essay is exactly that situation and its output is forty per cent below what the three fractions predict.

It also means that a circuit built with the fraction and then measured is not merely inaccurate but inaccurate in a direction that hides itself. An output that reads low is usually attributed to a resistor being out of tolerance, or to a supply that has sagged, and the loading explanation is several places down the list — even though it is typically an order of magnitude larger than either.

Where the deciding quantity actually lives

The reason a ratio cannot answer this question is that the divider has a property the ratio does not record, and it is worth naming because it turns up everywhere afterwards.

Looking back into the output of an unloaded divider, what is seen is the two resistances in parallel. That combination — not the ratio, not either resistance alone — is the quantity that decides everything about loading. A divider of two 10 kΩ resistors looks like a 6 V source behind 5 kΩ. A divider of two 1 MΩ resistors looks like a 6 V source behind 500 kΩ. The two are the same divider by every rule taught with the fraction and completely different objects by the only measure that matters when something is connected.

This is Thévenin’s observation, and it is not being derived here so much as pointed at: any network of sources and resistances, seen from a pair of terminals, behaves exactly like one source behind one resistance. What makes it more than a convenience is that it identifies the missing quantity. The divider’s fraction gives the source; the parallel combination gives the resistance; and the fraction alone is half a description.

A network solved, and checked: a three-section ladderNode 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 1.4e-16 of the largest current in the circuit, which is floating-point rounding and nothing else.a three-section laddernode a3.8462 Vnode b1.5385 Vnode out0.7692 Vcurrent law, rebuilt from the element laws1.41e-16 of the largest branch currentpower delivered against power dissipated2.26e-16 apart · 61.54 mWsolved, then checked — 7 elementsa linear network has no edge: this one is exact
Fig. 2 Three dividers in a row, solved together. Each section is two equal resistances, so each should halve what reaches it — and none of them does, because each is loaded by the two after it. The potentials come out as thirds rather than halves: 3.846 V, 1.538 V and 0.769 V from ten volts, which is a factor of a third at every step and not a factor of a half at any of them.

That ladder is the argument in its most compact form. Anyone building it from three applications of the divider fraction would predict five volts, two and a half, and one and a quarter. The solved answers are 3.846, 1.538 and 0.769 — a factor of a third each time. The fraction is wrong at the first step by twenty-three per cent, and the error compounds.

Two ways to get the right answer, and why the second one is better

The loaded divider has a closed form. Replace the lower resistance by its parallel combination with the load and apply the fraction again. It works, it is exact, and it is what most treatments offer.

It also does not scale. The ladder above cannot be done that way without working backwards from the far end, and a network with a bridge in it — the one in the previous essay — cannot be done that way at all, because no two of its resistors are purely in series or purely in parallel. The moment a circuit stops being a chain, the reduction technique runs out.

The nodal solve does not care. It writes one equation per node and solves them all at once, and a bridge, a ladder and a lone divider are the same problem at different sizes. Every number in this essay came out of it, which is also why they can be trusted: the solve checks itself twice before returning, once by rebuilding the branch currents from the element laws and summing them at each node, and once by requiring the resistors’ dissipation to equal the sources’ delivery.

The figures here take the closed form and the solve as two routes to the same answer, and require them to agree. That is not an abundance of caution. It is the only way to find out that a figure is right, because a plot of the wrong divider looks exactly like a plot of the right one.

What this generalises to

Loading is not a resistive-divider phenomenon. It is what happens whenever one part of a circuit is described without reference to what is attached to it, and the pattern recurs through the whole subject in forms that look unrelated:

  • A signal source has an internal resistance, and the voltage at its terminals falls as current is taken. That is the next essay, and it is the same arithmetic with the divider’s upper resistor playing the part of the source’s insides.
  • An amplifier’s gain is specified with a stated load, and changes with a different one.
  • A filter designed as a cascade of sections behaves as designed only if the sections do not load one another — which is why the filters elsewhere in this collection have followers between their sections, and why that decision is stated rather than hidden.
  • A probe placed on a circuit to measure it changes the circuit it is measuring, by exactly this mechanism, which is why an oscilloscope probe’s input resistance and capacitance appear on its label.
Impedance of a resistor in series with a capacitor, measured by driving itOne ampere is forced into the terminals at each frequency and the resulting voltage is the impedance. The minimum is 1.00e+3 Ω at 477 kHz.1k10k100k1M101001k10k100kfrequency (hertz)impedance magnitude (ohms)reactances cancel at 1.59 kHz1.00e+3 Ωsolved, then checked — one ampere in, 201 frequenciesnot a component value: what the pair does
Fig. 3 The same idea one step further on. What a resistor and capacitor look like from a pair of terminals, against frequency, computed by forcing one ampere in and reading the voltage. Below the corner the pair is a capacitor and above it a resistor — so the “output resistance” of a network is not generally a resistance at all, and the divider’s story continues into the next field.

The honest statement of the model

The divider fraction is not wrong. It is a statement about a particular network — one with nothing connected to its output — and it is exactly right about that network. What goes wrong is the silent extension of it to a network that has something connected, which is every network anybody actually builds.

Written out completely, the model reads: the output is the supply times the ratio of the lower resistance to the sum, provided that what is connected across the output has a resistance at least fifty times the parallel combination of the two. The proviso is not a footnote and it is not conservative. It is the boundary of the claim, it is computable from the divider’s own values, and it is the first instance on this site of the rule the whole collection runs on.

Where four of this site's models stop being trueIn order: the ideal operational amplifier at 1.42 kHz, a 10 V output at full amplitude at 7.96 kHz, Kirchhoff's laws on 10.0 cm at 3.97 MHz, the ideal 100 nF capacitor at 4.69 MHz. The fifth boundary is an amplitude rather than a frequency and cannot share this axis: a small-signal model is 1% wrong above 7.3 mV, at every frequency there is.101001k10k100k1M10M100M1G10Gfrequency (hertz)the ideal operational amplifier1.42 kHz — a gain of 100 from a 1 MHz part is 1% low herea 10 V output at full amplitude7.96 kHz — above this the output cannot move fast enoughthe ideal 100 nF capacitor4.69 MHz — 1.2 nH of lead makes it 10% wrong hereKirchhoff's laws on 10.0 cm3.97 MHz — the board is one degree long hereeach bar is where the model may be used; the rule at its end is the numbersolved, then checked — each boundary from its own modeland one that is not a frequency: 7.3 mV
Fig. 4 Where four other models on this site stop being true, for comparison. The divider’s boundary is a resistance rather than a frequency and so cannot share this axis — but it is the same kind of statement, arrived at the same way: solve the network the model is a simplification of, and find where the two answers part company by a stated amount.
Four networks the solver refusesEach has no answer, for a reason that is a fact about the circuit rather than about the arithmetic. The solver names the reason; it does not return a number.no path to grounda node whose potential nothing fixesrefused: floatingtwo sources in a loop5 V3 Vtheir currents are not determinedrefused: vloopa resistance of zero0 Ωan element with no law of its ownrefused: valuean output that is not a node?a response asked for where there is nothingrefused: nodesolved, then checked — each refusal was runa solver that never declines is untested
Fig. 5 What the same solver does with a network that has no answer at all. A divider whose lower arm has been replaced by a wire is not a divider with an extreme ratio; it is a constraint rather than an element, and the third panel here refuses it by name rather than dividing by zero somewhere quiet.

The choice that has no good answer

Since a divider is more faithful the lower its resistances are, an obvious question is why anybody would ever build one out of megohms. The answer is that the boundary drawn above is one of two, and they pull in opposite directions.

Lowering the resistances improves the loading behaviour and costs current. A two-to-one divider of 1 kΩ resistors across twelve volts draws six milliamperes continuously, does nothing useful with any of it, and dissipates seventy-two milliwatts as heat. The same divider in megohms draws six microamperes. For anything running from a battery, the second is the only tenable choice, and the loading problem then has to be solved elsewhere — by putting an amplifier with a very high input resistance immediately after the divider, which is precisely what such amplifiers are for.

There is a third consideration that decides the matter at high impedance, and it belongs to a later field: a divider made of megohms is loaded not only by whatever is deliberately connected to it but by stray capacitance, and stray capacitance loads it more at every increase in frequency. A megohm divider with five picofarads of stray capacitance across its lower arm has already lost a tenth of its output at sixty hertz — which is the frequency of the mains wiring in the room. The low-frequency loading rule and the capacitive one are the same statement about impedance, and at high resistance the second arrives first.

So there is no value of resistance that is simply correct. There is a range bounded below by the current that can be spared and above by what is connected and by the stray capacitance, and the range can be empty, at which point the divider is the wrong circuit and something active has to be used instead. Stating the boundary as a number is what makes that decidable rather than a matter of habit.

A note on what “one per cent” is doing

The boundary above was drawn at one per cent, and the choice deserves defending because an arbitrary-looking threshold is exactly the sort of thing that makes a computed number feel like a quoted one.

One per cent is not a tolerance anybody has to accept; it is a reporting threshold, chosen because it is small enough to be inside what a reader would call agreement and large enough to be well outside the noise of the computation. The solve here is exact to about fifteen digits, so the boundary could equally have been drawn at a part in a thousand or a part in a million; every such boundary on this site would simply move to a different, equally computable place.

What must not happen is the boundary being drawn where it is convenient. The figures state the fraction they used, the fraction is the same across a whole field wherever that is possible, and the solve underneath is the same one that produced the curve. A reader who prefers a tenth of a per cent can move the marker with one number and the argument does not change: a divider is ten times more demanding about its load at that threshold, needing a load about 500 times its own resistance, and the point that the ratio does not predict any of this is untouched.

Where this goes next

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. 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 What a network answers, and how the answer is checked A circuit has exactly one answer and a matrix finds it. The part that matters is not that the answer exists but that it can be checked twice, by routes that share no arithmetic — and that a circuit with no answer is refused by name rather than returned as a large plausible number. 101001k10k100k1M10M100M1G10Gfrequency (hertz)the ideal operational amplifier1.42 kHz — a gain of 100 from a 1 MHz part is 1% low herea 10 V output at full amplitude7.96 kHz — above this the output cannot move fast enoughthe ideal 100 nF capacitor4.69 MHz — 1.2 nH of lead makes it 10% wrong hereKirchhoff's laws on 10.0 cm3.97 MHz — the board is one degree long hereeach bar is where the model may be used; the rule at its end is the numbersolved, then checked — each boundary from its own modeland one that is not a frequency: 7.3 mV Every model has an edge Four assumptions this collection runs on, with the frequency at which each stops being true, on one axis. The ordering is not the one most readers would guess — an ordinary amplifier circuit runs out of model at 1.42 kHz, three thousand times sooner than a ten-centimetre circuit board does.