Circuit Response
every model has an edge, and the figure carries it
Frequency, which is the same solve

Three voltages that close on one, and the steady state they assume

Kirchhoff's voltage law drawn as a polygon in the complex plane. The three element voltages of a series circuit add head to tail to the source exactly — while their magnitudes add to five times it. And the whole picture is a statement about a settled circuit, which takes a computable number of cycles to arrive.
14 min read 4 figures Two routes to a numberEvery model has an edgeThe constants decide

A resistor, an inductor and a capacitor in series across a one-volt source. Kirchhoff’s voltage law says the three element voltages add to the source. At the frequency where the two reactances cancel they measure 1.000 V, 2.128 V and 2.128 V — a total, if the magnitudes are added, of 5.255 V from a one-volt supply.

Nothing is wrong. The law is about complex numbers and the meter is not.

Three element voltages closing on one source, at 1.59 kHzSolved at 1.59 kHz, which is 1.00× the frequency at which the two reactances cancel. The three phasors add head to tail to the 1 V source exactly; their magnitudes sum to 5.26 V, which is not the same statement.realimaginaryacross Racross Lacross Cthe source, 1 Vmagnitudes|v_R| = 1.000 V|v_L| = 2.128 V|v_C| = 2.128 Vsum 5.255 Vvector sum 1.000 Vsolved, then checked — one solve at 1.59 kHzsteady state only: 3 cycles to settle
Fig. 1 The same three voltages drawn as arrows in the complex plane, added head to tail. The chain closes exactly on the source: the residual is below 10−12 of a volt, which is arithmetic noise and not a measurement. The slider moves the frequency, and at resonance the inductor’s arrow and the capacitor’s arrow are equal and opposite — each more than twice the source, and cancelling.

What a phasor is, in this account

A phasor is usually introduced as a device: a rotating vector, a shorthand for a sinusoid, a notation to be learned. In a collection built on a complex-valued solver it is none of those. It is simply the number the solve returns.

Feed the network a source of one volt at complex frequency s = j2πf and solve. The answer at each node is one complex number, and that number is the whole steady-state description of what happens there: its magnitude is the amplitude of the sinusoid, its angle is the phase relative to the source. There is no conversion step, no reconstruction, and nothing to remember, because the representation and the computation are the same object.

That is why the arrows in the figure are not an illustration of the answer. They are the answer, plotted in the plane it lives in rather than as two separate curves against frequency. Both views show the same solve; the arrows make the addition visible and lose the frequency axis, and the Bode pair keeps the frequency axis and hides the addition.

Why the law still holds when the numbers look impossible

The element voltages sum to the source as complex numbers, and complex addition is head-to-tail addition of arrows. Three arrows pointing in different directions can each be long while their sum is short, in exactly the way three sides of a triangle can be long while the triangle closes.

The particular geometry here is fixed by what each element does. Across a resistor, voltage and current are in phase, so the resistor’s arrow points along the current. Across an inductor the voltage leads the current by 90°; across a capacitor it lags by 90°. Since the three elements are in series they share one current, so the two reactive arrows are at right angles to the resistive one and antiparallel to each other. Whatever their lengths, they subtract.

At the frequency where their magnitudes are equal, they cancel completely, the whole source appears across the resistor, and the current is at its maximum. That is resonance, seen from the arrows rather than from a curve, and the individual reactive voltages at that point are the current times their respective reactances — which is a number set by the components and not bounded by the source at all.

The number that surprises people

The quality factor of a series resonant circuit is the ratio of its reactance at resonance to its resistance, and it decides how large those cancelling arrows are. In the figure, an inductance of 10 mH and a capacitance of 1 µF give a reactance of 100 Ω at the resonant frequency of 1.59 kHz; with 47 Ω of resistance, the quality factor is 2.13.

The voltage across the inductor at resonance is therefore 2.13 times the source. With 4.7 Ω of resistance it would be 21.3 times. With 0.47 Ω it would be 213 times, and a one-volt signal generator would be producing two hundred volts across a component in a circuit that contains nothing capable of amplification.

This is not a paradox and it is not free energy — the energy sloshes between the inductor’s magnetic field and the capacitor’s electric field and only the resistor consumes any — but it is a genuinely practical hazard, and it is the mechanism behind several things that look unrelated. A power-factor correction capacitor resonating with the supply’s leakage inductance. A long cable resonating with an amplifier’s output inductance. A decoupling capacitor resonating with the inductance of the track that connects it, producing an impedance peak precisely where a smooth low impedance was intended.

The arrows make it obvious in a way the algebra does not: the sum being small says nothing whatever about the terms.

Why the arrows are drawn from the solve

A phasor diagram is normally constructed rather than measured: choose a current, draw the resistor’s voltage along it, draw the inductor’s at ninety degrees up and the capacitor’s at ninety degrees down, and close the figure with the source. Done that way the polygon closes because it was constructed to close, and the drawing demonstrates a convention rather than a fact.

Here the three arrows are read off a solved network. Each is the difference between two node potentials that came out of a matrix inversion, and nothing in that inversion knows that the three should add to the source — it only knows the current law at each node and each element’s own relation. The polygon closing is therefore a consequence, and its closing to within 10⁻¹² of a volt is a measurement of how well the solve satisfies a law it was not explicitly given.

The distinction matters more here than in most of the collection, because a phasor diagram is precisely the sort of figure that cannot be wrong when drawn by hand. It is worth having one on the site that could have been.

What happens at the two ends of the slider

The frequency slider runs from a quarter of resonance to four times it, and the arrows do something worth watching at both ends.

Well below resonance the capacitor’s arrow dominates: its reactance is large, so most of the source appears across it, and the circuit is essentially a capacitor with some resistance in series. The current is small and leads the source voltage by close to ninety degrees.

Well above resonance the inductor’s arrow dominates for the mirror-image reason, and the current lags by close to ninety degrees.

At resonance the two are equal and opposite, both several times the source, and the current is in phase with it.

The passage from one to the other is not gradual in the way the description suggests. The phase of the current swings through the whole hundred and eighty degrees in a band of width f₀/Q around the resonance, which for the circuit drawn is about seven hundred and fifty hertz out of sixteen hundred. That is the same f₀/Q that sets the amplitude bandwidth, and it is the reason the two arrows swap places so abruptly on the slider.

The assumption underneath the entire frequency axis

Every phasor statement is a statement about a steady state. The arrows describe a circuit in which the source has been running long enough that nothing is changing from cycle to cycle. Before that, the circuit is doing something else entirely: it is ringing at its own natural frequencies while also responding to the drive, and no arrow describes the sum.

The site’s rule applies to this model like any other, and the boundary is a duration rather than a frequency. It can be computed from the poles: the natural frequencies of this network have a real part of −R/2L, so any transient decays with a time constant of 2L/R, and the number of drive cycles needed for it to fall below one per cent of its initial size is

Q · ln(100) / π ≈ 1.47 Q

which for the figure’s quality factor of 2.13 is about three cycles. That is the number in the caption strip, and it is why the figure is honest about being a steady-state picture.

Three cycles is nothing, and that is exactly what makes this boundary treacherous rather than academic, because it scales with Q. A crystal filter with a quality factor of 50,000 needs about seventy-five thousand cycles, which at ten megahertz is seven milliseconds — an eternity in a circuit that is being asked to switch between channels. A high-quality resonator is by definition one that takes a long time to agree with its own frequency response.

The same statement in the other direction is the useful one for a designer: a circuit cannot be sharper in frequency than it is slow in time. Bandwidth and settling time are two readings of the same pole, and no arrangement of components separates them.

Impedance, drawn rather than derived

The addition in the figure also settles a question that trips readers who arrive from direct-current work: what it means to say a series combination has an impedance.

The impedance of the series chain is the sum of the individual impedances — as complex numbers, so the resistance adds along one axis and the two reactances add along the other, one positive and one negative. The magnitude of the result is the square root of the sum of squares, which is the reason a 500 Ω resistance and a 500 Ω reactance make 707 Ω rather than 1000 Ω, and the reason the answer falls to just the resistance when the two reactances match.

None of that needs a separate derivation once the arrows are on the page: the impedance triangle and the voltage triangle are the same triangle scaled by the current, because every element in a series chain carries the same current.

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. 2 The same chain, read as an impedance instead: one ampere is forced into the terminals at each frequency and the resulting voltage is plotted. The V-shaped minimum is where the two reactive arrows cancel, and the value at the bottom is the resistance alone.

Power, which the arrows also settle

There is a second thing the closed polygon explains, and it is the source of more confusion in practice than the voltages are.

The power delivered by the source is the product of the voltage and the current — as complex numbers, which means the answer has two parts. The real part is power in the ordinary sense: energy that leaves and does not come back, and in this circuit all of it ends up in the resistor. The imaginary part is energy that leaves the source, is stored in a field, and returns half a cycle later, having done nothing.

The arrows make the split visible without arithmetic. The current is along the resistor’s arrow; the component of the source voltage along that direction is the part that produces real power, and the component at right angles to it produces the part that goes back and forth. At resonance, the source voltage lies exactly along the current, the right-angle component is zero, and every watt leaving the source is consumed.

The name for the ratio between the two is the power factor, and it is the cosine of the angle between the voltage and the current arrows. It is worth noticing that it is a property of the load rather than of the supply, and that a load with a poor power factor draws more current than its real power requires — which is why industrial supplies are billed on it and why correction capacitors are fitted. The capacitor supplies the returning energy locally rather than letting it travel back down the distribution network each half cycle.

The convention, and why it has to be stated

One detail that is easy to skip and causes trouble later: the arrows in the figure are amplitudes in the sense of peaks, and half the literature draws them as root-mean-square values instead.

Both conventions are internally consistent and the diagrams are identical apart from a factor of √2. The trouble arises when a formula from one convention meets a number from the other — most commonly in power, where the real power of a sinusoid is half the product of peak voltage and peak current but the full product of the root-mean-square values. A factor of two in a power calculation is not a subtlety.

This site uses peak amplitudes throughout, because the solver returns the phasor that multiplies e^{jωt} and that is a peak by construction. Anywhere a power appears in this collection it is computed from those quantities with the factor made explicit rather than inherited. Stating the convention once is cheaper than checking it at every appearance, and it is the sort of thing that is invisible until two figures disagree by exactly two.

What the arrows cannot show

It is worth being explicit about the cost of the arrow view, since this collection generally prefers the picture that carries more.

An arrow diagram is drawn at one frequency. It shows the addition beautifully and it shows the frequency dependence not at all — which is why the figure here has a slider, and why the slider is the frequency rather than a component value. Moving it is the only way to see the two reactive arrows trade places as the frequency passes through resonance.

An arrow diagram also assumes a single frequency is present. Real signals are not single sinusoids, and the reason phasors are nevertheless useful is superposition: a linear network’s response to a sum of sinusoids is the sum of its responses to each. That is what makes a frequency response a complete description of a linear circuit and it is precisely what fails once anything in the circuit stops being linear — at which point new frequencies appear that were not in the input, and no phasor sum will account for them. The amplitude at which that happens is computable too, and it is the subject of a later essay.

A single-pole low-pass with its corner at 995 HzSolved at 209 frequencies. The straight-line sketch, drawn faintly, is 3.01 dB wrong at the corner and within a tenth of a decibel only below 152 Hz. The phase is already −5.7° a decade before the corner and −84° a decade after it.-60-40-200gain (decibels)the sketch: flat, then −20 dB/decade−3.01 dB-90-450101001k10k100kfrequency (hertz)phase (degrees)sketch within 0.1 dB below 152 Hzsolved, then checked — checked against a chain-matrix productthe sketch is 3.01 dB wrong at 995 Hz
Fig. 3 The other view of the same kind of solve: magnitude and phase against frequency, for a simpler network. Each point on these two curves is one arrow’s length and one arrow’s angle, and the two representations lose different things — the curves lose the addition, the arrows lose the axis.
Two poles at ζ = 0.3, recovered from the matrixThe poles are at -477.4 ± j1518 hertz. Their distance from the origin is the natural frequency to six digits; the cosine of their angle from the negative real axis is the damping ratio. The step response beside them follows.the step this produces00.50011.5001234σjωζ = 0.3000ω₀ = 1592 Hzsolved, then checked — poles by rooting the determinantnatural frequency recovered to 6 digits
Fig. 4 What the same circuit looks like off the imaginary axis. The arrows above are the response at one real frequency; these two points are the complex frequencies at which the network would respond with nothing driving it at all, and they are what decides how long the steady state takes to arrive.

Reading the closure as a check

There is one more use for the closed polygon, and it is the reason the figure is in this collection rather than in a textbook.

The chain closing on the source is not a fact that was drawn in; it is a fact that was tested. The three element voltages are computed independently from the solved node potentials, added, and required to equal the source to within a part in 1012. If the inductor’s contribution had been assembled with the wrong sign — which is one character in a matrix and produces a perfectly plausible response curve — the polygon would not close, and the assertion would stop the build.

This is the general shape of every check on this site. Kirchhoff’s laws are the only thing available that constrains a solution without going through the solve, so they are used as tests rather than being assumed as results. The polygon closing is Kirchhoff’s voltage law used as a test, in the same spirit as the current-law residual printed under the network figures in the previous field.

That the test is also the most legible thing in the figure is a piece of luck the subject offers and that this collection tries to take wherever it appears: the check and the explanation can be the same picture.

Where this goes next

00.200.400.600.8011001k10kfrequency (hertz)fraction of the source across the resistorhalf the power198.9 Hz measuredresonance 1.59 kHzsolved, then checked — half-power points by bisectionf₀/Q predicts 198.9 Hz — exactly Resonance, and the bandwidth it sets exactly The half-power bandwidth of a resonant circuit is f₀/Q — not approximately, but to every digit the arithmetic has, which is rare enough to be worth checking. What is not exact, and is drawn as though it were, is the idea that the band sits centred on the resonance. At a quality factor of one its middle is twelve per cent above. -60-40-200gain (decibels)the sketch: flat, then −20 dB/decade−3.01 dB-90-450101001k10k100kfrequency (hertz)phase (degrees)sketch within 0.1 dB below 152 Hzsolved, then checked — checked against a chain-matrix productthe sketch is 3.01 dB wrong at 995 Hz One solve, read four ways Reactance, phase, the corner frequency and the roll-off are not four ideas. They are four readings of one complex number, obtained from the same matrix that answers direct-current questions — and the straight-line sketch every engineer draws of them is itself a model, three decibels wrong exactly where it is read. the step this produces00.50011.5001234σjωζ = 0.3000ω₀ = 1592 Hzsolved, then checked — poles by rooting the determinantnatural frequency recovered to 6 digits Where the behaviour is written down Two numbers in the complex plane contain everything a second-order circuit will ever do. Their distance from the origin is the natural frequency, the cosine of their angle is the damping — and the fastest-settling circuit is not the critically damped one, which is the case the textbooks name.