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For the supplied linear meter model let ${\omega }_0^2=C/I$ and $2\gamma ={\Phi }_0^2/(RI)$. If the applied step is $i_s$, the settled deflection is ${\theta }_{\infty }={\Phi }_0{i}_s/C$. An underdamped response overshoots and rings about that value; a critically damped response reaches it monotonically without ringing; an overdamped response is also monotone from these initial conditions but has a slower final approach. The original plot uses the common final value and natural-frequency time scale.

The critical step response of a galvanometer has $\gamma ={\omega }_0$ and repeated homogeneous root $-{\omega }_0$. Apply $\theta (0)=\dot{\theta }(0)=0$ to the general constant-force solution to obtainIts derivative is ${\theta }_{\infty }{\omega }_0^{2}t{e}^{-{\omega }_{0}t}$, with the sign of the step. It is the fastest nonoscillatory response within this second-order family at fixed ${\omega }_0$, avoiding a needle that repeatedly crosses the reading or an unnecessarily slow overdamped settling.

For sinusoidal drive choose $e^{i\omega t}$ phasors. Writing $F_0={\Phi }_0{i}_0/I$, the response amplitude isFor positive $F_0$, choose the continuous lag branch from zero to $-\pi$ as frequency increases; an ordinary single-argument arctangent without the quadrant would give the wrong high-frequency phase.

For the critical galvanometer transfer function, the denominator factors as $({\omega }_0+i\omega {)}^2$. ThusFor $\omega \ll {\omega }_0$, the meter follows the quasi-static deflection ${\Phi }_0{i}_0/C$ with a small lag $\varphi \simeq -2\omega /{\omega }_0$. For $\omega \gg {\omega }_0$, its amplitude is $F_0/{\omega }^2$ and its lag approaches $-\pi$.

The stated components give $\gamma =12.5$ s⁻¹ and, if critical, ${\omega }_0=12.5$ rad/s, or $f_0=1.99$ Hz. At 50 Hz, $\omega =314.16$ rad/s andThe needle barely responds compared with its DC sensitivity and its mean deflection for a pure sinusoid is zero in this linear model. It is unsuitable for directly measuring 50-Hz amplitude or RMS current without an additional rectifier or another readout method. The calculation treats ${\Phi }_0$ as the stipulated torque/back-emf coefficient of the given equation; it does not infer an unprovided winding geometry.

Use $e^{i\omega t}$ phasors. The relation $V=L\dot{I}$ gives $Z_L=i\omega L$. The three branches form a parallel RLC circuit and share the source voltage. The source-current phasor and its real time dependence areConsequently its amplitude is $V_0\sqrt{R^{-2}+(\omega C-1/(\omega L){)}^2}$ and its phase relative to the voltage is ${\tan }^{-1}[R(\omega C-1/(\omega L))]$. At low frequency it is inductive and lags toward $-\pi /2$; at high frequency it is capacitive and leads toward $+\pi /2$. The ideal amplitude diverges at either extreme and has a finite minimum between them; real component losses and source electrical impedance limit those asymptotes.

Zero phase requires cancellation of electrical susceptances:The inductor and capacitor currents are nonzero, equal in magnitude and opposite in phase, while the resistor supplies the in-phase source current. energy oscillates between the reactive elements; dissipation occurs in the shunt resistor. A source-current maximum would describe a different, series-resonant circuit.

The specified ratio isIt exceeds $1/\sqrt{2}$ when $|\omega C-1/(\omega L)|<1/R$. Define $b=1/(RC)$ and ${\omega }_0=1/\sqrt{LC}$. The two positive endpoints are ${\omega }_{\pm }=[\sqrt{b^2+4{\omega }_0^2}\pm b]/2$, soThe endpoints have geometric mean ${\omega }_0$. Increasing the parallel resistance reduces dissipative loading and narrows this band, in contrast with increasing the resistance of a series RLC circuit. Equivalently, $Q={\omega }_0{U}_{\max }/P_{\mathrm{loss}}$, with $U_{\max }=C{V}_0^2/2$ and $P_{\mathrm{loss}}=V_0^2/(2R)$. The parallel RLC bandwidth here belongs to the explicitly defined current ratio, not a half-maximum rule applied to a source-current peak that does not exist.

In the $e^{i\omega t}$ phasor convention, the total series RL circuit impedance is $R+i\omega L$. The resistor voltage follows the current, so its transfer function relative to the applied voltage isAt the stated condition $R=\omega L$,Thus the resistor-voltage amplitude is $\boxed{V_{R,0}=V_{\mathrm{in},0}/\sqrt{2}}$ and it lags by $45^{\circ }$. For a cosine input it is $V_R(t)=V_{\mathrm{in},0}\cos (\omega t-\pi /4)/\sqrt{2}$.

Complex impedance. In sinusoidal steady state write physical voltage and electric current as real parts of $\tilde{V}{e}^{i\omega t}$ and $\tilde{I}{e}^{i\omega t}$. The electrical impedance $Z=\tilde{V}/\tilde{I}$ is a ratio of phasors; its magnitude gives the voltage-to-current amplitude ratio and its argument gives their phase difference. It is measured in ohms.

For an ideal capacitor, $q=Cv$ and $i=dq/dt=Cdv/dt$. Consequently $\tilde{I}=i\omega C\tilde{V}$, givingWith this time convention the voltage lags the electric current by $\pi /2$.

Bridge output. The original diagram has $R_1$ and $C_1$ in parallel in the upper-left arm, and $R_2$ and $C_2$ in series in the lower-left arm. The point labelled $C$ is connected to the upper supply node. Set the lower supply node to zero, so $V_C=V_i$. Assuming the measured output draws negligible electric current, the two voltage dividers giveTheir difference is thereforeThe reactive-arm electrical impedances are

Balance. Setting $V_{AB}=0$ and cross-multiplying gives $R_3{Z}_2=R_4{Z}_1$. For this Wien bridge,The right-hand side must be the real number $R_4/R_3$. Equating imaginary and real parts givesPositive capacitances require $R_4/R_3>R_2/R_1$. In the equal-component case $R_1=R_2=R$ and $C_1=C_2=C$, this reduces to

Band-pass output and phase. Preserve the requested voltage polarity: $V_{AC}=V_A-V_C$. Since $V_C=V_i$,Put $u=\omega RC>0$. The equal-component ratio is $Z_2/Z_1=2+i(u-u^{-1})$, so the Wien bridge band-pass response isAt finite positive frequency, its imaginary part vanishes exactly when $u-u^{-1}=0$. ThusThe magnitude $|H|=[9+(u-u^{-1}{)}^2{]}^{-1/2}$ is maximal there and tends to zero at both frequency extremes, as expected for a band-pass filter.

A convenient continuous, unwrapped wave phase isIt decreases from $-\pi /2$ as $\omega \to 0^{+}$, passes through $-\pi$ at ${\omega }_r$, and tends to $-3\pi /2$ as $\omega \to \infty$. A principal-value plot instead jumps from $-\pi$ to $+\pi$ at resonance; both represent the same phasor. Reversing the output to $V_{CA}$ would shift the phase by $\pi$ and make the resonant ratio $+1/3$, but that is a different output polarity from the one printed.

The original diagram places the output across the $10\Omega$ resistor, in series with an ideal $1\mathrm{mH}$ inductor. Treat the voltages as phasors with time dependence $e^{i\omega t}$. The electrical impedances are $Z_R=R$ and $Z_L=i\omega L$, so the same electric current flows through both:This is an RL low-pass filter, with corner angular frequencyThe frequency response tends to one for $\omega \ll {\omega }_c$, has magnitude $1/\sqrt{2}$ at the corner, and tends to ${\omega }_c/\omega$ for $\omega \gg {\omega }_c$. The requested sketch uses a logarithmic horizontal axis and a linear voltage-gain axis:

At the two specified angular frequencies,The relative attenuation by a linear filter is the ratio of these gains:Thus the noise amplitude is attenuated by a factor of 10 relative to the signal. A squared-voltage ratio would decrease by a factor of $100$; the question's voltage transfer function gives the amplitude comparison.

For electric potential difference and electric current varying as sinusoids, choose the phasor conventionThe electrical impedance isIts modulus gives the voltage-to-current amplitude ratio, and its complex argument gives the voltage-minus-current phase difference. For ideal components,These follow from Ohm's law $V=RI$, the capacitor law $I=CdV/dt$, and the inductor law $V=LdI/dt$. In phasor notation a time derivative becomes multiplication by $i\omega$, turning the linear ordinary differential equations of the electrical circuit into algebraic equations. Electrical impedances add in a series circuit; their reciprocals, the electrical admittances, add in a parallel circuit. This computes both amplitudes and phase differences in the steady response to sinusoids.

A peak amplitude phasor is the complex amplitude that reconstructs a real single-frequency sinusoid without an extra amplitude multiplier:Its modulus is the sinusoid's peak amplitude, and its complex argument is the wave phase relative to the chosen cosine reference. This is the default convention in phasor. MIT's sinusoidal notes explicitly describe the cosine reference and phase/amplitude parameters.^[ref]

For an original example, take $v(t)=3\cos \omega t-4\sin \omega t$ volts. Expanding the real part shows that $\hat{V}=3+4i$ volts, with peak amplitude $5$ volts and wave phase $\arctan (4/3)\approx 53.13^{\circ }$. The corresponding root mean square phasor is $(3+4i)/\sqrt{2}$ volts, whose magnitude is about $3.536$ volts. The physical signal is the same under both conventions.

The peak phasor versus positive-frequency Fourier coefficient distinction follows fromThus the positive-frequency coefficient in this two-exponential expansion is $\hat{V}/2$, not $\hat{V}$. A real sinusoid requires its conjugate negative-frequency term as well. This algebra does not assign ordinary Fourier-transform function values to an infinite-duration sinusoid; those require the corresponding distributional interpretation.

A common time-origin change multiplies both voltage and current peak phasors by the same unit-modulus factor, leaving their relative phase and impedance ratio unchanged. Peak-versus-RMS normalization in complex electrical power is still essential when an amplitude convention changes rather than just the time origin.

Peak-versus-RMS normalization in complex electrical power determines the coefficient multiplying the product of voltage and conjugate current phasors. A peak amplitude phasor reconstructs a sinusoid directly, while a root mean square phasor requires a factor of $\sqrt{2}$. The DOE effective-value derivation relates these amplitudes by comparing heating from AC and DC currents.^[ref]

To track any combination, writewhere each reconstruction scale is $1$ for a peak phasor or $\sqrt{2}$ for an RMS phasor. A period average of the product leaves the real powerUsing the same positive-frequency and passive-sign convention, the corresponding complex electrical power isThus the coefficient is $1/2$ for two peak-normalized amplitudes, $1$ for two RMS-normalized amplitudes, and $1/\sqrt{2}$ for a mixed pair. The last case is why merely recognizing a product as a power calculation is not enough to choose its normalization.

For an original example, let the electric potential difference be represented by peak amplitude $V=100\mathrm{V}$ and the electric current by RMS amplitude $I=2e^{-i\pi /6}\mathrm{A}$, at the same frequency. ThenMultiplying the two numbers as though both were RMS phasors overestimates the complex electrical power by $\sqrt{2}$. Inserting a peak-pair factor $1/2$ instead underestimates it by $1/\sqrt{2}$. The time-domain reconstruction, $v(t)=100\cos \omega t$ and $i(t)=2\sqrt{2}\cos (\omega t-\pi /6)$, independently verifies the period average.

The mixed peak and RMS gain-calibration error also matters when source and receiver amplitude reports are combined: matching units does not establish matching amplitude conventions. This calculation concerns one sinusoidal component. A nonsinusoidal waveform requires its actual period average or an appropriately normalized harmonic decomposition, not one unspecified phasor product.

A phasor is a complex amplitude representing a sinusoid at one specified angular frequency: $a(t)=\mathrm{Re}(\hat{A}{e}^{\mathrm{i}\omega t})$. Its modulus gives the peak amplitude and its complex argument gives the wave phase in this convention. Both may instead use root mean square magnitudes if that normalization is stated.

In an 1880 paper, John William Strutt, known as Lord Rayleigh, studied the resultant of many oscillations with a common frequency and arbitrary oscillation phases. His large-number calculation distinguished a mean intensity from a definite resultant amplitude and used planar coordinates followed by a polar integral. The resulting amplitude density is the Rayleigh distribution in a different scale convention. This is a large-number description, not an exact law for every finite collection of oscillations.^[ref]

To see the connection in a chosen model, take independent oscillation phases ${\Theta }_j$ with uniform distribution on $[0,2\pi )$ and equal component amplitude $a$. The phasor sum has coordinatesBoth have mean zero, variance $n{a}^2/2$ and zero covariance. The multivariate central limit theorem gives an approximately circular Gaussian random vector for large $n$, with $\sigma =a\sqrt{n/2}$. Its amplitude $\sqrt{X_n^2+Y_n^2}$ is therefore approximately Rayleigh. For every finite $n$, that amplitude is bounded by $na$, whereas a Rayleigh distribution has an unbounded tail. At $n=1$ it is exactly $a$. This simple endpoint check prevents the approximation from becoming a false exact identity.

A root mean square phasor represents a single-frequency sinusoid by a complex amplitude whose magnitude is its root mean square value. With the convention $v(t)=V_{\mathrm{pk}}\cos (\omega t+\varphi )$ and $\omega >0$, defineIndeed, the period average of $v^2$ is $V_{\mathrm{pk}}^2/2$, so $|\underline{V}|=V_{\mathrm{rms}}$. The factor $\sqrt{2}$ in reconstruction distinguishes this convention from the peak-amplitude phasor convention in the parent article. The DOE handbook derives the same effective value by comparing the heating from alternating current and direct current.^[ref]

Using RMS phasors for both electric potential difference and electric current, the passive-sign convention gives complex electrical powerTo see the normalization directly, multiplying the two reconstructed sinusoids and averaging over a period leaves $P=\mathrm{Re}(\underline{V}{\underline{I}}^{\ast })$. Peak-amplitude phasors would instead require the prefactor $1/2$. This peak-versus-RMS normalization in complex electrical power must be stated before combining a source model with measured amplitudes.

For an original example, let $\underline{V}=10e^{i0}\mathrm{V}$ and $\underline{I}=2e^{-i\pi /3}\mathrm{A}$. The electric current lags the electric potential difference by $60^{\circ }$, givingThe corresponding time-domain peaks are $10\sqrt{2}\mathrm{V}$ and $2\sqrt{2}\mathrm{A}$. Treating those peaks as RMS phasor magnitudes would falsely double the computed complex electrical power. This representation assumes a sinusoid at one common frequency; a nonsinusoidal signal can have a root mean square value without being represented by one RMS phasor.

An ideal voltage source drives a series resistor feeding a parallel inductor and capacitor. In the $e^{i\omega t}$ convention, the branch voltage ratio is the title expression. It is a band-pass response with unit peak at $1/\sqrt{LC}$, phase from $+\pi /2$ through zero to $-\pi /2$, and half-power bandwidth $1/(RC)$. At resonance the input current vanishes while equal and opposite inductor/capacitor currents remain finite. Reactive branch-current magnitudes must be combined as phasors, not added as positive scalars.

A series resistor $R$ and capacitor $C$ feed a parallel pair with the same values. Measuring the parallel pair's voltage relative to the return gives this unloaded transfer function for phasors proportional to $e^{i\omega t}$. The voltage divider applied to $Z_s=R+1/(i\omega C)$ and $Z_p=R/(1+i\omega RC)$ proves the formula. The amplitude peaks at $\omega =1/(RC)$ with ratio $1/3$ and zero relative phase. At low and high positive frequencies, the transfer function tends respectively to $i\omega RC$ and $-i/(\omega RC)$. Reversing output polarity changes its sign, while loading changes the response itself.

Consider a stipulated series circuit containing an ideal resistor of electrical resistance $R=100\Omega$ and an ideal capacitor of capacitance $C=1\mu \mathrm{F}$. The same electric current passes through both. Their terminal electric potential differences add by Kirchhoff's voltage law, soA quick derivation uses $i=Cd{v}_C/dt$: taking the derivative of the complex exponential multiplies the capacitor voltage phasor by $\mathrm{i}\omega$, hence ${\hat{V}}_C=\hat{I}/(\mathrm{i}\omega C)$. Dimensionally, $1/(\omega C)$ has units $s/F=V/A=\Omega$, matching $R$.

At frequency $f=1000\mathrm{Hz}$, the angular frequency is $\omega =2\pi f$. The signed capacitive electrical reactance, modulus and complex argument of the total electrical impedance areApply $V_{\mathrm{rms}}=1.00\mathrm{V}$ across the complete series circuit. Its root mean square electric current isWith voltage as the zero-phase reference, the current phasor has complex argument $+57.858092^{\circ }$: electric current leads. The phasor drops have moduli $V_{R,\mathrm{rms}}=0.532018\mathrm{V}$ and $V_{C,\mathrm{rms}}=0.846733\mathrm{V}$; these add as complex numbers, not as positive magnitudes. Their squared magnitudes sum to $(1.00\mathrm{V}{)}^2$.

The mean absorbed power uses the root mean square values:The ideal capacitor has zero cycle-averaged absorbed power; all the mean absorbed power becomes Joule heating in the resistor.^[ref] To see why the phase factor appears, average $p(t)=v(t)i(t)$ over a full cycle with $v(t)=\sqrt{2}{V}_{\mathrm{rms}}\cos \omega t$ and $i(t)=\sqrt{2}{I}_{\mathrm{rms}}\cos (\omega t-\varphi )$. The cosine addition formula reduces the average to $V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$; the term at $2\omega$ averages to zero. Independently, $p_C=d(C{v}_C^2/2)/dt$, and the capacitor energy returns to its starting value after each cycle.

At fixed applied $V_{\mathrm{rms}}$, the limit $\omega \to 0^{+}$ gives $|Z|\to \infty$, $\varphi \to -\pi /2$ and $I_{\mathrm{rms}}\to 0$: the ideal capacitor blocks steady electric current. The limit $\omega \to \infty$ gives $Z\to R$, $\varphi \to 0$ and $I_{\mathrm{rms}}\to V_{\mathrm{rms}}/R$. These are limits of the ideal lumped electrical circuit; unwanted lead inductance and component losses can invalidate that model at high frequency. At $\omega RC=1$, the equal legs of the impedance right triangle give $|Z|=\sqrt{2}R$ and $\varphi =-\pi /4$ without numerical evaluation. Measuring both the magnitude and the sign of the phase difference distinguishes this capacitor response from an equally large positive electrical reactance.

The sinusoidal steady state of a stable linear time-invariant system driven by a sinusoid is its persistent response at the driving frequency, after the relevant transient response has decayed. A phasor calculation finds that response; it does not generally satisfy the circuit's switch-on initial conditions. The late-time assumption is explicit in Berggren's treatment of sinusoidal circuit analysis.^[ref]

Using peak phasors, write the source as $v_s(t)=\mathrm{Re}({\hat{V}}_s{e}^{i\omega t})$. If $H(i\omega )$ is the relevant transfer function, its particular solution isThe complete response adds a homogeneous solution determined by the initial conditions. Approaching the displayed expression requires the modes affecting the measured output to decay; an unchanged amplitude and a fixed phase shift then characterize its frequency response.

For an original turn-on example, drive a series RC circuit with $v_s(t)=V_0\cos \omega t$ for $t\ge 0$, measure the capacitor's electric potential difference, and take $v_C(0)=0$. With time constant $\tau =RC>0$, Kirchhoff's voltage law gives the first-order linear differential equationPut $x=\omega \tau$. Substitution verifies the sinusoidal particular solution $V_0(\cos \omega t+x\sin \omega t)/(1+x^2)$. Adding $A{e}^{-t/\tau }$ and imposing the initial condition fixes $A=-V_0/(1+x^2)$, givingIts sinusoidal part has amplitude $V_0/\sqrt{1+x^2}$ and lags by $\arctan x$. The exponential settling bound for RC sinusoidal turn-on, relative to that amplitude, isWith chosen $R=1\mathrm{k\Omega }$, $C=1\mu \mathrm{F}$, $V_0=1\mathrm{V}$ and $\omega =1000rad/s$, we have $\tau =1\mathrm{ms}$ and $x=1$. At turn-on the phasor approximation predicts 0.5 volt while the capacitor is initially at zero. After 5 milliseconds, their difference is about 3.37 millivolts, or 0.476 percent of the sinusoidal amplitude. A 1 percent settling time threshold for this specific switch-on is about $4.259\tau$.

A useful measurement compares a long, low-frequency square-wave response with sinusoidal responses at several frequencies. An oscilloscope can identify the exponential decay and then measure amplitude ratios and phase shifts. MIT's laboratory uses that sequence to connect the time constant to the sinusoidal response; probe loading belongs in the measured circuit model.^[ref]

Undamped resonant forcing has no bounded sinusoidal steady state. For an ideal series LC circuit, let $q$ be electric charge and ${\omega }_0=1/\sqrt{LC}$. At resonance, the equationhas the particular solution $q_p(t)=V_{0}t\sin ({\omega }_{0}t)/(2L{\omega }_0)$. Its growing envelope prevents a bounded attracting sinusoidal response. This ideal counterexample explains why decay of the relevant homogeneous solutions is part of the physical steady-state assumption.

A steady periodic thermal state repeats each drive period $P$ after the startup transient has decayed. It is time dependent, unlike a time-independent steady state. For angular frequency $\omega =2\pi /P$, its single-frequency part has the phasor representation $\theta (x,t)=\mathrm{Re}\{\hat{\theta }(x)e^{i\omega t}\}$. Here $\theta$ is the oscillating temperature about the mean profile. Repeating the fitted amplitudes and wave phases over successive cycles checks that startup drift is small; merely completing one drive cycle is insufficient.

A thermal diffusion wave is a periodically forced solution of the heat equation, with spatially decreasing amplitude and varying wave phase. In a semi-infinite rod $x\ge 0$, assume constant thermal diffusivity, negligible side losses, a nearly uniform temperature across each cross-section, and the boundary conditions $\theta (0,t)={\Theta }_0\cos \omega t$ and $\theta (x,t)\to 0$ as $x\to \infty$. A steady periodic thermal state then satisfiesChoose the square root in the complex numbers with $a>0$ so that the amplitude decays into the rod. Substitution of the phasor into the heat equation proves the formula: $i\omega \hat{\theta }=\alpha {\hat{\theta }}^{\prime \prime }$. Taking the real part gives the displayed cosine. In this convention $a$ is the spatial attenuation slope and $b$ is the spatial lag slope, both in ${\mathrm{m}}^{-1}$; $\gamma$ combines them into one complex number. If a complex wavenumber $K$ is defined by $e^{i\omega t-iKx}$ instead, $K=b-ia$.

With $\Delta x=x_2-x_1>0$, let $A_j=|\hat{\theta }(x_j)|$ and define a positive downstream lag $\Delta \varphi$. Thenand either measurement determines thermal diffusivity:The logarithm uses an amplitude ratio, not a ratio of squared amplitudes. The lag is in radians and must be tracked continuously across cycles: a measured phase difference modulo $2\pi$ leaves an integer-cycle ambiguity. Use closer sensor spacings or vary frequency continuously to resolve that ambiguity. Equal attenuation and lag slopes provide a useful diagnostic of the no-side-loss model.

The apparent phase velocity is $\omega /b=\sqrt{2\alpha \omega }$, which depends on frequency. A thermal diffusion wave therefore does not give a material-independent travel speed for heat; it is a forced pattern in a dissipative diffusion equation. The ideal heat equation still has infinite propagation speed, and the observed lag is not the arrival time of a sharp front.

Undamped resonant forcing has no bounded sinusoidal steady state because the forcing continually adds energy to a mode that has no dissipation. For a series LC circuit driven from rest by $V_0\cos \omega t$, let $q$ be its electric charge and ${\omega }_0=1/\sqrt{LC}$. When $\omega \ne {\omega }_0$, imposing $q(0)=\dot{q}(0)=0$ on the linear ordinary differential equation givesBoth frequencies persist: the undamped homogeneous solution never disappears. Taking the limit $\omega \to {\omega }_0$ at fixed time produces the growing resonant response described in sinusoidal steady state. Its factor proportional to time cannot be removed by adding a bounded homogeneous solution.

For an original numerical example, choose $L=10\mathrm{mH}$, $C=1\mu \mathrm{F}$ and $V_0=1\mathrm{V}$. Then ${\omega }_0=10^4\mathrm{rad}{\mathrm{s}}^{-1}$. At exactly ten resonant periods, $t=20\pi /{\omega }_0\approx 6.283\mathrm{ms}$, the resonant solution has $q=0$ butAt twenty periods the electric current doubles and the stored magnetic energy quadruples. These are ideal-model values; a real source and winding have finite electrical resistance and ratings.

With positive series resistance $R$, the resonant phasor calculation instead gives current amplitude $V_0/R$. The DOE handbook explicitly identifies resistance as the remaining series electrical impedance at resonance.^[ref] For weak damping, decay takes a time of order $2L/R$, which grows as resistance decreases. These noncommuting damping and observation-time limits at resonance explain why an infinite steady-state amplitude is a failed limiting calculation, not the actual finite-time response. The late-time transient assumption underlying sinusoidal analysis must be checked.^[ref]