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For a specified photodetector measurement, coronagraph throughput $T(\theta )$ is the fraction of a source's light transmitted into that measurement at angular separation $\theta$ from the blocked source, relative to a stated reference configuration. Total transmitted power and power inside a small image opening can give different fractions; one must specify which is used. Throughput usually depends on wavelength and separation. A lower stellar residual is useful only in relation to the retained companion signal and noise. Guyon's worked designs plot transmitted off-axis light against separation to expose that tradeoff. ^[ref].

The cosmic microwave background is nearly isotropic thermal electromagnetic radiation. Its spectrum connects radio receiver measurements with the universe's thermal history: redshift preservation of a thermal photon spectrum cools a blackbody spectrum as space expands. The 1965 Princeton analysis described this relation.^[ref]

A radiometer cannot label a received photon as cosmic, atmospheric or instrumental. Its measured power combines the incoming radiation with the receiver response. The useful question is which comparisons and controls can isolate a sky contribution. This makes the Penzias-Wilson microwave background experiment an example of both observational cosmology and measurement calibration.

Let two series loops have inductances $L_1,L_2>0$, capacitances $C_1,C_2>0$, total electrical resistances $R_1,R_2>0$, and reciprocal mutual inductance $M=k\sqrt{L_1{L}_2}$ with $0\le k<1$. For a steady sinusoid of angular frequency $\omega >0$, choose complex peak amplitudes with time convention $e^{\mathrm{i}\omega t}$. An ideal source of peak electric potential difference $V$ drives loop 1; loop 2 has no independent source. ThenEliminating $I_1$ proves the response formula. Reversing one coil changes the sign of $M$ and the relative wave phase, but leaves this amplitude response unchanged. The measured sense ratio is $|R_s{I}_2/V|$, with $R_s$ included in $R_2$, not added a second time.

For identical loops at their uncoupled resonant frequency ${\omega }_0=1/\sqrt{LC}$,Writing $x={\omega }_{0}M/R$ makes this $x/[R(1+x^2)]$. Its derivative has the sign of $1-x^2$, so response at this fixed frequency is largest when $x=1$, equivalently $k=1/Q$. Too little coupling transfers little excitation; excessive coupling reduces the response at the original centre frequency. This is a fixed-frequency optimum, not a universal optimum over every possible receiver and drive frequency.

For the chosen values, $k=1/Q=0.063246$ gives $M=0.31623\mathrm{mH}$. At source peak $V=0.100\mathrm{V}$, the secondary peak electric current is $0.500\mathrm{mA}$ and the sense peak electric potential difference is $25.0\mathrm{mV}$. The primary peak electric current is also $0.500\mathrm{mA}$, so source mean power is $25.0\mu \mathrm{W}$: each loop dissipates $12.5\mu \mathrm{W}$, checking conservation of energy. Here $V$ is the equivalent ideal source behind $R_1$; a generator's terminal reading must be converted using its measured Thévenin resistance.

In the lossless identical-loop limit, the normal modes have equal or opposite electric currents, making their effective inductances $L+M$ or $L-M$. ThusAt $k=0.20$ these are $45.944$ and $56.270\mathrm{kHz}$. Finite electrical resistance broadens and shifts the driven peaks, so the exact response formula, rather than these lossless values, generates the plot. Changing a capacitor moves the uncoupled tuning through $f_0\propto C^{-1/2}$: increasing it by $21\%$ lowers $f_0$ by the factor $1/1.1$. The model explains why tuning and coupling must be adjusted together; it supplies no transatlantic propagation or detector-threshold prediction.

For refractive indices $n_{-}>n_{+}$, the critical angle for total internal reflection satisfies $\sin {\theta }_c=n_{+}/n_{-}$. Above it, the transmitted normal wavenumber is imaginary and the transmitted field is an evanescent wave.

The critical angle for total internal reflection is the boundary between outgoing refraction and an evanescent wave, rather than an angle at which the external electric field abruptly disappears. Write $n_1=n_{-}$ and $n_2=n_{+}$. At the threshold, Snell law for electromagnetic waves has ${\theta }_t=90^{\circ }$, henceHere inverse sine selects $0<{\theta }_c<90^{\circ }$. For $n_1=1.5$ and $n_2=1$, $\sin {\theta }_c=2/3$ and ${\theta }_c\approx 41.8103149^{\circ }$. The threshold is independent of polarization of an electromagnetic wave under the stated assumptions.

A short proof using phase matching at a planar wave interface reveals all three cases at once. Choose a Cartesian coordinate system with boundary $z=0$, incident medium $z<0$, and transmitted medium $z>0$. With vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the shared tangential wavevector component is $q=n_1{k}_0\sin \theta$. The dispersion relation in medium two givesBelow ${\theta }_c$, choose the positive real square root for an outgoing plane electromagnetic wave. At ${\theta }_c$, $k_{2z}=0$: there is a grazing wave, with zero normal average power but no exponential confinement in $z$. Above ${\theta }_c$, choose $k_{2z}=i\kappa$ with $\kappa >0$, so the wave amplitude decays into the second half-space.

If $n_1\le n_2$, no nongrazing angle of incidence produces this threshold. As $n_2/n_1\to 1$ from below, ${\theta }_c\to 90^{\circ }$; as that ratio tends to zero, ${\theta }_c\to 0$. These limits explain why a larger refractive index contrast permits confinement over a wider range of internal angles.

In NIST's 2001 comparison, the Laser-Optimized Cryogenic Radiometer held its cavity near $5.4\mathrm{K}$ and controlled the heat sink separately. Opening the beam reduced heater power at fixed cavity temperature; an AC bridge read the temperature. Tests used stabilized, spatially filtered lasers at $488$, $514$ and $633\mathrm{nm}$, with horizontal input and a movable photodiode transfer detector in the cavity's optical plane. The paper contrasts this with the High Accuracy Cryogenic Radiometer's optical heating followed by two electrical heating cycles and interpolation.^[ref]

The two laboratories compared calibrations of the same transfer detectors, and checked their spatial response. Alignment differences and transported dust limited that comparison; its agreement did not prove universal equivalence. Its correction tables retain window transmission, cavity radiometric absorptance, alignment, electrical readout and optical-electrical heating non-equivalence. The heat sink was controlled, but §3 does not specify its numerical setpoint: the $5.0\mathrm{K}$ reference in the preceding example is stipulated.^[ref]

For a reproducible substitution measurement, record the beam's wavelength and entrance boundary, receiver and reference temperatures, heater terminal voltage and electric current, dark readings before and after illumination, and the settling rule. Check beam collection at the aperture, window transmission, radiometric absorptance, deposited lead heating, and sensitivity to beam position. Repeatability alone misses a stable correction error. Photometric calibration can subsequently use this radiant flux scale, but photometry still needs the appropriate spectral visual weighting; the cold receiver does not itself measure luminous flux.

The current density $\mathbf{J}(\mathbf{r},t)$ describes the amount and direction of local electric charge transport per area per time. Its SI unit is $A/{\mathrm{m}}^2$. Through a small area element with unit normal $\mathbf{n}$, the contribution to electric current is $dI=\mathbf{J}\cdot \mathbf{n}dA$: the dot product selects transport through the surface. A current density tangent to the surface contributes zero. If current density is uniform and perpendicular to a wire's transverse slice, its magnitude $j=|\mathbf{J}|$ satisfiesHere $A$ is the wire cross-sectional area. For nonuniform current density, $I/A$ is only the average normal component, and the surface integral is needed. ^[ref].

Consider two separate uniform-flow cases, each carrying electric current $I=2\mathrm{A}$. A wire cross-sectional area of $1\mathrm{m}{\mathrm{m}}^2=10^{-6}{\mathrm{m}}^2$ gives $j=2\times 10^{6}A/{\mathrm{m}}^2$; doubling the wire cross-sectional area to $2\mathrm{m}{\mathrm{m}}^2$ gives $j=10^{6}A/{\mathrm{m}}^2$. Equal electric current can therefore mean different local transport intensity. This distinction matters for Joule heating, whose local power per volume depends on current density and electric field.

For a toy electron transport model, stipulate mobile electron number density $n=8\times 10^{28}{\mathrm{m}}^{-3}$ and drift velocity ${\mathbf{v}}_d=-10^{-4}m/s\hat{\mathbf{x}}$, with the positive unit normal $\hat{\mathbf{x}}$. Each electron has electric charge $-e$, where the elementary charge $e=1.602176634\times 10^{-19}\mathrm{C}$ is exact in SI. These are defined model inputs, rather than measured parameters of a particular material. ^[ref] The current density and its magnitude areFor wire cross-sectional area $A=1\mathrm{m}{\mathrm{m}}^2$ and constant transport over time $10\mathrm{s}$, this givesThe two minus signs make conventional current point opposite to the electron drift velocity. A small drift velocity can carry an appreciable electric current because many charge carriers contribute; it does not specify the propagation speed of an electrical disturbance.

The motivating observation is that a chemical reaction can produce a signed electrical reading even when its reacting materials occupy separate vessels. UC San Diego's teaching demonstration places a copper strip in $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ $\mathrm{CuS}{\mathrm{O}}_4$ and a zinc strip in $0.1\mathrm{mol}{\mathrm{L}}^{-1}$ $\mathrm{ZnS}{\mathrm{O}}_4$, using two $400\mathrm{mL}$ beakers with about $300\mathrm{mL}$ of chemical solution in each. The strips connect to a voltage meter. Initially there is no ionic connection; the demonstration specifies a zero reading. A filter-paper salt bridge containing aqueous $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$ then connects the chemical solutions. With the meter's positive lead on copper and negative lead on zinc, its reported readings are about $+1.1\mathrm{V}$; exchanging the leads reverses the reading. These are observations reported by the university, not data generated for this article. See UC San Diego, demonstration 9B: apparatus, bridge control and meter polarity.

The controls answer different questions. Exchanging only the meter leads checks the sign convention; it does not reverse the chemical reaction or swap the anode and cathode. Interrupting only the salt bridge tests the need for an ionic return path. Its absence prevents sustained cell electric current, but a very high-input-resistance meter may still show a transient or unstable reading while interfaces store electric charge. The zero reading specified for the university's apparatus is therefore not a universal theorem for every meter. Reconnecting the same salt bridge distinguishes a reversible interruption from a disconnected wire or exhausted chemical solution. This interpretation follows from electric charge conservation in a model allowing interfacial electric charge storage; see IUPAC Gold Book: integral capacitance of an electrode for the capacitance associated with electrode charge.

To distinguish terminal electric potential difference from delivered power, compare the nearly open-circuit reading with readings across a known resistor load. In the nearly open-circuit condition, a meter with input electrical resistance much larger than the cell's effective internal electrical resistance draws little electric current. With a load, measure its terminal electric potential difference $V$ and infer $I=V/R$ using Ohm's law; track the readings over time and repeat the nearly open-circuit reading afterwards. Keep the temperature, chemical solution compositions, immersed electrode areas, bridge geometry and electrical contacts comparable. This is a model-based extension of the documented demonstration, not a claim that UC San Diego published a load sweep. A finite-current drop can reflect ionic electrical resistance, changes of interfacial reaction rates, or depletion near an electrode; a single voltage reading cannot identify which dominates. The interfacial and composition contributions are distinguished in IUPAC Gold Book: overpotential and ^[ref].

For a deliberately simplified calculation, model the cell by a constant electromotive force $\mathcal{E}$ in series with a constant internal electrical resistance $r$, and a passive load $R>0$. Take $I$ positive from the cathode through the load to the anode. ThenThe nearly open-circuit limit $R\to \infty$ gives $V\to \mathcal{E}$ but $I\to 0$ and $P_{\mathrm{load}}\to 0$. A straight $V$-versus-$I$ line is a prediction of constant $r$, not a general law for electrochemistry. See Ling, Sanny and Moebs, University Physics II, §10.1: internal resistance, terminal voltage and load.

For the chosen values, $I=5.50\mathrm{mA}$ and $V=0.550\mathrm{V}$. Holding these toy parameters fixed for $600\mathrm{s}$ gives transported electric charge $Q=It=3.30\mathrm{C}$ and an extent of reaction $\xi =Q/(2F)\simeq 17.1\mu \mathrm{mol}$, assuming all the electric current serves the two specified half-reactions. The same $\xi$ counts zinc consumed and copper deposited, rather than equal masses. The energy budget isThe load's electrical work also becomes Joule heating when the load is a resistor. Here $\mathcal{E}Q$ accounts for the reaction's available reversible electrical work; a full cell thermal budget would also require its thermodynamic entropy contribution. Real measurements need uncertainty estimates, a specified meter input electrical resistance and calibration, and allowance for changing composition and unwanted chemical reactions. The toy value $r=100\Omega$ is not a measured property of the university's cell.

A comparison using a logarithm of two powers is expressed as $10{\log }_{10}(P_2/P_1)\mathrm{dB}$. Positive loss instead uses $10{\log }_{10}(P_{\mathrm{in}}/P_{\mathrm{out}})\mathrm{dB}$. For wave amplitudes, a factor twenty is appropriate only when the corresponding power ratio equals the square of the amplitude ratio.

Destructive interference occurs when coherent contributions partially or completely cancel their complex amplitudes. For two equal contributions, a phase difference of $\pi$ modulo $2\pi$ gives zero resultant wave amplitude. The missing local light intensity is consistent with conservation of energy because the full spatial interference pattern redistributes power.

Take a stipulated pair of effective rods with common length $L$, linear thermal expansion coefficient ${\alpha }_L$, heat capacity $C$, projected illuminated area $A$, and thermal conductance $G$ to an environment at constant temperature $T_a$. One reference represents the symmetric polished-rod assembly. Write ${\theta }_b=T_b-T_a$ and ${\theta }_r=T_r-T_a$. Let their effective radiometric absorptances be $a_b>a_r$ for a fixed incident optical spectrum, and let $E(t)$ be the common irradiance. A spatially uniform, linear heat-balance model givesHere $G$ includes a linear approximation to the relevant losses; changing the surface treatment can also change those losses in a real device. We stipulate equal $C$ and $G$ to isolate the response to absorbed illumination, with thermal coupling from the burner neglected. Small, mechanically unconstrained extensions then give a differential displacementThe time constant $\tau$ describes thermal memory. A common imposed temperature shift cancels from the extension difference under the matched-coefficient assumptions; mismatched materials, geometry or thermal coupling can spoil that cancellation.

For a step from darkness to constant irradiance $E_0$, with initially equal rod temperatures, the first-order linear differential equation givesChoose $L=0.20\mathrm{m}$, ${\alpha }_L=20\times 10^{-6}{\mathrm{K}}^{-1}$, $A=0.0020{\mathrm{m}}^2$, $a_b=0.9$, $a_r=0.1$, $C=5.0\mathrm{J}{\mathrm{K}}^{-1}$, $G=0.20\mathrm{W}{\mathrm{K}}^{-1}$ and $E_0=600\mathrm{W}{\mathrm{m}}^{-2}$. The calculated difference of absorbed powers is $0.96\mathrm{W}$, givingFor an illustrative ideal valve that closes at $\delta =g=10\mu \mathrm{m}$, the closing time isThis threshold is crossed only when $0<g<{\delta }_{\infty }$. These are chosen model parameters, not a fit to a historical valve.

After illumination stops at $t=t_1$, the displacement decays as $\delta (t)=\delta (t_1)e^{-(t-t_1)/\tau }$. With $t_1=180\mathrm{s}$, the model reopens about $16.3\mathrm{s}$ later. Thus a sun valve responds to the thermal history of illumination; it does not read a clock or directly measure the beacon's brightness.

A diffuse thermal emitter has emitted radiance independent of outgoing direction over the outward hemisphere. Its emitted radiant intensity from a planar patch still varies with viewing angle through projected area, as described by Lambert's cosine law. Diffuse emission does not mean equal power per solid angle from the patch. This approximation does not by itself impose a grey body spectrum or specify how incident radiation is reflected.

The angular distribution of electric dipole radiation measures squared projections of the dipole axis, so a few intensity measurements need not determine that axis or the total radiated power uniquely. For a real unit dipole axis $\mathbf{d}$, the radiation-zone formula is $I(\mathbf{n})=3P[1-(\mathbf{d}\cdot \mathbf{n}{)}^2]/(8\pi R^2)$. Equal intensities along $\hat{\mathbf{z}}$, $(\hat{\mathbf{x}}+\hat{\mathbf{z}})/\sqrt{2}$ and $(\hat{\mathbf{y}}+\hat{\mathbf{z}})/\sqrt{2}$ permit four axes, up to reversal:Three generally different radiated powers are compatible with the same intensity data. Additional directional measurements or polarization information can remove the ambiguity.

For strain $\gamma ={\gamma }_0\cos \omega t$ and stress $\sigma ={\gamma }_0(G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t)$, integrating supplied power $\sigma \dot{\gamma }$ over one period gives the displayed dissipated energy per unit volume. The storage term integrates to zero; passivity requires nonnegative loss modulus for positive frequency in this convention.

For a string occupying $x>0$, its small-slope force on the endpoint is $T{y}_x$. The driver force on the string has the opposite sign, so injected power is $-T{y}_x{y}_t$. A right-going traveling wave has nonnegative outgoing instantaneous power in this ideal model.

Use physical electric fields $\mathbf{E}(t)=\mathrm{Re}[\hat{\mathbf{E}}{e}^{-i\omega t}]$ and current densities $\mathbf{J}(t)=\mathrm{Re}[\hat{\mathbf{J}}{e}^{-i\omega t}]$, with peak complex amplitudes. After the initial transient has decayed, substitution into the Drude model equation gives the complex conductivityHere the mobile species is electrons. With the alternative $e^{+i\omega t}$ convention, the denominator is $1+i\omega \tau$ and the imaginary part changes sign; the physical response is identical.^[ref]

For $E_x(t)=E_0\cos \omega t$ with $\omega >0$, the real-time current density isThus a positive argument of complex conductivity in the $e^{-i\omega t}$ convention means that current density lags the electric field. At $\omega \tau =1$, its real part and imaginary part are both ${\sigma }_0/2$, its modulus is ${\sigma }_0/\sqrt{2}$, and the lag is $\pi /4$.

For $\omega \tau \ll 1$, complex conductivity tends to the positive constant ${\sigma }_0$: current density nearly follows the electric field. For $\omega \tau \gg 1$ within this model,The current density amplitude decreases and its response becomes predominantly inertial. Average power absorbed per volume is $\overline{p}=\frac{1}{2}\mathrm{Re}\sigma |\hat{\mathbf{E}}{|}^2\ge 0$. These limits describe the material response to a specified local electric field, not the complete electromagnetic response of a finite specimen with contacts. The Drude response has its pole at $\omega =-i/\tau$, consistent with the decaying causal response of a first-order relaxation equation.

The simple Drude model does not predict the microscopic momentum relaxation time. It neglects electronic band structure and Fermi-Dirac statistics; a useful semiclassical extension replaces $m$ by an appropriate band effective mass, but neither a single scalar mass nor one constant momentum relaxation time describes every material. Several charge carriers, energy-dependent scattering, and transitions between electronic bands can spoil the simple Drude frequency response. A successful low-angular frequency fit does not justify extrapolation to arbitrarily large angular frequencies.

An electrical calorimetry experiment asks whether a measured electrical work input accounts for a measured warming. The University of Iowa's demonstration uses an insulated flask containing water, an immersed heater, a thermometer, electrical meters and a timer. Its procedure compares the duration, electrical readings and temperature rise. The following low-power measurement design and synthetic readout develop that idea; they are not measurements reported by the demonstration. ^[ref]

Use a covered calorimeter with weighed water, an immersed resistor, a calibrated thermometer and gentle stirring. Put an ammeter in series with the heater branch and a high-input-electrical resistance voltmeter across the immersed resistor's terminals. Record $V(t)$, $I(t)$, temperature $T(t)$ and room temperature $T_a(t)$ before, during and after switching the heater on. Measure the electric potential difference at the heater, rather than at the supply: lead and contact electrical resistance outside the receiving boundary can consume power that never reaches the water. Account for voltmeter loading if the ammeter also includes the sensing branch's electric current.^[ref]

The electrical readout gives operating electrical resistance $R(t)=V(t)/I(t)$ when $I(t)\ne 0$, and power $P(t)=V(t)I(t)$. To test Ohm's law, obtain several pairs at the same temperature; one ratio alone cannot establish a linear response. For small electrical resistance, use four-terminal sensing to exclude lead drops and reverse electric current to test stable offsets from the thermoelectric effect. Use brief low-power resistance measurements if the intended result is the unheated value: the test itself causes Joule heating.

For a covered assembly at approximately constant pressure, with negligible evaporation, negligible endpoint mechanical energy changes and no phase transition, takeHere $c_w$ is the water's specific heat capacity at constant pressure, $C_{\mathrm{cal}}$ is the participating vessel, heater and sensor heat capacity, $Q_{\mathrm{loss},\mathrm{out}}$ is signed outward heat leakage, and $W_{\mathrm{stir},\mathrm{on}}$ is stirring work entering the assembly. The storage term approximates its enthalpy change; for a rigid fixed-volume receiver use the corresponding heat capacity at constant volume. Calibrate $C_{\mathrm{eff}}$ independently if the purpose is to test electrical input against thermal storage. Using the same heating run to define $C_{\mathrm{eff}}=E_{\mathrm{el}}/\Delta T$ and then claiming it verifies the equality would be circular.

For the synthetic trace, stipulate $V=5\mathrm{V}$, $I=1\mathrm{A}$, $R=5\Omega$, $C_{\mathrm{eff}}=500J/K$, room temperature $20{}^{\circ }\mathrm{C}$, and loss coefficient $G=0.5W/K$, with negligible stirring input. The heater runs for $120\mathrm{s}$, beginning at room temperature. The calorimeter heat-loss correction model isIt predicts $E_{\mathrm{el}}=600\mathrm{J}$, $\Delta T\approx 1.13080\mathrm{K}$, storage $565.398\mathrm{J}$ and outward loss $34.602\mathrm{J}$. Neglecting leakage undercounts deposited energy by about $5.77\%$; assuming no loss predicts $1.20\mathrm{K}$. After switching off, the excess temperature undergoes exponential decay with time constant $C_{\mathrm{eff}}/G=1000\mathrm{s}$. A cooling slope can estimate $G$ only with known $C_{\mathrm{eff}}$ and a credible single-temperature model.

Controls separate causes of a temperature rise. Check the electrical meters' zeros and thermometer calibration; keep water mass, immersion, lid and stirring fixed between runs. Record a heater-off stirred baseline to estimate sensor drift and stirring input. Repeat cooling runs over the same temperature range to estimate environmental heat transfer; a baseline taken only at room temperature cannot measure warming-dependent leakage. Vary electric current while keeping $R$ and initial thermal conditions comparable: corrected deposited energy should scale as $I^2$ for equal durations. An independent calorimeter calibration or known reference heat capacity makes this a quantitative test rather than a qualitative observation.

Finite sensor response, incomplete thermal equilibrium, changing $R(T)$ or heat capacity, evaporation, and imperfect loss corrections limit the inference. A final temperature alone cannot distinguish smaller electrical input from larger heat leakage or a larger participating heat capacity. Agreement tests the stated assembly and its energy accounting; it does not show that all electrical work in arbitrary devices becomes heat.

Historically, the Royal Society's archive description of James Prescott Joule's manuscript dated 13 October 1840 records a coil of wire in a jar of water, a thermometer and a galvanometer, and summarizes his inference that warming depends on electrical resistance and the square of electric current intensity. This electrical experiment precedes Joule's paddle-wheel experiment; the input here is electrical work, whereas the paddle supplies mechanical work. The catalogue, rather than the inaccessible manuscript images, supports this account. ^[ref]

An electrical insulator has sufficiently small transport of electric charge for its operating conditions. Insulation is a practical requirement involving electric potential difference, temperature, geometry and duration, rather than a universal claim of zero electrical conductivity. Electric polarization can be substantial in an electrical insulator; dielectric breakdown can end the insulating regime.

For an idealized uniform slab with small constant electrical conductivity $\sigma$, the Ohmic conductor model predicts leakage electrical resistance $R=d/(\sigma A)$ in parallel with capacitance $C=ϵA/d$. An isolated capacitor then hasThis is a derived model of slow discharge, not an assertion that every leakage mechanism obeys Ohm's law. Its time scale matches charge relaxation. The fixed-charge comparison in dielectric capacitor response needs an observation time short compared with $RC$, but long enough for electric polarization to settle. During a maintained voltage, leakage consumes power $V^2/R$ as Joule heating; delayed electric polarization can add dielectric loss under changing electric fields.^[ref]

Large relative permittivity alone does not rule out either mechanism.^[ref]

Electrical reactance $X$ is the signed imaginary part of electrical impedance $Z=R+\mathrm{i}X$, measured in ohms. It describes the part of the terminal electric potential difference in phase quadrature with the electric current in a sinusoidal steady state. In the $e^{\mathrm{i}\omega t}$ convention, an ideal inductor has $X_L=\omega L>0$ and an ideal capacitor has $X_C=-1/(\omega C)<0$ for $\omega >0$. A large $|X|$ reduces electric current at fixed applied voltage, but an ideal purely reactive component returns the energy it receives over each cycle. This is why electrical reactance and electrical resistance have the same units yet describe different physical behavior. ^[ref].

In a series RLC circuit, electrical impedances add because the same electric current passes through every component. For ideal positive inductance $L$, capacitance $C$, and total electrical resistance $R_{\Sigma }>0$,Here $\varphi$ is voltage wave phase minus current wave phase. Below ${\omega }_0=1/\sqrt{LC}$, $X<0$ and electric current leads terminal voltage; above it, $X>0$ and electric current lags. At ${\omega }_0$, the two signed electrical reactances cancel, leaving $Z=R_{\Sigma }$ and the maximum electric current at fixed terminal voltage. This is the resonant series RLC response. Electrical resistance does not cancel: the toy circuit gives $f_0={\omega }_0/(2\pi )=1591.55\mathrm{Hz}$ and $I_{\mathrm{rms}}=9.091\mathrm{mA}$ there. These numbers follow directly from the stipulated component values. ^[ref].

Cancellation concerns terminal voltage, not the disappearance of stored energy. In the toy circuit at resonance, the individual inductor and capacitor voltages are each $0.9091\mathrm{V}$ in root mean square magnitude, with opposite wave phases, while their sum is zero. Their peak stored magnetic energy and capacitor energy are each $0.8264\mu \mathrm{J}$, attained at different times. With smaller positive $R_{\Sigma }$, either component voltage can exceed the applied voltage. This ability to select a frequency while sustaining internal energy exchange connects electrical reactance to tuning and frequency response.

For an ideal inductor or capacitor, the absorbed instantaneous power equals the rate of change of stored energy:Positive power stores energy; negative power returns it. Over a periodic steady state, each stored energy returns to its initial value, so its cycle-averaged absorbed power is zero. An ideal resistor instead has $p_R=R{i}^2\ge 0$, transferring energy into Joule heating. More generally, the average terminal power of this series RLC circuit isZero average power in an ideal reactive component does not imply zero instantaneous power, zero electric current, or zero stored energy. Real coils and capacitors have losses, so zero ideal reactive loss is a model assumption rather than a claim about every component. ^[ref].

An explicit lumped model makes the cancellation understandable. Treat the receiver as having uniform temperature $T$, heat capacity $C$, heater power $P_h$ deposited in the receiver, net parasitic input $P_{\mathrm{par}}$, and outward heat transfer $Q_{\mathrm{out}}(T,T_b)$ to a reference at temperature $T_b$. With incident radiant flux $P_{\mathrm{in}}$ and radiometric absorptance $a$, conservation of energy givesThe dark state has no admitted beam. Compare settled dark and light states, denoted $d$ and $l$; each has $dT/dt=0$. Subtracting their balances giveswhere $\Delta Q_{\mathrm{out}}=Q_{\mathrm{out},l}-Q_{\mathrm{out},d}$ and $\Delta P_{\mathrm{par}}=P_{\mathrm{par},l}-P_{\mathrm{par},d}$. At identical $T$ and $T_b$, with unchanged surroundings and parasitic heating, both correction differences vanish. Thus $P_{\mathrm{abs}}=\Delta P_h$, and $P_{\mathrm{in}}=\Delta P_h/a$. Notice the signs: extra stray thermal radiation admitted with the beam also reduces heater power, so it must be subtracted from that reduction before attributing the result to the intended beam.

Measure the heater's own terminal voltage $V_h$ and electric current $I_h$: for steady electrical heating, $P_h=V_h{I}_h=I_h^2{R}_h$, where $R_h$ is its electrical resistance. A supply voltage that includes wire voltage drops counts Joule heating in the leads too; only the portion deposited in the receiver belongs in this balance. Changes in that deposited lead heat belong in $\Delta P_{\mathrm{par}}$. Optical-electrical heating non-equivalence adds a further instrument-specific correction when the uniform-temperature model is insufficient.^[ref]

An electrical substitution radiometer measures radiant flux by replacing absorbed optical power with calibrated Joule heating. At fixed temperature, feedback control holds the receiver equally warm with the beam blocked or admitted: after corrections, absorbed optical power equals the reduction in supplied electrical heating. Incident radiant flux additionally requires radiometric absorptance and entrance-path corrections. This connects radiometry to electrical standards.^[ref]

Picture a small absorbing cavity connected to a colder reference block by a radiometer thermal link. A heater supplies Joule heating, a thermometer senses the receiver, and feedback control adjusts the heater. When light supplies some of the heating, the controller supplies less. The readout is the change in heater power, rather than the beam's brightness or its number of photons. NIST's institutional description gives a concrete receiver with a conical interior, absorbing coating and weak Kapton support, operated with a receiver heater and thermometer.^[ref]