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A spark gap can rapidly discharge a charged capacitor through an inductor. After excitation, the stored energy runs down through Joule heating, radiation and other losses. A key held down can permit repeated discharges: each discharge starts a damped train of rapid oscillations. The spark repetition frequency, the electrical oscillation frequency within each train, and the rate of keyed symbols are different quantities. Repetition does not make each train a maintained sinusoid.

An original lumped model replaces the conducting spark gap and other losses by a constant total electrical resistance $R>0$ in an isolated series RLC circuit. For electric charge $q$ on the capacitor, the Kirchhoff cycle law givesFor $Q>1/2$, its characteristic roots are $-\alpha \pm \mathrm{i}{\omega }_d$, with ${\omega }_d=\sqrt{{\omega }_0^2-{\alpha }^2}$, proving the stated damped form. The electric current $\dot{q}$ has the same exponential decay scale. Its wave envelope falls by $\exp (-2\pi \alpha /{\omega }_d)$ per cycle; the energy envelope falls twice as fast in the exponent. A maintained sinusoid instead requires continuing input power balancing losses.

With $L=5\mathrm{mH}$, $C=2\mathrm{nF}$ and $R=100\Omega$, the model gives $f_0={\omega }_0/(2\pi )=50.329\mathrm{kHz}$, Q factor $Q=15.811$, amplitude time constant $1/\alpha =100\mu \mathrm{s}$, and damped frequency ${\omega }_d/(2\pi )=50.304\mathrm{kHz}$. These are chosen component values. A real spark gap switches nonlinearly, an antenna radiates, and coupling exchanges energy between circuits; one constant-$R$ free decay does not model the complete transmitter. Rapid damping also broadens the frequency spectrum, which helps explain why short trains make sharp receiver selection difficult.

For a long homogeneous circular wire of radius $a$, uniform volumetric heating $H$, constant thermal conductivity $\kappa$ and negligible end losses, the steady heat equation is $\kappa {\rho }^{-1}(\rho T^{\prime }{)}^{\prime }+H=0$. Regularity at the axis excludes a logarithm term, giving $T=T_s+H(a^2-{\rho }^2)/(4\kappa )$. If the surface obeys Newton's law of cooling, $-\kappa T^{\prime }(a)=\gamma (T_s-T_{\mathrm{air}})$, then $T_s=T_{\mathrm{air}}+Ha/(2\gamma )$. Here $\gamma$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ and $H$ has units $\mathrm{W}{\mathrm{m}}^{-3}$. For uniform electric current $I$ and constant electrical conductivity $\sigma$, Joule heating gives $H=I^2/(\sigma {\pi }^2{a}^4)$.

The magnetic energy stored in a linear solenoid of constant self-inductance $L$ isin the long-solenoid approximation. The first formula is an inductor relation; the second identifies its storage with magnetic energy in the interior magnetic field, under the vacuum-core approximation. This agreement joins the electrical circuit description to the spatial magnetic field description. The expression $B^2/(2{\mu }_0)$ is the vacuum magnetic energy density. ^[ref].

A short derivation starts from the terminal electric potential difference $v_L=LdI/dt$, with positive electric current entering the terminal at higher potential. The absorbed power is $v_{L}I$, and integrating from zero electric current givesThe result describes recoverable storage in the inductor; Joule heating in the wire is a separate expenditure of energy. ^[ref].

For the long-coil example, $L\approx 4.9348\mathrm{mH}$ and $I=0.50\mathrm{A}$ give $U\approx 6.1685\times 10^{-4}\mathrm{J}=0.61685\mathrm{mJ}$. Substituting the ideal $B\approx 1.256637\mathrm{mT}$ and volume $A\ell \approx 9.8175\times 10^{-4}{\mathrm{m}}^3$ into $B^{2}A\ell /(2{\mu }_0)$ gives the same value. Doubling electric current quadruples stored magnetic energy. In a real coil, reducing electric current can return this energy to an electrical circuit, or release it as Joule heating in a closed RL circuit.

Superconductivity is a phase of matter that supports steady electric current with zero bulk electrical resistivity and exhibits the Meissner effect: equilibrium screening of a sufficiently weak, static magnetic field from the interior. It occurs below a material-dependent superconducting transition temperature, provided the electric current and applied magnetic field are also sufficiently small. ^[ref] Bulk expulsion means that a sample much thicker than its London penetration depth has an almost field-free interior; the magnetic field need not vanish at its surface. Conventional BCS theory explains superconductivity through correlated electrons forming Cooper pairs. ^[ref]

The two observations answer different questions. A resistance-versus-temperature experiment for superconductivity asks whether a steady electric current requires a measurable electric potential difference. A Field-cooled Meissner experiment asks whether a magnetic field already present in the material is expelled when it enters superconductivity. An ideal perfect conductor with zero electrical resistivity can preserve that pre-existing magnetic field; zero electrical resistance alone therefore does not establish the Meissner effect.

The usefulness follows directly from Joule heating, $P=I^{2}R$: maintaining a steady electric current through a segment with zero electrical resistance produces no resistive heating in that segment. This makes persistent electric currents attractive for producing steady magnetic fields. The beauty is that one phase has both a transport property and an equilibrium magnetic response, and the second cannot be inferred just by making an ordinary electrical resistance very small. The London equations encode this extra magnetic response. ^[ref]

If a wanted response changes sign on reversing a controlled setting while an offset remains unchanged, the difference of the reversed readings isolates the response. For $V_{\pm }=\pm RI+V_0$, the estimator $(V_{+}-V_{-})/(2I)$ cancels $V_0$. The method requires offset stability between readings. It does not cancel a common current-scale calibration error, and it does not remove even-in-current Joule heating. A useful reversal test specifies which physical effects change sign, then tests the remaining effects independently.

The chosen steady-state model isHere $U$ is the terminal voltage drop along positive electric current through the module. Its Seebeck effect term opposes the cooling supply. The cold Peltier effect term removes heat; Joule heating and returning heat conduction reduce the available cooling. At zero electric current, $Q_c=Q_h=-K\Delta T$: the passive leak warms c and removes heat from h, consistent with the stated signs. Subtracting the two face equations cancels heat conduction and gives the electrical power, an independent conservation of energy check.

The factor one half is an ideal-model result, not a universal rule for every contact. To see its origin, let $x$ run from cold to hot along a uniform equivalent leg of length $L$, cross-sectional area $A$, electrical resistivity $\rho$ and thermal conductivity $\kappa$. With constant coefficients, insulated sides and no Thomson effect, steady conservation of energy givesThe Fourier's law heat current in the positive $x$ direction is $-\kappa A{T}^{\prime }$. At $x=0$ it equals $-K\Delta T-I^{2}R/2$, and at $x=L$ it equals $-K\Delta T+I^{2}R/2$, using $K=\kappa A/L$ and $R=\rho L/A$. Adding the local transported heat current $\alpha TI$ gives the two face equations. Summing equivalent legs preserves this accounting. Nonuniform material properties, side losses or asymmetric contact dissipation require a revised model.

For an original toy module, stipulate $\alpha =0.040V/K$, $R=2.0\Omega$, $K=0.40W/K$, $T_c=280\mathrm{K}$ and $T_h=300\mathrm{K}$. At $I=3.0\mathrm{A}$,The coefficient of performance is defined here only for refrigeration with $Q_c>0$ and $P_{\mathrm{in}}>0$. The hot-side assembly must remove $37.0\mathrm{W}$, not just the extracted $16.6\mathrm{W}$. These are stipulated calculations, not measurements or a commercial module specification.

At the same fixed face temperatures but $I=-3.0\mathrm{A}$, the same boundary definitions give $Q_c=-50.6\mathrm{W}$, $Q_h=-35.0\mathrm{W}$, $U=-5.2\mathrm{V}$ and $P_{\mathrm{in}}=15.6\mathrm{W}$. The formerly cold face is now heated; the formerly hot reservoir supplies $35.0\mathrm{W}$. Still $Q_h-Q_c=15.6\mathrm{W}$. Reversal changes the linear terms, while $I^{2}R$ and the leak at fixed temperatures remain unchanged. Holding the old labels fixed makes this check meaningful.

The corresponding steady entropy production of the module is the outgoing reservoir entropy minus the incoming reservoir entropy:The Peltier effect terms cancel because they carry the same entropy $\alpha I$ at each end. This cancellation explains the reversible coupling; it does not make finite-resistance operation reversible. In the toy example the two positive contributions are approximately $0.06214$ and $0.001905W/K$, giving ${\dot{S}}_{\mathrm{gen}}=0.06405W/K$ for either current direction. In cooling operation, combining this inequality with the energy balance gives the Carnot refrigerator limit $\mathrm{COP}\le T_c/(T_h-T_c)$, equal to $14$ for these reservoir temperatures. A coefficient of performance above one would describe transported heat per electrical work, not the creation of energy.

At fixed face temperatures, differentiation of the thermoelectric cooler energy balance givesThus current chosen for maximum cooling power is not automatically current for maximum coefficient of performance. Cooling exists only when $(\alpha T_c{)}^2>2RK\Delta T$, and then only between the two positive rootsAt equality the parabola merely touches $Q_c=0$. For the toy module the interval is approximately $0.767<I<10.433\mathrm{A}$, with $I_{\ast }=5.60\mathrm{A}$ and $Q_{c,\max }=23.36\mathrm{W}$. At $12\mathrm{A}$, the object is heated even in the nominal cooling polarity: $Q_c=-17.6\mathrm{W}$. This mathematical interval supplies no permission to exceed an actual device's current or temperature ratings.

The fixed-face constraint matters. With fixed $T_h$ but a freely cooling object, $T_c$ must be solved from its heat load and leaks rather than held at $280\mathrm{K}$ while claiming an arbitrary temperature drop. For zero load and optimal current, the same model givesHere $Z$ has units ${\mathrm{K}}^{-1}$ and is a module parameter. The toy model gives $Z=0.0020{\mathrm{K}}^{-1}$ and $T_{c,\min }\approx 241.62\mathrm{K}$ when $T_h=300\mathrm{K}$. This is an ideal zero-load limit with stipulated constant properties; cooling a real load requires $Q_c$ above zero and reduces the attainable drop. Contact thermal resistance, a warming heat sink, external leaks and temperature-dependent coefficients further change the result. Electrical control is convenient, but conservation of energy still requires a thermal path that removes the extracted heat plus the electrical input.

Use the A-positive terminal order of Seebeck effect for each thermocouple. Its thermocouple measuring junction has temperature $T_{h,i}$ and its reference connections have temperature $T_{c,i}$. Define the difference of Seebeck coefficients as ${\alpha }_i(T)=S_{A,i}(T)-S_{B,i}(T)$. Connect the B reference end of one thermocouple to the A reference end of the next at a common local temperature, so the terminal electric potential differences add:The first A reference terminal is the positive meter input; the final B reference terminal is the negative input. Following this connected path and adding the individual Seebeck effect integrals proves the result. Matched meter leads with common endpoint temperatures cancel as in an isothermal terminal block. The manufacturer's sum formula also allows unequal individual contributions.^[ref]

If all pairs share $T_h,T_c$ and the same nearly constant difference $\alpha$ of Seebeck coefficients throughout that temperature interval, thenThis $\alpha$ is the pair's relative Seebeck coefficient, not either material's individual coefficient or a universal device constant. As an original constructed example, choose $N=20$, $\alpha =40\mu \mathrm{V}{\mathrm{K}}^{-1}$ and $\Delta T=5\mathrm{K}$. Each thermocouple contributes $0.20\mathrm{mV}$ andReversing the meter connections changes the sign. Exchanging $T_h$ and $T_c$ reverses the integrals and hence the electromotive force; this reversible sign response does not make a detector with heat conduction a reversible thermodynamic process. Reversing the wiring of one pair instead gives $(N-2)\alpha \Delta T$, because one contribution is subtracted rather than added. For this example the result is $3.6\mathrm{mV}$, and at equal junction temperatures every contribution vanishes.

The open-circuit condition matters. If the array's total electrical resistance is $R_{\mathrm{pile}}$ and the meter input has electrical resistance $R_m$, the elementary voltage divider model gives $V_m=V_{\mathrm{oc}}{R}_m/(R_m+R_{\mathrm{pile}})$. Thus $R_m\gg R_{\mathrm{pile}}$ is needed to suppress voltmeter loading. Appreciable electric current also permits Joule heating and the thermoelectric heat transfer to disturb the assumed junction temperatures.

An authored teaching experiment uses an adjustable steady-voltage supply, a digital multimeter, fixed resistors, and a $10\mathrm{k\Omega }$ potentiometer or resistance decade box. Its original circuit uses a $10\mathrm{V}$ supply and $3.3$, $6.8$, and $4.7\mathrm{k\Omega }$ resistors; the student compares $8.2$ and $2.2\mathrm{k\Omega }$ electrical loads. The readout is load voltage, recorded against prediction and deviation. The student then removes the electrical load to measure the open-circuit voltage, disconnects the supply and substitutes a wire to measure the Thévenin resistance with the multimeter's resistance mode, and restores the supply for an independent method: adjust the variable electrical load until the voltmeter reads half the unloaded voltage. Finally the student builds the equivalent and repeats both load measurements. Its tables are blank recording forms, not reported observations. ^[ref]

The half-voltage inference assumes negligible voltmeter loading and positive $R_{\mathrm{th}}$. A finite voltmeter input electrical resistance $R_m$ actually gives an apparent unloaded readingWhen the same voltmeter remains connected across a variable electrical load $R_L$, the network sees $R_p=R_L{R}_m/(R_L+R_m)$. Setting its reading to $V_m/2$, rather than to the true $V_{\mathrm{oc}}/2$, givesThus the uncorrected half-reading method underestimates positive $R_{\mathrm{th}}$. In the original $12\mathrm{V}$ example, a stipulated $R_m=10\mathrm{k\Omega }$ gives $V_m=120/17\mathrm{V}\approx 7.059\mathrm{V}$ and half-reading load $R_L=20/17\mathrm{k\Omega }\approx 1.176\mathrm{k\Omega }$, instead of $\frac{4}{3}\mathrm{k\Omega }$. These are model predictions for a deliberately loading meter, not the specification or observed readout of the source experiment's instrument.

Finite readings can also estimate the terminal line without a physical short. For two distinct load settings with corrected delivered electric currents $I_1,I_2$ and terminal voltages $V_1,V_2$,If the voltmeter remains across each electrical load, the delivered electric current is $I_k=V_k(1/R_{L,k}+1/R_m)$, including the meter branch. Omitting it biases the inferred slope.

Useful additional controls are to measure each resistor separately before assembly, verify terminal polarity, repeat an earlier load setting to detect supply drift, and compare a third setting with the inferred Thévenin terminal response. These are proposed controls, not reported outcomes of the source experiment. Closely spaced electric currents make the slope estimate sensitive to voltage resolution and measurement error; widely separated settings may trigger source current limiting or Joule heating that changes electrical resistance. Lead and contact electrical resistance, meter input uncertainty and rounding limit the inference. An ohmmeter injects its own test electric current; its reading on a powered network is not the suppressed-network ratio. Measurements with independent sources physically removed do not characterize an active dependent electrical source whose control or supply has also been disconnected.

Thévenin's theorem replaces the terminal behavior of a linear resistive circuit by an ideal voltage source in series with a Thévenin resistance, whenever its terminal electric potential difference is a well-defined affine function of the delivered electric current. Fix terminals $a,b$, write $V=\varphi (a)-\varphi (b)$, and take $I$ positive out of the network at $a$ into the electrical load, returning through $b$. ThenThe open-circuit voltage $V_{\mathrm{oc}}$ is the intercept at $I=0$; $-R_{\mathrm{th}}$ is the slope. Thus two parameters predict what happens when many different electrical loads are connected. The Wheatstone bridge detector loading calculation uses precisely this replacement to find the detector reading even when the bridge's two voltage dividers interact through its detector.

For example, a $12\mathrm{V}$ ideal voltage source with $2\mathrm{k\Omega }$ above and $4\mathrm{k\Omega }$ below a voltage divider junction supplies $V_{\mathrm{oc}}=8\mathrm{V}$ and $R_{\mathrm{th}}=\frac{4}{3}\mathrm{k\Omega }$ at that junction relative to the lower terminal. Attaching a $4\mathrm{k\Omega }$ electrical load reduces its voltage to $6\mathrm{V}$ and draws $1.5\mathrm{mA}$. The equivalent explains why an unloaded divider ratio alone cannot predict a loaded output.

The equivalence concerns the chosen terminals, including the power $VI$ delivered there. It does not identify internal branch electric currents or internal Joule heating. At open circuit the original divider still draws $2\mathrm{mA}$ from its $12\mathrm{V}$ supply, while the equivalent's series branch carries no electric current. ^[ref]

The Thomson effect is distributed reversible heating or cooling when an electric current flows through a homogeneous material with a temperature gradient. It is distinct from junction Peltier effect heating and even-in-current Joule heating.

Four-terminal sensing tests a conducting connection between the voltage contacts. A magnetometer instead tests the sample's magnetic response. These measurements provide complementary constraints: the first is sensitive to a continuous path, while the second also depends on screening geometry and preparation history.

For transport, define $V_{ab}$ between the inner contacts and use the reversed-electric current estimator from four-terminal sensing. Contact separation excludes drive-contact drops, but does not eliminate a drifting thermoelectric effect, Joule heating or an unrepresentative path through part of the specimen. Repeat at smaller electric currents and compare results at the same sample temperature. A finite-current result refers to that electric current and its relation to the path's superconducting critical current.

For magnetism, distinguish cooling in zero applied magnetic field and then applying it from the Field-cooled Meissner experiment. Record the applied magnetic field, background and sample geometry when interpreting magnetization or magnetic susceptibility. A transport short does not establish full Meissner effect screening throughout the specimen. Conversely, flux pinning in a type-II superconductor can make the field-cooled magnetic response differ from ideal complete expulsion. Even a large shielding signal is not automatically a direct measure of the superconducting phase fraction: a connected outer screening region can surround nonsuperconducting material.

For $R_1,R_2,R_L>0$ and $V_s\ne 0$, define the fractional shortfall in output magnitude from the unloaded value byThis is a loading bias relative to the unloaded target, not the uncertainty of an instrument's measurement of the connected circuit. The equality with the signed ratio also holds for negative $V_s$, because the output retains its sign. The Thévenin resistance $R_{\mathrm{th}}=R_1\parallel R_2$ summarizes the divider's susceptibility: the relevant comparison is $R_L/R_{\mathrm{th}}$, not merely a large absolute $R_L$.

Set $a=R_1/R_2>0$ and $x=R_L/R_2>0$. ThenThe derivative proves that increasing the load electrical resistance always reduces this error. For equal divider arms, $e=1/(1+2x)$: $x=1$ gives $33\frac{1}{3}\%$, $x=10$ gives $100/21\%=4.7619\dots \%$, and $x=100$ gives $100/201\%=0.497512\dots \%$. Thus a “ten times the lower resistor” rule does not guarantee one-percent agreement with the unloaded target.

A quantitative design requirement follows without a rule of thumb. For a specified $0<ϵ<1$,For equal $10\mathrm{k\Omega }$ arms, $R_{\mathrm{th}}=5\mathrm{k\Omega }$, so one-percent loading error requires $R_L\ge 495\mathrm{k\Omega }$. Reducing both divider electrical resistances reduces loading but increases unloaded supply electric current and Joule heating: $P=V_s^2/(R_1+R_2)$. The stipulated equal-arm example dissipates $5\mathrm{mW}$ in total when unloaded. This tradeoff explains why the ratio alone does not determine a useful divider.

The watt is the SI unit of power, one joule per second. Joule heating at $12.5\mathrm{W}$ transfers $125\mathrm{J}$ of energy over a time of $10\mathrm{s}$ if that power stays constant. ^[ref].

The waveguide cutoff frequency separates oscillatory axial propagation from exponential decay for a particular normal mode. For the vacuum-filled rectangular electromagnetic waveguide, the dispersion relation isFor a transverse electric mode of a rectangular waveguide, $m,n$ are nonnegative integers but not both zero. For a mode with $B_z=0$ and nonzero $E_z$, both indices must be positive. The lowest cutoff for $a>b$ is consequently ${\omega }_{10}=\pi c/a$, or ordinary frequency $f_c=c/(2a)$.

A square cross-section instead gives equal ${\mathrm{TE}}_{10}$ and ${\mathrm{TE}}_{01}$ cutoffs. The absence of a transverse electromagnetic mode is specific to this hollow pipe, rather than all electromagnetic waveguides.^[ref]

For $\omega >{\omega }_{mn}$, choose real $\beta >0$ for propagation toward $+z$. For $0<\omega <{\omega }_{mn}$, a field decaying into a semi-infinite guide hasA finite section can contain both growing and decaying axial solutions set by its two ends; the single decaying solution is not a complete finite-section transmission calculation. At exact cutoff $\beta =0$, the two axial solutions become a constant and a linear function of $z$: there is no finite axial wavelength for a traveling sinusoid.

In the chosen $a=30\mathrm{mm}$ guide at $f=0.8f_c=3.99723277\mathrm{GHz}$, the ${\mathrm{TE}}_{10}$ branch has $\alpha =20\pi {\mathrm{m}}^{-1}$. Its amplitude decay length is $1/\alpha \simeq 15.9155\mathrm{mm}$; after $50\mathrm{mm}$, its normalized wave amplitude is $e^{-\pi }\simeq 0.0432139$. This is geometric evanescence even with ideal walls, not Joule heating. A single decaying branch has zero cycle-averaged axial Poynting vector: substituting $\beta =i\alpha$ makes $B_x$ a quarter-cycle out of phase with $E_y$. Coupling across a finite section requires the complete boundary-value solution.

For a finite undirected graph without self-loops, give each edge a symmetric nonnegative weight $w_{ij}=w_{ji}$. In the weighted graph, set $(A_w{)}_{ij}=w_{ij}$ for an edge and zero otherwise, and $(D_w{)}_{ii}={\sum }_j{w}_{ij}$ with all other entries zero. The Weighted graph Laplacian isEach edge appears once in the quadratic form. For the path graph $A-B-C$ with weights $w_{AB}=2$ and $w_{BC}=3$,The Weighted graph Laplacian is still a positive semidefinite matrix. Its kernel consists of vectors constant on each connected component of a graph after zero-weight edges are removed: an edge of weight zero imposes no constraint. In an electrical network, taking the weights to be electrical conductances makes $x^T{L}_{w}x$ the total power dissipated by Joule heating when $x$ gives the voltages at the vertices. This physical interpretation explains why stronger connections penalize disagreement more heavily.^[ref]

A real voltmeter has finite input electrical resistance $R_d$; a galvanometer also has a finite winding electrical resistance. Away from balance, its electric current changes both intermediate electric potentials. This is voltmeter loading when the detector is a voltmeter. Model the detector as a passive resistor, with no offset or injected electric current. Thévenin's theorem supplies the full response without pretending the connected bridge remains two independent voltage dividers.

With the detector removed, the equivalent source is $V_{\mathrm{oc}}$. Replace the ideal supply by a wire to calculate the source electrical resistance seen between $B,C$: nodes $A,D$ merge, making $R_1,R_2$ parallel and $R_3,R_x$ parallel. ConsequentlySince the attenuation factor is strictly positive for $R_d>0$, the null remains $R_2{R}_3=R_1{R}_x$. A smaller detector electrical resistance reduces the measured electric potential difference off balance but does not shift the ideal null. Its electric current sensitivity instead depends on $1/(R_{\mathrm{th}}+R_d)$; detector gain and resolution also matter in deciding how closely a null can be located.

In the plotted example, $R_x=1.100\mathrm{k\Omega }$ gives $V_{\mathrm{oc}}=-119.048\mathrm{mV}$ and $R_{\mathrm{th}}=1.02381\mathrm{k\Omega }$. A detector with $R_d=1.000\mathrm{k\Omega }$ instead reads $V_d=-58.824\mathrm{mV}$. Using that loaded reading in the unloaded voltage divider formula would underestimate the deviation of $R_x$ from $1\mathrm{k\Omega }$. Balancing the bridge avoids this particular systematic measurement error.

The local derivative makes the sensitivity explicit. For $R_1=R_2=R_3=R$ and $R_x=R+\delta R$, the first-order Taylor expansion isAt the plotted null, the open-circuit slope is $-1.25mV/\Omega$, compared with $-0.625mV/\Omega$ for $R_d=1\mathrm{k\Omega }$. Increasing the supply improves this ideal sensitivity but also increases Joule heating. A finite source electrical resistance requires recomputing $R_{\mathrm{th}}$; the displayed parallel-pair expression assumes an ideal supply.