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Thermal boundary resistance describes a temperature drop between the two sides of a material interface carrying heat flux. Efficient thermal conduction within each material does not guarantee efficient heat transfer between them. The solid–helium case is Kapitza resistance; the broader concept also includes solid–solid interfaces. ^[ref]

Let $T_s$ and $T_l$ be the temperatures extrapolated to the interface from its solid and liquid sides, each treated as internally in thermal equilibrium. For uniform normal heat flux $q>0$ through area $A$, define the finite-drop, area-specific thermal resistance and the resistance of the whole interface byThe macroscopic jump represents an unresolved interfacial region; it is not a claim that one point has two equilibrium temperatures. At nonzero heat flux, the two sides are outside mutual thermal equilibrium. A measured difference between distant thermometers also contains bulk temperature gradients, which must be measured or bounded before assigning all of it to the interface.

An energy balance explains why a temperature jump can coexist with continuous heat flux. For a stationary interfacial region with no local work or heat source,The first law of thermodynamics therefore requires $q_{\mathrm{in}}=q_{\mathrm{out}}=q$ in a steady state. It does not require $T_s=T_l$. For a heated sample with electrical power $P=VI$, an independently determined outward parasitic loss $P_{\mathrm{leak}}$ and no other inputs,Here $V$ is the heater's electric potential difference and $I$ its electric current. Subtracting a lead loss means subtracting power before dividing by area; $q$ is not the total heating power. Unknown losses and stored internal energy prevent that simple steady inference.

For a one-dimensional comparison, put an interface between two slabs with the same area $A$, positive constant thermal conductivities $k_s,k_l$ and lengths $L_s,L_l$. Neglect side losses. Integrating Fourier's law across each slab givesThus, for a linear interface with constant $r_B$, the series result isThese slab formulas are a chosen thermal conduction model, not a bulk transport law for moving liquid helium. If the interface law is nonlinear, substitute its own relation between $q$ and $T_s-T_l$ into the preceding balance; a single constant resistance cannot describe every load. Increasing bulk thermal conductivity has diminishing benefit when the interfacial drop dominates. Increasing usable contact area reduces the whole-interface thermal resistance if the local interface law and the other conditions remain fixed.

Thermal electrical noise is the fluctuating electric potential difference or electric current produced by thermal fluctuations in a dissipative electrical circuit, even without an applied voltage. A resistor at thermal equilibrium is both a source of noise and a sink for electrical energy. Its electrical resistance describes the dissipation, and its absolute temperature sets the scale of the fluctuations. This connects a practical limit on signal-to-noise ratio to statistical mechanics: a warm, unbiased resistor is not electrically silent.

John Bertrand Johnson measured the effect using a vacuum-tube amplifier and thermocouple; his 1928 paper reported agreement with the Boltzmann constant. Harry Nyquist's companion paper calculated the effect from thermodynamics and statistical mechanics.^[ref][ref]

Thermal resistance measures the temperature drop required to drive a given rate of heat transfer. For the steady one-dimensional bar model in thermal conduction,Here $r$ relates the temperature drop to heat flux, while $R_{\mathrm{th}}$ relates it to the rate of heat transfer. The single bar above has $r=0.050{\mathrm{m}}^2\mathrm{K}{\mathrm{W}}^{-1}$ and $R_{\mathrm{th}}=5\mathrm{K}{\mathrm{W}}^{-1}$, so its $20\mathrm{K}$ drop drives $4\mathrm{W}$.

To see why series thermal resistances add, apply Fourier's law in each layer to obtain $\Delta T_i=\dot{Q}{R}_i$. With no accumulation or side losses the rate of heat transfer is the same through both, so adding the drops gives $T_h-T_c=\dot{Q}(R_1+R_2)$. This lets us size layered walls without solving a new differential equation for every layer. The structure matches Ohm's law: temperature difference plays the role of electric potential difference, the rate of heat transfer plays the role of electric current, and thermal resistance plays the role of electrical resistance. ^[ref]

The electric potential difference corresponding to thermal energy $k_{B}T$ per elementary charge $q$.

A thermocouple uses two different electrical conductors and the Seebeck effect to infer a measuring-junction temperature from a voltage and a known reference temperature. The reference material and the reference temperature are distinct choices: B specifies the comparison Seebeck coefficient, while $T_c$ specifies the lower limit of the integral. The two-material experiment in Seebeck effect gives the geometry, sign convention, and calibration limitations.

The useful measurement is a pair: the raw terminal electric potential difference and the thermocouple reference junction temperature. For the A-positive terminal order and homogeneous-wire assumptions of Seebeck effect, write the open-circuit electromotive force asHere $T_h$ is the thermocouple measuring junction temperature and $T_c$ is the common temperature of the reference connections. The subscript $h$ names the measuring end; it need not be hotter. A voltmeter senses the complete connected thermoelectric path through the wires and its leads, in the electrochemical potential convention already described under Seebeck effect. It does not measure an absolute temperature, the individual Seebeck coefficient of A, or an isolated contact electric potential.

A thermopile connects multiple thermocouples in series to turn a temperature difference into a larger electrical signal. Absorbed thermal radiation can heat its sensing junctions while reference junctions remain cooler; the signal depends on the thermal and optical design.

A thermopile is an array of thermocouples connected as a series circuit with their electromotive forces adding in the chosen terminal order. It measures a collective temperature difference through the Seebeck effect. An open-circuit electric potential difference requires no externally supplied electric current through the array: a high-input-electrical resistance voltmeter reads the signal. The heat that maintains the junction temperature difference still comes from somewhere, such as absorbed electromagnetic radiation.

In a membrane detector, an absorbing film heats the measuring junctions while reference junctions lie on the surrounding support. Actual junction temperatures can differ across the membrane.^[ref] This converts difficult-to-observe heat transfer into an accessible electrical signal; the series connection combines many small contributions without requiring every thermocouple to have a separate meter.

PASCO's documented lamp experiment connects a TD-8553 thermopile sensor to a millivoltmeter and places it level with a lamp filament along a meter stick. Lamp and detector axes are aligned. With the lamp off, readings at several positions establish an average ambient thermal radiation background. With the lamp on at a fixed supply setting, detector readings at different distances are compared after background subtraction. Between readings, a reflective shield prevents the detector warming; its shutter also protects the reference junction when idle. Moving the shutter can move the sensor, so fixed geometry matters. The manual's lamp limits must be followed.^[ref]

This protocol tests whether the background-corrected electric potential difference approximately follows the inverse-square law over a suitable distance range. It is a relative electromagnetic radiation measurement; those readings alone do not calibrate absorbed power in watts. The manual treats the filament as an approximate point source.^[ref]

For an absolute calibration, independently known absorbed power and a monitored reference temperature would be needed. A useful additional control is to repeat a fixed detector position before and after a distance sweep: disagreement reveals possible drift. These additions test the measurement chain rather than assuming that background subtraction fixes every error. A finite source, changing illuminated area, reflections, changing reference temperature, and insufficient settling can alter the measured dependence. In particular, the time constant in thermopile thermal response explains why a quick reading and a settled reading may differ. The detector infers its own absorber temperature difference; inferring the source's temperature additionally requires its emissivity, optical geometry and an appropriate thermal radiation model.

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.

To determine the Thévenin resistance, remove the external electrical load and set each independent electrical source to zero in the circuit model. A zero ideal voltage source enforces zero electric potential difference, so replace it by a short connection. A zero ideal current source enforces zero electric current, so replace it by an open connection. Retain every resistor, including any explicitly modeled supply electrical resistance. Turning an actual power supply's front-panel control to zero is not evidence that it physically becomes an ideal wire.

Retain every dependent electrical source and its controlling variable: its response to an electrical test source is part of the network's electrical resistance. Impose terminal voltage $v_t$ with an electrical test source and calculate $j_t$ entering the suppressed network at $a$. For a nonzero test yielding a finite ratio,Alternatively inject a known entering electric current and solve for $v_t$. The test electric current is opposite to the delivered-current convention of the original Thévenin terminal response. ^[ref]

For an original illustration, put a positive resistor $R$ across $a,b$ and a dependent electrical source in parallel that injects electric current $g{v}_t$ from $b$ toward $a$, where $g$ has units of electrical conductance. Its control is the same terminal voltage $v_t$. Kirchhoff's current law givesWith $R=1\mathrm{k\Omega }$ and $g=0.5\mathrm{mS}$, a $1\mathrm{V}$ test draws $0.5\mathrm{mA}$, giving $R_{\mathrm{th}}=2\mathrm{k\Omega }$. Removing the dependent electrical source would falsely give $1\mathrm{k\Omega }$. With $g=2\mathrm{mS}$, the model instead has $R_{\mathrm{th}}=-1\mathrm{k\Omega }$; with $g=1\mathrm{mS}$ it draws zero test electric current for any imposed voltage and has no finite ratio. An active physical realization requires a supply and has a limited operating range.

For a finite nonzero $R_{\mathrm{th}}$ and a well-defined short-circuit response, $I_{\mathrm{sc}}=V_{\mathrm{oc}}/R_{\mathrm{th}}$. Thus $R_{\mathrm{th}}=V_{\mathrm{oc}}/I_{\mathrm{sc}}$ is usable when $I_{\mathrm{sc}}\ne 0$. A source-free positive resistor has both $V_{\mathrm{oc}}=0$ and $I_{\mathrm{sc}}=0$, so this quotient cannot reveal its electrical resistance; an electrical test source can.

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 charge-to-mass experiment infers the magnitude of an electron's charge-to-mass ratio by combining electric and magnetic deflection of cathode rays. The central idea is to determine speed separately from magnetic curvature: a known speed turns the observed gyroradius into a measurement of $e/m$.

In 1897, J. J. Thomson used two grounded slits in an evacuated glass tube, then aluminium plates $5\mathrm{cm}$ long, $2\mathrm{cm}$ wide and $1.5\mathrm{cm}$ apart. A battery supplied their electric potential difference. Deflected cathode rays produced a phosphorescence patch on the glass; an external scale and movable luminous pointer recorded its position in darkness. Greater evacuation reduced gas electrical conductivity that screened the electric field; reversing plate polarity reversed deflection. External coils supplied the magnetic field, calibrated by induction against larger coils. He adjusted equal electric and magnetic deflections. ^[ref]

His comparison included air, hydrogen, carbon dioxide, and aluminium versus platinum cathodes, supporting a common carrier. He explicitly noted uncorrected magnetic bending outside the plates. His tables give $m/e$, the reciprocal of the magnitude used here. The inferred ratio alone could not separate electric charge from mass. ^[ref]

To connect this readout with the Lorentz force, derive a simple nonrelativistic model in modern SI notation. Let the uniform deflecting region have length $\ell$, followed by a field-free drift of length $L$. Assume a narrow speed distribution, small deflection angles, negligible collisions, and matching electric and magnetic field profiles. For an electron with speed $v$, the electric acceleration has magnitude $eE/m$, the transit time is approximately $\ell /v$, and the subsequent drift adds exit transverse velocity times $L/v$. Thus the displacement magnitudes measured separately with only electric field or only magnetic field areThe electric expression follows from constant transverse acceleration; the magnetic expression uses its small-deflection approximation. Equal displacement magnitudes then giveThis is an ideal derivation of the comparison principle, not a numerical reanalysis of the historical tube. Different exterior magnetic field profiles invalidate cancellation of the common geometrical factor. The comparison determines $e/m$ through deflection and the known apparatus dimensions; it does not require a downstream semicircle.

For a second, deliberately different teaching geometry, combine a velocity selector with a separate region containing only a uniform magnetic field $B_a$ perpendicular to the selected velocity. Here $B_s$ belongs to the velocity selector, while $B_a$ belongs to the analyzer: they need not be equal. The magnetic Lorentz force supplies the centripetal acceleration, soIf the analyzer turns the beam through a semicircle before it hits a screen in the entry plane, the separation $D$ between entrance and impact is the diameter: $r=D/2$. For a shorter observed arc, the screen displacement is not generally a radius; fit the appropriate circle geometry. This two-stage velocity selector and curvature apparatus is an original ideal teaching model, not a claim that Thomson's Fig. 2 used a separate selector and semicircular analyzer.

For a worked example, stipulate $E=2.00\times 10^3\mathrm{V}{\mathrm{m}}^{-1}$, $B_s=1.00\times 10^{-3}\mathrm{T}$, $B_a=2.00\times 10^{-4}\mathrm{T}$, and $D=0.1136\mathrm{m}$. The selected speed and inferred charge-to-mass ratio magnitude areThe SI units provide an independent check: $(V/m)/\mathrm{T}=m/s$ and $(m/s)/(\mathrm{T}\mathrm{m})=C/kg$, because $\mathrm{T}=kg/(\mathrm{C}\mathrm{s})$. The charge-to-mass ratio of the electron is negative; this calculation gives its magnitude. The bending direction gives the sign of the electric charge once the magnetic field direction is known. Scaling both $e$ and $m$ by the same positive factor preserves every displayed trajectory, so an independent electric charge measurement is required to determine mass separately.

The curvature formula assumes nonrelativistic momentum $p=mv$. At higher speeds, the relativistic cyclotron orbit gives $p=\gamma mv=e{B}_{a}r$, hence $e/m=\gamma v/(B_{a}r)$, where $\gamma$ is the Lorentz factor; the ideal straight-through selection $v=E/B_s$ still follows from vanishing Lorentz force. The stipulated speed is about $0.0067$ times the speed of light, making $\gamma -1$ about $2.2\times 10^{-5}$. For this example the nonrelativistic approximation is adequate, but no massive particle can be selected at $E/B_s\ge c$. Finite slit widths, collisions with residual gas, uncertain magnetic field calibration, and fields outside the ideal boundaries limit a real inference; repeat at several selected speeds and analyzer strengths to test consistency of $e/m$ rather than trusting one spot position.

A titration curve plots a measured or calculated property against added titrant volume or amount of substance. In an acid-base titration, plotting pH exposes gradual changes and a steep endpoint region. A potentiometric titration uses an electrode electric potential difference; a calibrated glass electrode and reference electrode provide operational pH. Finite additions bracket a transition rather than measuring a continuous curve. A chemical equilibrium fit can estimate the equivalence point; identifying the steepest sampled interval alone has a resolution set by the dose spacing and instrument response.

The change of output electric current per change of input electric potential difference, evaluated at an operating point with the other terminal conditions specified. In a small-signal model, $i_{\mathrm{out}}=g_m{v}_{\mathrm{in}}$ describes this contribution; transconductance has the dimensions of electrical conductance.

An electronic amplifier converting input electric current to output electric potential difference.

A tunnel diode is a p-n junction with sufficiently heavy semiconductor doping on both sides to make a thin depletion region and permit substantial interband tunnelling. The neutral regions are degenerate semiconductors. Its distinctive forward current-voltage characteristic contains a peak followed by a valley: over part of the curve, increasing the applied electric potential difference decreases the electric current. The Esaki tunnel-diode experiment made this behavior experimentally concrete.

The useful quantity is the local slope, not the ratio of total voltage to total electric current. With conventional electric current $I$ entering the positive-voltage terminal, defineNegative differential resistance means $g_d<0$, hence $r_d<0$ where defined. A forward operating point can still have $V>0$, $I>0$ and absorbed power $VI>0$. A small voltage increase then gives a negative current increment; this does not make the total electric current negative or turn the device into an unlimited source of energy. The external supply sets the operating point and supplies the energy available for an electronic amplifier.

An original idealized construction explains the shape. Put the n-type semiconductor on the left and the p-type semiconductor on the right. Take $q>0$ as the elementary charge and $V=V_p-V_n$. Increasing forward bias lowers the right-hand electron energy levels relative to the left. Denote the two reservoir electrochemical potentials by ${\mu }_n,{\mu }_p$, so that ${\mu }_n-{\mu }_p=qV$. At zero temperature, occupied left-hand conduction band states extend up to ${\mu }_n$; empty right-hand valence band states start at ${\mu }_p$. For transfer conserving electron energy, the available interval isIt is nonempty only when its upper endpoint exceeds its lower endpoint. Both an occupied initial quantum state and an empty final quantum state must exist at the same energy. The Pauli exclusion principle prevents transfer into an already occupied final quantum state.

For the original diagram, choose arbitrary energy units with $E_{c,n}=-0.5$, ${\mu }_n=0$, ${\mu }_p=-u$ and $E_{v,p}=0.5-u$, where $u$ is $qV$ in those units. The available interval has widthThus $L(0.25)=0.25$, $L(0.50)=0.50$, $L(0.70)=0.30$, and $L(1.15)=0$. Initially forward bias opens a transfer window; further forward bias closes it by moving the right-hand valence band edge below the left-hand conduction band edge. This is a state-counting illustration, not a formula for a measured electric current.

For reverse bias, the transfer direction changes: occupied right-hand valence band states can feed empty left-hand conduction band states. In the same construction, $u=-0.70$ opens the interval $0<E<0.70$ for this reverse transfer. A narrow depletion region allows a large reverse quantum tunnelling electric current at small electric potential difference, whereas the ordinary Shockley diode equation predicts only a small reverse saturation electric current before diode reverse breakdown. At zero biasing, equal reservoir occupations give zero net electric current; quantum tunnelling does not require a net equilibrium flow.

A finite-temperature refinement replaces sharp occupation cutoffs with the Fermi-Dirac distribution. An illustrative weak-transfer expression isHere $\mathcal{W}\ge 0$ describes allowed initial/final quantum states and their coupling, and $f_n,f_p$ are Fermi-Dirac distributions with the respective electrochemical potentials. The occupation difference follows by subtracting the reverse transfer factor from the forward one: $f_n(1-f_p)-f_p(1-f_n)=f_n-f_p$. An overlap interval alone does not determine $\mathcal{W}$: transmission probability, density of states, crystal momentum selection and phonons can matter. A complete germanium calculation is more demanding than multiplying two constant density of states factors. At larger forward bias, ordinary carrier injection and diffusion supply a rising current branch even after direct band overlap is lost.

Consider two identical magnetic layers separated by a nonmagnetic layer, between reservoirs that impose the same terminal electric potential difference on both spin channels. Neglect spin-flip scattering and assume each layer obeys Ohm's law. In a magnetic layer, define positive channel electrical resistances $a,b>0$. Let the spacer contribute $r\ge 0$ to each channel. These are circuit parameters, not measured iron or chromium resistivities.

For parallel layer magnetizations, one channel meets $a,r,a$ and the other $b,r,b$. For opposite layer magnetizations, each meets one $a$, one $b$, and the spacer. The two channels share the same terminal electric potential difference and run in parallel. Writing $x\parallel y=xy/(x+y)$ givesThe inequality is algebra, given the specified network. It shows how a difference between spin-channel responses can yield a difference between magnetic configurations even though all the constituent resistors are ohmic.

For chosen $a=1\Omega$, $b=4\Omega$, $r=1\Omega$, the parallel channels have total electrical resistances $3$ and $9\Omega$, while the opposite-alignment channels each have $6\Omega$. Thus $R_P=2.25\Omega$ and $R_{AP}=3.00\Omega$. At a common terminal electric potential difference $1\mathrm{V}$, the total electric currents are $4/9\mathrm{A}$ and $1/3\mathrm{A}$, respectively.

Reuse the velocity selector analysis in the Thomson charge-to-mass experiment, but now send positive ions with $q=ze$ into a separate magnetic analyzer. The selector fields $E_s$ and $B_s$ give $v_0=E_s/B_s$. If the analyzer field $B_a$ is perpendicular to the selected velocity, its gyroradius givesfor a semicircular return to the entrance plane. The displayed impact separation $D$ is a diameter. This formula is linear in $r$ at fixed selected speed; the fixed-electric potential difference model above instead gives $m/q\propto r^2$. They impose different experimental conditions.

For a chosen example, take $E_s=2.00\times 10^{4}V/m$, $B_s=0.200\mathrm{T}$, and $B_a=0.500\mathrm{T}$. Thus $v_0=1.00\times 10^{5}m/s$. Approximating an ion mass by $24\mathrm{u}$ and setting $q=e$ gives $r\approx 49.75\mathrm{mm}$ and $D\approx 99.50\mathrm{mm}$. Here $\mathrm{u}$ denotes the unified atomic mass unit. These are stipulated teaching inputs, not a reported Aston measurement or an exact magnesium atomic mass.

A finite accepted speed range broadens the radius distribution. At fixed $m/q$ and $B_a$, $\delta r/r=\delta v/v$. At fixed electric charge and mass, $K=m{v}^2/2$ gives $\delta v/v\simeq \delta K/(2K)$ for a small kinetic energy spread. A one-percent kinetic energy spread therefore produces roughly a half-percent radius spread before other broadening is included. Its width convention must be specified before converting that statement into a mass resolving power.

The volt is the SI unit of electric potential and electric potential difference: $1\mathrm{V}=1J/C$, one joule per coulomb. A positive electric charge of $2\mathrm{C}$ moved through a potential increase of $3\mathrm{V}$ gains $\Delta U=q\Delta \Phi =6\mathrm{J}$ of potential energy; the electric field does $-6\mathrm{J}$ of work. ^[ref].

An oscillator whose frequency is adjusted by an input electric potential difference. An approximately linear control characteristic provides a model for frequency modulation.

A voltage divider uses two resistors connected in series to obtain a fraction of an applied electric potential difference. Let the upper electrical resistance be $R_1>0$, the lower one $R_2>0$, and the input be $V_{\mathrm{in}}=V_s$ relative to a common electrical reference potential. The output is the junction's electric potential difference relative to that same reference, across $R_2$. The familiar ratio assumes zero output electric current: no external electrical load draws from the junction. In a steady direct current model with fixed resistors obeying Ohm's law and an ideal voltage source,Here $V_0$ is the open-circuit voltage; $R_1,R_2$ correspond to $R_a,R_b$ in the header formula. With no junction branch, Kirchhoff's current law makes the same electric current pass through both resistors, and Ohm's law gives the two voltage drops. For $V_s\ne 0$, division gives $V_0/V_s=R_2/(R_1+R_2)$; if $V_s=0$, the predicted output is zero but this ratio is undefined.

An original stipulated example has $R_1=R_2=10\mathrm{k\Omega }$ and $V_{\mathrm{in}}=10\mathrm{V}$. With zero output electric current, $I=0.5\mathrm{mA}$ and $V_0=5\mathrm{V}$. An attached $10\mathrm{k\Omega }$ electrical load instead gives $V_{\mathrm{out}}=10/3\mathrm{V}$; a $1\mathrm{M\Omega }$ load gives $V_{\mathrm{out}}=1000/201\mathrm{V}=4.975124378\dots \mathrm{V}$. These are exact model predictions, not measured readings or specifications of a particular multimeter or oscilloscope.

The divider is useful for scaling a voltage or converting a change in a sensor's electrical resistance into an electric potential difference. Its beauty is also its trap: multiplying both electrical resistances by the same positive factor preserves the unloaded ratio, but changes the effect of any fixed electrical load. A divider therefore supplies a predictable ratio under specified loading conditions; it is not a source that maintains that voltage for arbitrary output electric currents. A Wheatstone bridge compares two divider junctions, while Thévenin's theorem describes the divider's response to many different loads.