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The ratio of output to input signal electric potential difference for specified source and load conditions. A signed voltage gain records whether the output reverses polarity; its value can depend on frequency.

An electrical circuit that uses diodes and capacitors to produce an output electric potential difference larger in magnitude than the input peak. The attainable output depends on the circuit arrangement, electric current drawn by the load, capacitance and input frequency; an ideal unloaded result does not specify the output under load.

An ideal voltage source prescribes an electric potential difference; its electric current is determined by the connected electrical circuit. A dependent electrical source can prescribe a voltage controlled by another circuit variable instead.

A voltmeter measures an electric potential difference between two contacts; its finite input electrical resistance can cause voltmeter loading.

Voltmeter loading is the change in an electric potential difference caused by the electric current drawn by the measuring voltmeter.

For a loaded voltage divider in steady direct current, a voltmeter connected across the lower resistor places its finite input electrical resistance $R_m$ in parallel with the lower arm's electrical resistance $R_2$ and any intended electrical load of electrical resistance $R_L$. With positive, fixed electrical resistances, an upper resistor of electrical resistance $R_1$, and an ideal voltage source supplying electric potential difference $V_{\mathrm{in}}$, Kirchhoff's current law and Ohm's law giveHere $V_m$ is the connected circuit's electric potential difference across the voltmeter; omit $1/R_L$ if there is no intended electrical load. The parallel connection and need for large input electrical resistance are standard voltmeter operating principles.^[ref]

Voltmeter loading is a bias relative to the target electrical circuit with the voltmeter absent and any intended electrical load still present. Voltage divider loading error instead compares the attached-load output with the wholly unloaded output. An accurate readout of $V_m$, supported by measurement calibration, can still differ from either target: calibration accuracy concerns the indication of the connected electric potential difference, whereas voltmeter loading changes that electric potential difference itself.

The von Klitzing constant is $R_K=h/e^2=25812.807459\dots \Omega$. The current SI fixes the Planck constant $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and elementary charge $e=1.602176634\times 10^{-19}\mathrm{C}$ exactly, so $R_K$ is exact; the displayed decimal is truncated. It differs from the historical conventional value $R_{K\text{-}90}=25812.807\Omega$.^[ref]

As an original ideal example, $N=2$ and electric current $I=1\mu \mathrm{A}$ give $|V_H|=I{R}_K/2\simeq 12.9064\mathrm{mV}$. Changing the Hall bar's width changes the transverse electric field required for this electric potential difference, while the ideal ratio $|V_H|/I$ remains $R_K/2$.

The Wheatstone bridge has four resistor arms and two diagonals. Here the supply holds node $A$ at constant electric potential difference $V_s>0$ relative to node $D$, our electrical reference potential. The left path is $A\to R_1\to B\to R_2\to D$; the right path is $A\to R_3\to C\to R_x\to D$. The unknown is $R_x$, and $R_3$ is adjustable. A voltmeter or galvanometer connects $B$ to $C$. We define its positive reading as $V_B-V_C$; exchanging its leads reverses the sign. All four electrical resistances are positive, the supply is ideal, and the calculation describes steady electric currents at fixed temperature.

Charles Wheatstone described the instrument in his 1843 Bakerian lecture while explicitly crediting Samuel Hunter Christie. The account from King's College London's collection identifies that attribution; the familiar name should not be read as an assertion that Wheatstone alone originated the bridge.^[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.

Use the equal-component Wien bridge with upper-left parallel arm and lower-left series arm. Let $A$ be the left arm junction, $C$ the upper input node, and $V_i$ the upper-minus-lower input electric potential difference. With $u=\omega RC$, the unloaded transfer ratio $H=(V_A-V_C)/V_i$ is the displayed expression. Its band-pass filter magnitude is $[9+(u-u^{-1}{)}^2{]}^{-1/2}$, maximal at $u=1$ with $H=-1/3$. One continuous wave phase is $\varphi =-\pi -\arctan [(u-u^{-1})/3]$, tending from $-\pi /2$ at low positive frequency to $-3\pi /2$ at high frequency. Reversing the measured output polarity multiplies $H$ by $-1$ and shifts its phase by $\pi$ without changing its magnitude.

For an electron starting with negligible kinetic energy and accelerated through a constant positive electric potential difference $V$, electrical work supplies $K=eV$. In the ideal single-electron bremsstrahlung model, conservation of energy bounds any one emitted photon:The endpoint corresponds to essentially all available kinetic energy going into one photon. It is a limiting bound, not a claim that every electron produces an endpoint photon. Neglect target recoil, initial kinetic energy, and losses before the emitting interaction. If the accelerating voltage varies, its largest value sets the corresponding ideal upper endpoint; a mean voltage does not substitute for that value.

For a calculated constant-$V$ example, $50\mathrm{kV}$ gives $E_{\max }=50\mathrm{keV}$ and ${\lambda }_{\min }=0.0248\mathrm{nm}$. Doubling $V$ halves ${\lambda }_{\min }$. Producing a $0.0200\mathrm{nm}$ photon requires at least $V=hc/(e\lambda )\simeq 62.0\mathrm{kV}$ in this model. A continuous frequency spectrum spans lower photon energies; the cutoff calculation predicts neither its shape nor the efficiency of photon emission.