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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.

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.

Southern Illinois University's ET 304A guide, Lab 4: Voltage Measurements and Meter Loading Effects, documents an instructive test: build equal-arm voltage dividers on a $20\mathrm{V}$ supply, then measure across the lower arm with a digital multimeter, analog multimeter, and oscilloscope. Starting with two $10\mathrm{k\Omega }$ resistors, repeat with equal pairs from $220\mathrm{k\Omega }$ to $4.7\mathrm{M\Omega }$. It requires separate recorded readings, unloaded predictions, and predictions including each instrument's input electrical resistance. The guide obtains meter resistances from specification sheets and scope resistance from the input-panel marking. Equal ratios control the unloaded target while changing susceptibility to voltmeter loading. The provided tables are blank: the guide specifies a procedure, not measured outcomes.^[ref]

Preparation and accuracy checks are made concrete by Berkeley's Lab 1: Introductory Experiments and Linear Circuits I. Its different divider uses nominal $470\mathrm{k\Omega }$ and $10\mathrm{k\Omega }$ resistors and a $24\mathrm{V}$ supply on a breadboard. Measure isolated resistors first, record actual supply voltage, recalculate with these measured values, then record the junction voltage with the digital multimeter. It explicitly compares actual values with resistor tolerance and the final voltage with a range-dependent accuracy bound from the instrument manual. This is an uncertainty check, not an independent measurement calibration.^[ref]

For a controlled comparison, additionally record the selected instrument range, probe factor, and input coupling; connect each instrument separately to the same electrical reference potential, and monitor the input rather than treating the supply label as an exact voltage. These are proposed controls for interpreting the documented procedures. A voltage indication and a resistance indication from the same digital multimeter do not independently validate its measurement calibration; a traceable reference comparison would be needed for that. Resistor tolerance describes component variation, while measurement uncertainty describes what is known about measured values. Using nominal arms alone can confuse mismatch with loading: equal nominal arms each within one-percent tolerance can give unloaded ratios from $0.495$ to $0.505$.

The inference to test is specific: once actual arm values, input voltage, and meter input electrical resistance are accounted for, the readout should approach the loaded voltage divider prediction within the remaining measurement uncertainty. A discrepancy can instead reflect contacts, changing temperature, source droop, or an input that is not a fixed passive resistor. A source with positive series electrical resistance $R_s$ can be modeled by replacing $R_1$ with $R_1+R_s$ only when $V_s$ denotes its ideal internal voltage; when the actual voltage across the divider is measured, that terminal value is already $V_{\mathrm{in}}$. During settling, or with appreciable input capacitance and changing signals, the steady direct current calculation is insufficient. Neither institutional procedure supplies measured values for the original $10\mathrm{V}$ example above.

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 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.

An alternating-current electrical circuit comparing a parallel resistor-capacitor arm with a series resistor-capacitor arm. Let the upper-left arm have electrical impedance $Z_1=R_1/(1+i\omega R_1{C}_1)$, the lower-left arm $Z_2=R_2+1/(i\omega C_2)$, and the upper-right and lower-right arms have electrical resistances $R_3,R_4$. For input between the upper and lower nodes and an unloaded difference measurement between the arm junctions, voltage divider balance requires $Z_2/Z_1=R_4/R_3$. Separating real and imaginary parts givesThe balance angular frequency is positive for positive components. Equal $R_1=R_2=R$ and $C_1=C_2=C$ require $\omega =1/(RC)$ and $R_4=2R_3$. Reversing which reactive arm is upper changes the divider ratios, so the connections must always be specified.

A wirewound resistor uses resistive metal wire wound on an insulating former. Suitable alloys and heat-resistant construction provide stable electrical resistance and substantial electrical dissipation; parasitic inductance and interturn capacitance limit the frequency range over which it behaves as a pure resistor.