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In a Wheatstone bridge, the galvanometer connects the two comparison junctions. Its purpose is to locate a null measurement, rather than to measure the branch electric current through the unknown resistor. Using Wheatstone bridge detector loading and Thévenin's theorem, a settled moving-coil galvanometer with winding electrical resistance $R_g$ obeysHere $V_{\mathrm{oc}}$ is the bridge's open-circuit voltage and $R_{\mathrm{th}}$ its Thévenin resistance. For a functioning passive detector with finite positive electrical resistances, exact $I_g=0$ implies $V_{\mathrm{oc}}=0$, preserving Wheatstone bridge balance. Away from balance, the detector loads the junctions; its electrical resistance cannot be ignored.

For the illustrative $R_g=100\Omega$, $S_I=3000rad/A$ and a chosen $R_{\mathrm{th}}=1900\Omega$, a $1.00\mathrm{mV}$ bridge open-circuit voltage produces $0.500\mu \mathrm{A}$ and $0.0859^{\circ }$. That lies inside the stipulated $0.10^{\circ }$ zero tolerance. The corresponding resolvable open-circuit voltage scale is about $1.16\mathrm{mV}$, not arbitrarily small. Actual limits also depend on the zero stability and calibration.

Use the detector checks in the slide-wire resistance bridge experiment: perturb the comparison to either side of the proposed balance and require opposite signs, while confirming the supply is on. A broken connection, a jammed pointer or an inactive supply can also give a zero reading. Conversely, a healthy bridge detector reads zero while the bridge arms still carry electric current. A zero deflection alone proves neither a broken conducting path nor infinite sensitivity.

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.

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.