= Wheatstone bridge detector loading
{c}
{title2=$V_d=V_{\mathrm{oc}}R_d/(R_{\mathrm{th}}+R_d)$}
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. Consequently
$$
R_{\mathrm{th}}=\frac{R_1R_2}{R_1+R_2}+\frac{R_3R_x}{R_3+R_x},\qquad
I_d=\frac{V_{\mathrm{oc}}}{R_{\mathrm{th}}+R_d},\qquad
V_d=I_dR_d=V_{\mathrm{oc}}\frac{R_d}{R_{\mathrm{th}}+R_d}.
$$
Since the attenuation factor is strictly positive for $R_d>0$, the null remains $R_2R_3=R_1R_x$. 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.
\Image[/wheatstone-bridge-detector-voltage-and-loading.png]
{title=Finite detector resistance reduces bridge voltage sensitivity while preserving the null}
{id=wheatstone-bridge-detector-voltage-and-loading}
{description=Calculated curves use $V_s=5\,\mathrm V$ and $R_1=R_2=R_3=1\,\mathrm{k\Omega}$. All cross zero at $R_x=1\,\mathrm{k\Omega}$; the inset resolves their different slopes. These curves are a model, not measured data.}
{source=wheatstone-bridge-detector-voltage-and-loading.py}
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> is
$$
V_d=-\frac{V_s}{4R}\frac{R_d}{R+R_d}\,\delta R+O((\delta R)^2).
$$
At the plotted null, the open-circuit slope is $-1.25\,\mathrm{mV/\Omega}$, compared with $-0.625\,\mathrm{mV/\Omega}$ 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.
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