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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.

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]