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A useful short-interval approximation replaces a battery by an ideal electromotive force $\mathcal{E}$ in series with a positive internal electrical resistance $r$. This is the Thévenin terminal response with open-circuit voltage $\mathcal{E}$ and Thévenin resistance $r$. The model describes the terminals without asserting that there is a separate physical resistor hidden inside the cell. Both $\mathcal{E}$ and $r$ can change with composition, temperature, electric current and history.^[ref]

For a resistive electrical load $R>0$, take $I>0$ to mean discharge: conventional current leaves the positive terminal. Kirchhoff's voltage law and Ohm's law giveFor the stipulated $\mathcal{E}=3\mathrm{V}$, $r=1\Omega$, $R=4\Omega$,An open-circuit voltage of $3\mathrm{V}$ therefore does not promise $3\mathrm{V}$ at this electrical load. Increasing the load electrical resistance reduces electric current and brings $V_L$ closer to $\mathcal{E}$, as in Thévenin load response.

The signed relation also explains charging within this model. If an external source drives electric current into the positive terminal, $I<0$ and $V_L=\mathcal{E}+|I|r$. The terminal electric potential difference then exceeds $\mathcal{E}$. This algebra is not evidence that every battery is rechargeable.^[ref]

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 positive resistive electrical load obeys Ohm's law, $V_L=I{R}_L$. Its resistive load line intersects the Thévenin terminal response atprovided $R_{\mathrm{th}}+R_L\ne 0$. For the Thévenin divider example, $R_L=1\mathrm{k\Omega }$ gives $I=24/7\mathrm{mA}$ and $V_L=24/7\mathrm{V}$; $R_L=4\mathrm{k\Omega }$ gives $I=1.5\mathrm{mA}$ and $V_L=6\mathrm{V}$. At $R_L=R_{\mathrm{th}}=\frac{4}{3}\mathrm{k\Omega }$, the voltage is $4\mathrm{V}$, half the open-circuit voltage.

For this positive $R_{\mathrm{th}}$, increasing $R_L$ reduces the drawn electric current and raises $V_L$ toward $V_{\mathrm{oc}}$. This explains voltmeter loading: a voltmeter itself is an electrical load. The equivalence also applies to a nonlinear electrical load through intersection with its terminal relation, but an arbitrary load need not give a unique operating point. If negative $R_{\mathrm{th}}$ cancels positive $R_L$, the displayed formula is singular; a real active circuit's limits and dynamics must then be considered.

An authored teaching experiment uses an adjustable steady-voltage supply, a digital multimeter, fixed resistors, and a $10\mathrm{k\Omega }$ potentiometer or resistance decade box. Its original circuit uses a $10\mathrm{V}$ supply and $3.3$, $6.8$, and $4.7\mathrm{k\Omega }$ resistors; the student compares $8.2$ and $2.2\mathrm{k\Omega }$ electrical loads. The readout is load voltage, recorded against prediction and deviation. The student then removes the electrical load to measure the open-circuit voltage, disconnects the supply and substitutes a wire to measure the Thévenin resistance with the multimeter's resistance mode, and restores the supply for an independent method: adjust the variable electrical load until the voltmeter reads half the unloaded voltage. Finally the student builds the equivalent and repeats both load measurements. Its tables are blank recording forms, not reported observations. ^[ref]

The half-voltage inference assumes negligible voltmeter loading and positive $R_{\mathrm{th}}$. A finite voltmeter input electrical resistance $R_m$ actually gives an apparent unloaded readingWhen the same voltmeter remains connected across a variable electrical load $R_L$, the network sees $R_p=R_L{R}_m/(R_L+R_m)$. Setting its reading to $V_m/2$, rather than to the true $V_{\mathrm{oc}}/2$, givesThus the uncorrected half-reading method underestimates positive $R_{\mathrm{th}}$. In the original $12\mathrm{V}$ example, a stipulated $R_m=10\mathrm{k\Omega }$ gives $V_m=120/17\mathrm{V}\approx 7.059\mathrm{V}$ and half-reading load $R_L=20/17\mathrm{k\Omega }\approx 1.176\mathrm{k\Omega }$, instead of $\frac{4}{3}\mathrm{k\Omega }$. These are model predictions for a deliberately loading meter, not the specification or observed readout of the source experiment's instrument.

Finite readings can also estimate the terminal line without a physical short. For two distinct load settings with corrected delivered electric currents $I_1,I_2$ and terminal voltages $V_1,V_2$,If the voltmeter remains across each electrical load, the delivered electric current is $I_k=V_k(1/R_{L,k}+1/R_m)$, including the meter branch. Omitting it biases the inferred slope.

Useful additional controls are to measure each resistor separately before assembly, verify terminal polarity, repeat an earlier load setting to detect supply drift, and compare a third setting with the inferred Thévenin terminal response. These are proposed controls, not reported outcomes of the source experiment. Closely spaced electric currents make the slope estimate sensitive to voltage resolution and measurement error; widely separated settings may trigger source current limiting or Joule heating that changes electrical resistance. Lead and contact electrical resistance, meter input uncertainty and rounding limit the inference. An ohmmeter injects its own test electric current; its reading on a powered network is not the suppressed-network ratio. Measurements with independent sources physically removed do not characterize an active dependent electrical source whose control or supply has also been disconnected.

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]

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.