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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.

To determine the Thévenin resistance, remove the external electrical load and set each independent electrical source to zero in the circuit model. A zero ideal voltage source enforces zero electric potential difference, so replace it by a short connection. A zero ideal current source enforces zero electric current, so replace it by an open connection. Retain every resistor, including any explicitly modeled supply electrical resistance. Turning an actual power supply's front-panel control to zero is not evidence that it physically becomes an ideal wire.

Retain every dependent electrical source and its controlling variable: its response to an electrical test source is part of the network's electrical resistance. Impose terminal voltage $v_t$ with an electrical test source and calculate $j_t$ entering the suppressed network at $a$. For a nonzero test yielding a finite ratio,Alternatively inject a known entering electric current and solve for $v_t$. The test electric current is opposite to the delivered-current convention of the original Thévenin terminal response. ^[ref]

For an original illustration, put a positive resistor $R$ across $a,b$ and a dependent electrical source in parallel that injects electric current $g{v}_t$ from $b$ toward $a$, where $g$ has units of electrical conductance. Its control is the same terminal voltage $v_t$. Kirchhoff's current law givesWith $R=1\mathrm{k\Omega }$ and $g=0.5\mathrm{mS}$, a $1\mathrm{V}$ test draws $0.5\mathrm{mA}$, giving $R_{\mathrm{th}}=2\mathrm{k\Omega }$. Removing the dependent electrical source would falsely give $1\mathrm{k\Omega }$. With $g=2\mathrm{mS}$, the model instead has $R_{\mathrm{th}}=-1\mathrm{k\Omega }$; with $g=1\mathrm{mS}$ it draws zero test electric current for any imposed voltage and has no finite ratio. An active physical realization requires a supply and has a limited operating range.

For a finite nonzero $R_{\mathrm{th}}$ and a well-defined short-circuit response, $I_{\mathrm{sc}}=V_{\mathrm{oc}}/R_{\mathrm{th}}$. Thus $R_{\mathrm{th}}=V_{\mathrm{oc}}/I_{\mathrm{sc}}$ is usable when $I_{\mathrm{sc}}\ne 0$. A source-free positive resistor has both $V_{\mathrm{oc}}=0$ and $I_{\mathrm{sc}}=0$, so this quotient cannot reveal its electrical resistance; an electrical test source can.

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]

With the passive sign convention applied to the network, its entering electric current is $J=-I$, and the same relation becomes $V=V_{\mathrm{oc}}+J{R}_{\mathrm{th}}$. The electrical load uses the passive sign convention with entering electric current $I$, so it absorbs power $VI$. A positive delivered electric current therefore produces a voltage drop for positive $R_{\mathrm{th}}$; changing the reference direction without changing the equation would give the wrong sign.

The reason is linearity of Ohm's law, Kirchhoff's current law, Kirchhoff's voltage law, and the constitutive relations of any linear dependent electrical sources. Solve once at $I=0$ with the independent electrical sources present. Subtract this solution from a solution with prescribed delivered electric current $I$: the fixed independent electrical sources cancel, leaving a homogeneous linear network whose terminal voltage change is proportional to $I$. This is a short superposition principle argument for an affine function; the fixed offset need not vanish.

The qualification about a well-defined response matters. An ideal voltage source alone has $R_{\mathrm{th}}=0$. An ideal nonzero current source alone constrains $I$ and does not have a finite open-circuit voltage or a representation $V=V_{\mathrm{oc}}-I{R}_{\mathrm{th}}$ with finite coefficients. Linear dependent electrical sources can give negative resistance, and singular ideal networks may fail to have a unique finite response. Neither a finite nor a strictly positive passive $R_{\mathrm{th}}$ follows merely from calling an electrical circuit linear. ^[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.

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.