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A steady electrical circuit described by Ohm's law, Kirchhoff's current law, Kirchhoff's voltage law, and ideal independent electrical sources or linear dependent electrical sources. Fixed sources make the terminal response an affine function when the circuit equations admit a unique voltage for each prescribed terminal electric current.

In the pictured original linear resistive circuit, the ideal voltage source keeps its upper node at $12\mathrm{V}$ relative to $b$. Let $R_1=2\mathrm{k\Omega }$ connect that node to $a$, and $R_2=4\mathrm{k\Omega }$ connect $a$ to $b$. With the external electrical load removed, the voltage divider givesTo find the Thévenin resistance, Thévenin source suppression replaces the ideal $12\mathrm{V}$ voltage source by a wire. Both resistors now join $a$ to $b$ in parallel, givingThis is the electrical resistance looking into the original network with its independent electrical source suppressed; the external electrical load is excluded.

An independent check uses Kirchhoff's current law at $a$, without suppressing anything:The Thévenin terminal response thus follows directly from the original topology. Shorting $a$ to $b$ in this ideal calculation gives $I_{\mathrm{sc}}=6\mathrm{mA}$: the lower resistor has zero voltage and the upper resistor limits the electric current. All numerical values here and in the figures are stipulated model values, not laboratory data.

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.

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]