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A prescribed voltage source or current source used to probe a terminal response. For a suppressed linear resistive circuit, the ratio of terminal voltage to entering electric current determines the Thévenin resistance when finite.

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.

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]