= Thévenin source suppression
{c}
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 $j_t$ \i[entering] the suppressed network at $a$. For a nonzero test yielding a finite ratio,
$$
R_{\mathrm{th}}=\frac{v_t}{j_t}.
$$
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>. <Thevenin and Norton Equivalents, University of Houston teaching notes>[Thevenin and Norton Equivalents, pp. 5–7, “Test Source Method for RTH” and Notes: suppress independent sources, retain dependent sources]{ref}
For an original illustration, put a positive <resistor> $R$ across $a,b$ and a <dependent electrical source> in parallel that injects <electric current> $gv_t$ from $b$ toward $a$, where $g$ has units of <electrical conductance>. Its control is the same terminal voltage $v_t$. <Kirchhoff's current law> gives
$$
j_t=\frac{v_t}{R}-gv_t,\qquad
R_{\mathrm{th}}=\frac{1}{R^{-1}-g}.
$$
With $R=1\,\mathrm{k\Omega}$ 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 $I_{\mathrm{sc}}\ne0$. 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.
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