= Thévenin load response
{c}
{title2=$V_L=V_{\mathrm{oc}}R_L/(R_{\mathrm{th}}+R_L)$}
A positive resistive <electrical load> obeys <Ohm's law>, $V_L=IR_L$. Its <resistive load line> intersects the <Thévenin terminal response> at
$$
I=\frac{V_{\mathrm{oc}}}{R_{\mathrm{th}}+R_L},\qquad
V_L=\frac{V_{\mathrm{oc}}R_L}{R_{\mathrm{th}}+R_L},
$$
provided $R_{\mathrm{th}}+R_L\ne0$. 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 $R_L=R_{\mathrm{th}}=\tfrac43\,\mathrm{k\Omega}$, the voltage is $4\,\mathrm V$, half the <open-circuit voltage>.
\Image[/thevenin-terminal-load-lines-and-voltage.png]
{title=Load-line intersections and load voltage for the 12 V divider}
{id=thevenin-terminal-load-lines-and-voltage}
{description=Left: the <Thévenin terminal response> falls from $8\,\mathrm V$ at zero delivered <electric current> to zero voltage at $6\,\mathrm{mA}$; three positive <resistive load lines> select operating points. Right: both the original loaded <voltage divider> calculation and its equivalent give the same voltage curve, approaching $8\,\mathrm V$ as $R_L$ increases. Calculated values only.}
{source=thevenin-terminal-load-lines-and-voltage.py}
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.
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