= Thévenin divider example
{c}
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> gives
$$
V_{\mathrm{oc}}=12\,\mathrm V\,\frac{4\,\mathrm{k\Omega}}{2\,\mathrm{k\Omega}+4\,\mathrm{k\Omega}}=8\,\mathrm V.
$$
To 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, giving
$$
R_{\mathrm{th}}=\frac{R_1R_2}{R_1+R_2}=\frac43\,\mathrm{k\Omega}.
$$
This 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:
$$
\frac{12\,\mathrm V-V}{2\,\mathrm{k\Omega}}=\frac{V}{4\,\mathrm{k\Omega}}+I,
\qquad V=8\,\mathrm V-\left(\frac43\,\mathrm{k\Omega}\right)I.
$$
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.
Back to article page
