= Solution
The original diagram has an ideal $5\,\mathrm V$ <electric battery> feeding series <resistors> of $8\,\Omega$ above the tapping point and $2\,\Omega$ below it. The long battery bar is positive, so terminal $A$ is above terminal $B$ in <electric potential>. With an <open circuit> at the output, the internal <electric current> is $5/(8+2)=0.5\,\mathrm A$. The <voltage divider> therefore gives
$$
\boxed{\mathcal E_0=V_A-V_B=(0.5\,\mathrm A)(2\,\Omega)=1.0\,\mathrm V.}
$$
For the <Thévenin resistance>, suppress the ideal voltage source by replacing it with a <short circuit>. The two <resistors> are then in parallel as seen from $A,B$:
$$
\boxed{R_{\rm Th}=\frac{8\times2}{8+2}\,\Omega=1.6\,\Omega.}
$$
By <Thévenin's theorem>, the terminal <voltage> under an outward load <electric current> $I$ is $V_A-V_B=\mathcal E_0-IR_{\rm Th}$. Consequently
$$
\boxed{I_{\rm sc}=\frac{\mathcal E_0}{R_{\rm Th}}=0.625\,\mathrm A,}
$$
flowing through an external <short circuit> from $A$ to $B$. Directly shorting the original terminals bypasses the $2\,\Omega$ <resistor> and leaves $5\,\mathrm V$ across $8\,\Omega$, independently confirming the result.
Back to article page
