 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 givesFor 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 $A,B$:By Thévenin's theorem, the terminal voltage under an outward load electric current $I$ is $V_A-V_B={\mathcal{E}}_0-I{R}_{\mathrm{Th}}$. Consequentlyflowing 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.