= Solution
Take B as reference and define positive source current $I$ upward into the upper internal node; the source symbol does not specify an arrow, so this definition fixes its sign. With A open, no current flows through the upper series resistor. The shunt resistor therefore has voltage $IR$, and the battery's longer plate is on the right, adding $+V$ toward A. Thus
$$
\boxed{V_{\rm Th}=V_A-V_B=IR+V.}
$$
Deactivate the independent sources: open the ideal current source and short the ideal voltage source. The two resistors then lie in series between A and B, giving
$$
\boxed{R_{\rm Th}=2R.}
$$
The <Thévenin equivalent> is a source $IR+V$ in series with $2R$, positive toward A. If $I$ is instead defined downward, replace $I$ by $-I$; the resistance is unchanged.
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