Past exam of the physics 2 paper of the University of Oxford / 2013 / 5 / 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. ThusDeactivate 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, givingThe 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.