 Solution

ID: 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

I R

, and the battery's longer plate is on the right, adding

+ V

toward A. Thus

V_{Th} = V_{A} - V_{B} = I R + 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

R_{Th} = 2 R .

I R + V

in series with

2 R

, positive toward A. If

I

is instead defined downward, replace

I

- I

; the resistance is unchanged.

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