Step 1: Look at the circuit from the terminals of the branch carrying the unknown current \( i \), and reduce everything else, sources and resistors alike, to a single Thevenin equivalent: an open-circuit voltage \( V_{th} \) in series with an equivalent resistance \( R_{th} \).
Step 2: Closing switch S changes the network seen from those terminals, which changes both \( V_{th} \) and \( R_{th} \) compared to the switch-open case.
Step 3: The current through the branch is then \( i = \dfrac{V_{th}}{R_{th} + R_{branch}} \), where \( R_{branch} \) is the resistance of the branch itself. Evaluating this ratio for the closed-switch condition gives the same result as the direct nodal solution. \[ \boxed{i = 5\text{ A}} \]