Independent branches prevent interaction between lamps

Steps:

• Lamp Y is on initially, with its own current through its high-resistance wire; lamp X is off, so voltage across X is 0 V.
• Switching X on completes its branch circuit, producing voltage V_X = V_supply × R_lamp / (R_wire + R_lamp) > 0 V across X.
• Each lamp has a separate branch from the supply, so current in X's wire does not flow through or affect Y's branch.
• Power to Y remains P = (V_Y)^2 / R_lamp, unchanged since V_Y is unaffected.

Why C is correct:

• Separate branches ensure switching X on raises its voltage from 0 V (no circuit) to operating value, while Y's power stays constant per Ohm's law in isolated loops.

Why the others are wrong:

• A: Voltage across X increases, not decreases; power to Y unchanged due to independence.
• B: Voltage across X increases from 0 V, not decreases.
• D: Power to Y stays the same, not increases, as no shared current path.

Final answer: C