Lamp X shunts high-resistance wires, reducing voltage drop to Y Steps:

• Only Y on: Current flows through high-R wires in series with Y, large voltage drop across wires, low V_Y.
• Circuit setup: High-R transmission wires in series with Y; X (with switch) connected in parallel across the high-R wires.
• Both on: X provides low-resistance path parallel to high R, effective series resistance drops to R || r.
• Result: Total circuit resistance decreases, current rises, voltage drop across effective wires falls, V_Y = I × r increases. Why C is correct:
• Voltage increases per Ohm's law (V_Y = V × r / (R_eff + r), R_eff < R so V_Y rises); power stays the same as Y operates at rated condition with minimal change in I^2 r due to circuit balance. Why the others are wrong:
• A: Voltage does not decrease; shunting reduces drop.
• B: Voltage does not decrease; power does not increase beyond balance.
• D: Duplicate of C, invalid choice.

Final answer: C