Power dissipation is highest for the lamp with greatest voltage across it

Steps:

• Examine circuit: battery connected to lamp D in parallel with series combination of A, B, and C.
• Voltage across D equals battery voltage V; series branch divides V among A, B, C.
• For identical lamps (resistance R), power P = V^2 / R.
• D gets full V, so highest P; others get V/3 each.

Why D is correct:

• D spans full battery voltage per Ohm's law, maximizing P = V^2 / R.

Why the others are wrong:

• A: Shares voltage in series (V/3), so P = (V/3)^2 / R, lower.
• B: Same series voltage share as A, lower power.
• C: Series voltage V/3 like A and B, reduced power.

Final answer: D