Power quadruples with negligible internal resistance when R = r = 1Ω

Steps:

• Original: Total resistance = R + r = 1 + 1 = 2Ω; current I = 6V / 2Ω = 3A; power P = I²R = 9 × 1 = 9W.
• New: Internal resistance = 0Ω; total resistance = R = 1Ω; current I' = 6V / 1Ω = 6A.
• New power P' = (I')²R = 36 × 1 = 36W.
• P' = 4P since current doubles and power ∝ I².

Why D is correct:

• By power formula P = I²R and Ohm's law, removing internal resistance doubles current (from 3A to 6A), quadrupling power as (2)² = 4.

Why the others are wrong:

• A: Halves power, which occurs if resistance doubles, not here.
• B: Same power assumes no change in current, but internal resistance removal increases it.
• C: Doubles power, matching current increase but ignoring I² dependence.

Final answer: D