Power loss in battery internal resistance decreases with larger load

Steps:

• Original total resistance: 0.5Ω + 5Ω = 5.5Ω; current I₁ = 10V / 5.5Ω ≈ 1.82A.
• Original power in internal resistance: P₁ = I₁² × 0.5Ω ≈ 1.65W.
• New total resistance: 0.5Ω + 50Ω = 50.5Ω; current I₂ = 10V / 50.5Ω ≈ 0.20A.
• New power in internal resistance: P₂ = I₂² × 0.5Ω ≈ 0.02W, which is less than P₁.

Why B is correct:

• Power in internal resistance follows P = I²r; increasing load resistance R raises total resistance, lowering current I and thus P.

Why the others are wrong:

• A: Current I = ε / (r + R) decreases as R increases from 5Ω to 50Ω.
• C: Identical to B, but question specifies B as correct option.
• D: Total power = εI; I decreases, so total power drops from ≈18.2W to ≈2.0W.

Final answer: B