Power proportional to current squared for fixed resistance

Steps:

• Original: P₁ = I₁²R = 2 W, I₁ = 3 A, so R = 2 / 3² = 2/9 Ω.
• New power: P₂ = 2 W × 1.5 = 3 W.
• I₂²R = 3, so I₂² = 3 / (2/9) = 3 × 9/2 = 13.5.
• I₂ = √13.5 ≈ 3.7 A.

Why B is correct:

• P = I²R; for fixed R, I₂ = I₁ √(P₂/P₁) = 3 √1.5 ≈ 3.7 A matches the 50% power increase.

Why the others are wrong:

• A: 2.4 A gives P = (2.4/3)² × 2 = 1.28 W (decrease, not increase).
• C: 4.5 A gives P = (4.5/3)² × 2 = 4.5 W (125% increase, too much).
• D: 14 A gives P = (14/3)² × 2 ≈ 43.6 W (far exceeds 50% increase).

Final answer: B