Energy from constant voltage charging over two hours Steps: - Current is constant at 0.50 A for the first 1.0 hour (3600 s), transferring charge Q₁ = 0.50 × 3600 = 1800 C; energy E₁ = 5.0 × 1800 = 9000 J. - After 1.0 hour, current decreases slowly from 0.50 A to 0 A over the next hour; average current ≈ 0.30 A (higher than linear 0.25 A due to slow decrease); Q₂ ≈ 0.30 × 3600 = 1080 C; E₂ ≈ 5.0 × 1080 = 5400 J. - Total energy E = E₁ + E₂ ≈ 9000 + 5400 = 14400 J. - Best estimate from graph description is 14000 J. Why C is correct: - Matches V × total Q, where Q is estimated from initial constant phase plus gradual decrease phase per the problem's timing and "slowly decreases" description. Why the others are wrong: - A: Underestimates by ignoring most of the charge transfer. - B: Captures only the first hour, neglecting second-hour contribution. - D: Overestimates by assuming constant 0.50 A over full 2 hours. …