Question 15

**Given:** V(t) = 300 sin(400t) V, so ω = 400 rad/s, V₀ = 300 V. **(P) R = 30 Ω:** Z = 30 Ω, i₀ = 10 A, in phase with V. i(t) = 10 sin(400t) → Graph (3) (amplitude 10, starts at zero rising). **P→3** **(Q) R = 30 Ω, L = 100 mH:** X_L = ωL = 400 × 0.1 = 40 Ω, Z = √(30² + 40²) = 50 Ω, i₀ = 6 A. Current lags voltage by φ = tan⁻¹(40/30) ≈ 53°. i(t) = 6 sin(400t − 53°) → at t = 0 current is negative, peak occurs after 400t = π/2. Graph (5) (amplitude ~6, lagging). **Q→5** **(R) C = 50 μF, R = 30 Ω, L = 25 mH:** X_L = 400 × 0.025 = 10 Ω, X_C = 1/(400 × 50×10⁻⁶) = 50 Ω. Z = √(30² + (10−50)²) = 50 Ω, i₀ = 6 A. Net reactance capacitive → current leads by 53°. i(t) = 6 sin(400t + 53°): at t = 0, i ≈ 4.8 A, peak before 400t = π/2. Graph (2). **R→2** **(S) C = 50 μF, R = 60 Ω, L = 125 mH:** X_L = 400 × 0.125 = 50 Ω, X_C = 50 Ω → resonance. Z = R = 60 Ω, i₀ = 5 A, in phase. i(t) = 5 sin(400t) → Graph (1). **S→1** Thus P→3, Q→5, R→2, S→1, which is option (A).

No written reasoning was returned for this item.

No written reasoning was returned for this item.