A series LCR circuit has R = 10 Ω, L = 0.10 H, C = 10 μF, driven by a 50 V (rms) AC source. Find (a) the resonant frequency, (b) the current at resonance, (c) the impedance at 100 Hz (below resonance). Step 1: Resonant frequency f₀ = 1/(2π√(LC)) = 1/(2π√(0.10 × 10×10⁻⁶)) = 1/(2π√(10⁻⁶)) = 1/(2π × 10⁻³) ≈ 159 Hz. Step 2: At resonance Z = R = 10 Ω, so I_rms = V_rms/R = 50/10 = 5.0 A (maximum current; voltage and current in phase). Step 3: At f = 100 Hz, ω = 2π×100 = 628 rad/s. X_L = ωL = 628 × 0.10 = 62.8 Ω. X_C = 1/(ωC) = 1/(628 × 10×10⁻⁶) = 1/0.00628 ≈ 159 Ω. Z = √(10² + (62.8 − 159)²) = √(100 + 9260) ≈ 96.7 Ω. Current = 50/96.7 ≈ 0.52 A — far below the 5 A at resonance, and capacitive (X_C > X_L).