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**Worked Example: Series RL Circuit with AC Source**\n1. **Given:**\n - AC supply voltage V = 120 V (rms)\n - Frequency f = 60 Hz\n - Inductor L = 0.1 H\n - Resistor R = 30 Ω\n2. **Find:**\n - Inductive reactance X_L\n - Total impedance Z\n - Current I\n - Reactive power Q\n3. **Step‑by‑Step:**\n - Calculate X_L = 2πfL = 2π × 60 × 0.1 ≈ 37.7 Ω\n - Total impedance Z = √(R² + X_L²) = √(30² + 37.7²) ≈ 48.7 Ω\n - Current I = V / Z = 120 / 48.7 ≈ 2.46 A\n - Phase angle φ = arctan(X_L / R) = arctan(37.7 / 30) ≈ 52°\n - Reactive power Q = V × I × sin(φ) = 120 × 2.46 × sin(52°) ≈ 200 VAR\n4. **Result:** The circuit draws about 2.46 A of current, but only about 200 VAR of it is reactive; the rest is real power used by the resistor.