• Tags: inductor, reactance, lagging
• Difficulty: Moderate

For V = V_0 sin(omegat) across a pure inductor L: the back-EMF equals the applied voltage, so L(dI/dt) = V_0 sin(omegat). Integrating: I = -(V_0/(omegaL))cos(omegat) = (V_0/(omegaL))sin(omegat - pi/2). Current lags voltage by pi/2 (or 90 degrees). ELI: Voltage (E) Leads current (I) in an Inductor (L). Inductive reactance X_L = omegaL = 2pifL (units: ohm). X_L increases linearly with frequency — an inductor opposes rapid changes in current. At DC (f=0): X_L = 0 (short circuit). At very high f: X_L -> infinity (open circuit). Average power P = V_rmsI_rms*cos(90) = 0. The inductor stores energy in its magnetic field during one quarter cycle and returns it during the next — no net energy dissipation.