• 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$\frac{dI}{dt}$ = $V_0$ sin(omegat). Integrating: I = -($\frac{V_0}{omega*L}$)cos(omegat) = ($\frac{V_0}{omega*L}$)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_{rms}$$I_{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.