• Tags: capacitor, reactance, leading
• Difficulty: Moderate

For V = V_0 sin(omegat) across a pure capacitor C: Q = CV, so I = C(dV/dt) = CV_0omegacos(omegat) = (V_0/(1/(omegaC)))sin(omegat + pi/2). Current leads voltage by pi/2. ICE: Current (I) leads voltage (E) in a Capacitor (C). Capacitive reactance X_C = 1/(omegaC) = 1/(2pifC). X_C decreases with frequency — a capacitor passes high frequencies and blocks DC. At DC (f=0): X_C -> infinity (open circuit). At very high f: X_C -> 0 (short circuit). This is opposite to inductor behavior. Average power = 0 (wattless). The capacitor stores energy in its electric field and returns it. Phasor diagram: I_C leads V_C by 90 degrees. JEE trick: for a capacitor, "current leads" can be remembered from I = CdV/dt — the current responds to the rate of change of voltage.