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

For V = $V_0$ sin(omegat) across a pure capacitor C: Q = CV, so I = C$\frac{dV}{dt}$ = C*$V_0$omegacos(omegat) = ($\frac{V_0}{1/(omega*C}$))sin(omegat + pi/2). Current leads voltage by pi/2. ICE: Current (I) leads voltage (E) in a Capacitor (C). Capacitive reactance $X_C$ = $\frac{1}{omega*C}$ = $\frac{1}{2*pi*f*C}$. $X_C$ 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.