A \( 200 \, \Omega \) resistor and \( 15 \, \mu\text{F} \) capacitor are in series with a \( 220 \, \text{V} \), \( 50 \
**Capacitive reactance** X_C =1/(ω C)=1/(2π f C) (Ω), C capacitance (F), current leads voltage by 90°, I_rms = V_rms/X_C = V_rms ω C, I_peak = V_peak ω C, impedance Z = X_C for pure C. For 45 μF, 60 Hz, X_C=1/(2π×60×45×10⁻⁶)=58.9 Ω, V_rms=110 V, I_rms=1.867 A, I_peak=2.64 A. X_C = (1/ω C) = (1/314 × 15 × 10⁻⁶) ≈ 212.3 Ω . Z = √(200² + 212.3²) ≈ 291.5 Ω . RMS current: I = (V/Z) = (220/291.5) ≈ 0.755 A . Voltage across capacitor: V_C = I X_C = 0.755 × 212.3 ≈ 160.3 V . Applying X_L = ωL, X_C = 1/ωC, Z
Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance
