A \( 12 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) source. What is the c
**Capacitor average power** zero over complete cycle because P=V I =½ V_peak I_peak sin2ωt average zero, energy stored in field, not dissipated, unlike resistor. For 15 μF, 60 Hz, X_C=176.8 Ω, V_rms=110 V, I_rms=0.622 A, illustrating lower C higher X_C. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 12 × 10⁻⁶ F . X_C = (1/314 × 12 × 10⁻⁶) ≈ 265.3 Ω . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²), I_rms = V_rms/Z, P = V_rms I_rms cosφ, V_s/V_p = N_s/N_p, calculation gives 265.3 Ω, consistent with phasor analysis
Ref: NCERT > Physics Book > Alternating Currents > AC Through Capacitor - Capacitive Reactance