Practice question
Question
A \( 13 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \)
source. What is the capacitive reactance?
Explanation
**Resistor in AC** behaves as DC, no reactance, impedance Z=R, current follows voltage exactly, average power over cycle V_rms I_rms, for 200 V rms, 80 Ω, P=500 W? Actually 200²/80=500 W, peak current √2×2.5=3.535 A, average power ½ V_peak I_peak. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 13 × 10⁻⁶ F . X_C = (1/314 × 13 × 10⁻⁶) ≈ 245.1 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 245.1 Ω, consistent with phasor analysis
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.