Practice question
Question
A \( 50 \, \mu\text{F} \) capacitor is connected to a \( 220 \, \text{V} \), \( 50 \, \text{Hz} \) AC
source. What is the peak current?
Explanation
**RMS value** I_rms = I_peak/√2, V_rms = V_peak/√2 for sinusoidal AC, significance rms gives equivalent DC value producing same heating power P = I_rms² R, average power over cycle, instruments measure rms, average over full cycle zero, half-cycle average 2 I_peak/π, peak = √2 rms. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 50 × 10⁻⁶ F . X_C = (1/314 × 50 × 10⁻⁶) ≈ 63.7 Ω . RMS current: I = (V/X_C) = (220/63.7) ≈ 3.45 A . Peak current: i_m = √(2) I = 1.414 × 3.45 ≈ 4.88 A . Applying X_L =
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