Practice question
Question
A \( 60 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC
supply. What is the peak current in the circuit?
Explanation
**Power factor** cos φ = R/Z, 0≤cos φ≤1, at resonance cos φ=1 maximum power, pure L or C cos φ=0 zero average power, current wattless because energy stored returned. Instantaneous power when current maximum in pure L: I=I_peak, V=0 because V leads 90°, so P= V I =0 at that instant, average zero. Capacitive reactance: X_C = (1/ω C) , where ω = 2π f . f = 60 Hz , C = 60 × 10⁻⁶ F . ω = 2 × 3.14 × 60 = 376.8 rad/s . X_C = (1/376.8 × 60 × 10⁻⁶) = 44.24 Ω . RMS current: I = (V/X_C)
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