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Question

A \( 45 \, \mu\text{F} \) capacitor is connected to a \( 110 \, \text{V} \), \( 60 \, \text{Hz} \) AC
source. What is the peak current?

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Explanation

**Inductive reactance calculation** for 60 mH, 50 Hz, X_L=2π×50×0.06=18.85 Ω, V_rms=220 V, I_rms=220/18.85=11.67 A, for 70 mH, 50 Hz, X_L=21.99 Ω, I=110/21.99=5 A, showing X_L ∝ f L, higher f or L increases opposition. X_C = (1/ω C) , ω = 2π × 60 = 376.8 rad/s . C = 45 × 10⁻⁶ F . X_C = (1/376.8 × 45 × 10⁻⁶) ≈ 59 Ω . RMS current: I = (V/X_C) = (110/59) ≈ 1.864 A . Peak current: i_m = √(2) I = 1.414 × 1.864 ≈ 2.64 A . Applying X_L = ωL, X_C = 1/ωC, Z = √(R² + (X_L - X_C)²),

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