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Question

A \( 35 \, \mu\text{F} \) capacitor is connected to a \( 230 \, \text{V} \), \( 50 \, \text{Hz} \) AC
source. What is the peak current?

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Explanation

**Transformer principle** alternating current in primary creates changing flux linking secondary, emf induced e = -N dΦ/dt, flux same through both, so V∝N. For N_p=550, N_s=1100, V_p=110 V rms, V_s= V_p×N_s/N_p=220 V, step-up factor 2, efficiency <100% due to copper, iron, hysteresis losses. X_C = (1/ω C) , ω = 2π × 50 = 314 rad/s . C = 35 × 10⁻⁶ F . X_C = (1/314 × 35 × 10⁻⁶) ≈ 91 Ω . RMS current: I = (V/X_C) = (230/91) ≈ 2.527 A . Peak current: i_m = √(2) I = 1.414 × 2.527 ≈ 3.57 A . Applying X_L = ωL, X_C

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