A capacitor of capacitance 100 μF is connected to an AC source of voltage 200 V and frequency 50 Hz. The reactance is
Q1. A capacitor of capacitance 100 μF is connected to an AC source of voltage 200 V and frequency 50 Hz. The reactance is
Answer: 31.84 Ω
Explanation: Capacitive reactance Xc = 1 / (2πfC). With f = 50 Hz and C = 100 μF = 1e-4 F, Xc = 1 / (2π × 50 × 1e-4) ≈ 31.84 Ω.