For an ideal capacitor in AC, capacitive reactance:
A1/(ωC)
BC / ω
CωC
Dω²C
Answer & Solution
Correct answer: A. 1/(ωC)
X_C = 1/(ωC).
Current leads voltage by 90°.
Reactance decreases with frequency — capacitors pass high frequencies, block DC.
Keep practising Electromagnetic Induction
Browse all Electromagnetic Induction questions for JEE Advanced, or explore more Physics topics.
Related questions
Faraday + Lenz — true or false?Inductive reactance increases linearly with frequency. At f = 50 Hz, X_L = 100 Ω. At f = 2Phase difference between voltage and current in pure capacitor in AC:For an AC generator with EMF ε(t) = ε₀ sin(ωt), the RMS EMF:In AC circuit with only inductor, average power consumed over one full cycle:A capacitor C in series with inductor L (LC circuit) oscillates with frequency:In an LR circuit with V = 12 V, R = 4 Ω, L = 2 H, current after t = 0.5 s:Energy density of magnetic field B in vacuum: