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A parallel plate capacitor has a capacitance C = 200 pF. It is connected to 230 V ac supply with an angular frequency 300 rad/s. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are

Asked in JEE Main 1st Feb 1st Shift 2024 · Capacitor on an a.c. source

Answer: (3) 13.8 μA and 13.8 μA

Step-by-step solution

X_C = 1/(ω C) = 1/(300 × 200 × 10⁻¹²) = 1.667 × 10⁷ Ω

Iᵣₘₛ = Vᵣₘₛ/X_C = 230/(1.667 × 10⁷) = 1.38 × 10⁻⁵ A = 13.8 μA

i_d = i_c at every instant, so the displacement current is the same 13.8 μA

Why the other options are wrong

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