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 :

Solution

Physics sir · 4 steps

Find the current from Vω C. The displacement current between the plates must exactly match it.

1. Capacitive reactance. For a capacitor on ac, X_C = 1/(ω C), so the rms current is Iᵣₘₛ = Vᵣₘₛ/X_C = Vᵣₘₛ ω C.

   Iᵣₘₛ = 230 × 300 × 200 × 10⁻¹² = 1.38 × 10⁻⁵ A

2. Conduction current. That is 13.8 μA, flowing in the connecting wires.
3. Displacement current. No charge crosses the gap. The changing electric field there carries a displacement current I_d = ε₀ dΦ_E/dt = dq/dt, which is the same as the current in the wires. This keeps the current continuous around the loop.
4. So. Both currents are 13.8 μA, which is option (b).

The idea to keep. Displacement current inside a capacitor always equals the conduction current in its wires. Work out one and you have the other.

Answer from the NTA final answer key. Solution by PrepWiser.