An alternating voltage of amplitude 40 V and frequency 4 kHz is applied directly across the capacitor of 12 μ F. The maximum displacement current between the plates of the capacitor is nearly :

Solution

Physics sir · 4 steps

Displacement current between the plates equals the current in the wires, so find C dV/dt at its peak.

1. Displacement current. Between the plates no charge flows, but the changing field gives a displacement current I_d = ε₀ dΦ_E/dt = CdV/dt. It is exactly equal to the conduction current in the leads.

   V = V₀ sin ω t ⇒ I_d = C V₀ ω cos ω t, Iₘₐₓ = C V₀ ω

2. Angular frequency. ω = 2π f = 2π × 4000 = 8000π ≈ 2.51 × 10⁴ rad s⁻¹.
3. Put in numbers. Iₘₐₓ = 12 × 10⁻⁶ × 40 × 8000π = 3.84π ≈ 12.06 A.
4. So. The maximum displacement current is nearly 12 A, option (d).

The idea to keep. For a capacitor on AC, peak displacement current is V₀ ω C, the same as the peak conduction current. Always use ω = 2π f, not f.

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