Test: Displacement & Conduction Current - Electrical Engineering (EE) MCQ

Answer: a
Explanation: The conduction current density is given by, Jc = σE. To get conductivity, σ = J/E = 100/4 = 25 units.

Answer: b
Explanation: The displacement current density is given by, Jd = dD/dt.
Jd = d(20sin 0.5t)/dt = 20cos 0.5t (0.5) = 10cos 0.5t.

Answer: d
Explanation: Jd = dD/dt = εdE/dt in time domain. For frequency domain, convert using Fourier transform, Jd = εjωE. The magnitude of
Jd = εωE = ε(2πf)E. On substituting, we get 4 ampere.

Answer: b
Explanation: When Jd = Jc , we get εωE = σE. Thus εo(2∏f) = σ. On substituting conductivity as one and permittivity as 2, we get f = 9GHz.

Answer: c
Explanation: Jc /Jd is a standard ratio, which is referred to as loss tangent given by σ /ε ω. The loss tangent is used to determine if the material is a conductor or dielectric.

Answer: b
Explanation: If loss tangent is less, then σ /ε ω <<1. This implies the conductivity is very poor and the material should be a dielectric. Since it is specifically mentioned very less, assuming the conductivity to be zero, the dielectric will be lossless (ideal).

Answer: b
Explanation: The electric and magnetic fields will be out of phase by 45 in good conductors. This is because their intrinsic impedance is given by η = √(ωμ/σ) X (1+j). In polar form we get 45 out of phase.

Answer: c
Explanation: In free space, ε = ε0 and μ = μ0. The relative permittivity and permeability will be unity. Since the free space will contain no charges in it, the conductivity will be zero.

Answer: b
Explanation: The loss tangent is given by tan 2θn, where θn = 20. Thus the loss tangent will be tan 40.

Answer: d
Explanation: The intrinsic impedance is given by η = √(μo/εo) ohm. Here εo = 8.854 x 10^-12and μo = 4π x 10^-7.
On substituting the values, we get η = 377 ohm.