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# Test: Displacement And Conduction Current

## 10 Questions MCQ Test Electromagnetic Theory | Test: Displacement And Conduction Current

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This mock test of Test: Displacement And Conduction Current for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Displacement And Conduction Current (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Displacement And Conduction Current quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Displacement And Conduction Current exercise for a better result in the exam. You can find other Test: Displacement And Conduction Current extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

### Find the conductivity of a material with conduction current density 100 units and electric field of 4 units.

Solution:

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

QUESTION: 2

### Calculate the displacement current density when the electric flux density is 20sin 0.5t.

Solution:

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.

QUESTION: 3

### Find the magnitude of the displacement current density in air at a frequency of 18GHz in frequency domain. Take electric field E as 4 units.

Solution:

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.

QUESTION: 4

Calculate the frequency at which the conduction and displacement currents become equal with unity conductivity in a material of permittivity 2.

Solution:

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.

QUESTION: 5

The ratio of conduction to displacement current density is referred to as

Solution:

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.

QUESTION: 6

If the loss tangent is very less, then the material will be a

Solution:

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).

QUESTION: 7

In good conductors, the electric and magnetic fields will be

Solution:

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.

QUESTION: 8

In free space, which of the following will be zero?

Solution:

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.

QUESTION: 9

If the intrinsic angle is 20, then find the loss tangent.

Solution:

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

QUESTION: 10

The intrinsic impedance of free space is given by

Solution:

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.