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

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


Test Description

10 Questions MCQ Test - Test: Displacement & Conduction Current

Test: Displacement & Conduction Current for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Displacement & Conduction Current questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Displacement & Conduction Current MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Displacement & Conduction Current below.
Solutions of Test: Displacement & Conduction Current questions in English are available as part of our course for Electrical Engineering (EE) & Test: Displacement & Conduction Current solutions in Hindi for Electrical Engineering (EE) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt Test: Displacement & Conduction Current | 10 questions in 10 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study for Electrical Engineering (EE) Exam | Download free PDF with solutions
Test: Displacement & Conduction Current - Question 1

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

Detailed Solution for Test: Displacement & Conduction Current - Question 1

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

Test: Displacement & Conduction Current - Question 2

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

Detailed Solution for Test: Displacement & Conduction Current - Question 2

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.

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Displacement & Conduction Current - 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.

Detailed Solution for Test: Displacement & Conduction Current - Question 3

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.

Test: Displacement & Conduction Current - Question 4

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

Detailed Solution for Test: Displacement & Conduction Current - Question 4

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.

Test: Displacement & Conduction Current - Question 5

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

Detailed Solution for Test: Displacement & Conduction Current - Question 5

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.

Test: Displacement & Conduction Current - Question 6

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

Detailed Solution for Test: Displacement & Conduction Current - Question 6

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).

Test: Displacement & Conduction Current - Question 7

In good conductors, the electric and magnetic fields will be

Detailed Solution for Test: Displacement & Conduction Current - Question 7

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.

Test: Displacement & Conduction Current - Question 8

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

Detailed Solution for Test: Displacement & Conduction Current - Question 8

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.

Test: Displacement & Conduction Current - Question 9

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

Detailed Solution for Test: Displacement & Conduction Current - Question 9

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

Test: Displacement & Conduction Current - Question 10

The intrinsic impedance of free space is given by

Detailed Solution for Test: Displacement & Conduction Current - Question 10

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.

Information about Test: Displacement & Conduction Current Page
In this test you can find the Exam questions for Test: Displacement & Conduction Current solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Displacement & Conduction Current, EduRev gives you an ample number of Online tests for practice

Top Courses for Electrical Engineering (EE)

Download as PDF

Top Courses for Electrical Engineering (EE)