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Test: Loss Tangent - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test Electromagnetic Fields Theory (EMFT) - Test: Loss Tangent

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

The loss tangent refers to the

Detailed Solution for Test: Loss Tangent - Question 1

Answer: a
Explanation: The loss tangent is the tangent angle formed by the plot of conduction current density vs displacement current density. It is the ratio of Jc by Jd. It represents the loss of power due to propagation in a dielectric, when compared to that in a conductor.

Test: Loss Tangent - Question 2

Calculate the conduction current density when the resistivity of a material with an electric field of 5 units is 4.5 units.

Detailed Solution for Test: Loss Tangent - Question 2

Answer: c
Explanation: The conduction current density is the product of the conductivity and the electric field. The resistivity is the reciprocal of the conductivity. Thus the required formula is Jc = σ E = E/ρ = 5/4.5 units.

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Test: Loss Tangent - Question 3

At high frequencies, which parameter is significant?

Detailed Solution for Test: Loss Tangent - Question 3

Answer: b
Explanation: The conduction current occurs in metals and is independent of the frequency. The attenuation and phase constant highly depend on the varying frequency. The displacement current occurs due to dielectrics and is significant only at very high frequencies.

Test: Loss Tangent - Question 4

Find the loss tangent of a material with conduction current density of 5 units and displacement current density of 10 units.

Detailed Solution for Test: Loss Tangent - Question 4

Answer: b
Explanation: The loss tangent is the ratio of Jc by Jd. On substituting for Jc = 5 and Jd = 10, the loss tangent, tan δ = 5/10 = 0.5. It is to be noted that it is tangent angle, so that the maxima and minima lies between 1 and -1 respectively.

Test: Loss Tangent - Question 5

The loss tangent is also referred to as

Detailed Solution for Test: Loss Tangent - Question 5

Answer: c
Explanation: The loss tangent is the measure of the loss of power due to propagation in a dielectric, when compared to that in a conductor. Hence it is also referred to as dissipation factor.

Test: Loss Tangent - Question 6

The loss tangent of a wave propagation with an intrinsic angle of 20 degree is

Detailed Solution for Test: Loss Tangent - Question 6

Answer: b
Explanation: The angle of the loss tangent δ is twice the intrinsic angle θn. Thus tan δ = tan 2θn = tan 2(20) = tan 40.

Test: Loss Tangent - Question 7

The expression for the loss tangent is given by

Detailed Solution for Test: Loss Tangent - Question 7

Answer: a
Explanation: The conduction current density is Jc = σ E and the displacement current density is Jd = jωεE. Its magnitude will be ωεE. Thus the loss tangent tan δ = Jc /Jd = σ/ωε is the required expression.

Test: Loss Tangent - Question 8

Find the loss angle in degrees when the loss tangent is 1.

Detailed Solution for Test: Loss Tangent - Question 8

Answer: c
Explanation: The loss tangent is tan δ, where δ is the loss angle. Given that loss tangent tan δ = 1. Thus we get δ = tan-1(1) = 450.

Test: Loss Tangent - Question 9

The complex permittivity is given by 2-j. Find the loss tangent.

Detailed Solution for Test: Loss Tangent - Question 9

Answer: a
Explanation: The loss tangent for a given complex permittivity of ε = ε’ – jε’’ is given by tan δ = ε’’/ ε’. Thus the loss tangent is 1/2.

Test: Loss Tangent - Question 10

The intrinsic angle of the wave with a loss angle of 60 is

Detailed Solution for Test: Loss Tangent - Question 10

Answer: d
Explanation: The angle of the loss tangent δ is twice the intrinsic angle θn. Thus tan δ = tan 2θn. We get θn = δ/2 = 60/2 = 30 degrees.

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