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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Test: Singly Excited Magnetic System - Electrical Engineering (EE) MCQ

Test: Singly Excited Magnetic System - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test - Test: Singly Excited Magnetic System

Test: Singly Excited Magnetic System for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Singly Excited Magnetic System questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Singly Excited Magnetic System MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Singly Excited Magnetic System below.
Solutions of Test: Singly Excited Magnetic System questions in English are available as part of our course for Electrical Engineering (EE) & Test: Singly Excited Magnetic System 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: Singly Excited Magnetic System | 10 questions in 30 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: Singly Excited Magnetic System - Question 1
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For a toroid to extract the energy from the supply system, the flux linkages of the magnetic field must be ________

Detailed Solution for Test: Singly Excited Magnetic System - Question 1

dWelec = idφ = eidt, where dWelec = differential electrical energy to coupling field, and if the flux linkages are either constant or zero, i.e, dφ = 0, then dWelec = 0.

Test: Singly Excited Magnetic System - Question 2
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The energy stored in a magnetic field is given by ____________ where L = self-inductance and Rl=reluctance.

Detailed Solution for Test: Singly Excited Magnetic System - Question 2

We know that Wfld = 1/2 φi and L = φ/i, thus Wfld = 1/2 Li2.

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Test: Singly Excited Magnetic System - Question 3
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For a linear electromagnetic circuit, which of the following statement is true?

Detailed Solution for Test: Singly Excited Magnetic System - Question 3

Wfld = Wfld1 = 1/2 φ*i = 1/2 F*∅.

Test: Singly Excited Magnetic System - Question 4
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The electromagnetic force developed in any physical system acts in such a direction as to tend to _____________
Detailed Solution for Test: Singly Excited Magnetic System - Question 4

fe = (∂Wfld1 (i,x))/∂x = (∂Wfld1 (F,x))/∂x, the positive sign before ∂Wfld1 indicates that force fe acts in a direction as to tend to increase the co-energy at constant mmf.

Test: Singly Excited Magnetic System - Question 5
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The electromagnetic torque developed in any physical system, and with magnetic saturation neglected, acts in such a direction as to tend to ____________
Detailed Solution for Test: Singly Excited Magnetic System - Question 5

fe = 1/2 ∅2 dRl/dx, Te = -1/2 ∅2 dRl/dθ = 1/2 i2 dL/dθ.

Test: Singly Excited Magnetic System - Question 6
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Magnetic stored energy density for iron is given by ______
Detailed Solution for Test: Singly Excited Magnetic System - Question 6

Magnetic stored energy density for iron is given as
wfld = Wfld/((Length of the magnetic path through Iron)*(Iron area normal to the magnetic flux)) = 1/2 (F∅)/(length*Area) = 1/2 F/length ∅/area = 1/2 H*B
Also, H = B/μ,thus wfld = 1/2 B2/μ.

Test: Singly Excited Magnetic System - Question 7
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When a current of 5A flows through a coil of linear magnetic circuit, it has flux linkages of 2.4 wb-turns. What is the energy stored in the magnetic field of this coil in Joules?
Detailed Solution for Test: Singly Excited Magnetic System - Question 7

Wfld = 1/2 φ*i = 1/2*2.4*5 = 6 Joules.

Test: Singly Excited Magnetic System - Question 8
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The electromagnetic force and/or torque, developed in any physical system, acts in such a direction as to tend to ____________
Detailed Solution for Test: Singly Excited Magnetic System - Question 8

fe = -(∂Wfld (φ, x))/∂x = -(∂Wfld (∅,x))/∂x and Te = -(∂Wfld(φ,θ))/∂θ = -(∂Wfld (∅,θ))/∂θ
The negative sign before ∂Wfld indicates that fe acts in a direction as to tend to decrease the stored energy at constant mmf.

Test: Singly Excited Magnetic System - Question 9
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Consider a magnetic relay with linear magnetization curve in both of its open and closed position. What happens to the electrical energy input to the relay, when the armature moves slowly from open position to closed position?
Detailed Solution for Test: Singly Excited Magnetic System - Question 9

For the above mentioned case, Wfld = Wmech and Wfld = Welec/2 hence, option “c” is the correct answer.

Test: Singly Excited Magnetic System - Question 10
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Electromagnetic force and/or torque developed in any physical system, acts in such a direction as to tend to ____________
Detailed Solution for Test: Singly Excited Magnetic System - Question 10

fe = (∂Wfld1(i,x))/∂x = (∂Wfld(i,x))/∂x.

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