This set of Electrical Machines Multiple Choice Questions & Answers (MCQs) focuses on “Singly Excited Magnetic Systems”.
1. For a toroid to extract the energy from the supply system, the flux linkages of the magnetic field must be ________
a) zero
b) changing or varying
c) constant
d) any of the mentioned
View Answer
Explanation: 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.
2. Magnetic stored energy density for iron is given by ______
a) 1/2 B/μ
b) 1/2 B2 μ
c) 1/2 ∅2 Rl
d) 1/2 B2/μ
View Answer
Explanation: 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.
3. The energy stored in a magnetic field is given by ____________ where L=self-inductance and Rl=reluctance.
a) 1/2 Li2
b) 1/2 (mmf*Rl)2
c) 1/2∅Rl
d) 1/2 φ2i
View Answer
Explanation: We know that Wfld=1/2 φi and L=φ/i, thus Wfld=1/2 Li2.
4. 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?
a) 6
b) 12
c) 1.2
d) 2.4
View Answer
Explanation: Wfld = 1/2 φ*i = 1/2*2.4*5 = 6 Joules.
5. For a linear electromagnetic circuit, which of the following statement is true?
a) Field energy is less than the Co-energy
b) Field energy is equal to the Co-energy
c) Field energy is greater than the Co-energy
d) Co-energy is zero
View Answer
Explanation: Wfld=Wfld1=1/2 φ*i=1/2 F*∅.
6. The electromagnetic force and/or torque, developed in any physical system, acts in such a direction as to tend to ____________
a) decrease the magnetic stored energy at constant mmf
b) decrease the magnetic stored energy at constant flux
c) increase the magnetic stored energy at constant flux
d) increase the magnetic stored energy at constant current
View Answer
Explanation: 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.
7. The electromagnetic force developed in any physical system acts in such a direction as to tend to _____________
a) decrease the co-energy at constant mmf
b) increase the co-energy at constant flux
c) decrease the co-energy at constant flux
d) increase the co-energy at constant mmf
View Answer
Explanation: 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.
8. 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?
a) Welec=Wfld
b) Welec=Wmech
c) Welec=Wmech/2+Wfld/2
d) Welec=0
View Answer
Explanation: For the above mentioned case, Wfld=Wmech and Wfld=Welec/2 hence, option “c” is the correct answer.
9. The electromagnetic torque developed in any physical system, and with magnetic saturation neglected, acts in such a direction as to tend to ____________
a) decrease both the reluctance and inductance
b) increase both the reluctance and inductance
c) decrease the reluctance and increase the inductance
d) increase the reluctance and decrease the inductance
View Answer
Explanation: fe=1/2 ∅2 dRl/dx, Te=-1/2 ∅2 dRl/dθ = 1/2 i2 dL/dθ.
10. Electromagnetic force and/or torque developed in any physical system, acts in such a direction as to tend to ____________
a) increase both the field energy and co-energy at constant current
b) increase the field energy and decrease the co-energy at constant current
c) decrease both the field energy and co-energy at constant current
d) decrease the field energy and increase the co-energy at constant current
View Answer
Explanation: fe = (∂Wfld1(i,x))/∂x = (∂Wfld(i,x))/∂x.
Sanfoundry Global Education & Learning Series – Electrical Machines.
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