Quiz: Electrical Engineering
Exam: SSC -JE
Topic: Miscellaneous
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. In the circuit given below, determine Vab.
(a)2.5 V
(b)3.5 V
(c)7 V
(d)5V
Q2. The rotor power output of a three-phase induction motor is 15 KW and the corresponding slip is 4%. The rotor ohmic losses are
(a)600 W
(b)625 W
(c)650 W
(d)700 W
Q3. A current i=5+14.14 sin(ωt+45^0) is passed through a center-zero PMMC and a moving-iron instrument respectively, the respective readings are:
(a)-5 and 15
(b)5 and √125
(c)-5 and 19.14
(d)5 and 10
Q4. The type of single-phase induction motor having the highest power factor at full load is:
(a)Shaded pole type
(b)Split-phase type
(c)Capacitor-start type
(d)Capacitor-run type
Q5. In house wiring, Black and Green wires indicate
(a)Earth and neutral respectively
(b)Phase and neutral respectively
(c)Phase and earth respectively
(d)Neutral and earth respectively
Q6. The commutator in a d.c. machine acts as
(a) a mechanical inverter
(b) a mechanical rectifier
(c) current controller
(d) either (a) or (b)
Q7. Speed control is possible for _______ and not possible for __________
(a) induction motor, synchronous motor
(b) induction motor, differential motor
(c) synchronous motor, synchronous-induction motor
(d) dc motor, induction motor
Q8. Which of the following can be done using a synchronous motor but not by induction motor?
(a) Power factor improvement
(b) Supplying mechanical load
(c) Power factor improvement and supply mechanical load
(d) None of the mentioned
Q9. The torque in an induction motor varies as _______________
(a) square of voltage
(b) proportion to voltage
(c) inversely proportion to voltage
(d) none of the above
Q10. For satisfactory performance of 3-phase 480V, 60 Hz induction motor, the supply voltage at 50 Hz should be equal to
(a)480 V
(b)350 V
(c)400 V
(d)420 V
SOLUTIONS
S1. Ans.(b)
Sol. Nodal analysis at point a: (Va-1)/2+Va/2-3=0
∴2Va=7
⇒Va=3.5=V_ab
S2. Ans.(b)
Sol. rotor power output of 3 phase I/M=P_g-P_cu
Where, Pg=air gap power and Pcu=rotor cu loss=sPg
According to question: Pg-Pcu=15
∴Pg-sPg=15
⇒Pg=15/(1-s)=15/(1-0.04)=15.625 KW
∴Rotor ohmic loss= sPg=0.04×15.625=0.625 KW=625 W
S3. Ans.(b)
Sol. Reading of Centre-zero PMMC =IDC=5
Reading of moving iron=Irms=√(5^2+〖(14.14/1.414)〗^2 )=√125
S4. Ans.(d)
Sol. capacitor-run type motor will have high power factor in which capacitor will be connected in running condition.
S5. Ans.(d)
Sol. In house wiring, Black and Green wires indicate neutral and earth respectively.
S6. Ans.(d)
Sol. Commutator acts as a reversing switch.
S7. Ans.(a)
Sol. Speed can be adjusted for an induction motor while it cannot be altered for a synchronous motor operating at normal speed. Synchronous motor rotates at synchronous speed.
S8. Ans.(a)
Sol. An over-excited synchronous motor has a leading power factor. This makes it useful for power factor correction of industrial loads.
Induction motor is a singly excited machine and it always needs exciting current to set up the flux in the machine.
S9. Ans.(a)
Sol. Torque in induction motor varies as square of the voltage.
S10. Ans.(c)
Sol. For satisfactory performance, the V / f ratio must be constant.
