# Quiz: Electrical Engineering 30 April 2020

Quiz: Electrical Engineering
Exam: UPPSC-AE
Topic: Miscellaneous
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. If the fault current is 2000 A, the relay setting 50% and C.T. ratio 200/5, the plug setting multiplier will be
(a) 25
(b) 20
(c) 35
(d) 50

Q2. For very long high voltage transmission lines, the system of overcurrent protection used is
(b) pilot-wire system
(c) distance protection
(d) none of the above

Q3. Balanced earth-fault protection is generally provided for
(a) small-size alternators
(b) medium-size alternators
(c) large-size alternators
(d) none of the above

Q4. A wire has a resistance of 24 Ω. It is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is
(a) 144 Ω
(b) 6 Ω
(c) 3Ω
(d) 24Ω

Q5. A 400 W and 200 W bulb both meant for operation at 220 V are connected in series. When connected to a 220 V supply, the power consumed by them is
(a) 133.33 W
(b) 139 W
(c) 800W
(d) 350 W

Q6. An electric fan and a heater are marked as 100 W, 220 V and 1000 W, 220 V respectively. The resistance of heater is
(a) zero
(b) greater than that of fan
(c) less than that of fan
(d) equal to that of fan

Q7. The resistor values in delta network that is equivalent to a wye containing three 60 Ω resistors is …………
(a) 360 Ω each
(b) 240 Ω each
(c) 180 Ω each
(d) 120 Ω each

Q8. Two electric lamps of 80 W each are connected in parallel. The power consumed by the combinations is
(a) 20 W
(b) 60 W
(c) 80 W
(d) 160 W

Q9. A house is served by a 220 V line. In a circuit protected by a fuse marked 9 A, the maximum number of 60 W lamps in parallel that can be turned on is
(a) 22
(b) 66
(c) 33
(d) 11

Q10. The armature of a d.c. machine is made of ………
(a) silicon steel
(b) wrought iron
(c) cast steel
(d) soft iron

Solution

S1. Ans.(b)
Sol. Pick-up value = Rated secondary current of C.T. × Current setting
= 5 × 0.5 = 2.5 A
Fault current in relay coil = 2000 × 5/200 = 50 A
∴Plug setting multiplier = 50/2.5 = 20

S2. Ans.(c)
Sol. The time-graded protection system gives an unduly long-time delay in fault clearance at the generating station end particularly when there are more than four or five sections. The pilot-wire system becomes too expensive owing to the greater length of pilot wires required.

S3. Ans.(a)
Sol. In small-size alternators, the neutral ends of the three-phase windings are often connected internally to a single terminal. Therefore, it is not possible to use Merz-Price circulating current principle. Under these circumstances, it is considered sufficient to provide protection against earth-faults only by the use of balanced earth-fault protection scheme.
S4. Ans.(b)
Sol. According to the conditions of the problem. Each parallel path has a resistance of 12 Ω.
∴ Effective resistance between two points on the diameter
= 12‖12┤=(12×12)/(12+12)=144/24=6Ω

S5. Ans.(a)
Sol. RT=R1+R2 or RT/V2 =R1/V2 +R2/V2
Or 1/PT =1/P1 +1/P2
∴ PT=(P1 P2)/(P1+P2 )=(400×200)/(400+200)=133.33W

S6. Ans.(c)
Sol.
S7. Ans.(c)
Sol. RD=3RY=3 × 60 = 180 Ω
S8. Ans.(d)
Sol. 1/RT =1/R1 +1/R2
or V^2/RT =V^2/R1 +V^2/R^2
or PT=P1+P2=40+40= 80 W

S9. Ans.(a)
Sol. Let n be the required number of 90 W lamps connected in parallel.
Total power, P = 90 n.
Now, P = V I or 90 × n = 220 × 9
n = (220×9)/90 = 22

S10. Ans.(a)
Sol. It is desired that armature should provide a path of low reluctance for the flux. For this reason, armature is made of material having high permeability e.g. silicon steel

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