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UPRVUNL’21 EE: Daily Practices Quiz 01-Sep-2021

UPRVUNL JE Recruitment 2021: 

A total of 196 vacancies were released by the UPRVUNL Recruitment 2021. The selection process will be based on the written examination which will be divided into two parts – Part I and Part II. Part I will have 150 objective-type questions from the syllabus for the Diploma Engineering in the relevant branch. While Part II will have 50 objective-type questions comprising General Knowledge, General Hindi, and Reasoning.

UPRVUNL’21 EE:  Daily Practices Quiz 01-Sep-2021

 

Each question carries 1 mark.

Negative marking: 1/4 mark

Total Questions: 06

Time: 08 min.

Q1. If both the number of turns and core length of an inductive coil are doubled, then its self-inductance will be
(a) halved
(b) doubled
(c) quadrupled
(d) unaffected

Q2. An alternating current given by i =14.14 sin⁡〖(ωt+π/6)〗 has an r.m.s. value of
(a) 1.96 A
(b) 7.07 A
(c) 10.0 A
(d) 14.14 A

Q3. A 3-phase induction motor runs at super synchronous speed. For self-excitation the machine
(a) Draws reactive power from the mains
(b) Draws active power from the mains
(c) Feeds reactive power to the mains
(d) None of the above

Q4. A 3-phase induction motor has a starting torque of 200 Nm when switched on-directly to supply. If an auto-transformer with 50 % tapping is used for starting, the starting torque would be
(a) 50 Nm
(b) 100 Nm
(c) 200 Nm
(d) 400 Nm

Q5. The emf induced per phase in the rotor winding of 3-phase induction motor is 200 V at standstill. Under full-load condition, this emf would be normally
(a) 200 V
(b) 100 V
(c) 8 V
(d) 0.4 V

Q6. A 2KVA transformer has iron loss of 100 watts and full-load copper loss of 150 watts. The maximum efficiency of the transformer would occur when the total loss is-
(a) 300 W
(b) 250 W
(c) 200 W
(d) 175 W

SOLUTIONS
S1. Ans.(b)
Sol. Self-inductance(L)=(μ_0 μ_r AN^2)/l
If both the number of turns and core length of an inductive coil are doubled,
Then L^’=(μ_0 μ_r A(2N)^2)/2l=2L.

S2. Ans.(c)
Sol. i_rms=i_m/√2=14.14/1.414=10 A

S3. Ans.(a)
Sol. If a 3-phase induction motor is running at super synchronous speed then:
Slip is negative
It goes in to generating region
To provide the reactive power for the machine, it will draw reactive power from a source.
NOTE: Induction generator is not a self-excited machine. Therefore, when running as a generator, the machine takes reactive power from the AC power line and supplies active power back into the line. Reactive power is needed for producing rotating magnetic field. The active power supplied back in the line is proportional to slip above the synchronous speed.

S4. Ans.(a)
Sol. For 50 % tapping, x=0.5
∴T_auto=x^2×T_DOL=(0.5)^2×200=50 Nm

S5. Ans.(c)
Sol. At standstill, s=1
∴E_2^’=sE_2=1×200=200 V
For full-load condition, slip varies from 2% to 5 %.
∴E_2^’=sE_2=(0.02 to 0.05)×200
So possible values lie between: 4 V to 10 V.
So, option© is correct.
NOTE: At standstill, induction motor slip s is 1, hence the frequency of induced EMF in the rotor of the induction motor is same as that of supply frequency and reduces with increase in speed (due to the reduction in slip).

S6. Ans.(c)
Sol. → for maximum efficiency
copper loss = Iron loss
Total loss = 2 × iron loss
= 2 × 100
= 200 watts

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