RVUNL Recruitment 2021: The RVUNL exam date is released for a total of 1075 vacancies.
There are two parts for the RVUNL test, Part A and Part B.
Part A will have 60% weightage while Part B has 40% weightage out of the total marks in the online written exam.
RVUNL’21 EE: Daily Practices Quiz 28-Aug-2021
Each question carries 2 marks.
Negative marking: 0.5 mark
Total Questions: 06
Time: 08 min.
Q1. A fuse 10 A current rating of fuse element and it can be blown out at minimum fusing current of 15 A. The fusing factor will be:
Q2. The advantage of PMMC instrument is that it:
(a) Has low torque to weight ratio
(b) Can be used on both AC and DC
(c) Has high torque to weight ratio of moving parts
(d) Is free from friction error
(e) Both b and c
Q3. For measurement of which of the following is Low Power Factor (LPF) wattmeter is NOT suitable?
(a) Load test on induction motor
(b) Open circuit test on single phase transformer
(c) Measurement of power in resistive loads
(d) Measurement of power in inductive loads
(e) None of the above
Q4. Which of the following statement is true for no-load current of the transformer?
(a) has high magnitude and low power factor
(b) has high magnitude and high-power factor
(c) has small magnitude and high-power factor
(d) has small magnitude and low power factor
(e) None of the above
Q5. When supply voltage to a 3-phase squirrel cage induction motor is reduced by 20%, the maximum torque will decrease by
(a) 10 %
(c) 36 %
(d) 40 %
(e) 4 %
Q6. A 200/400 V, 10 kVA, 50 Hz single-phase transformer has, at full-load, a copper loss
of 120 W. If it has an efficiency of 98% at full-load, determine the iron losses.
(a) 84 W
(b) 117 W
(c) 92 W
(d) 106 W
(e) 204 W
Sol. Fusing factor=(minimum fusing current)/(rated current)=15/10=1.5
Note: the value of fusing factor is always more then 1.
Sol. PMMC instruments:
High torque to weight ratio
Less friction errors
PMMC instruments (i.e., D’Arsonval meters) are only used for measuring the Direct Current (DC) current.
Sol. LPF wattmeter:
Works with low pf system
Resistive loads have unity pf.so, it will give large error in such system.
Sol. The no load current is the current drawn by the transformer winding when no external load is connected to the secondary of the transformer that is the secondary circuit is open. Theoretically this current should be zero. But actually, it is not zero but a small factor of the full load current.
Since no-load current lags voltage by the angle of nearly 90⁰, power factor being equal to cosine of the angle between current and voltage, it will be equal to value which is near to 0. Thus, power factor will be low.
Sol. T_max α V^2
According to question, voltage is reduced by 20%, i.e., V_2=0.8 V_1
∴T_(max2/T_max1 =(V_2/V_1 )^2=0.64)
So, T_max2 is reduced by 36 %.
Sol. Output at full load at unity pf = 10×1 = 10kW
Input = output / η = 10 / 0.98 = 10.204 kW
Total losses (iron loss+Cu loss) = 10.204-10=0.204 Kw=204W
Iron loss = 204 – 120 = 84 W