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RVUNL’21 EE: Daily Practices Quiz 12-Aug-2021

Each question carries 2 marks.
Negative marking: 0.5 mark
Total Questions: 06
Time: 06 min.

Q1. A network has 10 independent loops and 7 nodes. The number of branches in the network will be….
(a) 10
(b) 12
(c) 14
(d) 16
(e) 18

Q2. Which of the following is INCORRECT with regard to PMMC instrument?
(a) No hysteresis errors
(b) Very high accuracy
(c) High torque to weight ratio
(d) Non-uniform scale
(e) All of the above

Q3. Super tension cables are manufactured for the voltage….
(a) Up to 11 KV
(b) Above 132 KV
(c) From 22 KV to 33 KV
(d) From 33 KV to 66 KV
(e) Less than 1 KV

Q4. An isolation transformer has primary to secondary turns ratio of
(a) 1:1
(b) 1:2
(c) 2:1
(d) can be any ratio
(e) None of the above

Q5. If a voltmeter when connected to the rotor of an induction motor gives 150 oscillations per minute and stator frequency is 50 Hz. Then the slip of induction motor will be:
(a) 2 %
(b) 4 %
(c) 2.5 %
(d) 5 %
(e) 3 %

Q6. In a 220 KV system, the line to ground capacitance is 0.02 µF and capacitance is 8 H. what will be the value of resistance to be used across the contacts of the circuit breaker to eliminate the re-striking voltage if an instantaneous magnetizing current of 6 A is interrupted.
(a) 10.6 kΩ
(b) 10 KΩ
(c) 1 KΩ
(d) 100 KΩ
(e) 10 Ω

SOLUTIONS

S1. Ans.(d)
Sol. L=B-(N-1)
Where; B=number of branches=Loops and N=Nodes
∴10=B-(7-1)
⇒B=10+6=16.

S2. Ans.(d)
Sol. For PMMC instrument:
Ө α I ⇒uniform scale
the scale in the instrument can be divided properly
It generates no losses because of hysteresis.
It uses less power
It is not influenced by the stray magnetic field.
High accuracy

S3. Ans.(c)
Sol. Type of cable   Operating voltage
Low tension         <1 KV
High tension        1-11 KV
Super tension       11-33 KV
Extra high tension  <66 KV

S4. Ans.(a)
Sol. For isolation transformer = 1:1

S5. Ans.(d)
Sol. f_r=150/60=2.5 Hz
We know, f_r=sf
∴s=f_r/f=2.5/50×100=5 %

S6. Ans.(b)
Sol. Resistance to be connected across the contacts to eliminate the re-striking voltage is R=0.5√(L/C)
∴ R=0.5 √(8/(0.02×10^(-6) )) = 10 KΩ.

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