NHPC-JE’21 EE: Daily Practices Quiz 22-Oct-2021

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NHPC-JE’21 EE: Daily Practices Quiz

NHPC-JE’21 EE: Daily Practices Quiz 22-Oct-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06
Time: 06 min.

Q1. Differential relays are used for protection of
(a) Alternators
(b) Transformers
(c) Feeders
(d) All of the above

Q2. In a synchronous motor, during hunting when the rotor speed exceeds the synchronous speed then damper bar develop
(a) Induction generator torque
(b) Harmonic
(c) DC motor torque
(d) Synchronous motor torque

Q3. Power consumed by a balanced 3-phase; 3-wire load is measured by two wattmeter methods. The first wattmeter reads twice that of the second. Then what will be the load impedance angle in radian?
(a) (π/6)
(b) (π/3)
(c) (π/2)
(d) (π/4)

Q4. Single or One Wattmeter method can only be used for
(a) Balanced three-phase load
(b) Imbalanced two-phase load
(c) Balanced one-phase load
(d) Imbalanced one-phase load

Q5. At any power factor of the load the efficiency of transformer will be maximum when:
(a) copper loss is equal to core loss
(b) copper loss is equal to eddy current loss
(c) copper loss is less then core loss
(d) copper loss is greater than core loss

Q6. A synchronous machine with low value of short circuit ratio has:
(a) higher stability limit
(b) good voltage regulation
(c) good speed regulation
(d) lower stability limit

SOLUTIONS

S1. Ans.(d)
Sol. A differential relay operates when phasor difference of two or more electrical quantities exceeds a predominant amount.
It is used for protection of generators, generator-transformers unit, feeder, transformer, large motors and bus-bars.

S2. Ans.(a)
Sol. When hunting occurs, the difference in the speed of stator and rotor poles develops an induced emf in the damper winding, which acts in such a way to suppress the rotor oscillation. The induced emf generated develop induction torque in the synchronous motor.

S3. Ans.(a)
Sol. W1=2W2
W1 :W2=2:1
Let W1=2
W2=1
Then,
tanθ= √3 ((W1-W2)/(W1+W2 ))
=√3 ((2-1)/(2+1))
=√3×1/3=1/√3
⇒θ=tan^(-1) (1/√3)=30°=(π\/6)_radian

S4. Ans.(a)
Sol.
I. one wattmeter → for 3-ϕ balanced load
II. Two wattmeters → 3-ϕ balanced or unbalanced star, Delta connected load
III. Three wattmeter →3-ϕ, 4 wire system

S5. Ans.(a)
Sol. At any power factor of the load, the efficiency of the transformer will be maximum when the copper loss is equal to core loss.

S6. Ans.(d)
Sol. Xd in pu = 1/SCR
SCR = 1/ Xd in per unit
Thus, the short circuit ratio is equal to the reciprocal of per unit value of direct axis synchronous reactance. The more the value of Xd, the lesser will be the short circuit ratio.
For the small value of the short circuit ratio (SCR), the synchronizing power is small. As the synchronizing power keeps the machine in synchronism, a lower value of the SCR has a low stability limit. In other words, a machine with a low SCR is less stable when operating in parallel with the other generators.