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NHPC-JE’21 EE: Daily Practices Quiz 28-Sep-2021

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NHPC-JE’21 EE: Daily Practices Quiz

NHPC-JE’21 EE: Daily Practices Quiz 28-Sep-2021

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

Q1. The rotor used in alternators of Hydroelectric station is:
(a) Salient pole rotor
(b) Cylindrical rotor
(c) Non-salient pole rotor
(d) Round rotor with ac excitation

Q2. Swamping resistance is used to compensate error due to
(a) Stray magnetic field
(b) Temperature variations
(c) Large supply variations
(d) None of the above

Q3. Which of the following is an Integrating instrument?
(a) Watt-hour meter
(b) Ammeter
(c) Voltmeter
(d) Wattmeter

Q4. Superposition theorem is essentially based on the concept of
(a) Reciprocity
(b) Linearity
(c) Duality
(d) Non-linearity

Q5. A transformer is working at its full load and its efficiency is also maximum. Its iron
loss is 1000 watts, its copper loss at half load will be:
(a) 250 watts
(b) 300 watts
(c) 400 watts
(d) 500 watts

Q6. Two coupled coils with L1=L2=0.8 H, have a coupling coefficient of K = 0.8. The
turn ratio N1/N2 is
(a) 2: 1
(b) 1: 2
(c) 1: 1
(d) 4: 3


S1. Ans.(a)
Sol. As we know, N_s=120f/P
The hydraulic turbine in hydroelectric plant operates at low speed. Therefore, to achieve the required power frequency, number of poles are need to be increased which is efficiently possible in salient pole rotor.

S2. Ans.(b)
Sol. The ammeter is a sensitive device which is easily affected by the surrounding temperature. The variation in temperature causes the error in the reading. This can reduce by swamping resistance. The resistance having zero temperature coefficient is known as the swamping resistance. It connects in series with the ammeter.

S3. Ans.(a)
Sol. integrating instrument are those instruments which measures the electrical quantity over a period.

S4. Ans.(b)
Sol. Superposition theorem is used in the linear circuits having more than one
independent voltage or current sources.

S5. Ans.(a)
Sol. The iron loss at full load is 1000 Watts and maximum efficiency is obtained at full
For maximum efficiency: copper loss=iron or core loss=1000 watts
∴ Full load copper loss = I_fl^2.R=1000 Watts
∴ Half load copper loss = (I_fl/2)^2×R =1000/4=250 Watts

S6. Ans.(c)
Sol. The inductance of a coil is given by:
L=(N^2 μ_0 A)/l (For air core)
Or, L1/L2 =(N_1^2)/(N_2^2 )
∴ N1/N2 =√(L1/L2 )=√(0.8/0.8)=1

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