NHPC-JE’21 EE: Daily Practices Quiz 19-Oct-2021

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NHPC-JE’21 EE: Daily Practices Quiz

NHPC-JE’21 EE: Daily Practices Quiz 19-Oct-2021

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

Q1. Pelton wheels are most efficient under the conditions of……….
(a) High pressure and high flow
(b) Low pressure and low flow
(c) Low pressure and high flow
(d) High pressure and low flow

Q2. In a power station,
(a) Reserve capacity = Plant capacity + Max. demand
(b) Reserve capacity = Plant capacity – Average demand
(c) Reserve capacity = Plant capacity – Max. demand
(d) Reserve capacity = Plant capacity + Average demand

Q3. In a 3-phase system, the line losses are
(a) directly proportional to cos ϕ
(b) inversely proportional to cos ϕ
(c) inversely proportional to cos² ϕ
(d) none of the above

Q4. The breaking capacity of a 3-phase circuit breaker is given by
(a) √3 × service voltage × rated symmetrical current
(b) 3 × service voltage × rated symmetrical current
(c) 2 × service voltage × rated symmetrical current
(d) None of the above

Q5. Tellegen’s theorem is based on the principle of law of _________________.
(a) Conservation of charge
(b) Conservation of mass
(c) Conservation of velocity
(d) conservation of energy

Q6. A D.C. Generator has 6 poles, A brush shift of 6° actual means a brush shift of
(a) 6° electrical
(b) 18° electrical
(c) 30° electrical
(d) 2° electrical

SOLUTIONS

S1. Ans.(d)
Sol. Pelton wheels are suitable for low discharge and high head(pressure)
Pelton wheel is an example of impulse turbine.

S2. Ans.(c)
Sol. Reserve capacity = Plant capacity – Max. demand

S3. Ans.(c)
Sol. Line current, IL=P/(√3 VL cos⁡ϕ )
Since line losses are proportional to the square of IL, it follows that line losses are proportional to 1/cos² ϕ.

S4. Ans.(a)
Sol. It is current (rms) that a circuit breaker is capable of breaking at given recovery voltage and under specified conditions.
Breaking capacity = √3 × service voltage × rated symmetrical current

S6. Ans.(b)
Sol. (θele= P/2×θmech )
= (6/2)×6
= 18° electrical

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