NHPC-JE’21 EE: Daily Practices Quiz 20-Sep-2021

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NHPC-JE’21 EE: Daily Practices Quiz

NHPC-JE’21 EE: Daily Practices Quiz 20-Sep-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) High pressure and low flow
(c) Low pressure and low flow
(d) Low pressure and high flow

Q2. Which of the following is an example of impulse turbine?
(a) Pelton wheel
(b) Tyson turbine
(c) Kaplan turbine
(d) All of the above

Q3. Corona loss can be reduced by the used of hollow conductors because,
(a) The current density is reduced
(b) Eddy current in the conductor is eliminated
(c) Of better ventilation in the conductor
(d) For a given cross-section, the radius of the conductor increases

Q4. A voltage source having an open-circuit voltage of 150 V and internal resistance of 75 Ω, is equivalent to a current source of
(a) 2 A in parallel with 75 Ω
(b) 2 A in series with 75 Ω
(c) 2 A in series with 37.5 Ω
(d) 1 A in parallel with 150 Ω

Q5. …………. opposes the change in circuit voltage.
(a) Resistance
(b) Inductance
(c) Capacitance
(d) Inductive reactance

Q6. A dc series motor drawing an armature current of Ia is operating under saturated magnetic conditions, torque developed in the motor is proportional to
(a) 1/Ia
(b) 1/Ia^2
(c) Ia^2
(d) Ia

SOLUTIONS

S1. Ans.(b)
Sol. Pelton wheels are most efficient under the conditions of high pressure and low flow. Pelton wheels are the preferred turbine for hydro-power where the available water source has relatively high hydraulic head at low flow rates. Thus, it is most suitable to being fed by a hydro source with a low ratio of flow to pressure, (meaning relatively low flow and/or relatively high pressure). It is an example of impulse turbine.
NOTE:
Kaplan & Francis: Reaction turbine
Pelton: Impulse turbine

S2. Ans.(a)
Sol. Pelton wheels are most efficient under the conditions of high pressure and low flow. It is an example of impulse turbine.

S3. Ans.(d)
Sol. Pc α 1/d
So, by using hollow conductor d increase and corona loss decreases.

S4. Ans.(a)
Sol. Using source transformation (i.e., voltage source to current source transformation)
Isc=Voc/Ri =150/75=2 A
So, given circuit is equivalent to a current source of 2 A in parallel with 75 Ω.

S5. Ans.(c)
Sol. Capacitor resists a sudden change in the voltage.

S6. Ans.(d)
Sol. We know Torque Tα ΦIa, As the motor is operating under saturation condition so torque will be independent of flux i.e., T α Ia.

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