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NHPC-JE’21 EE: Daily Practices Quiz 06-Oct-2021

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NHPC-JE’21 EE: Daily Practices Quiz

NHPC-JE’21 EE: Daily Practices Quiz 06-Oct-2021

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

Q1. A diesel power plant is most suitable as:
(a) Baseload power plant
(b) Stand by power plant
(c) Peak load power plant
(d) None of the above

Q2. The efficiency of a d.c. motor when developing maximum mechanical power will
(a) 100%
(b) 50%
(c) less than 50%
(d) more than 50%

Q3. Running the machine at no-load is not advisable for
(a) DC shunt motor
(b) Induction motor
(c) DC series motor
(d) Synchronous motor

Q4. The potential difference V across and current I flowing through an instrument in
an a.c circuit is given by;
V = 5 cos ωt volts; I = 2 sin ωt amperes. The power dissipated in the instrument is:
(a) 0 W
(b) 10 W
(c) 10/(√2 W)
(d) 2.5 W

Q5. A 40 Ω electric heater is connected to a 200 V, 50 Hz supply. The peak value of
the electric current in the circuit is approximately
(a) 7 A
(b) 2.5 A
(c) 5 A
(d) 10 A

Q6. An a.c source is in series with R and L. If the respective potential drops are 200 V
and 150 V, the applied voltage is
(a) 125 V
(b) 350 V
(c) 200 V
(d) 250 V


S1. Ans.(b)
Sol. A diesel power plant is most suitable as a standby power plant.
During power outages, emergency backup electrical generators powered by diesel
engines provide reliable, immediate, and full-strength electric power, and unique start

S2. Ans.(c)
Sol. Under the condition of the maximum power output of d.c. motor, half of the input power
Is wasted in the armature circuit. In fact, if we take into account other losses (iron and
mechanical), the efficiency will be well below 50%.

S3. Ans.(c)
Sol. At no-load, the armature current in a d.c. the series motor is very small and so is the
magnetic flux ϕ due to series excitation. Hence the speed of the motor rises to an
excessively high value (N ∝ 1/ϕ). This is dangerous for the machine which may be
destroyed due to centrifugal forces set up in the rotating parts. Hence the answer is d.c.
series motor.

S4. Ans.(a)
Sol. Power, P=Vrms IrmsCos ϕ
Here, V=5 cos ωt
And I=2 cos (ωt–90)
⇒ϕ=Өv-ӨI=90⁰⇒ P= 0

S5. Ans.(a)
Sol. Vm=Vrms×√2=200√2 V
∴Im=Vm/R=(200√2)/40=7 A

S6. Ans.(d)
Sol. VR=200V
VL=150 V
∴ Vs=√(VR^2+VL^2 )
=250 V

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