2019 SSC Junior Engineer Electrical Online Exam 2019 CPWD/CWC/MES Electrical Engineering Set 1.10


91. A magnetic circuit is applied with a current that changes at a rate of 5 A/sec. The circuit has an inductance of 2H, then the self-induced EMF is ?




... Answer is D)
Given Inductance L = 2 H
Rate of change of current di/dt = 5 A/sec
Self induced EMF = − (Rate of change of current × Inductance) = −L(di/dt)
= −(5 × 2) = −10V


92. What would the total loss of the 2kVA transformer corresponding to maximum efficiency be, provided the transformer has iron loss of 150W and full-load copper loss of 250W ?




... Answer is B)
The condition for maximum efficiency is,
Full load copper loss = Iron loss + Full load copper loss
Total loss P = Pe + Pi = 150 + 150 = 300 W .


93. Is it possible to have current in a transmission line, under no-load conditions ?




... Answer is B)
During no load condition the current that is flowing is only charging current for line capacitance, it increase capacitive var in the system. And since the line is under no load the line inductance will be less .
There fore the capacitive var becomes greater than inductive var during no load or light load condition.
Due to this phenomenon the receiving end voltage becomes greater than sending end voltage.
This effect is also called ferranti effect.This happens mainly in long transmission line.


94. Which of the following losses is together called iron losses ?




... Answer is B)
Iron Losses Iron losses are caused by the alternating flux in the core of the transformer/Motor as this loss occurs in the core it is also known as Core loss. Iron loss is further divided into hysteresis and eddy current loss.


95. What is the reason behind using a centrifugal switch in a single phase induction motor ?




... Answer is C)
Perhaps the most common use of centrifugal switches is within single-phase, split-phase induction motors. Here, the switch is used to disconnect the starting winding of the motor once the motor approaches its normal operating speed.


These induced voltages set up currents in the rotor bars. As a result, a rotor field is created which reacts with the stator field to develop the torque which causes the rotor to turn. As the rotor accelerates to the rated speed, the centrifugal switch disconnects the starting winding from the line.


96. The Thevenin's resistance as seen through the terminal A and B is ?




... Answer is B)
Simple Question try to solve and comment the answer.


97. If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is ?




... Answer is D)
Copper loss of a transformer at half full load = 400

If I is full load current than half load current = I/2

Copper loss of a transformer at full load = $(I)^2$ R

Copper loss of a transformer at half full load $(I/2)^2$ R= 400

= 1/4*$(I)^2$ R = 400

= $(I)^2$ R = 400*4 =1600 W



98. A 200 V d.c. machine has Ra = 0.5 Ω and its full-load Ia= 20A. Determine the induced e.m.f. when the machine acts as motor ?




... Answer is A)
Given data,
V = 200 v
Ra = 0.5 ohm
I a = 20 A
Eb =? Now,
Eb = V – Ia*Ra
Eb= 200 – 20 × 0.5
Eb= 200 – 10
Eb = 190 V .


99. An alternator has 9 slots/pole. What is the value of pitch factor if each coil spans 8 slot pitches ?




... Answer is A)
Coil Span = 8

Pole Pitch = 9 slots/pole

Slot angle (α) = 180° / Pole Pitch = 180°/9 = 20°

Pitch Factor = cos(α/2) = cos (20/2)

Pitch Factor = cos 10°.



100. If admittance Y = a + jb, then a= ?




... Answer is C)
Admittance is given by
Y = G + j B
Where ‘G’ is conductance, and B is susceptance.
Now comparing with given equation we get ‘a’  conductance. .








1 2 3 4 5 6 7 8 9 10

टिप्पणियाँ