2019 SSC Junior Engineer Electrical Online Exam 2019 CPWD/CWC/MES Electrical Engineering Set 1.7


61. The Norton's current in the circuit shown below is ?




... Answer is C)
For calculating Norton's current we have short R ohm resistance and calculate current.

when we short R resistance then no current flow in 150 ohm

only 30 ohm resistance work, so current drow from 360v source.

I =360/30 = 12 A,



62. In a capacitor start single phase induction motor the current in the ?




... Answer is D)
Ideal requirement to have maximum starting torque is the current through starting winding should lead current through main winding by 90°.

Normally current through main winding lag supply voltage by an angle θ.

In order to get maximum starting torque the current through starting winding should lead supply voltage by ( 90 - θ )°.



63. motor is referred to as a universal motor ?




... Answer is D)
The universal motor is a type of electric motor that can operate on either AC or DC power and uses an electromagnet as its stator to create its magnetic field.

It is a commutated series-wound motor where the stator's field coils are connected in series with the rotor windings through a commutator.



64. Find H=__A/m at the center of a circular coil of diameter 1 m and carrying a current of 2A ?




... Answer is #)
Given I =2A, Radius R = 1/2 m, H =?


B = (μ0 I)/2R Magnetic field intensity at centre of circular coil = I/2R.

circular coil H = I/2R

H = 2/2(1/2) = 2 A/m

.


65. If the field of the synchronous motor is under excited the power factor will be ?




... Answer is C)
An overexcited synchronous motor operate at leading power factor, under-excited synchronous motor operate at lagging power factor and normal excited synchronous motor operate at unity power factor.

Synchronous generator supplies leading power factor when it is under excited and lagging power factor when it is over excited.



66. A synchronous motor runs at 600 rpm, which of the following case is true ?




... Answer is C)
Synchronous speed NS = 120f/P

When

P = 12 and F = 60 then Ns = ?

NS = 120f/P = (120 × 60)/12 = 600 RPM



67. With the current direction marked in the circuit shown, the net voltage applied is ?




... Answer is D)
Use KVL..
Net voltage = –V2 + V1 = – (V2 – V1)


68. Identify the machine shown in the circuit. ?




... Answer is A)


69. Prevention of interference with neighbouring telephone lines can be done by ?




... Answer is C)
The transmission lines transmit bulk power at relatively high vol.
... on the true speech currents in the neighboring telephone wires and set up distortion while ... There are various ways that can reduce the telephone interference.
... Transposition of lines reduces the induced voltages to a considerable extent.


70. reflected light _ _ _ incident light ?




... Answer is B)
Reflection factor = luminous flux reflected by a body / luminous flux received by the body


The ratio of the total amount of radiation, as of light, reflected by a surface to the total amount of radiation incident on the surface.
In other words, it is the fraction of radiant energy that is reflected from a surface.









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