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DFCCIL Executive Electrical 2018 Solved Question Paper Part One-Machines

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DFCCIL Executive Electrical 2018 Solved Question Paper PDF Part One-Machines 

Questions-1.A transformer have 400 W as iron loss at a full load,The iron loss at half full load will be____

A.400 W
B.100 W
C.200 W
D.800 W

Solution-Iron losses are of two types namely hysteresis loss and eddy current loss. They depend on the frequency and the constructional properties of the material which are considered to be constant from no load to full load. They are calculated during no-load test on the transformer.So, the iron loss at full load will also be 400 W.

Questions-2.A two pole alternator running at 1500 r.p.m. its angular velocity will be____

A.192 rad/s
B.157 rad/s
C.212 rad/s
D.118 rad/s

Solution-The frequency of an AC machine is expressed by the formula
F = (P * N) / 120
where F = Frequency, in hertz
P = the number of poles of the generator rotor
N = the speed of the generator rotor (in RPM)
Angular velocity=2πF rad/s
=2*3.14*25=157 rad/s

Questions-3.In a split-phase induction motor ,The two stator winding’s_____

A.Have equal R/XL ratio
B.Draw only the in phase currents
C.Are mutually displaced by 90⁰ electrically
D.Draw equal current

Solution-The Split Phase Motor is also known as a Resistance Start Motor. It has a single cage rotor, and its stator has two windings known as main winding and starting winding. Both the windings are displaced 90⁰ degrees in space.

Questions-4.Core lamination in transformer decreases____

A.Eddy current loss
B.Hysteresis loss
C.Copper loss
D.Leakage reactance

Solution-Eddy Current Loss is directly proportional to the square of lamination thickness. If more lamination are used for a given rotor axial length, the thickness of lamination decreases which result in decrease in Eddy current loss. Thus the use of thin lamination reduces the Eddy Current Loss. Normally the thickness of lamination is in between 0.4-0.5 mm. Further reduction in the thickness results into reduction of loss but at an increased cost.

Questions-5.A three phase induction motor running at 4 % slip ,if the input to rotor is 1000 W,the mechanical power develop by the motor will be___

A.960 W
B.9600 W
C.96 W
D.0.96 W

Solution-The formula is given by
Mechanical power = (1 – S) × Power input.
Power input = 1000 W
S = 4%
S = 4/100 = 0.04
Doing the substitution we have
(1 – 0.04) = 0.96
= 0.96 × 1000
= 960 W

Questions-6.Armature of DC machine is placed on the rotor to_____

A.Reduce the losses
B.Save iron
C.Support commutation
D.Decrease armature reaction

Solution-The armature winding of a DC machine is placed on the rotor to improve commutation i.e. to convert the alternating voltage produced in the winding into direct voltage at the brushes.

Questions-7.A washing machine generally employs a___motor.

A.Shaded pole
B.Resistance split phase
C.Single phase series

Solution-The Split Phase Motor is also known as a Resistance Start Motor.This type of motors are cheap and are suitable for easily starting loads where the frequency of starting is limited. This type of motor is not used for drives which require more than 1 KW because of the low starting torque. The various applications are as follows

  • Used in the washing machine, and air conditioning fans.
  • The motors are used in mixer grinder, floor polishers.
  • Blowers, Centrifugal pumps
  • Drilling and lathe machine.

Questions-8.An open circuit test on a transformer gives____

A.Friction losses
B.Iron losses
C.Total losses
D.Copper losses

Solution-Open circuit test or no load test on a transformer is performed to determine ‘no load loss (core loss/Iron losses)’ and ‘no load current I0‘.

Questions-9.In a 6 pole DC machine,90 mechanical degrees corresponds to how many electrical degrees____


Solution-One Degree Mechanical=P/2 Electrical Degree (Here P=Total number of poles)
90=6/2 Electrical Degree
Electrical Degree=90*3=270

Questions-10.A_____motor runs at the highest speed when load is removed.

A.Cumulatively compound
D.Differential compound

Solution-We know the relation, N ∝ Eb
For small load current (and hence for small armature current) change in back emf Eb is small and it may be neglected. Hence, for small currents speed is inversely proportional to ɸ. As we know, flux is directly proportional to Ia, speed is inversely proportional to Ia. Therefore, when armature current is very small the speed becomes dangerously high. That is why a series motor should never be started without some mechanical load.

Questions-11.The magnetic flux path in transformer must have _____

A.High resistance
B.Low resistance
C.High reluctance
D.Low reluctance

Solution-The lower the reluctance, the easier it is for magnetic flux to flow through the core material.

Questions-12.Leakage flux in a transformer ____

A.Helps in transfer of energy
B.Produces mutually induced EMF
C.Is minimized by interleaving the primary and secondary winding’s
D.Is negligible at full load

Solution-Leakage fluxes of transformer may be minimized by

  • Reducing the magnetizing current to the minimum
  • Reducing the reluctance of the iron core to the minimum
  • Reducing the number of primary and secondary turn to the minimum
  • Sectionalizing and interleaving the primary and secondary winding’s

Questions-13.The primary of a ___should never energized when its secondary is open circuited

A.Potential transformer
B.Current transformer
C.Auto transformer
D.Power transformer

Solution-Current transformers are intended to be used as proportional current devices. Therefore a current transformers secondary winding should never be operated into an open circuit, just as a voltage transformer should never be operated into a short circuit.

Very high voltages will result from open circuiting the secondary circuit of an energized CT so their terminals must be short-circuited if the ammeter is to be removed or when a CT is not in use before powering up the system.

Questions-14.For a simplex wave wound generator,EMF generated path given by

φ: Flux/pole in Weber
Z:Total number of armature conductor
P:Number of generator poles
N:Armature rotation in revolution per minutes(rpm)


Solution-Average e.m.f generated /conductor = dΦ/dt volt (n=1)
Now, flux cut/conductor in one revolutions Φ = ΦP Wb
No.of revolutions/second = N/60
Time for one revolution, dt = 60/N second
Hence, according to Faraday’s Laws of Electromagnetic Induction,
E.M.F generated/conductor dΦ/dt =(φPN)/60

For a simplex wave-wound generator
No.of parallel paths = 2
No.of conductors (in series) in one path = Z/2
E.M.F. generated/path =(φZPN)/120

Questions-15.Three phase wound rotor motors are commonly called as ____motors.

C.Slip ring

Solution-A wound rotor motor is a variation of the three-phase induction motor, designed to provide high starting torque for loads with high inertia, while requiring very low current. Wound rotor motors are also referred to as “slip ring motors.”

Questions-16.Running the mechanic as no load is NOT recommended for a/an____

A.Induction motor
B.DC shunt motor
C.Synchronous motor
D.DC series motor

Solution-We know the relation, N ∝ Eb
For small load current (and hence for small armature current) change in back emf Eb is small and it may be neglected. Hence, for small currents speed is inversely proportional to ɸ. As we know, flux is directly proportional to Ia, speed is inversely proportional to Ia. Therefore, when armature current is very small the speed becomes dangerously high. That is why a series motor should never be started without some mechanical load.

Questions-17.___Is a type of single phase motor which have lowest speed.

C.Shaded pole

Solution-The simplest (and least-expensive) type of AC single-phase motor is the Shaded-Pole Motor. Due to low starting-torque, its use is limited to applications that require less than ¾ horsepower and most commonly range in size from 1/20-to-1/6 horsepower and synchronous speeds of 1800 or 3600 RPM.

Questions-18.A capacitor start motor has a____

A.High starting torque
B.Low efficiency
C.Low power factor
D.High power factor

Solution-Capacitor induction motors are widely used for heavy-duty applications requiring high starting torque. Examples are refrigerator compressors, pumps, and conveyors.

Questions-19.The open circuit characteristics of a DC generator also called as____characteristics.

D.No load saturation

Solution-Open circuit characteristic is also known as magnetic characteristic or no-load saturation characteristic. This characteristic shows the relation between generated emf at no load (E0) and the field current (If) at a given fixed speed.

Questions-20.The direction of rotation of field in a three phase induction motors depends on the____

A.Supply voltage
B.Number of poles
C.Supply frequency
D.Phase sequence of supply voltage

Solution-The direction of rotation of the motor depends on the phase sequence of supply lines, and the order in which these lines are connected to the stator. Thus interchanging the connection of any two primary terminals to the supply will reverse the direction of rotation.

Questions-21.The function of dummy coils in DC mechanic is to_____

A.Improve commutation
B.Reduce machine cost
C.Mechanically balance the armature
D.Increase efficiency

Solution-Dummy coils are used with wave-winding and are resorted to when the requirements of the winding are not met by the standard armature punchings available in armature-winding shops. These dummy coils do not influence the electrical characteristics of the winding because they are not connected to the commutator. They are exactly similar to the other coils except that their ends are cut short and taped. They are there simply to provide mechanical balance for the armature because an armature having some slots without winding’s would be out of balance mechanically.

Questions-22.To obtain greater efficiency,the slip of an induction motor should be

B.Very High

Solution-Slip can be defined as the difference between the flux speed (Ns) and the rotor speed (N).
Speed of the rotor of an induction motor is always less than its synchronous speed.To obtain greater efficiency,the slip of an induction motor should be low or less than one.

Questions-23.The current drawn by a 220 V DC motor of armature resistance 0.5 ohm and back emf is 180 V is

A.0.8 A
B.80 A
C.8 A
D.18 A

Solution-V = 220, Eb = 180, Ra = 0.5
Eb = V – (Ia*Ra)
Ia=40/0.5=80 A

Questions-24.The magnetic flux density on the surface of an iron face is 1.5 T,Which is typical saturation value of ferromagnetic material.Find the force density on the iron face.

A.0.59×10⁶ N/m²
B.0.89×10⁶ N/m
C.0.59×10⁶ N/m
D.0.89×10⁶ N/m²

Solution-Force density=1/2 B²/µ
=1/2(1.5)²/ 4π×10⁻⁷
=0.89×10⁶ N/m²

Click Here To Download DFCCIL Executive EE 2018 Solved Questions Part One PDF

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