HPCL Assistant Maintenance Technician 2019 Solved Question Paper Part Two

HPCL Assistant Maintenance Technician 2019

Attempt Free HPCL Assistant Maintenance Technician 2019 Here. Read The Important Electrical MCQ From Below.

HPCL Assistant Maintenance Technician Electrical 2019 Solved Question Paper Part Two

Here we are providing HPCL assistant maintenance technician electrical solved question paper part one which exam held on 24th April 2019.It is useful for upcoming RRB JE CT 2,UPSSC JE and others exam

Candidate can download full official question paper and part one solution PDF from post end.

Question-26.In a star connected three-phase balanced system, if the line current is 125A .then the phase current is

A) 250A

B) 72.16A

C) 216.5A

D) 125A

Answer- D) 125 A

Solution-In star connection the line current is equal to phase current i.e IL=Iph ,so phase current =125A

Question-27.The mutual inductance between two unity-coupled coils of 12H and 12H is

A) 12 H

B) 1H

C) 144H

D) 24H

Answer- A) 12 H

Solution-Given, K=1, L1=12H And L2=12H

So, M=k√L₁×L₂=1×√12×12 =12H

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Question-28.If the magnetic flux linking a conductor changes ,then it induces

A) emf

B) both, current and emf

C) neither emf nor current

D) current

Answer- A) emf

Solution-According to farady’s first law of electromagnetic induction when changing nature of flux linking with any conductor ,then emf is induces.

Question-29.The flow of electric current in a metallic conductor is only due to:

A) neutral

B) positive ions and electrons

C) positive ions

D) electrons

Answer- D) electrons

Solution-

The flow of current in a metallic conductor is due to- Electron

The flow of current in a semi-conductor is due to-electron and holes

The flow of current in a gases is due to-positive and negative ions

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Question-30.Damper winding is used in a synchronous machine to

A) reduce temperature rise

B) improve pf

C) improve efficiency

D) eliminate hunting

Answer- D) eliminate hunting

Solution-Due to sudden change in load in a synchronous motor hunting is produce.For eliminating hunting effect in a synchronous motor damper winding is used.

Question-31.To operate 6V bulb in a 240V supply ,the no of bulbs required is

A) 32

B) 42

C) 40

D) 30

Answer- C) 40

Solution-The no of bulbs required is=240/6=40.

Question-32.A transformer has 300W as iron loss at full load.The iron loss at half full load is:

A) 150W

B) 0W

C) 300W

D) 600W

Answer- C) 300W

Solution-In a transformer the iron loss is constant at any load condition.

So iron loss at full load=iron loss at half full load=300W.

HPCL Assistant Maintenance Technician 2019

Question-33.The field poles and armature of a DC machine is laminated to

A) reduce the weight of the machine

B) reduce armature reaction

C) reduce eddy currents

D) decrease the speed

Answer- C) reduce eddy currents

Solution-The field poles and armature of a DC machine laminated to reduce eddy currents.

Question-34.The synchronous speed of a three-phase induction motor having 10 poles , 50 Hz.

A) 1500 rpm

B) 600 rpm

C) 1200 rpm

D) 6000 rpm

Answer- B) 600 rpm

Solution-Ns=120×f/P=120×50/10=600 rpm

Question-35.If the full scale current of a meter is 50µA, then its sensitivity is-

A) 2000 Ω/V

B) 20000 Ω/V

C) 0 Ω/V

D) 200Ω/V

Answer- B) 20000 Ω/V

Solution-S=1/full scale deflection current =1/50µA=20000 Ω/V

Question-36.The mechanical power developed by the synchronous motor is independent of:

A) Torque angle

B) applied stator voltage

C) field excitation

D) speed

Answer- D) speed

Solution-The mechanical power developed by the sysnchronous motor is independent of speed-

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P=VE/Xs sinδ ,from this equation we can see that mechanical power developed by the synchronous motor is dependent on

1.torque angle

2.field excitation

3.applied stator voltage

Question-37.The power factor of an AC circuit can be expressed as

A) R/X_L

B) Z/R

C) X_L/R

D) R/Z

Answer- D) R/Z

Solution-Power factor of an AC circuit can be expressed as following way-

1.R/Z

2.COS

3.active power/apparent power

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Question-38.Which of the following is used to measure high values of alternating current with a dynamometer

A) Potential transformer

B) Shunt

C) Multiplier

D)Current transformer

Answer- D) Current transformer

Solution-Current transformer is a device which is used to measure high value of alternating current with a dynamometer. Here primary winding of current transformer is directly connected with current carrying conductor and secondary winding is connected to measuring instrument ( dynamometer).

HPCL Assistant Maintenance Technician 2019

Question-39.The form factor of a sine wave is

A)1.11

B)1.414

C)2

D)1.5

Answer- A) 1.11

Solution-The form factor of a sine wave=r.m.s value /average value=0.707Vm/0.637Vm=1.11

Question-40.A current of 3 A through a coil sets up flux linkage of 15Wb-turn.The induction of the coil is

A)15H

B)5H

C)1H

D)3H

Answer- B) 5H

Solution-

L=Nφ/i=15/3=5H

Question-41.An electric heater is marked with 1000W,200V .The resistance of the coil is

A)40 Ω

B)400 Ω

C)4 Ω

D)5Ω

Answer- A) 40 Ω

Solution-

R=V^²/P=200^²/1000=40 Ω

HPCL Assistant Maintenance Technician 2019

Question-42.The temperature rise of a transformer is directly proportional to

A) leakage reactance

B) reactive power

C) material used

D) apparent power

Answer- D) apparent power

Solution-The temperature rise of a transformer is directly proportional volt-ampere(VI).Because losses of transformer is always given in Kva or apparent power

Question-43.The contact resistance of the circuit breaker should be

A) 20 µΩ

B) 4 Ω

C) 20 kΩ

D) 40Ω

Answer- A) 20 µΩ

Solution-The contact resistance of the circuit breaker should be very less it is 20 µΩ.

HPCL Assistant Maintenance Technician 2019

Question-44.The open circuit test on transformer is conducted on

A) HV winding

B) LV winding

C) both,LV and HV winding

D) either LV or HV winding

Answer- B) LV winding

Solution-The open circuit test of a transformer is conducted on no load condition ,it is used to find iron loss at no load, in open circuit test the primary winding of transformer is LV and all instrument should be placed on this side and secondary winding is HV and it is open circuit .

HPCL Assistant Maintenance Technician 2019

Question-45.Which of the following devices will obey Ohm’s law?

A) Incandescent bulb

B) Rectifier

C) Arc lamps

D) Zener diode

Answer- A) Incandescent bulb

Solution-only here incandescent bulb has unity power factor so it obey Ohm’s law

Question-46.A wattmeter measures ………………power.

A) reactive

B) apparent

C) both,real and reactive