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
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
Answer- A) 12 H
Solution-Given, K=1, L1=12H And L2=12H
So, M=k√L₁×L₂=1×√12×12 =12H
Question-28.If the magnetic flux linking a conductor changes ,then it induces
B) both, current and emf
C) neither emf nor 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:
B) positive ions and electrons
C) positive ions
Answer- D) electrons
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
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
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:
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.
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
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
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
Answer- D) speed
Solution-The mechanical power developed by the sysnchronous motor is independent of speed-
P=VE/Xs sinδ ,from this equation we can see that mechanical power developed by the synchronous motor is dependent on
3.applied stator voltage
Question-37.The power factor of an AC circuit can be expressed as
Answer- D) R/Z
Solution-Power factor of an AC circuit can be expressed as following way-
3.active power/apparent power
Question-38.Which of the following is used to measure high values of alternating current with a dynamometer
A) Potential 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).
Question-39.The form factor of a sine wave is
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
Answer- B) 5H
Question-41.An electric heater is marked with 1000W,200V .The resistance of the coil is
Answer- A) 40 Ω
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
Answer- A) 20 µΩ
Solution-The contact resistance of the circuit breaker should be very less it is 20 µΩ.
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 .
Question-45.Which of the following devices will obey Ohm’s law?
A) Incandescent bulb
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.
C) both,real and reactive
Answer- D) real
Solution-A wattmeter measures always average active power
Question-47.The unit of inductance is
Answer- B) henry
Solution-The unit of inductance is henry.It is represented by ‘H’.
Question-48.The metric unit of illumination is
Answer- D) lux
Solution-The unit of illumination is=Lux or 1 lux= lumen/m²
Question-49.A 50 Hz sysnchronous motor runs at a speed of 500 rpm .The number of salient poles on the rotor is
Answer- D) lux
Solution-Given, f=50Hz , Ns=500 rpm ,p=?
Question-50.If the applied voltage of a DC machine is 220V ,then back emf for Maximum power developed is
Solution-In maximum power developed condition, Eb=V/2=220/2=115V.
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