# SAIL OCTT Electrical 2019 Solved Question Paper Part One

SAIL Rourkela Operator-Cum-Technician Trainee(OCTT) Electrical 2019 Solved Question Paper PDF Part One

Questions-1.The purpose of ground wire placed at the top of towers in transmission lines is that

(A) It protects the human body from electric shock

(B) It protects the line against short circuits

(C) It protects the transformer against short circuits

(D) It protects the line against over voltages

Solution-The ground wire placed in top of the transmission line to protect the line against

A) Over voltages

B) Indirect lightning strokes

Questions-2.Sag is independent of

(A) Length of span

(B) Line voltage

(C) Tension in the conductor

(D) None of these

Solution-The sag of a transmission line depends on following factor

Sag = WtL²/2T

Here ,W=weight of the conductor L=length of the conductor T=tension of the conductor

Questions-3.A mho relay is used for protection of

(A) Protection of a transformer against external faults.

(B) Long Transmission lines

(C) Protection of a transformer against all the internal faults and external faults.

(D) Medium length lines

Solution-A mho relay is a voltage controlled directional relay.It is least affected by switching resistance so use for long transmission line.It is also known as the admittance relay.

Questions-4.In coal-fired thermal power stations, what are the electrostatic precipitators used for

(A) To remove dust particles settling on the bus bar conductors in the station yard.

(B) To condense steam by electrostatic means.

(C) To keep the air heaters clean.

(D) To collect the dust particles from the flue gases.

Solution-Electrostatic precipitatotrs is a device which is used to collect the dust particles from the flue gases.this is installed at top of the chimney to collect dust before leaving the flue gases in atmosphere.

Question-5.In shunt generator inter-pole winding carries

(B) Armature current

(C) Shunt field current

(D) None of these

Solution-In shunt generator inter-pole winding are connected in series with armature,Which carry armature current.

Question-6.In the circuit shown in figure, if i= 2A, then the value of the battery voltage V will be (A) 1V

(B) 0.5V

(C) 2V

(D) 0.25V

Solution- At node 2

(V1-V)/0.5+(V1/1)+(V1/1)=0

-2+2V1=0

V1=1

At node 1

(V-V1)/0.5=2

V-1/0.5=2  (From above V1=1)

V=1+1=2

Hence net voltage should be =2V.

Questions-7.Which of the following combination of 3-phase transformers can be successfully operated in parallel

(A) Δ-Υ and Δ-Υ

(B) Υ-Υ and Δ-Υ

(C) Δ-Δ and Δ-Υ

(D) Υ-Δ and Δ-Δ

Solution-Here only Δ-Υ and Δ-Υ this group is only connected in same vector group.The following same group of transformers can be operated in parallel operation Questions-8.In a circuit containing R, L and C power loss can take place in

(A) R only

(B) L only

(C) C only

(D) All the above

Solution-In a circuit containing R,L and C power loss can take place in only in R due to I²R losses.

Questions-9.Find the node voltage VA (A) 6 V

(B) 12 V

(C) 4.28 V

(D) 3 V

Solution-

Applying the KCL at Node VA

(VA-12/49)+(VA/24)+(VA-6/80)=0

After solve we get 0.0746 VA=0.319

VA=0.319/0.0746=4.25 V

Questions-10.A single binary digit is called

(A) Byte

(B) Bit

(C) Data

(D) Logic

Solution-A bit is the smallest unit of data in a computer. A bit has a single binary value, either 1 or 0.

Questions-11.A spring controlled moving iron voltmeter draws a current of 1mA for full scale value of 100 V. If it draws a current of 0.5 mA, the meter reading is

(A) 25 V

(B) 50 V

(C) 100 V

(D) None of these

Solution-In case of moving iron instruments.

Therefore,If it draws a current of 0.5 mA, the meter reading is 25V.

Questions-12.Magnetic core losses arises due to

(A) Eddy currents

(B) Hysteresis losses

(C) Magnetic saturation

(D) All of these

Solution-Magnetic core losses arises due to following

1.Eddy current losses

2.Hysteresis loss

3.Magnetic saturation

4.Core loss

5.Magnetic vibration

Questions-13.The earthing transformer is used

(A) To avoid the harmonics in the transformers

(B) To provide artificial neutral earthing where the neutral points of the 3–phase system are not accessible

(C) To improve the current capacity of neutral

(D) To increase the neutral current to ground in case of fault

Solution-In cases where the neutral point of three phase system is not accessible like the system connected to the delta connected side of a electrical power transformer, an artificial neutral point may be created with help of a zigzag connected earthing transformer.

Questions-14.An ac current is given as

i = 10+10 sin 314t

The average and rms values of the current, respectively, are.

(A) 10A, 12.2A

(B) 10A, 17.07A

(C) 16.36A, 12.2A

(D) 16.36A, 17.7A

Solution- In this given equation i = 10+10 sin 314t

Here average value=10A

And R.M.S value =(√10)²+(10/√2)²=10+2.2=12.2A

Questions-15.A ceiling fan uses

(A) Split phase motor

(B) Capacitor start capacitor run motor

(C) Capacitor start motor

(D) Universal motor

Solution-Ceiling fan uses capacitors start capacitors run motor or permanent split capacitors motors.

Questions-16.In a Kelvin’s double bridge, two set of readings are taken when measuring a low resistance, one with the current in one direction and the other with direction of current reversed. This is done to

(A) Eliminate the effect of contact resistance

(B) Eliminate the effect of resistance of leads

(C) Correct for changes in battery voltage

(D) Eliminate the effect of thermo-electric effects

Solution-To Eliminate the effect of thermo-electric effects in a kelvin’s double bridge,two set of readings are taken when measuring a low resistance,one with the current in one direction and the other with direction of current reversed.

Questions-17.Megger is an instrument for

(A) Measuring current

(B) Measuring voltage

(C) Testing insulation

(D) Measuring power

Solution-Megger is an instrument which is used for measuring insulation resistance or testing insulation.

Questions-18.The material used for fuse wire should have the following characteristics

(A) Low melting point, high conductivity

(B) High melting point, low conductivity

(C) Highly malleable and coercive

(D) High resistance, low melting point

Solution-The material used for fuse wire should have the following properties

1.Low melting point

2.High conductivity

3.Low resistance

4.Low deteriorate due to oxidation

5.Low resistivity

Questions-19.A Kaplan turbine is

(A) Inward flow, impulse turbine

(B) Outward flow, reaction turbine

(C) A high lead mixed flow turbine

(D) Low head axial flow turbine

Solution-Kaplan turbine-low head axial flow turbine

Pelton wheel-impulsive type and high head

Questions-20.Kirchoff’s law is applicable to

(A) Passive networks only

(B) A.c. netwoks only

(C) d.c. netwoks only

(D) Both a.c. and d.c. circuits

Solution-Kirchhoff’s current law (1st Law)-The current flowing into a node (or a junction) must be equal to current flowing out of it.

Kirchhoff’s voltage law (2nd Law)-The sum of all voltages around any closed loop in a circuit must equal zero

These law is applicable for both a,c and dc circuits.

Questions-21.The nominal ratio of a current transformer is

(A) Primary winding current / secondary winding current

(B) Number of Primary winding turns / Number of secondary winding turns

(C) Rated Primary winding current / Rated secondary winding current

(D) Number of Secondary winding turns / Number of Primary winding turns

Solution-Nominal ratio of current transformer is the ratio of rated primary winding current to the rated secondary winding current.

Questions-22.Lap winding is most suitable for

(A) Low voltage, low current machine

(B) High voltage, high current machine

(C) Low voltage, high current machine

(D) High voltage, low current machine

Solution-Lap winding is used for low voltage and high current machine only

(A) Fleming’s rule

(B) Thumb’s rule

(D) Lenz’s law

Solution-The induced emf due to negative sign indicate lenz’s law.According to this law it state that the induced emf always opposes the cause that produced it.

Questions-24.When two transformers of different KVA ratings are connected in parallel, they divide the total load in proportion to their respective KVA ratings only when their

(A) KVA ratings are identical

(B) Efficiencies are equal

(C) Per unit impedance’s are equal

(D) Equivalent impedance’s are equal

Solution-The total load of two transformer are divided in proportion to their respective KVA ratings only when their per unit impedance are equal. And it is directly proportional to each other.

Questions-25.The ratio RMS value/ Average value is the

(A) Peak value

(B) Peak factor

(C) Mean value

(D) Form factor

Solution-Form factor is the ratio of RMS value to the average value. For full sine wave form factor should be 1.11 and for have sine wave it is 1.57

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