SSC JE electrical 2019 question paper with solution SET-1

Ques.71. An electric motor in which both the rotor and stator fields rotate with the same speed is called

  1. Synchronous motor
  2. DC motor
  3. Universal motor
  4. Induction motor

Answer.1. Synchronous Motor


In the synchronous motor, the stator and rotor rotate at the same speed.


Ques.72. Which of the following lamps employs a bimetallic strip?

  1. Fluorescent lamp
  2. Mercury vapor lamp
  3. Incandescent lamp
  4. Sodium vapor lamp

Answer.1. Fluorescent lamp


A fluorescent lamp is a hot cathode low-pressure mercury-vapor lamp.

Construction of Fluorescent lamp

  • It is a low-pressure mercury-vapor lamp. Due to low pressure, the lamp is in the form of a long tube, coated inside with phosphor.
  • The tube contains a small amount of mercury and a small quantity of argon gas at a pressure of 2.5 mm of mercury.
  • At each end of the tube, the electrodes are of spiral form made of tungsten and coated with an electron-emitting material.
  • A choke is connected in series with the tube filament. ft provides a voltage impulse for starting the lamp and acts as a ballast later on when the lamp is running.
  • The filament is connected to a starter switch which is a small cathode glow lamp with a bimetal strip at the electrodes.

Fluorescent lamp

Working of Fluorescent lamp

  • At the time of starting, when both the lamp and the glow starters are cold, the mercury is in the form of globules.
  • When supply is switched on, the glow starter terminals are open-circuited and full supply voltage appeared across these terminals, due to the low resistance of electrodes and choke coil.
  • The small quantity of argon gas gets ionized, which establishes an arc with a starting glow.
  • This glow warms up the bimetallic strip, and the strip gets bend thus glow starts gets short-circuited. Hence, the two electrodes come in series and are connected across the supply voltage.
  • Now, the two electrodes get heated and start emitting electrons due to the flow of current through them. These electrons collide with the argon atoms present in the long tube discharge that takes place through the argon gas.
  • So, in the beginning, the lamp starts conduction with argon gas as the temperature increases, the mercury changes into vapor form and takes over the conduction of current.


Ques.73. Let T be the net torque developed by the rotor runs at ω rad/s. What is the mechanical power developed?

  1. Pmech = (2πωT)/60
  2. Pmech = (ωT)/60
  3. Pmech = ωT
  4. Pmech = (2πωT)

Answer.3. pmech = ωT


By the term, torque is meant the turning or twisting moment of a force about an axis. It is measured by the product of the force and the radius at which this force acts. Consider a pulley of radius r meter acted upon by a circumferential force of F Newton which causes it to rotate at N r.p.m.


The angular speed of the pulley is

ω = 2πN/60 rad/sec

Work is done by this force in one revolution

= Force × distance = F × 2πR Joule

The power developed = Work Done/Time

= (F × 2πR)/60/N

= (F × R) × (2πN)/60

The power developed = T × ω watt or P = T ω Watt

pmech = (ωT)


Ques.74. If an alternator is operating at unity power factor, then its terminal voltage is

  1. Equal to the induced EMF, with zero voltage regulation
  2. Less than the induced EMF, with negative regulation
  3. Less than the induced EMF, with positive voltage regulation
  4. Greater than the induced EMF, with negative voltage regulation

Answer.3. Less than the induced EMF, with positive voltage regulation


Unity Power Factor Load (Cross-Magnetizing)

Consider a purely resistive load connected to the alternator, having a unity power factor. As induced e.m.f Eph drives a current Iph and load power factor is unity, Ephand Iph is in phase with each other.

If φf is the main flux produced by the field winding responsible for producing Ephthen Eph lags φf by 90°.

Now current through armature Ia produces the armature flux say to,, So flux φa, and la are always in the same direction.

Unity Power factor

It can be seen from the phasor diagram that there exists a phase difference of 90°between the armature flux and the main flux. From the waveforms, it can be seen that the two fluxes oppose each other on the left half of each pole while assisting each other on the right half of each pole. Hence average flux in the air gap remains constant but its distribution gets distorted. Hence such distorting effect of armature reaction under unity p.f. the condition of the load is called the cross magnetizing effect of armature reaction. Due to such distortion of the flux, there is a small drop in the terminal voltage of the alternator

Voltage Regulation:-

The difference in voltage at the receiving end of a transmission line between conditions of no-load and full-load is called voltage regulation and is expressed as a percentage of the receiving end voltage. Mathematically,

% Voltage regulation =  (Vs − VR) ⁄ VR × 100

  • The value of the regulation not only depends on the load current but also on the power factor of the load.
  • For lagging and unity p.f. conditions there is always a drop in the terminal voltage hence regulation values are always positive.
  • While for leading capacitive load conditions, the terminal voltage increases as load current increases. Hence the regulation is negative in such cases.


Ques.75. In the circuit shown, find the equivalent resistance between A and B.

In the circuit shown, find the equivalent resistance between A and B.

  1. 75 ohm
  2. 100 ohm
  3. 150 ohm
  4. 50 ohm

Answer.1. 75 ohm


The resistance R3 &  R4 are connected in series

= (30 + 70) = 100Ω

In the circuit shown, find the equivalent resistance between A and B.

Now three resistor i.e 100Ω, 100Ω & 50Ω is connected in parallel

1/Rp = 1/100 + 1/100 + 1/50

Rp = 100/4 = 25 Ω

In the circuit shown, find the equivalent resistance between A and B.

Resistance 50Ω and 25Ω are connected in series.

Rtotal = (50 + 25)Ω

Rtotal = 75Ω


Ques.76. The below symbol which is used in single line diagrams represents


  1. Potential transformer
  2. Power transformer
  3. Current transformer
  4. Circuit breaker

Answer.1. Potential transformer


The single-line diagram (also called one-line diagram) is a pictorial representation of connections of various components of the power system, using standard symbols for components and lines for the transmission lines connecting them.

The symbol given in the question is of Potential transformer.

single-line-diagram of power system


Ques.77. P = ρQHg is the water power equation. Head ‘H’ is measured in meter ‘g’ gravity constant is measured in meter/second square, then the measuring unit of ‘Q’, the flow rate of water is:

  1. Meter per hour
  2. Square meter per second
  3. Meter per second
  4. Cubic meter per second

Answer.4. Cubic meter per second


P = ρ QHg is the water power equation.


H = heead = meter

g = gravity =m/s2

ρ =density  = (kg/m3)

P = power = watt = N-m/sec

Q = water flow rate = m3/sec


Ques.78.  In a capacitor start single phase induction motor the current in the:

  1. The main winding leads the voltage
  2. Supply lines lead the voltage
  3. Starting winding leads the voltage
  4. Starting winding lags the voltage

Answer.4. Starting winding leads the voltage


In the capacitor start, single-phase induction motor the capacitor is connected in series with the starting auxiliary winding. In this manner, the current in the starting winding may be made to lead the line voltage. Since the running winding current lags the line voltage, the phase displacement between the two currents is made to approximately 90° on starting.

Placing the capacitor in the auxiliary winding circuit to produce a greater phase difference between the current in the main and the auxiliary windings. Due to greater phase difference capacitor Start motors have very high starting torque for a single-phase AC motor.


Ques.79. An electric heater is connected across 230V and it draws a current of 2A. Then the resistance offered by the heater is?

  1. 115 ohm
  2. 29 ohm
  3. 290 ohm
  4. 11.5 ohm

Answer.1. 115 ohm



Voltage V = 230 V

Current I = 2 A

Resistance R = V/I

R = 230/2 = 115 Ω


Ques.80. A coil is wound with 50 turns and a current 8A produces a flux of 200μWb. Calculate the inductance of the coil.

  1. 0.125 mH
  2. 1.25 mH
  3. 0.25 mH
  4. 2.5 mH

Answer.2. 1.25 mH


The inductance of the coil is given by the relation

L = Nφ/I


N = number of turns = 50

φ = flux = 200μWb

I = current = 8 A

L = 50 × 200 ×  10−6 ⁄ 8

L = 1.25 mH

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