SSC JE 2014 Electrical question paper with solution(Evening-shift)

Ques 91. The domestic load that has UPF is

  1. Fan
  2. Mixer
  3. Tube
  4. Filament Lamp
UPF or unity power factor: When voltage and current are in the same phase i.e displaced by zero degrees simply cos(Φ)=1. Filament lamp is highly resistive in nature, therefore, its power factor is approximately equal to 1

 

Ques 92. An industrial consumer has a daily load pattern of 2000 kW, 0.8 lag for 12 hours and 1000 kW UPF for 12 hours. The load factor is

  1. 0.5
  2. 0.75
  3. 0.6
  4. 2.0

Load factor: The ratio of average load to the maximum demand during a given period is known as the load factor.

Load factor = Average load × time / Maximum Demand × time

= 2000 × 0.8 × 12 + 1000 × 1 × 12/ 2000 × 24 = 0.65

 

Ques 93. Dielectric loss is proportional to

  1. (Frequency)1/2
  2. (frequency)
  3. (frequency)2
  4. (frequency)3
Dielectric loss is the dissipation of energy through the movement of charge in an al alternating electromagnetic field as polarisation switches direction. The loss is found to be proportional to the frequency and to the square of the applied voltage.

 

Ques 94. Which of the following applications needs frequent starting and stopping of an electric motor?

  1. Air-conditioner
  2. Lifts and Hoists
  3. Grinding mills
  4. Paper mills
Lifts and hoists need frequent starting and stopping of an electric motor.

 

Ques 95. In a CE transistor, Vcc = 12V and the zero signal collector current is 1mA. Determine the operating point when collector Load Rc is 6 KΩ

  1. 6 V, 1 mA
  2. 6 V, 2mA
  3. 12 V, 1 mA
  4. 12 V, 2 mA

Collector-emitter voltage Vce is given by

Vce = Vcc – IcRc

Vce = 12 – 6×1

Vce = 6V

Therefore operating voltage Q = (6V, 1 mA)

 

Ques 96.  An AC supply of 230 V is applied to half-wave rectifier through the transformer of turns ratio 10: 1 as shown in the figure. Determine the peak inverse voltage across the diode

Numerical96

  1. 37.6 V
  2. 32.5 V
  3. 23.0 V
  4. 14.54 V
Given V1 = 200V

N2/N1 = 1/10

V2 =?

Secondary voltage V2

V2 = V1 x N2/N1

= 230 x1/10 = 23 V

Peak Inverse voltage of diode (PIV) = √2 x 23 = 32.52 volts

 

Ques 97. The potential barrier existing across PN junction

  1.  Prevents flow of minority carriers
  2.  Prevents flow of majority carriers
  3. Prevents total recombination of holes and electrons
  4.  Prevents neutralization of acceptor and donor

The potential barrier in the PN junction is the barrier that does not allow charge flow across the junction normally. This barrier is created by the charge present in the space charge region.

When a p-type material is brought in contact with an n-type material charge flow across the junction takes place due to the concentration gradient between the two sides (n and p-type).

After some time the immobile ions are created near the contact which creates an electric field that opposes the flow of current. This resistance to the flow of charge is known as barrier potential.

 

Ques 98. The technique of adding a precise amount of time between the trigger point and the beginning of the scope sweep in a CRO is known as

  1.  Free running sweep
  2.  Delayed sweep
  3. Triggered sweep
  4.  Non-sawtooth sweep

The technique of adding a precise amount of time between the trigger point and the beginning of the scope sweep in a CRO is known as the Delayed sweep. A useful sweep range is from one second to 100 nanoseconds, with appropriate triggering and (for analog instruments) sweep delay.

A well-designed, stable trigger circuit is required for a steady display. The chief benefit of a quality oscilloscope is the quality of the trigger circuit.

 

Ques 99. In a CRO, a sinusoidal waveform of a certain frequency is displayed. The value of the quantity that can be made out by observation is

  1.  RMS value of the sine wave
  2. Average value of the sine wave
  3. Form factor of the sine wave
  4. Peak- peak value of the sine wave
CRO is voltage-dependent instruments and can be used for the measurement of the voltages at any frequency within the range of the CRO. CRO measures Peak to Peak value of the sine wave because from the peak value of the AC signal the RMS and other parameters can be determined such as distortion.

 

Ques 100.  In a Cathode Ray Tube, the focusing anode is located

  1. After accelerating anode
  2. Between pre-accelerating and accelerating anode
  3. Before pre-accelerating anode
  4. Just after electron-guns

The electron gun assembly consists of six parts i.e an indirectly heated cathode, a control, a grid surrounding the cathode, a focusing anode, an accel­erating anode, and a pre accelerating anode. And the focusing anode is located between the pre-accelerating and accelerating anode.

The pre accelerating anode and accelerating anode are at a higher potential than the focusing anode, therefore, these anodes create an electric field between them and act as a lens with a focus on the screen.

 

For SSC JE 2017  Electrical paper with complete solution Click Here

For SSC JE 2015  Electrical paper with complete solution Click Here

For SSC JE 2013 Electrical paper with complete solution Click Here

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