SSC JE 2014 Electrical question paper with solution(Morning-Shift)

Ques 31. Bundled conductor in EHV transmission system provides

  1. Increase corona loss
  2. Increased line reactance
  3. Reduced Line capacitance
  4. Reduced voltage gradient

Advantage of Bundle Conductor

  • Bundled conductors per phase reduce the voltage gradient in the vicinity of the line. Thus reduces the possibility of the corona discharge. (Corona effect will be observed when the air medium present between the phases charged up and start to ionize and acts as a conducting medium. This is avoided by employing bundled conductors).
  • Bundled conductor lines will have higher capacitance to neutral in comparison with single lines. Thus they will have higher charging currents which help in improving the power factor.
  • Bundled conductor lines will have higher capacitance and lower inductance than ordinary lines they will have higher Surge Impedance Loading  (Z=(L/C)1/2). Higher Surge Impedance Loading (SIL) will have higher maximum power transfer capability. 
  • With an increase in self GMD or GMR inductance per phase will be reduced compared to single conductor line. This results in lesser reactance per phase compared to an ordinary single line. Hence lesser loss due to reactance drop.

 

Ques 32. Welding is injurious to eye because of

(i) Infrared radiation

(ii) Ultraviolet radiation

Among the above two choose the correct one from the following choices

  1. Both are wrong
  2. (i) alone is correct
  3. (ii) alone is correct
  4. Both are right

Welding arcs give off radiation over a broad range of wavelengths – from 200 nm (nanometers) to 1,400 nm (or 0.2 to 1.4 µm, micrometers). This includes ultraviolet (UV) radiation (200 to 400 nm), visible light (400 to 700 nm), and infrared (IR) radiation (700 to 1,400 nm).

An injury like flash burn occurs when you are exposed to bright ultraviolet (UV) light. It can be caused by all types of UV light, but welding torches are the most common source. That’s why it is sometimes called ‘welder’s flash’ or ‘arc eye

 

Ques 33. The rated speed of a given D.C. shunt motor is 1050 rpm. To run this machine at 1200 r.p.m. the following speed control scheme will be used

  1. Varying frequency
  2. Armature circuit resistance control
  3. Field resistance control
  4. Ward-Leonard Control

In field resistance control, the lower the field current in a shunt (or separately excited) dc motor, the faster it turns and the higher the field current, the slower it turns.

Since an increase in field current causes a decrease in speed, there is always a minimum achievable speed by field circuit control. This minimum speed occurs when the motor’s field circuit has the maximum permissible current flowing through it.

If a motor is operating at its rated terminal voltage, power, and field current, then it will be running at rated speed, also known as base speed.

Field resistance control can control the speed of the motor for speeds above base speed but not for speeds below the base speed.

To achieve a speed slower than base speed by field circuit control would require excessive field current, possibly- burning up the field windings.

 

Ques 34. After closing the switch ‘s’ at t = 0, the current i(t) at any instant ‘t’ in the network shown in the figure:

Numerical 33

  1. 10 – 10 e-100t
  2. 10 + 10 e100t
  3. 10 – 10 e100t
  4. 10 + 10 e-100t

Time constant τ  in series LR circuit is

τ =L/R = 0.01/1 = 0.01

Under steady state condition, L is short-circuited then

Io = V/R =10A

Now the instantaneous current i(t) is

solution 34

 

Ques 35. To increase the range of a.c ammeter you would use:

  1.  A condenser across the meter
  2.  Current transformer
  3.  A potential transformer
  4. An inductance across the meter

Instrument transformers are used in conjunction with an ammeter and voltmeter to extend the range of meters. In dc circuit shunt and multipliers are used to extend the range of measuring instruments. The shunt is used to extend the range of the ammeter whereas the multiplier is used to extend the range of voltmeters. But in case of AC circuits instruments, it is not convenient to use shunt and multiplier to extend the range of the ammeter and voltmeter.

The current transformer is used in conjunction with the current measuring device (such as an ammeter, wattmeter, energy meter, etc.). The current transformer is used with low-range ammeters to measure currents in a high voltage alternating current circuit. Its primary winding consists of one or more turns of thick wire connected in series with the line whose current is to be measured.

 

Ques 36. The voltage across 5-H inductor is

Numerical 36

Find the energy stored at t = 5 s. Assume zero initial current

  1. 312.5 kJ
  2. 0.625 kJ
  3. 3.125 kJ
  4. 156 .25 kJ

Energy stored in inductor “E” = 1/2(LI2)

Voltage across inductor E(l) = L di(t)/dt

or  i(t) = 1/L di(t)/dt

solution 36

= 30/5 x 125/3 = 250A

Energy stored in an inductor = (5 x 2502)/2 = 156.25 kJ

 

Ques 37. The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter with 1,000 turns of wire carrying current of 10 A is

  1. 1.15 J
  2. 0.015 J
  3. 0.15 J
  4. 0.5 J

Inductance in solenoid is given as

L =μoN2A/l

Where μο =vacuum permeability = 4π x 10-7 wb/m

N = Number of turns

A = area of cross-section

l = length of the solenoid

L =(μoN2πr2)/l

solution 37

Energy stored =  LI2/2

E = (29.43 x 10-4 x 102) /2

=0.14715 J ≅ 0.15 J

 

Ques 38. In a power plant if the maximum demand on the plant is equal to the plant capacity then

  1. Load factor will be nearly 60%
  2. Plant reserve capacity will be zero
  3. Diversity Factor will be unity
  4. Load factor will be unity

The plant capacity factor is the measure of the reserve capacity of the plant. A power station must be designed in such a way that it has some reserve capacity for meeting the increased load demand in the future. Therefore the installed capacity of the plant is always greater than the maximum demand on plants.

Reserve capacity = Plant capacity – Max. Demand

If the plant capacity is equal to Max. Demand then the Reserve capacity will be zero.

 

Ques 39. The least expensive fractional horsepower motor is ________ 

  1. Ac series motor
  2. Shaded pole Motor
  3. Capacitor start motor
  4. Split phase motor

Shaded pole motor is small in size and less expensive to manufacture. These motors are also called as fractional KW motors because most of these motors are constructed in fractional kilo-watt capacity.

 

Ques 40. Which of the following condition is NOT mandatory for alternators working in parallel?

  1. The alternators must have the same phase sequence
  2. The terminal voltage of each machine must be the same
  3. The machines must have equal kVA ratings
  4. The alternators must operate at the same frequency

There are five conditions that must be met before when two alternators running in parallel.

    1. Equal line voltage
    2. Same frequency
    3. Same phase sequence
    4. Same phase angle
    5. Same waveform

We can use 2 alternators of 6 MVA and 4 MVA each instead of using single 10 MVA alternator because it is economical than using a single alternator of the same rating.

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