SSC JE Basic Electrical Questions (2009 – 2018) Solved | MES Electrical

Ques.111. What is the average value of a DC voltage having a peak value of 25 V? (SSC-2018, Set-4)

  1. 0
  2. 17.73
  3. 22.5
  4. 25

Answer.4. 25

Explanation

Practically no DC source provides an absolutely constant voltage that means that it is not alternating in nature. Hence the average value of DC voltage or current is equal to Peak or maximum value of DC voltage or current.

DC volatge

Average Voltage of DC = Maximum voltage of DC = 25 V

 

Ques.112. The waveform given in the figure below is an example of the waveform (SSC-2018, Set-4)

Sawtooth waveform

  1. Sinusoidal
  2. Rectangular
  3. Square
  4. Sawtooth

Answer.4. Sawtooth

Explanation:-

A sawtooth waveform is a special case of the triangular wave. First, it rises at a linear rate and then rapidly declines to its negative peak value. The positive ramp is of relatively long duration and has a smaller slope than the short ramp. Sawtooth waves are used to trigger the operations of electronic circuits. In television sets and oscilloscopes they are used to sweep the electron beam across the screen creating the image.

Sawtooth waveform

 

Ques.113. The inductive reactance of a circuit is 60 Ohms when it is supplied with a 50 Hz supply. What will be the value of inductive reactance (in Ohms) of the circuit, if it is supplied with a 60 Hz supply? (SSC-2018, Set-4)

  1. 72
  2. 86
  3. 94
  4. 105

Answer.1. 72

Explanation

The value of inductive reactance can be found in the formula.

Inductive reactance XL = 2πfL

Given

XL1 = 60Ω

f1 = 50 Hz

f2 = 60 Hz

XL2 =?

XL1  ⁄  XL2  =  f1 ⁄ f2

60  ⁄  XL2 = 50/60

XL2 = 72 Ω

 

Ques.114. Calculate the Impedance (in Ω) 0f the series RLC circuit given in the figure below? (SSC-2018, Set-4)

ques.34

  1. 130
  2. 120
  3. 13
  4. 15

Answer.1. 130

Explanation:-

Inductive Reactance = 2.π.f.L 

Where

f = frequency = 50 Hz

L = Inductance = 1/π 

XL = 2.π.f.L = 2 × π × 50 × 1/π

XL = 100 Ω

Capacitive Reactance =  1 ⁄ 2.π.f.C

Where C = 200/π μF = (200 × 10−6) ⁄ π

XC = 1 2.π.f.C = 1 2 × π × 50 × 200/π × 10−6 

XC = 50 Ω

The impedance of an RLC series circuit is given as

[latex]\begin{array}{l}Z = \sqrt {{R^2} + {{({X_L} – {X_C})}^2}} \\\\Z = \sqrt {{{120}^2} + {{(100 – 50)}^2}} \\\\Z = 130\Omega \end{array}[/latex]

 

Ques.115. Calculate the value of the quality factor of a series RLC circuit having resistance, inductance, and capacitance of 30 Ohms, 27 mH, and 0.03 mF respectively. (SSC-2018, Set-4)

  1. 0.86
  2. 1
  3. 2.6
  4. 3

Answer.2. 1

Explanation

Given

Resistance R = 30 Ω

Inductance L = 27mH = 27 × 10−3 H

Capacitance C = 0.03 mF = 0.03 × 10−3 H

The Quality factor of the RLC circuit is given as

[latex]\begin{array}{l}Q = \dfrac{1}{R}\sqrt {\dfrac{L}{C}} \\\\Q = \dfrac{1}{{30}}\sqrt {\dfrac{{27 \times {{10}^{ – 3}}}}{{0.03 \times {{10}^{ – 3}}}}} \\\\Q = \dfrac{1}{{30}}\sqrt {900} \\\\Q = 1\end{array}[/latex]

 

Ques.116. What is the bandwidth (in kHz) of a series RLC circuit having resistance, inductance, and capacitance of 80 Ohms, 2 mH, and 0.01 mF respectively? (SSC-2018, Set-4)

  1. 10
  2. 20
  3. 40
  4. 80

Answer.3. 40

Explanation

img.4 2

Given Resistance “R” = 80 Ω
Capacitance “C” = 0.01 mF
Inductance “L” = 2 mH = 2 × 10-3

The bandwidth of the series RLC circuit is given as

B.W = R/L

B.W = 80 ⁄ (2 × 10-3)

B.W = 40 Hz

 

Ques.117. Calculate the time taken (in seconds) by the capacitor to get fully charged in a series RC circuit having resistance and capacitance of 40 kilo-Ohms and 0.01 mF respectively. (SSC-2018, Set-4)

  1. 2
  2. 4
  3. 1
  4. 3

Answer.1. 2

Explanation

The time taken by the capacitor to get fully charged in the series RC circuit is 5 × Time constant

The time constant of an RC circuit is given as

T = RC

T = 40 × 103 × 0.01 × 10−3

T = 0.4 sec

Hence the time is taken by the capacitor to get them fully charged 

Tfull = 5 × 0.4

Tfull = 2 sec

 

Ques.118. What is the value of line voltage (in kV) of a 3-phase star connected system having a phase voltage of 3.3 kV? (SSC-2018, Set-4)

  1. 2.1
  2. 3.3
  3. 5.7
  4. 6.8

Answer.3. 5.7

Explanation

In a star connected system, the phase voltage is 1/√3 times the line voltage.and the phase current is equal to the line current of the system.

VPH = VLine /√3

VLine = VPH × √3

VLine = 3.3 × 103 × √3

VLine = 5.71 kV

 

Ques.119. A 3-phase star connected system is supplied by a line voltage of 440 V. The value of phase current is 50 A. What is the power (in kW) consumed by the system, if the current lags the voltage by 45 degrees? (SSC-2018, Set-4)

  1. 8.95
  2. 24
  3. 26.94
  4. 47

Answer.3. 26.94

Explanation

Given

Line voltage VL = 440 V

Phase current Iph = 50 A

Phase angle φ = 45°

In star connection the line current and phase current are equal i.e Iph = IL = 50A

The power in 3-phase star connection is given as

P = √3 × VL × IL × cos45°

= √3 × 440 × 50 × cos45°

P = 26940.3 watt

Power = 26.94 kW

 

Ques.120.  What is the magnitude of reactive power (in kVAR) of a balanced 3-phase delta connected system having a line voltage of 400 V and a line current of 50 A and the phase difference between the voltage and current is 53.13 degrees? (SSC-2018, Set-4)

  1. 0.2771
  2. 2.771
  3. 27.71
  4. 277.1

Answer.3. 27.71

Explanation

Given

Line Voltage VL = 400 V

Line current IL = 50A

Phase angle φ = 53.13

The reactive power of a three-phase AC circuit is given as

Q = √3 × VL × IL × sinφ

Q = √3 × 400 × 50 × sin 53.13°

Q = √3 × 400 × 50 × 0.8

Q = 27712.8 VAR

or

Q = 27.71 KVAR

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