SSC JE Electrical Previous Year Question Paper 2018-SET 6

Ques. 31. Determine the peak value (in A) of the current used by a motor rated at 220 V, 30 A.

  1. 21.2
  2. 30
  3. 42.4
  4. 51.9

The r.m.s. value or effective value or virtual value of alternating current is 0.707 times the peak value of alternating current or voltage for the half-cycle as well as for full cycle.

IRMS = 0.707 IPeak

Ipeak = IRMS ⁄ 0.707 

Ipeak = 30 ⁄ 0.707 = 42.4 Ampere

Ipeak = 42.4 Ampere

 

Ques.32. What will be the frequency (in Hz) of a sinusoidal wave, when the time period is 20 ms?

  1. 50
  2. 60
  3. 40
  4. 30

As we know that the frequency is inversely proportional to the time period

F = 1/t = F = 1 ⁄ 20 × 10−3

F = 50 Hz

 

Ques.33. Determine the instantaneous value of the alternating current (in A), which is represented by i(t) = 10 sin(25t − 15)A, if the value of t is 3?

  1. 9.36
  2. 7.87
  3. 8.66
  4. 4.42

Let the expression be as follows:-

I(t) = Im sin(ωt ± φ)

Where φ represent the concerned phase-shift

Im = Maximum value of the current

Time t = 3 sec

I(t) = 10 sin(25t − 15)A

I(t) = 10 sin(25 × 3 − 15)

I(t) = 10sin(75 − 15)

I(t) = 10sin(60)

I(t) = 8.66 A

 

Ques.34. At resonance frequency, the impedance of a series resonant circuit is

  1. Zero
  2. Maximum
  3. Infinite
  4. Minimum

In the RLC series circuit the total impedance of the series LCR circuit is given as

Z2 = R2 +  (X1 – X2)2

where X1 is inductive reactance

and X2 is capacitive reactance.

At a particular frequency (resonant frequency), we find that X1=Xbecause resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase. Therefore, the phase angle between voltage and current is zero and the power factor is unity.

Thus, at the resonant frequency, the net reactance is zero because of X1=X2. The circuit impedance Z becomes minimum and is equal to the resistance R. “Since the impedance is minimum, the current will be maximum”.

Hence electrical resonance is said to take place in a series LCR circuit when the circuit allows maximum current for a given frequency of alternating supply at which capacitive reactance becomes equal to the inductive reactance.

 

Ques.35. Determine the value of the phase angle (in degrees) in a series RC circuit having a resistance of 20 ohms and the capacitive reactance of 11.56 ohms when supplied by a frequency of 50 Hz.

  1. 60
  2. 30
  3. 90
  4. 45

The phase angle of series RC circuit is given as

φ = tan−1XC/R

φ = tan−111.56/20

φ = tan−10.57

φ = 30°

 

Ques.36. A 3-phase delta connected system is supplied by a line voltage of 200 V. The value of phase current is 30 A. What is the total power consumed (in W) by the system, if the current lags the voltage by 30 degrees?

  1. 16.32
  2. 14.45
  3. 15.59
  4. 16.69

The power consumed by the three-phase circuit is

P = 3 Vph Iph cosφph

In delta connected system

Line voltage is equal to the phase voltage i.e

VL = VPH = 200 V

Phase current IPH = 30 A

phase angle φ = 30°

∴ P = 3 × 200 × 30 × cos30°

P = 15.58 watts

 

Ques.37. What will be the capacitive susceptance (in Siemens) of the circuit given below?

ques.37

  1. 0.126
  2. 1.26
  3. 12.6
  4. 126.1

Susceptance (B) is an expression of the ease with which alternating current ( AC ) passes through capacitance or an Inductance.

Capacitive Susceptance BC is given as

BC = 1/XC Siemens

Capacitive reactance XC = 1/ωC = 1/2πfC

XC = 1/2 × π × 50 × 0.4 × 10−3

XC = 7.96Ω

BC = 1/XC = 1/7.96

BC = 0.126 Siemens

 

Ques.38. Determine the reactive power (in kVAR) of a 3-phase delta connected system having a line voltage of 400 V, line current of 40 A and the phase difference between the voltage and current is 45 degrees.

  1. 14.4
  2. 15.8
  3. 18.7
  4. 19.6

Given

Line Voltage VL = 400 V

Line current IL = 40A

Phase angle φ = 45

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

Q = √3 × VL × IL × sinφ

Q = √3 × 400 × 40 × sin 45°

Q = √3 × 400 × 40 × 0.707

Q = 19592.9 VAR

or

Q = 19.59 KVAR

 

Ques.39. Determine the quality factor of a series RLC circuit, when the value of capacitance is 0.02 mF, the value of inductance is 8 mH and the value of resistance is 40 ohms

  1. 0.5
  2. 0.4
  3. 0.8
  4. 1.3

Given

Resistance R = 40 Ω

Inductance L = 8mH = 8 × 10−3 H

Capacitance C = 0.02 mF = 0.02 × 10−3 H

The Quality factor of the RLC circuit is given as

$\begin{array}{l}Q = \dfrac{1}{R}\sqrt {\dfrac{L}{C}} \\\\Q = \dfrac{1}{{30}}\sqrt {\dfrac{{27 \times {{10}^{ – 3}}}}{{0.03 \times {{10}^{ – 3}}}}} \\\\Q = \dfrac{1}{{40}}\sqrt {400} \\\\Q = 0.5\end{array}$

 

Ques.40. What will be the time (in seconds) taken by the inductor reach its maximum steady-state value in a series RL circuit, when the value of the inductance is 0.8 H and the value of the resistance is 10 Ohms?

  1. 0.2
  2. 0.6
  3. 0.4
  4. 0.8

The time taken by the RL circuit to reach steady state is 5 times constant of RL circuit i.e

Total time = 5τ

Also, the time constant of the RL circuit is given as τ = R/L

Total time = 5 × R/L

= (5 × 0.8) ⁄ 10

= 0.4 sec

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