SSC JE Electrical Previous Year Question Paper 2018-SET 5

Ques.31. Determine the total impedance (in ohms) of a series RLC circuit having a resistance of 10 Ohms, capacitive reactance of 2 ohms and the inductive reactance of 6 ohms, connected in a series across a 200 V, 50 Hz supply.

  1. 8.64
  2. 10.77
  3. 12.21
  4. 14.65

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

Z2 = R2 +  (XL – XC)2

where

R = Resistance = 10Ω

XL is inductive reactance = 6Ω

and XC is capacitive reactance = 2Ω

Ans.36

Z2 = 102 + (6 − 2)2

Z2 = 116

Z = 10.77Ω

 

Ques.32. What is the RMS value of the alternating voltage (in V) having a peak value of 1000 V?

  1. 1414
  2. 707
  3. 141.4
  4. 70.777

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.

VRMS = 0.707 VPeak

VRMS = 0.707 × 1000

VRMS = 707 volt

 

Ques.33. Calculate the quality factor of the resonant series RLC circuit, if its bandwidth is 30 Hz.

  1. 30
  2. 15
  3. 2/30
  4. 1/30✓

Resonant frequency (f0) = 1 Hz

Bandwidth (BW) = 30 Hz

Quality factor (Q) = 1/30

 

Ques.34. The line current for a balanced delta connected system is 10 A. Calculate the value of phase current (in A)

  1. 10
  2. 17.32
  3. 5.77
  4. 11

In delta connected system

Line voltage is equal to the phase voltage i.e

VL = VPH

and the line current is √3 times the phase current

IL = √3 × Iph

Iph = 10/√3

Iph = 5.77A

 

Ques.35. For the circuit shown below, the voltage (in V) across the 4mF capacitor is____

ques.35

  1. 37.3
  2. 3.76
  3. 0.373
  4. 37

Inductive Reactance = 2.π.f.L 

Where

f = frequency = 50 Hz

L = Inductance = 0.15 H 

XL = 2.π.f.L = 2 × π × 50 × 0.15

XL = 47.1 Ω

Capacitive Reactance =  1 ⁄ 2.π.f.C

Where C = 4mF = (4 × 10−3)F

XC = 1 ⁄ 2.π.f.C = 1 ⁄ 2 × π × 50 × 4 × 10−3

XC = 0.8 Ω

The impedance of an RLC series circuit is given as

$\begin{array}{l}Z = \sqrt {{R^2} + {{({X_L} – {X_C})}^2}} \\\\Z = \sqrt {{5^2} + {{(47.1 – 0.8)}^2}} \\\\Z = 46.56\Omega \end{array}$

Now the circuit current I is

I = V/Z = 220/46.56 = 4.725 Ω

Voltages across the capacitor

VC = I × XC = 4.725 × 0.8

VC = 3.77 V

 

Ques. 36. The maximum voltage induced in the coil is 200 V and the rotation angle of the coil is 45 degrees with respect to the coil. Find the instantaneous value of the sinusoidal waveform produced.

  1. 200 sin45°
  2. 200 cos45°
  3. 200/√2 sin45°
  4. 200/√2 cos45°

Sine Wave Instantaneous Values

The instantaneous are those value which is not constant. They are only momentary. They are that particular value only for an instant. 

The relationship of the sinusoidal waveform to the trigonometric sine function is useful in determining the instantaneous value of a sinusoidal voltage or current waveform at any electrical degree point. The relationship of instantaneous voltage values to the sine function is expressed mathematically by

VI = Vmax. sinθ

Where, Vmax = the maximum voltage induced in the coil 

θ = ωt, is the rotational angle of the coil with respect to time.

VI = 200 × sin45 = 200 sin45°

 

Ques.37. For a pure sinusoidal waveform, the form factor and crest factors are______ and ______ respectively.

  1. 1.414, 1.11
  2. 1.11, 1.414
  3. 11.1, 14.14
  4. 14.14, 11.1

Form and crest factors

The form and crest factors are used to compare the shape of a sinusoidal wave with that of a non-sinusoidal wave.

Form Factor

In electrical, the form factor of an alternating current waveform (signal) is the ratio of the RMS (root mean square) value to the average value (mathematical mean of absolute values of all points on the waveform).

The form factor is the ratio of R.M.S value to the Average value

R.M.S value of sine wave = 0.707

The average value of sine wave = 0.637

Therefore Form factor = 0.707/0.637

=1.109 ≅ 1.11

Crest Factor

The Crest Factor is the ratio of the Maximum value of sin wave to the RMS value of sin wave

Crest factor = Maximum value ⁄ RMS value

The maximum value of sin wave = 1

RMS value of the sin wave = 0.707

Crest factor = 1/0.707 = 1.414

 

Ques.38. A sinusoidal voltage is applied across a series RC circuit is given by 40 sinωt V. The current flowing in the circuit is 20 sin(ωt + 45°) A. Determine the value of the average power (in W)

  1. 282.84
  2. 286.64
  3. 288.04
  4. 292.24

Let the expression be as follows:-

I(t) = Im sin(ωt ± φ) = 20 sin(ωt + 45°)

Where φ represent the concerned phase-shift

Im = Maximum value of the current = 20 A or IRMS = Im/√2 = 20/√2

φ = 45°

Similarly

V(t) = Vm sin(ωt ± φ)

Vm = 40 V or VRMS = 40/√2

Average power consumed is

PAVG = VRMS. IRMS. Cosφ

PAVG = (40/√2) × (20/√2) × Cos45°

PAVG = 800/2√2 = 282.84 watt

PAVG = 282.84 watts

 

Ques.39. A constant voltage of 60 V is applied at t = 0 across a series R-L circuit is shown in the figure. Determine the current (in A) in the circuit at t = 0

Ques.39

  1. 4
  2. 3
  3. 0
  4. 2

The current flow in an inductor does not change instantaneously. Since the initial current in the inductor is zero. So at t = 0 current flow in the RL circuit is Zero.

Hence the correct answer is zero.

Why the Inductor Opposes the Sudden change in the current.

An inductor is a device that temporarily stores energy in the form of a magnetic field. It is usually a coil of wire. One of the basic properties of electromagnetism is that when you have current flowing through the wire it creates a small magnetic field around it.

The current first start to flow through the inductor a magnetic field start to expand then after some time the magnetic field becomes constant then we have some energy stored in the magnetic field.

Once a constant magnetic field is generated in the Inductor, it will not change any further. As magnetic flux = N x I (Turns × Current), Inductor will draw a constant current to maintain the magnetic field.

When the current stop flowing the magnetic field start to collapse and the magnetic energy turned back into electric energy.

So when the current flowing through the inductor changes, the magnetic field also changes in the inductor and emf (electromotive force) is induced in the inductor as per Faraday’s law of electromagnetic induction.

Because the inductance of the inductor results in a back emf, an inductor in a circuit opposes changes in the current in that circuit.

The inductor attempts to keep the current the same as it was before the change occurred. If the battery voltage in the circuit is increased so that the current rises, the inductor opposes this change and the rise is not instantaneous. If the battery voltage is decreased, the inductor causes a slow drop in the current rather than an immediate drop. Therefore, the inductor causes the circuit to be “sluggish” as it reacts to changes in the voltage.

According to Lenz’s law, the direction of the electromotive force(emf) opposes the change of current that created it.  V= -L dI/dt (rate of change of current). when t tends to zero..means i is changing instantaneously in small time t = 0 ..E will be infinite which is not possible..hence it does not change instantaneously

So the inductor opposes the sudden change of current through them.

 

Ques.40. Determine the value of the inductor (in mH) connected in parallel with a capacitance of 5 F having a quality factor 6. Assume the resonant frequency of 5 rad /s

  1. 8
  2. 7
  3. 9
  4. 8.5

The equation of resonance frequency for parallel RLC circuit is given as

fr = 1 ⁄ 2π√LC Hz

Note:-  The equation of resonance is the same for both series and the parallel circuit

The above equation can be also written in rad/sec as

ωr = 2πfr = 1 ⁄ √LC rad/sec

Squaring the above equation

52 = 1 ⁄ 5L……….(Given C = 5F &  ω= 5 rad/sec)

L = 0.008H = 8 mH

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