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

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.

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

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.

Answer.1. 72

Explanation

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

Inductive reactance X_L = 2πfL

Given

X_L1 = 60Ω

f₁ = 50 Hz

f₂ = 60 Hz

X_L2 =?

X_L1  ⁄  X_L2  =  f₁ ⁄ f₂

60  ⁄  X_L2 = 50/60

X_L2 = 72 Ω

Answer.1. 130

Explanation:-

Inductive Reactance = 2.π.f.L 

Where

f = frequency = 50 Hz

L = Inductance = 1/π 

X_L = 2.π.f.L = 2 × π × 50 × 1/π

X_L = 100 Ω

Capacitive Reactance =  1 ⁄ 2.π.f.C

Where C = 200/π μF = (200 × 10⁻⁶) ⁄ π

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

X_C = 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]

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]

Answer.3. 40

Explanation

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

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 

T_full = 5 × 0.4

T_full = 2 sec

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.

V_PH = V_Line /√3

V_Line = V_PH × √3

V_Line = 3.3 × 10³ × √3

V_Line = 5.71 kV

Answer.3. 26.94

Explanation

Given

Line voltage V_L = 440 V

Phase current I_ph = 50 A

Phase angle φ = 45°

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

The power in 3-phase star connection is given as

P = √3 × V_L × I_L × cos45°

= √3 × 440 × 50 × cos45°

P = 26940.3 watt

Power = 26.94 kW

Answer.3. 27.71

Explanation

Given

Line Voltage V_L = 400 V

Line current I_L = 50A

Phase angle φ = 53.13

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

Q = √3 × V_L × I_L × sinφ

Q = √3 × 400 × 50 × sin 53.13°

Q = √3 × 400 × 50 × 0.8

Q = 27712.8 VAR

or

Q = 27.71 KVAR