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

Answer.3. 4

Explanation

Given

Resistance R = 15 ohms

Potential Difference V = 60 V

Current I =?

According to the Ohm’s Law

I = V/R = 60/15 = 4A

I = 4A

Answer.4. 64

Explanation

Given
Resistance = 16 ohms
Potential  difference V = 32 V

∴ Power dissipated by the resistor

P = V² ⁄ R = (32)² ⁄ 16

P = 64 Watt

Answer.1.

Explanation:-

The quality factor is defined as the ratio of Resonant Frequency to the Bandwidth.

[latex]\begin{array}{l}{\text{Quality Factor = }}\dfrac{{{\omega _o}}}{{B.W}}\\\\{\text{Resonant Frequency }}{\omega _o} = \dfrac{1}{{\sqrt {LC} }}\\\\Bandwidth = \dfrac{R}{L}\\\\{\text{Q Factor = }}\dfrac{{\frac{1}{{\sqrt {LC} }}}}{{\dfrac{R}{L}}}\\\\{\text{Q Factor = }}\dfrac{1}{R}\sqrt {\dfrac{L}{C}} \end{array}[/latex]

Answer.4. 100

Explanation

Given

The instantaneous value of Voltage V = V_msinωt = 20 sinωt = 20 V

The instantaneous value of voltage I = I_msin(ωt + φ) = 10 sin(ωt + 60) = 10 A

Phase angle φ = 60°

Average Power delivered by an AC circuit

P_avg = I_RMS.V_RMS.Cosφ

Where

I_RMS = I_m ⁄ √2 = 20 ⁄ √2

V_RMS = V_m ⁄ √2 = 10 ⁄ √2

[latex]\begin{array}{l}{P_{avg}} = \dfrac{{20}}{{\sqrt 2 }} \times \dfrac{{10}}{{\sqrt 2 }} \times \cos 60^\circ \\\\{P_{avg}} = {\text{100 watt}}\end{array}[/latex]

Answer.4. Ampere/Meter

Explanation

MAGNETIC FIELD INTENSITY

Magnetic field Intensity (H) is also called as Magnetic field Strength, Magnetic Intensity, Magnetic field, Magnetic Force and Magnetization Force.

Magnetic Field Strength (H)  gives the quantitative measure of the strongness or weakness of the magnetic field.

Suppose that a current of I amperes flows through a coil of N turns wound on a toroid of length I meters. The MMF is the total current linked to the magnetic circuit i.e  IN ampere-turns. If the magnetic circuit is homogeneous and of the uniform cross-sectional area, the MMF per meter length of the magnetic circuit is termed as the magneticfield strength, magnetic field intensity, or magnetizing force. It is represented by symbol H and is measured in ampere-turns per meter (At/m).

[latex display=”true”]H = \dfrac{{NI}}{l}{\text{ AT/m}}[/latex]

Hence the magnetic field Intensity can be defined as the ratio of applied MMF to the length of the path that it acts over.

Note:- Magnetic field intensity, H, is independent of the medium. Its value depends only on the number of turns N and the current I flowing in the coil.

Magnetic Field Strength is equivalent to the Voltage gradient in an Electrical Circuit.

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.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 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

Answer.4. 314

Explanation

The average value of a sin wave for the complete cycle i.e 2π is given as

V_avg = 2⁄π V_peak

100 = 2⁄π V_peak

V_Peak = 157 V

Or

V_avg = 0.636 × V_Peak

V_Peak = 157

Now Peak to Peak Voltage is given as

V_PK − V_PK = 2 × V_Peak

V_PK − V_PK = 2 × 157 = 314 V

V_PK − V_PK = 314 V

Answer.1. 50

Explanation

If current flows for time t then the production of heat is given as

H = I²R.t

Where

I = current = 1 A

R = Resistance = 10 ohms

t = time = 5 sec

H = 1² × 10 × 5

H = 50 watt-sec or Joule