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SSC JE Electrical 2019 with solution SET-1

SSC JE Electrical 2019 with solution SET-1
 At f =  _____R – L – C series circuit operates at unity power factor

1/2π√LC
1/RC
1 ⁄ LC
1 ⁄ RLC

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SSC JE Electrical 2019 with solution SET-1
 Two coupled coils with L1 = L2 = 0.5H have a coupling coefficient of K = 0.75. The turn ratio N1 ⁄ N2 = ?

0.5
2
1
4

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

The self-inductance is given as
L = μN2A/I
L ∝ N2
where
N is the number of turns of the solenoid
A is the area of each turn of the coil
l is the length of the solenoid
and μ is the permeability constant
L1/L2 = N21/N22
0.5/0.5 = N21/N22
N1/N2 = 1
 

SSC JE Electrical 2019 with solution SET-1
________ is used to manufacture stay wire, earth wire, and structural components

Galvanized steel
Cadmium copper
Hard drawn copper
Nichrome

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Galvanized steel conductors do not corrode, and possess high resistance. Hence such Wires are used in telecommunications circuits, earth wires, guard wire, stray wire, etc.
 

SSC JE Electrical 2019 with solution SET-1
 Find the frequency of rotor induced EMF of a 3-phase, 440V, 50 Hz induction motor has a slip of 10%

5Hz
50Hz
2.5Hz
25Hz

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

Given Data
Voltage = 440 V
Frequency = 50 Hz
Slip = 10% =0.1
Now,
Rotor frequency = slip × frequency
= 0.1 × 50 = 5 Hz
 

SSC JE Electrical 2019 with solution SET-1
 4F2D is a/an number

Binary
Decimal
Hexadecimal
Octal

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

The hexadecimal number system is also called base-16, a numeration system in which all numbers are represented using the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F only. The system is base of 16. The hexadecimal numbers are used to represent binary numbers because of ease of conversion and compactness.

SSC JE Electrical 2019 with solution SET-1
In the two wattmeter method, the readings of the two wattmeters are 500W, 500W respectively. The load power factor in a balanced 3-phase 3-wire circuit is:

0.9
1
0.8
0.5

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

In two wattmeter method the phase angle is tanφ = √3(W1 − W2)/(W1 + W2) tanφ = √3(500 − 500)/(500 + 500) tanφ = 0° φ = tan−10° = 0° Power factor = cosφ PF = cos0° = 1

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