EE MCQ

SSC JE Electrical 2019 with solution SET-1
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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:

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

Related Question

SSC JE Electrical 2019 with solution SET-1
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Which motor is referred to as a universal motor?

Universal motors may be powered by either AC or DC power sources.

The universal motor is constructed in the same way as a series-wound DC motor. However, it is designed to operate with either AC or DC applied.

The series-wound motor is the only type of DC motor that will operate with AC applied.

The windings of shunt-wound motors have inductance values that are too high to allow the motor to function with AC applied.

The series-wound motors have windings that have low inductances (few turns of large diameter wire), and they, therefore, offer a low impedance to the flow of AC.

The universal motor is one type of AC motor that has concentrated or salient field windings. These field windings are similar to those of all DC motors.

SSC JE Electrical 2019 with solution SET-1
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Reflected Light/Incident Light =

The ratio of luminous flux leaving the surface (Reflected Light) to the luminous flux incident on it (Incident Light) is known as the reflection factor.

Reflection Factor = Reflected light/Incident light

The value of the Reflection factor is always less than 1.

SSC JE Electrical 2019 with solution SET-1
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For a three-layer stranded wire, the number of strands in the first, second, and third layer respectively are:

The total number of the conductor in stranded cable is given by

N = 3x2 − 3x + 1

Where

x = no. of layer

For layer 1

N1 = 3(1)2 − 3(1) + 1

N1 = 1

For layer 2

N2 = 3(2)2 − 3(2) + 1

N2 = 7

In layer 2 no. of conductor = N2 − N1 = 7 − 1 = 6

For layer 3

N3 = 3(3)2 − 3(3) + 1

N3 = 19

In layer 3 no. of conductor = N3 − N2 = 19 − 7 = 12

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

Given Data

Voltage = 440 V

Frequency = 50 Hz

Slip = 10% =0.1

Now,

Rotor frequency = slip × frequency

= 0.1 × 50 = 5 Hz