EE MCQ

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.

Related Question

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

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
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A coil is wound with 50 turns and a current 8A produces a flux of 200μWb. Calculate the inductance of the coil

The inductance of the coil is given by the relation

L = Nφ/I

Where

N = number of turns = 50

φ = flux = 200μWb

I = current = 8 A

L = 50 × 200 × 10−6 ⁄ 8

L = 1.25 mH

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

SSC JE Electrical 2019 with solution SET-1
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Two decimal coils A and B of 1000 turns each lies in the parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is:

Flux in coil A = 0.05 mWb = 5 × 10−5 wb = φA

No. of turns NA = NB = 1000

Flux linkage in a coil with B = Flux linkage in coil A × 80/100

= 0.8 × 5 × 10−5

= 4 × 10−5 wb =0.04 mwb

SSC JE Electrical 2019 with solution SET-1
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Please check out below figure.

From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below. Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R Here Current I = 1/2 A = 0.5 A Voltage V = 5 V R = V/I = 0.5/5 R = 10Ω