EE MCQ

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

Related Question

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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The laws involved in the expression e = −dφ/dt are

Faraday’s 1st laws of electromagnetic induction tell us about the condition under which an e.m.f. is induced in a conductor or coil a when the magnetic flux linking a conductor or coil changes.

Faraday’s 2nd laws of electromagnetic induction give the magnitude of the induced e.m.f in a conductor or coil and may be stated as:

The magnitude of the e.mf induced in a conductor or coil is directly proportional to the rate of change of magnetic flux linkages.

Suppose a coil has N turns and magnetic flux linking the coil increases (i.e. changes) from φ1 Wb to φ2 Wb in t seconds. Now, magnetic flux linkages mean the product of magnetic flux and the number of turns of the coil.

N = e dφ/dt

Lenz Law:- Lenz’s law states: the direction of the induced e.m.f. is such as to oppose the change producing it. Therefore, the magnitude and direction of induced e.m.f. should be written as :

N = −e dφ/dt

SSC JE Electrical 2019 with solution SET-1
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A magnetic circuit is applied with a current that changes at a rate of 5 A/sec. The circuit has an inductance of 2H, then the self-induced EMF is

Given

Inductance L = 2 H

Rate of change of current di/dt = 5 A/sec

Self induced EMF = − (Rate of change of current × Inductance) = −L(di/dt)

= −(5 × 2) = −10V

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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Split-phase, capacitor, and shaded-pole single-phase motors classified as:

Split-phase, capacitor, and shaded-pole single-phase motors classified as single phase motor.

The single-phase motor with a single winding is not self-starting, because if the single winding is fed with a.c., it simply produces a “pulsating field” which rises and falls with the alternating current, and this type of field produces no torque in the rotor.