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QUESTION: 1

Which of the following statements is/are correct regarding the generation of EMF in rotating electrical machines in the armature winding? EMF is generated ____________

Solution:

Any of the above mentioned methods will produce EMF in the armature windings.

QUESTION: 2

The EMF equation e = NωrΦsinωrt is applicable to ____________

Solution:

e = Nω_{r}Φsinω_{r}t-NdΦ/dtcosω_{r}t, for time invariant field flux dΦ/dt = 0 ⇒ e = Nω_{r}Φsinω_{r}t, is the general equation, and is applicable to both AC and DC systems.

QUESTION: 3

In the equation for RMS value of the generated EMF in a full pitched coil of an AC machine,E = Emax/√2 = √2πfrNΦ, fr depends on

Solution:

f_{r} is called the rotational or speed frequency, since its value depends upon the relative velocity between the flux density wave and the armature coil.

QUESTION: 4

In AC rotating machines, the generated or speed EMF

Solution:

e = E_{max}cos(ω_{r}t-π/2)

ψ = NΦcosω_{r}t

It reveals that the speed or generated EMF lags by 90° the flux that generates it, and is true when flux is time invariant and is sine distributed in space.

QUESTION: 5

In a short pitched coil, the coil pitch factor kp, is given by ____________

Solution:

The effect of short pitched coil is to reduce the generated EMF, and this reduction factor is cosε/2, and is referred to as coil pitch factor.

QUESTION: 6

Which of the following equations represents the RMS value of the generated EMF in a short-pitched N-turn armature coil of an AC machine ____________

Solution:

EMF induced = NΦω_{r}k_{p}sinω_{r}t,

E_{max} = NΦω_{r}k_{p} when sinω_{r}t = 1

RMS value, E = E_{max}/√2 = √2πf_{r}k_{p}NΦ

QUESTION: 7

The effect of short pitched coil on the generated EMF is _____________

Solution:

The effect of short pitched coil is to reduce the generated EMF and this reduction factor is cosε/2 and is referred to as coil pitch factor.

QUESTION: 8

A winding is distributed in the slots along the air gap periphery

(i) to add mechanical strength to the winding

(ii) to reduce the amount of conductor material required

(iii) to reduce the harmonics in generated EMF

(iv) to reduce the size of the machine

(v) for full utilization of iron and conductor materials

From these, the correct answer is

Solution:

In rotating electrical machines, the armature turns are usually distributed in slots rather than concentrated in single slot. This is essential from the view point of utilizing the armature periphery completely, and add mechanical strength.

QUESTION: 9

In an AC machine, the effect of distributing the turns in different slots, results in a further reduction of generated EMF by the factor kd. This factor is called ____________

Solution:

E = √2πk_{p}k_{d}f_{r}NΦ, k_{d} is the distribution factor.

QUESTION: 10

A polyphase induction motor of the slip ring or wound rotor type can be used ____________

Solution:

A polyphase induction motor can be used as a frequency converter (or changer) for changing the supply frequency f to other frequencies sf and (2-s)f at the slip ring terminals.

QUESTION: 11

The equation for slip speed is ____________

Solution:

The relative speed between rotor and rotating flux wave, i.e, (ω-ω_{r}) is referred to as the slip speed in rad/sec

slip speed = (ω-ω_{r}) rad/sec

QUESTION: 12

If the rotor of an induction motor is made to revolve in a direction opposite to the rotating flux wave, then RMS value of EMF induced in one phase of rotor E is proportional to ____________

Solution:

Relative velocity between rotor winding and rotating flux wave becomes

(ω+ω_{r}) = ω+ω(1-s) = ω(2-s) and

E = 2πf(2-s)N_{phr}k_{w}Φ

QUESTION: 13

In an alternator, frequency per revolution is equal to

Solution:

For a P-pole machine, in n rev/sec, P/2n cycles/second are generated and cycles per second is referred to as frequency f of the EMF wave. Here in this question n=1

⇒ frequency per revolution = P/2

⇒ number of pole pairs.

QUESTION: 14

A 6 pole machine is rotating at a speed of 1200rpm. This speed in mechanical rad/sec and electrical radians per second is respectively

Solution:

N = 1200rpm,P = 6 and f = PN/120 = 60 cycles per sec.

1cycle = 360° electrical degrees = 2π electrical radians

f in electrical radians per sec = 60∗2π electrical radians/sec = 120π elect radians/sec

speed in mechanical radians/sec = 2/P(speed in electrical rad/sec) = 2/P(120π)=40π mechanical rad/sec.

QUESTION: 15

The DC machines are designed with flat topped flux density waves because

Solution:

For the same value of peak flux density B_{p}, it is easy to see that average value of brush voltage could be more for a flat topped B-wave than for a sinusoidal flux density wave.

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