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Applied Mathematics for Electrical Engineering - 3130908
Complex Variables and Partial Differential Equations - 3130005
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Basic Electronics - 3110016
Mathematics-II - 3110015
Basic Civil Engineering - 3110004
Physics Group - II - 3110018
Basic Electrical Engineering - 3110005
Basic Mechanical Engineering - 3110006
Programming for Problem Solving - 3110003
Physics Group - I - 3110011
Mathematics-I - 3110014
English - 3110002
Environmental Science - 3110007
Software Engineering - 2160701
Data Structure - 2130702
Database Management Systems - 2130703
Operating System - 2140702
Advanced Java - 2160707
Compiler Design - 2170701
Data Mining And Business Intelligence - 2170715
Information And Network Security - 2170709
Mobile Computing And Wireless Communication - 2170710
Theory Of Computation - 2160704
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Semester - 8
Basic Electrical Engineering
(3110005)
BEE-3110005
Winter-2019
Question-4c-OR
BE | Semester-
1
Winter-2019
|
11-01-2020
Q4) (c)
7 Marks
Explain Generation of Rotating Magnetic Field in 3-phase Induction Motor with diagrams and equations.
Generation of Rotating Magnetic Field (RMF)
When stationary three phase winding coils are supplied by an alternating 3-phase supply then uniform Rotating Magnetic Field (or flux) of constant value is produced.
The principle of 3-phase, 2-pole stator having three identical winding coils are placed by 120° electrical apart. The sinusoidal flux due to three phase windings is shown in Fig. 2.
The directions of the positive fluxes are shown individually below at different positions.
Let us say that the maximum value of the flux due to any one of the three phases be
∅
m
. The resultant flux
∅
r
, at any instant is given by the resultant sum of the individual fluxes
∅
1
,
∅
2
, and
∅
3
, due to three phases.
We have considered the
1
6
th time period apart corresponding to points marked 0, 1, 2 and 3 in Fig. 1.
When
𝛉
θ
=
0
°
(at point 0),the Resultant flux,
We have,
∅
1
=
0
∅
2
=
-
3
2
∅
m
∅
3
=
3
2
∅
m
Now,
θ
∅
r
=
∅
2
2
+
∅
3
2
-
2
∅
2
∅
3
cos
θ
∅
r
=
-
3
2
∅
m
2
+
3
2
∅
m
2
-
2
-
3
2
∅
m
3
2
∅
m
cos
60
°
∅
r
=
3
4
∅
m
2
+
3
4
∅
m
2
+
3
4
∅
m
2
∅
r
=
3
4
+
3
4
+
3
4
∅
m
2
∅
r
=
3
+
3
+
3
4
∅
m
2
∅
r
=
9
4
∅
m
2
∅
r
=
3
2
∅
m
When
𝛉
θ
=
60
°
(at point 1),the Resultant flux,
We have,
∅
1
=
3
2
∅
m
∅
2
=
-
3
2
∅
m
∅
3
=
0
Now,
θ
∅
r
=
∅
1
2
+
∅
2
2
-
2
∅
1
∅
2
cos
θ
∅
r
=
3
2
∅
m
2
+
-
3
2
∅
m
2
-
2
3
2
∅
m
-
3
2
∅
m
cos
60
°
∅
r
=
3
4
∅
m
2
+
3
4
∅
m
2
+
3
2
1
2
∅
m
2
∅
r
=
3
4
+
3
4
+
3
4
∅
m
2
∅
r
=
3
+
3
+
3
4
∅
m
2
∅
r
=
9
4
∅
m
2
∅
r
=
3
2
∅
m
When
𝛉
θ
=
120
°
(at point 2),the Resultant flux,
We have,
∅
1
=
3
2
∅
m
∅
2
=
0
∅
3
=
-
3
2
∅
m
Now,
θ
∅
r
=
∅
1
2
+
∅
3
2
-
2
∅
1
∅
3
cos
θ
∅
r
=
3
2
∅
m
2
+
-
3
2
∅
m
2
-
2
3
2
∅
m
-
3
2
∅
m
cos
60
°
∅
r
=
3
4
∅
m
2
+
3
4
∅
m
2
+
3
4
∅
m
2
∅
r
=
3
4
+
3
4
+
3
4
∅
m
2
∅
r
=
3
+
3
+
3
4
∅
m
2
∅
r
=
9
4
∅
m
2
∅
r
=
3
2
∅
m
When
𝛉
θ
=
180
°
(at point 3),the Resultant flux,
We have,
∅
1
=
0
∅
2
=
3
2
∅
m
∅
3
=
-
3
2
∅
m
Now,
θ
∅
r
=
∅
2
2
+
∅
3
2
-
2
∅
2
∅
3
cos
θ
∅
r
=
-
3
2
∅
m
2
+
3
2
∅
m
2
-
2
-
3
2
∅
m
3
2
∅
m
cos
60
°
∅
r
=
3
4
∅
m
2
+
3
4
∅
m
2
+
3
4
∅
m
2
∅
r
=
3
4
+
3
4
+
3
4
∅
m
2
∅
r
=
3
+
3
+
3
4
∅
m
2
∅
r
=
9
4
∅
m
2
∅
r
=
3
2
∅
m
FOR UNDERSTANDING ONLY
Note:
Any above cases, any one vector has zero value and 60° is the angle between other two present vectors as per law of parallelogram, hence, in each above cases cos60 is considered. The value of θ is considered with sin function, as the waveforms are sinewaves.
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