GTHS KUMBO_Electrical Department_Third sequence examJan.

2012
1
Courses In
Electrical
Engineering
Volume V
POWER ELECTRONICS
THIRD SEQUENCE EXAM WITH SOLUTION
By
Jean-Paul NGOUNE
DIPET I (Electrotechnics), DIPET II (Electrotechnics)
M.sc. (Electrical Engineering)
Teacher in the Electrical Department, GTHS KUMBO, Cameroon.
GTHS KUMBO_Electrical Department_Third sequence examJan.2012
2
Exercice 1 : Battery charger
A battery having an emf of 60V and a negligible internal resistance is charged
by means of a Graetz bridge through a transformer whose primary is supplied by a
sinusoidal voltage 220V-50Hz. A resistor is placed in series with the battery such as
to limit the current to a maximal value of 30A.
1. Draw the diagram of the circuit so described.
2. Knowing that the transformer ratio of the transformer is m = 0.386, calculate
the effective value of the voltage at the secondary of the transformer.
3. Calculate the conducting angle of each diode in degree, and deduce the
duration of conduction in milliseconds.
4. Calculate the value of the resistor in series with the battery.
5. Calculate the average load current.
6. Calculate the charging time of the battery assuming that its capacity is 500Ah.
7. Determine the effective current that flows in the battery.
8. Determine the maximal reverse voltage across the diodes.
Exercise 2: Three phase rectification.
Consider the diagram of the figure below.
D1
D2
D3
RP
E
A
B
C
v1
v2
v3
u
All the diodes are perfect.
REPUBLIC OF CAMEROON
Peace Work Fatherland

GTHS KUMBO/ ELECT DPT

THIRD SEQUENCE EXAM
Class: F
3
7
Option: Electrotechnology
Duration: 03H
Coefficient: 4
Written paper
POWER ELECTRONICS
No document is allowed except the one given to
the candidates by the examiners.
AH Q
V E
Hz f
t t v
t t v
t t v
60
12
50
3
4
cos 2 30 ) (
3
2
cos 2 30 ) (
cos 2 30 ) (
3
2
1
=
=
=
|
.
|

\
|
=
|
.
|

\
|
=
=

GTHS KUMBO_Electrical Department_Third sequence examJan.2012

3
1. Draw in synchronism the waveforms of the voltages v1, v2, v3 and u.
2. Calculate the average and the effective values of the voltage u.
3. Determine the value of RP if the maximal value of the current 30.4A
4. Calculate the charging time of the battery.
5. Determine the power absorbed by the resistor RP
6. Calculate the maximal reverse voltage supported by each diode.
7. The battery is replaced by a resistor R having a value of 500 .
7.1Calculate the average load current.
7.2Calculate the effective load current; deduce the power absorbed by R.
8. Let the diodes be replaced by three perfect thyristors having a firing angle ,
where
3
2
0

; the load remaining R = 500 .
8.1Repeat question 1 above taking
3

= .
8.2Show that the expression of the average voltage across the load is:

cos
2
3 3
max
V U
av
=
8.3Draw the graph U
av
as a function of .
Exercise 3: Three phase Graetz bridge.
A three phase Graetz bridge is supplying two phases of the stator of a three
phase asynchronous motor whose windings are star connected. The bridge is
connected to the secondary of a three phase transformer 380/220V-50Hz. The
voltages at the secondary of that transformer are expressed as follows:
|
.
|

\
|
=
|
.
|

\
|
=
=
3
4
sin 2 127 ) (
3
2
sin 2 127 ) (
sin 2 127 ) (
3
2
1

t t v
t t v
t t v
1. Give the role of this circuit and precise how the secondary of the transformer is
connected, justify your answer.
2. Draw the diagram of the circuit described above.
3. Sketch the waveform of the load voltage and that of the voltage across the
diode D1.
GTHS KUMBO_Electrical Department_Third sequence examJan.2012
4
4. Calculate the average voltage across the load.
5. The stator windings of the motor are assumed to be pure resistance. Calculate
the value of the resistance measured between two phases of the stator,
knowing that the average load current is 70A. Deduce the resistance of one
winding of the stator.
6. Calculate the power consumed by the load.
Exercise 4: Control of a resistive load by thyristors and TRIACs
Three TRIACs are controlling a three phase balanced resistive load as show
by the figure below. The line voltage (between two phases) is 380V. The firing angle
of the TRIACs is equal to zero. The total power absorbed by the load is 15kW.
R
R
R
N
1
2
3
u
1. Calculate the effective value the current flowing through each TRIAC.
2. Calculate the maximal reverse voltage across each TRIAC.
3. The TRIACs are replaced by three thyristors; the firing angle still being equal
to zero.
3.1 Determine the effective value of the current flowing in each thyristor.
3.2 Calculate the maximal reverse voltage across each thyristor.
Subject Master: NGOUNE Jean-Paul, PLET
Electrotechnics, GTHS Kumbo.
GTHS KUMBO_Electrical Department_Third sequence examJan.2012
5
AKNOWLEDGEMENT
The exercises solved in this document are past Baccalaureat Technique
examination questions proposed by the Cameroon General Certificate of Education
Board (CGCEB) and the Office du Baccalaureat du Cameroun (OBC).
Exercise 1: Battery charger
1. Diagram of the circuit.
D1
D4 D3
D2
R
E
220V
50Hz
E = 60V; I
max
= 30A.
2. Effective voltage at the secondary of the transformer.
92 . 84 220 386 . 0
1 2
1
2
= = = = mU U
U
U
m
3. Conducting angle of each diode.
The angle at the beginning of the conduction (opening angle) is
1
, such that,
o
U
E
U
E
30
5 . 0
09 . 120
60
2 92 . 84
60
2

sin
1
2 2
1
=
~ = = = =

Then we can deduce the conducting angle

C
.
The duration of conduction can therefore be deduced.
ms
f
T T
t t
C
C c C
67 . 6
50 3
1
3
1
3 2 3
2
=

= = = = = =

4. Value of the resistor in series with the battery.
O =

= 2
30
60 120

2 2
I
E U
R
R
E U
I
o
C
120 30 2 180 2
1
= = =
GTHS KUMBO_Electrical Department_Third sequence examJan.2012
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5. Average load current.
It is known that,
( )
A I
E U
R
I
LAV
C LAV
086 . 13 60
3
14 . 3 2
2
3
120 2
2 14 . 3
1
cos

2
1
1 2
=
|
|
.
|

\
|

=
=

6. Charging time of the battery.

sec 31 min 12 38
086 . 13
500
h
I
Q
t t I Q
LAV
ch ch LAV
= = = = .
7. Effective current in the battery.
It is known that,
A I
T
t
I I I
Leff
C C
Leff
32 . 17
3
30
3
1
30
2
)
3
2
(
30

= = = =
= =

8. Maximal reverse voltage across the diodes.

V U U
D
120
max max
= =
Exercise 2: Three phase rectification.
Let us consider a three phase rectifier circuit made up
of 3 diodes with an R-E load.
D1
D2
D3
RP
E
A
B
C
v1
v2
v3
u
2

V
E < , the conduction of current in the load in not interrupted.
1. Waveforms of v1, v2 v3 and u in synchronism.
AH Q
V E
Hz f
t t v
t t v
t t v
60
12
50
3
4
cos 2 30 ) (
3
2
cos 2 30 ) (
cos 2 30 ) (
3
2
1
=
=
=
|
.
|

\
|
=
|
.
|

\
|
=
=

GTHS KUMBO_Electrical Department_Third sequence examJan.2012

7
The waveform of u is made up of the upper parts of v1,v2 and v3. The sketch of the
waveforms of v1, v2 and v3 is shown below.
v1 v2 v3 v1 v2 v3
0
-T/6 T/6 2T/6 3T/6 4T/6 5T/6 T t
v3
2. Average and effective load voltage.
| |
V U
V
V
d V U
LAV
LAV
104 . 35 2 30
28 . 6
3 3

2
3 3
sin
2

3
cos

2
3
3
3
3
3
= =
= = =

( )
V
V
U
V V
d
V
d V U
f
f
66 . 35
4
3 3
1
2
2 30
4
3 3
1
2

4
3 3
1
2

2
2 sin
4

3
cos 1
4

3
cos

2
3
2
3
3
2 3
3
2
2
3
3
2 2
= + = + =
|
|
.
|

\
|
+ =
(

+ = + = =

} }

3. Value of RP.
O ~

= 1
4 . 30
12 2 30

I
E V
RP
RP
E V
I
4. Charging time of the battery
LAV
ch ch LAV
I
Q
t t I Q = = with A
R
E U
I
LAV
LAV
104 . 23
1
12 104 . 35
=

=
Therefore, . sec 48 min 35 2 596 . 2
104 . 23
60
h h t
ch
= = =
5. Power absorbed by the resistor RP.
( ) W I RP P
f
79 . 559 66 . 23 1
2 2
= = =
GTHS KUMBO_Electrical Department_Third sequence examJan.2012
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6. Maximal reverse voltage supported by each diode.
V V V
D
48 . 73 6
max
= =
7. The battery is replaced by a resistor R = 500 .
Average load current.
mA
R
U
I
LAV
LAV
2 . 70
500
104 . 35
= = =
Effective current and power dissipated.
mA
R
U
I
f
f
32 . 71
500
66 . 35
= = =
W RI P
f
54 . 2
2
= =
8. The diodes be replaced by three perfect thyristors having a firing angle ,
where
3
2
0

; the load remaining R = 500 .
Waveforms of v1, v2, v3 and u (pending).
Expression of the verage load voltage:
| |

cos

2
3 3
sin
2

3
cos

2
3
3
3
3
3
}
+
+
+
+
= = = V
V
d V U
LAV
Graph of U
LAV
as function of .

cos cos

2
3 3
= = V U
LAV
We can deduce the following table of values:
0
6

3
2
U
LAV

2
3
2
0
2

The graph of U
LAV
as function of is drawn as follows.
0
pi/6 pi/3 pi/2 2pi/3
Uavmax
-Uavmax
Alpha
Uav
GTHS KUMBO_Electrical Department_Third sequence examJan.2012
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Exercise 3: Three phase Graetz bridge.
1. This circuit is used for the breaking of the asynchronous motor by injection of
direct current. The resistance of stator windings can also be easily determined
using this method. The secondary of the transformer is star connected.
2. Diagram of the circuit described.
R
R
R
v1
v2
v3
3. Waveform of the load voltage and that of the voltage across the diode D1
(PENDING).
4. Average load voltage.
V V U
LAV
21 . 297 3
14 . 3
2 127 3

3 3
=

= =

5. Resistance measured between two windings of the stator.

O = = = 24 . 4
LAV
LAV
LAV LAV
I
U
R RI U
We can then deduce the resistance of one winding of the stator.
O = = = 12 . 2
2
24 . 4
2
R
r
6. Power consumed by the load.
2
I R P = .
Let us first determine the effective load current I.
08 . 70
2
3 3
1
2
6
= + = =
R
V
R
U
I
f
; ( ) W P 51 . 20823 08 . 70 24 . 4
2
= =
Exercise 4: Control of a resistive load by thyristors and TRIACs
Three TRIACs are controlling a three phase balanced resistive load as show
by the figure below. The line voltage (between two phases) is 380V. The firing angle
of the TRIACs is equal to zero. The total power absorbed by the load is 15kW.
GTHS KUMBO_Electrical Department_Third sequence examJan.2012
10
R
R
R
N
1
2
3
u
1. Effective current flowing through each TRIAC.
Since the firing angle is 0, the load is assumed to be supplied with a three phase
sinusoidal voltage. Hence.
3
3
U
P
I UI P = = (Resistive load)
A I 7904 . 22
3 380
15000
= = .
2. Maximal reverse voltage across each TRIAC.
Because of the fact that the firing angle of the TRIACs is 0, they will conduct
throughout the period. Hence the maximal reverse voltage across each TRIAC is 0V
(the TRIAC is never blocked.
3. TRIACs are replaced by thyristors, the firing angle being still 0.
Effective value of current in each thyristor.
The current flows through each period during a half of period, hence.
V
I
I
Th
12 . 16
2
7904 . 22
2
= = =
Maximal reverse voltage across each thyristor.
V V
D
311 2 220
max
= =
END.
GTHS KUMBO_Electrical Department_Third sequence examJan.2012
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ABOUT THE AUTHOR
NGOUNE Jean-Paul is a teacher in the electrical
department at GTHS KUMBO. He is teaching this year
in the following subjects: Power electronics, Electrical
Machines, Digital and Analogue Circuits, Electricity-
Electronics, and Automation.
Any suggestion or critic is welcome
NGOUNE Jean-Paul, PLET, M.sc.(Electrical Engineering).
P.O. Box: 102 NSO, Kumbo, Cameroon.
Phone: (+237) 7506 2458.
Email : jngoune@yahoo.fr
Web site : www.scribd.com/jngoune