ele 2601 assign 248591238
TRANSCRIPT
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8/22/2019 Ele 2601 Assign 248591238
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Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
QUESTION 1
1.1 The inductance of the inductor:
mHL
L
fL
fLX
X
X
XRZ
Z
Z
ZIV
R
R
IRV
L
L
318,31
502
86.9
86.92
2
86.9
1216
16
15240
.
6.12
15189
.
222
222
1.2 The impedance of the circuit:
o
L
j
jXRZ
044,3816
86.96.12
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8/22/2019 Ele 2601 Assign 248591238
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Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
1.3 The overall power factor of the circuit:
Since the V is at the reference the current angle is the opposite of impedance:
lags
PF
Z
VI
7875,0
0445,380cos
0445,3816
240
QUESTION 2
2.1 Circuit Diagram
2.2 The total impedance of the circuit
o
L
jZ
fLX
792,71686,39
699,374.12
699,37
12
10120502
2
1
3
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8/22/2019 Ele 2601 Assign 248591238
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Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
o
o
o
oo
oo
E
o
C
ZZ
ZZZ
jZ
fCX
298,6103,32
864,23021,40
162,30834,1284
63,41375,32792,71686,39
63,41375,32792,71686,39
63,41375,32
507.212.24
507.21
101485021
2
1
21
12
2
6
2.3. The current drawn from the supply
Aj
Z
VI
o
o
o
E
82,0492,7
298,6475,7
298,6103,32
0240
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8/22/2019 Ele 2601 Assign 248591238
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Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
2.4 Voltage across each component
Vj
V
ZIV
Vj
V
ZIV
o
oo
T
o
oo
T
639,179157,162
928,47003,242
63,41375,32298,6475,7
928,269047,123
494,65652,296
792,71686,39298,6475,7
22
11
2.5 Power Factor
lags
PF
Z
VI
o
o
o
993,0
)298,60cos(
cos
298,6103,32
0240
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8/22/2019 Ele 2601 Assign 248591238
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Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
QUESTION 3
3.1 Values of currents I3 and I4:
A
Z
VI
A
Z
VI
jVEV
jVjE
jZ
j
jjZ
X
X
o
o
o
o
o
o
o
o
o
o
o
C
L
085.18913.6
988.60764.38
903,42268
097.47677.14
90259.18
903,42268
903,42268
46,182323,196
704,20720,631867168.203043,26038330
988.60764.38
9.338.18
90259.18
259.180
159.3009.110
9.11
10270502
1
159.30
1096502
4
2
4
3
2
3
12
1
4
3
6
3
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8/22/2019 Ele 2601 Assign 248591238
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Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
3.2 Current drawn from the supply:
3.3 Current I2:
A
III
o
oo
T
179,46257,22
70637.3906.37990.20
12
3.4 Power Factor of coil L2:
lags
PF
I
VZ
oo
o
o
435.0
179,4618cos
179,46257,22
1867
2
1
2
A
III
o
oo
T
906.37990.20
085.18913.6097.47677.14
43
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8/22/2019 Ele 2601 Assign 248591238
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Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
QUESTION 4
886.0742.3
33.103846.3
8.398.7
13.14330
8.398.7
90313.5310
56
386
33.132.128
13.53108.3982.12
868.3982.12
8.3982.12
8.398.7
0100
563086
'
j
j
jjZ
V
j
ZIV
A
Z
VI
jjjZ
o
o
o
o
o
o
oo
o
o
o
o
E
E
A
B
V
V
A
B
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8/22/2019 Ele 2601 Assign 248591238
8/9
Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
QUESTION 5
5.1.1 Total Power
kW
kWkW
WWP
kWW
kWW
T
40
1030
10
30
21
2
1
5.1.2 Power Factor
756.0
892.40cos
cos
892.40
866.0tan
8660.0
1030
10303
3tan
1
21
21
o
o
Pf
WW
WW
5.2 The advantages of a three phase system are:-
Higher power/weight ratio of alternators. A three phase transmission system requires less copper or aluminium to
transmit the same quantity of power of a specific distance than a single
phase system.
Three phase motors are self-starting due to the rotating magnetic fieldinduced by the three phases.
Three phase motors have better power factor compared to single phasemotors.
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8/22/2019 Ele 2601 Assign 248591238
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Surname & Initials : MABENGO NDStudent No : 48591238
Module Code : ELE2601
Assignment No : 2
5.3.1 cos3 LLT IVP
5.3.2 LLT IVS 3
5.3.3 sin3 LLT IVQ