Introduction to Electrical
Engineering
Lecture #11
AC CIRCUIT ELEMENTS AND LAWS
Dr. ARIUNBOLOR Purvee
11/28/2019
German-Mongolian Institute for Resources and Technology
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POLYPHASE SYSTEM
• Circuit or system in which AC sources operate at the
same frequency but different phases are known as
poly-phase.
• Three phase is the most economical poly-phase
system.
• Melting purposes need 48 phases supply.
POLYPHASE SYSTEM
• Three Phase System:
• A generator consists of three coils placed 120 apart.
• The voltage generated are equal in magnitude but, out of phase by 120.
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¾ Cycle
½ Cycle
¼ Cycle
Start
Three-phase Voltage Sources
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Three phase vs single phase circuits
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FARADAYS LAW
•
Three things must be present in order to produce
electrical current:
a) Magnetic field
b) Conductor
c) Relative motion
•
•
Conductor cuts lines of magnetic flux, a voltage is
induced in the conductor
Direction and Speed are important
THREE PHASE GENERATOR
Video: Three phase generation (7:52)
• https://www.youtube.com/watch?v=4oRT7PoXSS0&t=53s
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GENERATOR WORK
• The generator consists of a rotating magnet (rotor)
surrounded by a stationary winding (stator).
• Three separate windings or coils with terminals a-a’,
b-b’, and c-c’ are physically placed 120 apart around
the stator.
• As the rotor rotates, its magnetic field cuts the flux
from the three coils and induces voltages in the coils.
• The induced voltage have equal magnitude but out of
phase by 120.
PHASE SEQUENCE
van (t )  VM cos t
vbn (t )  VM cost  120
vcn (t )  VM cost  120
Van  VM 0
Van  VM 0
Vbn  VM   120
Vbn  VM   120
Vcn  VM   120
Vcn  VM   120
POSITIVE
SEQUENCE
NEGATIVE
SEQUENCE
Balanced Y-connected Voltage Source
• Balanced phase voltages are equal in magnitude and are out of phase
with one another by 120 degrees.
• Phase voltages sum up to zero.
• Two possible combinations:
abc or (+) sequence
acb or () sequence
Balanced Y-connected Voltage Source
• Balanced line voltages are equal in magnitude and are out of phase with one
another by 120 degrees.
• Line voltages sum up to zero.
• The magnitude of line voltages is 1.732 times the magnitude of the phase voltages
• Line Voltages lead their corresponding phase voltages by 30 degrees
Three-phase Voltage Sources
Y-connected Source
D-connected Source
Balanced Three-phase Load Configurations
A balanced load is one in which the phase impedances are equal in magnitude and in phase.
Y-connected Load
D-connected Load
Balanced Three-phase Load Configurations
Reminder: A Y-connected load consists of three impedances
connected to a neutral node, while a –connected load consists of
three impedances connected around a loop. The load is balanced
when the three impedances are equal in either case.
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THREE PHASE QUANTITIES
QUANTITY
SYMBOL
Phase current
I
Line current
IL
Phase voltage
V
Line voltage
VL
PHASE VOLTAGES and LINE VOLTAGES
• Phase voltage is measured between
the neutral and any line: line to
neutral voltage
• Line voltage is measured between any
two of the three lines: line to line
voltage.
PHASE CURRENTS and LINE CURRENTS
• Line current (IL) is the current in each line of the
source or load.
• Phase current (I) is the current in each phase of
the source or load.
LINE VOLTAGES, VL
Vab  Van  Vbn
Vbc  Vbn  Vcn
Vca  Vca  Van
Vab  3VM 30
Vbc  3VM   90
Vca  3VM 150
Van  VM 0
volt
Vbn  VM   120 volt
Vcn  VM 120 volt
LINE
VOLTAGE
(VL)
PHASE
VOLTAGE (V)
Vab  3 VM 30 volt
Vbc  3 VM   90 volt
Vca  3 VM 150 volt
PHASE DIAGRAM OF V L AND V 
Vca
Vcn
Vab
30°
120°
Vbn
Vbc
-Vbn
Van
PROPERTIES OF PHASE VOLTAGE
• All phase voltages have the same
magnitude,
Vca
Vcn
Vab
V = lVanl = lVbnl = lVcnl
30°
120°
• Out of phase with each other by 120
Vbn
Vbc
-Vbn
Van
PROPERTIES OF LINE VOLTAGE
• All line voltages have the same
magnitude,
VL = lVabl = lVbcl = lVcal
Vca
Vcn
• Out of phase with each other
by 120
Vab
30°
120°
Vbn
Vbc
-Vbn
Van
RELATIONSHIP BETWEEN V and VL
1.
Magnitude
VL  3 V
2.
Phase
Vca
Vcn
Vab
- VL LEAD their corresponding
30°
V by 30
VL  V  30
120°
Vbn
Vbc
-Vbn
Van
Three phase connections
In three phase circuit, star and delta connection can be arranged in four
different ways•
•
•
•
Star-Star connection
Star-Delta connection
Delta-Star connection
Delta-Delta connection
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Balanced Y-Y Connection
ZY = ZS + Zl + ZL
IL = I
Balanced Y-Y Connection
𝑉𝑎𝑛 = 𝑉𝑝00
𝑉𝑏𝑛 = 𝑉𝑝  − 1200
𝑉𝑐𝑛 = 𝑉𝑝1200
𝑉𝑎𝑏 = 𝑉𝑎𝑛 +𝑉𝑛𝑏 = 𝑉𝑎𝑛-𝑉𝑏𝑛 = 3𝑉𝑝 300
𝑉𝑏𝑐 = 𝑉𝑏𝑛 − 𝑉𝑐𝑛 = 3𝑉𝑝 − 900
𝑉𝑐𝑎 = 𝑉𝑐𝑛 − 𝑉𝑎𝑛 = 3𝑉𝑝  − 2100
Balanced Y-Y Connection
• Thus, the magnitude of the line voltages is times the magnitude of the
phase voltages 𝑉𝑝 , or
𝑉𝐿 = 𝑉𝑝 3
𝑉𝑝 = 𝑉𝑎𝑛 = 𝑉𝑏𝑛 = 𝑉𝑐𝑛
𝑉𝐿 = 𝑉𝑎𝑏 = 𝑉𝑏𝑐 = 𝑉𝑐
• Also the line voltages lead their corresponding phase voltages by 300
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Problem: Calculate the line currents in the three-wire Y-Y system as shown.
n
Van
Ia 
ZY
1100
Ia 
 6.81  21.8
16.15521.8
I b  I a   120
 6.81  141.8A
I c  I a   240
 6.81  261.8  6.8198.2A
PHASE VOLTAGE AND LINE VOLTAGE
• In D-D system, line voltages equal to phase voltages:
VL  Vφ
PROPERTIES OF PHASE CURRENT
• All phase currents have the same magnitude,
Iφ  I AB  I BC  ICA 
Vφ
ZΔ
• phase shifts are by 120 with each other
PROPERTIES OF LINE CURRENT
• All line currents have the same magnitude,
I L  Ia  I b  Ic
• Out of phase with each other by 120
RELATIONSHIP BETWEEN I  and I L
1.
Magnitude
IL  3 IF
2.
Phase
- VL LAG their corresponding V by 30
IL  IF  30
PHASE DIAGRAM OF IL AND I
Conclusions
1. Line voltage and phase voltage
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Conclusions
2. Peak voltage, rms
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Conclusions
3. Vector Diagram
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Problem:
A balanced delta connected load having an impedance 20-j15
 is connected to a delta connected, positive sequence
generator having Vab = 3300 V. Calculate the phase currents
of the load and the line currents.
Balanced ∆ − ∆ Connection
Given Quantities
 ZΔ  20  j15   25  36.87
 Vab  3300
Phase Currents
VAB
3300
I AB 

 13.236.87A
ZΔ
25  38.87
I BC  I AB  120  13.2 - 83.13A
I CA  I AB  120  13.2156.87A
Line Currents
I a  I AB 3  30


 13.236.87 3  30 A
 22.866.87
I b  I a   120  22.86 - 113.13A
I c  I a   120  22.86126.87A
Balanced Y-D Connection
Problem: A balanced abc-sequence Y-connected source withVan  10010 
V is connected to a delta-connected balanced load of 8+j4  per phase.
Calculate the phase and the line currents.
Solution: Given Quantities
• Balanced WYE source
• Van= 10010 V
• Balanced DELTA load
• ZD= 8+j4 
Phase Currents
VAB
I AB 
ZΔ
VAB= voltage across ZD
= Vab= source line voltage
 VAB  3 Van 30
VAB  173.240 V
173.240
 I AB 
 19.3613.43
8  j4
Phase Currents
I AB  19.3613.43
 I BC  I AB13.43  120
I BC  19.36  106.57 A
 I CA  19.3613.43  120
I CA  19.36133.43 A
Line Currents
I a  3 I AB   30
 3 (19.36) 13.43  30
I a  33.53   16.57 A
 I b  I a   120  33.53   136.57 A
 I c  I a   120  33.53 103.43 A
Balanced D-Y Connection
Balanced D-Y Connection
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Example A balanced Y-connected load with a phase impedance of 40+j 25 
is connected to a balanced, positive sequence D-connected source with a
line voltage of 210 V. Calculate the phase currents. Use Vab as reference.
𝑍𝑌 = 40 + 𝑗25 = 47.17 320
𝑉𝑎𝑛 =
𝑉𝑎𝑏 = 21000
𝑉𝑎𝑏
3
300 = 121.2 − 620
𝑉𝑎𝑛 121.2 − 620
0
=
=
2.57
−
62
𝐼𝑎 =
47.17320
𝑍𝑌
𝐼𝑏 = 𝐼𝑎 1200 = 2.57 −1780
𝐼𝑐 = 𝐼𝑎 1200 = 2.57 −1780
11/28/2019
¾ Cycle
½ Cycle
¼ Cycle
Start
Three-phase Voltage Sources
German-Mongolian Institute for Resources and Technology
53
Three phase vs single phase circuits
11/28/2019
German-Mongolian Institute for Resources and Technology
54