ENE 104
Electric Circuit Theory
Lecture 11:
Polyphase Circuits
(ENE) Mon, 26 Mar 2012 / (EIE) Wed, 4 Apr 2012
Week #12 : Dejwoot KHAWPARISUTH
http://webstaff.kmutt.ac.th/~dejwoot.kha/
Objectives : Ch12
Page 2
• single-phase and polyphase systems
• Y- and Δ- connected three-phase system
• per-phase analysis of three-phase systems
ENE 104
Polyphase Circuits
Week #12
Introduction:
Page 3
An example set of three voltages, each of which is 120o out of
phase with the other two. As can be seen, only one of the voltages
will be zero at a particular instant.
ENE 104
Polyphase Circuits
Week #12
Double-Subscript Notation:
Page 4
(a) The definition of the
voltage Vab.
(b) Vad = Vab + Vbc+ Vcd
= Vab + Vcd.
ENE 104
Polyphase Circuits
Week #12
Double-Subscript Notation:
Page 5
Van  100 0 V.
Vbn  100  120 V.
Vcn  100  240 V.
ENE 104
Polyphase Circuits
Week #12
Double-Subscript Notation:
Vab
Page 6
 Van  Vnb  Van  Vbn
 100 0  100  120 V.
 100  (50  j86.6)  173.2 30 V.
ENE 104
Polyphase Circuits
Week #12
Practice: 12.1
Page 7
Let 𝐕𝑎𝑏 = 100∠0° V, 𝐕𝑏𝑑 = 40∠80° V, and
𝐕𝑐𝑎 = 70∠200° V. Find (a) 𝐕𝑎𝑑 ; (b) 𝐕𝑏𝑐 ; (c) 𝐕𝑐𝑑
ENE 104
Polyphase Circuits
Week #12
Practice: 12.1
ENE 104
Page 8
Polyphase Circuits
Week #12
Practice: 12.2
Page 9
Refer to the circuit of figure below and let
𝐈𝑓𝑗 = 3 A, 𝐈𝑑𝑒 = 2 A, and 𝐈ℎ𝑑 = −6 A. Find (a),
𝐈𝑐𝑑 ; (b) 𝐈𝑒𝑓 ; (c) 𝐈𝑖𝑗
ENE 104
Polyphase Circuits
Week #12
Practice: 12.2
ENE 104
Page 10
Polyphase Circuits
Week #12
Single-Phase Three-Wire System:
Page 11
ENE 104
Polyphase Circuits
Week #12
Single-Phase Three-Wire System:
Page 12
Consider:
ENE 104
Since
Van = Vnb
Then
Van
Vnb
I aA 
 I Bb 
Zp
Zp
And
InN = IBb + IAa = IBb - IaA = 0
Polyphase Circuits
Week #12
Example:
Page 13
Effect of Finite Wire Impedance:
Analyze the system below and determine the power
delivered to each of the three loads as well as the power
lost in the neutral wire and each of the two lines.
ENE 104
Polyphase Circuits
Week #12
Example:
Page 14
 115 0  I1  50(I1  I 2 )  3(I1  I 3 )  0
(20  j10)I 2  100(I 2  I 3 )  50(I 2  I1 )  0
 115 0  3(I 3  I1 )  100(I 3  I 2 )  I 3
ENE 104
Polyphase Circuits
0
Week #12
Example:
Page 15
I1  11.24  19.83 A.
I 2  9.389  24.47 A.
I 3  10.37  21.80 A.
The average power drawn by each load:
P50  I1  I 2  50  206 W.
2
P100  I 3  I 2 100  117 W.
2
P20 j10  I 2  20  1763 W.
2
ENE 104
Polyphase Circuits
Week #12
Example:
Page 16
I1  11.24  19.83 A.
I 2  9.389  24.47 A.
I 3  10.37  21.80 A.
The loss in each of the wires:
PaA  I1 1  126 W.
2
PbB  I 3 1  108 W.
2
PnN  I 3  I1 1  3 W.
2
ENE 104
Polyphase Circuits
Week #12
Example:
Page 17
The transmission efficiency, :

ENE 104
total
power delivered to load

total power generated
2086

 89.8%
2086  237
Polyphase Circuits
Week #12
Practice: 12.3
Page 18
Modified Figure below by adding a 1.5 Ω
resistance to each of the two outer lines, and a
2.5 Ω resistance to the neutral wire. Find the
average power delivered to each of the three
loads.
ENE 104
Polyphase Circuits
Week #12
Practice: 12.3
ENE 104
Page 19
Polyphase Circuits
Week #12
Practice: 12.3
ENE 104
Page 20
Polyphase Circuits
Week #12
Three-Phase Y-Y Connection:
Page 21
A balanced three-phase system:
may be defined as having
Van  Vbn  Vcn
and
Van  Vbn  Vcn  0
A Y-connected three-phase four-wire source.
ENE 104
Polyphase Circuits
Week #12
Three-Phase Y-Y Connection:
Positive, or abc,
phase sequence.
ENE 104
Page 22
Negative, or cba,
phase sequence.
Polyphase Circuits
Week #12
Three-Phase Y-Y Connection:
Page 23
Line-to-Line Voltage:
Vab
Vab
ENE 104
 Van  Vbn
 V p 0  V p
 3 Vp
Polyphase Circuits
 120
30
Week #12
Three-Phase Y-Y Connection:
Page 24
Line-to-Line Voltage:
ENE 104
Vab  3 V p
30
Vbc  3 V p
 90
Vca  3 Vp
 210
Polyphase Circuits
Week #12
Three-Phase Y-Y Connection:
Page 25
Van
I aA 
Zp
I bB
Vbn

Zp
Van

 I aA
 120
Zp
 120
I cC  I aA  240
ENE 104
A balanced three-phase system, connected
Y-Y and including a neutral.
Therefore:
I Nn  I aA  I bB  I cC  0
Polyphase Circuits
Week #12
Example:
Page 26
Find the currents, voltages and the total power
dissipated in the load
ENE 104
Polyphase Circuits
Week #12
Example:
Page 27
“Per-phase”:
Van  200 0 V.
Vbn  200  120 V.
Vcn  200  240 V.
ENE 104
Van
I aA 
 2  60
Zp
Vab  3 V p
30
Vbc  3 V p
 90
Vca  3 Vp
 210
I bB  2  180
I cC  2  300
Polyphase Circuits
Week #12
Example:
Page 28
“Per-phase”:
v AN  200 2 cos(120t  0)
iAN  2 2 cos(120t  60)
ENE 104
Polyphase Circuits
Week #12
Example:
Page 29
even though phase voltages and currents have
zero value at specific instants in time, the
instantaneous power delivered to the total load is
never zero.
p A (t )  v AN i AN
 800 cos(120t ) cos(120t  60)
 400[cos(60)  cos(240t  60)]
 200  400 cos(240t  60)]
pB (t )  200  400 cos(240t  300)]
pC (t )  200  400 cos(240t  180)]
ENE 104
Polyphase Circuits
Week #12
Example:
Page 30
the instantaneous power absorbed by the total
load is therefore
p(t )  p A (t )  pB (t )  pC (t )  600W
ENE 104
Polyphase Circuits
Week #12
Practice: 12.4
Page 31
A balanced three-phase three-wire system has a
Y-connected load. Each phase contains three
loads in parallel: −j100 Ω, 100 Ω, and 50 + j50 Ω.
Assume positive phase sequence with 𝐕𝑎𝑏 =
400∠0° V. Find (a) 𝐕𝑎𝑛 ; (b) 𝐈𝑎𝐴 ; (c) the total power
drawn by the load
ENE 104
Polyphase Circuits
Week #12
Practice: 12.4
ENE 104
Page 32
Polyphase Circuits
Week #12
Example:
Page 33
A balanced three-phase system with a line
voltage of 300V is supplying a balanced Yconnected load with 1200W at a leading PF of
0.8. Find the line current and the per-phase load
impedance.
The phase voltage is
The per-phase power is
ENE 104
V p  3003 V .
1200
3
Polyphase Circuits
 400 W .
Week #12
Example:
Page 34
1
P  Vm I m cos(   )  Veff I eff cos(   )
2
V p  3003 V .
Pp  1200
3  400 W .
The line current is
Pp
400
IL 
 300
 2.89 A.
V p cos(   )
 0.8
3
The phase impedance is
Zp 
ENE 104
Vp
IL

300
3
2.89
Polyphase Circuits
Week #12
Example:
Page 35
The PF is 0.8 leading, so the impedance phase
angle is cos 1 (0.8)  36.9
thus
ENE 104
Z p  60  36.9 
Polyphase Circuits
Week #12
Practice: 12.5
Page 36
A balanced three-phase three-wire system has a
line voltage of 500 V. Two balanced Y-connected
loads are present. One is a capacitive load with
7 − j2 Ω per phase, and the other is an inductive
load with 4 + j2 Ω per phase. Find (a) the phase
voltage; (b) the line current; (c) the total power
drawn by the load; (d) the power factor at which
the source is operating.
ENE 104
Polyphase Circuits
Week #12
Practice: 12.5
ENE 104
Page 37
Polyphase Circuits
Week #12
Example:
Page 38
A balanced 600-W lighting load is added (in
parallel) to the previous example. Determine the
new line current.
I 2  2.89
per-phase circuit
ENE 104
Polyphase Circuits
Z p  60  36.9 
Week #12
Example:
Page 39
The amplitude of the lighting current is
determined by
P  Veff I eff cos(   )
200  3003 I1 cos 0
So
I1  1.155 assume
I1  1.155 0 A.
I 2  2.89  36.9 A.
And the line current is
I L  I1  I 2  3.87  26.6 A.
ENE 104
Polyphase Circuits
Week #12
Practice: 12.6
Page 40
Three balanced Y-connected loads are installed on a
balanced three-phase four-wire system. Load 1
draws a total power of 6kW at unity PF, load 2
requires 10kVA at PF=0.96 lagging, and load 3
needs 7kW at 0.85 lagging. If the phase voltage at
the load is 135 V, if each line has a resistance of
0.1 Ω, and if the neutral has a resistance of 1 Ω, find
(a) the total power drawn by the load; (b) the
combined PF of the load; (c) the total power lost in
the four lines; (d) the phase voltage at the source;
(e) the power factor at which the source is operating.
ENE 104
Polyphase Circuits
Week #12
Practice: 12.6
ENE 104
Page 41
Polyphase Circuits
Week #12
Practice: 12.6
ENE 104
Page 42
Polyphase Circuits
Week #12
The Delta (∆) Connection:
VL
Page 43
 Vab
 Vbc
 Vca
Vp
 Van
 Vbn
 Vcn
A balanced D-connected load is present on a three-wire
three-phase system. The source happens to be Yconnected.
Where VL  3V p
ENE 104
and
Polyphase Circuits
Vab  3 V p
30
Week #12
The Delta (∆) Connection:
Page 44
The phase current
I AB
I BC
I CA
Vab

Zp
Vbc

Zp
Vca

Zp
I p  I AB  I BC  I CA
ENE 104
the line current:
I aA  I AB  I CA
I L  I aA  I bB  I cC
and
Polyphase Circuits
I L  3I p
Week #12
The Delta (∆) Connection:
Page 45
An example with an inductive load
I p  I AB  I BC  I CA
I L  I aA  I bB  I cC
I aA  I AB  I CA
I L  3I p
ENE 104
Polyphase Circuits
Week #12
∆-Connected:
Page 46
Pp
 V p I p cos 
 VL I p cos 
IL
 VL
cos 
3
total power,
P  3Pp
 3VL I L cos 
ENE 104
Polyphase Circuits
Week #12
Y-connected:
Page 47
Pp
 V p I p cos 
 V p I L cos 
VL

I L cos 
3
total power,
P  3Pp
 3VL I L cos 
ENE 104
Polyphase Circuits
Week #12
Practice: 12.7
Page 48
Each phase of a balanced three-phase Δ-
connected load consists of a 0.2 H inductor in
series with the parallel combination of a 5 𝜇F
capacitor and a 200 Ω resistance. Assume zero
line resistance and a phase voltage of 200 V at
𝜔 = 400 rad/s. Find (a) the phase current; (b)
the line current; (c) the total power absorbed by
the load
ENE 104
Polyphase Circuits
Week #12
Practice: 12.7
ENE 104
Page 49
Polyphase Circuits
Week #12
Example 12.5:
Page 50
Determine the amplitude of the line current in a
three-phase system with a line voltage of 300 V
that supplies 1200 W to a ∆-connected load at a
lagging PF of 0.8
Pp
 V p I p cos 
 VL I p cos 
IL
 VL
cos 
3
I L  2.89 A., Z p  180 36.9
ENE 104
Polyphase Circuits
Week #12
Example 12.6:
Page 51
Determine the amplitude of the line current in a
three-phase system with a line voltage of 300 V
that supplies 1200 W to a Y-connected load at a
lagging PF of 0.8
Pp
I L  2.89 A., Z p  60 36.9
ENE 104
Polyphase Circuits
 V p I p cos 
 V p I L cos 
VL

I L cos 
3
ZD
ZY 
3
Week #12
Practice: 12.8
Page 52
A balanced three-phase three-wire system is
terminated with two Δ-connected load in
parallel. Load 1 draws 40kVA at lagging PF of
0.8, while load 2 absorbs 24 kW at a leading PF
of 0.9. Assume no line resistance, and let
𝐕𝑎𝑏 = 440∠30° V. Find (a) the total power drawn
by the load; (b) the phase current 𝐈𝐴𝐵1 for the
lagging load; (c) 𝐈𝐴𝐵2 ; (d) 𝐈𝑎𝐴 .
ENE 104
Polyphase Circuits
Week #12
Practice: 12.8
ENE 104
Page 53
Polyphase Circuits
Week #12
Power Measurement:
Page 54
wattmeter
ENE 104
Polyphase Circuits
Week #12
Power Measurement:
Page 55
P  V2  I cosangV2  angI 
ENE 104
Polyphase Circuits
Week #12
Practice: 12.9
Page 56
Determine the wattmeter reading in Figure
below, state weather or not the potential coil had
to be reversed in order to obtain an upscale
reading, and identify the device or devices
absorbing or generating this power. The (+)
terminal of the wattmeter is connected to: (a) x;
(b) y; (c) z.
ENE 104
Polyphase Circuits
Week #12
Practice: 12.9
ENE 104
Page 57
Polyphase Circuits
Week #12
Practice: 12.9
ENE 104
Page 58
Polyphase Circuits
Week #12
Power Measurement:
Page 59
The wattmeter in a Three-Phase System:
ENE 104
Polyphase Circuits
Week #12
Power Measurement:
Page 60
The wattmeter in a Three-Phase System:
T
1
PA   v Ax iaAdt
T 0
v Ax  v AN  vNx
vBx  vBN  v Nx
vCx  vCN  v Nx
T
1
P  PA  PB  PC   (v Ax iaA  vBx ibB  vCxicC )dt
T 0
ENE 104
Polyphase Circuits
Week #12
Power Measurement:
Page 61
The wattmeter in a Three-Phase System:
v Ax  v AN  vNx
vBx  vBN  v Nx
vCx  vCN  v Nx
T
1
P  PA  PB  PC   (v Ax iaA  vBx ibB  vCxicC )dt
T 0
T
T
1
1
  (v AN iaA  vBN ibB  vCN icC )dt   v Nx (iaA  ibB  icC )dt
T 0
T 0
T
1
  (v AN iaA  vBN ibB  vCN icC )dt
T 0
ENE 104
Polyphase Circuits
Week #12
Power Measurement:
Page 62
The Two-Wattmeter Method:
P1  VAB  I aA cosangVAB  angI aA 
P2  VCB  I cC cosangVCB  angI cC 
ENE 104
Polyphase Circuits
Week #12
Power Measurement:
Page 63
The Two-Wattmeter Method:
in the case of a
balance load:
we can find PF
ENE 104
P1
 VAB  I aA cosangVAB  angI aA 
P2
 VCB  I cC cosangVCB  angI cC 
 VL I L cos30   
 VL I L cos30   
Polyphase Circuits
Week #12
Power Measurement:
Page 64
The Two-Wattmeter Method:
in the case of a
balance load:
we can find PF
P1 cos30   

P2 cos30   
P2  P1
tan   3
P2  P1
ENE 104
Polyphase Circuits
Week #12
Example 12.7:
Page 65
The balanced load is fed by a balanced threephase system having Vab  230 0 and positive
phase sequence. Find the reading of each
wattmeter and the total power drawn by the
load.
Vab  230 0
Vbc  230  120
Vca  230 120
ENE 104
Polyphase Circuits
Week #12
Example:
Page 66
Vab  230 0
Vca  230 120
Vac  230  60
Van
I aA 

4  j15
P1
ENE 104
230
3
 30
4  j15
 8.554  105.1
 Vac  I aA cosangVac  angI aA 
 230  8.554 cos 60  105.1  1389 W
Polyphase Circuits
Week #12
Example:
Page 67
Vbc  230  120
I aA  8.554  105.1
I bB  8.554  134.9
P2
ENE 104
 Vbc  I bB cosangVbc  angI bB 
 230  8.554 cos 120  134.9   512.5 W
Polyphase Circuits
Week #12
Practice: 12.10
Page 68
For the circuit of Figure below, let the loads be
𝐙𝐴 = 25∠60°Ω, 𝐙𝐵 = 50∠ − 60°Ω, 𝐙𝐶 = 50∠60°Ω,
𝐕𝐴𝐵 = 600∠0° V rms with (+) phase sequence, and
locate point x at C. Find (a) 𝑃𝐴 ; (b) 𝑃𝐵 ; (c)𝑃𝐶
ENE 104
Polyphase Circuits
Week #12
Practice: 12.10
ENE 104
Polyphase Circuits
Page 69
Week #12
Practice: 12.10
ENE 104
Polyphase Circuits
Page 70
Week #12
Example: Final 2/47
Page 71
เมื่อ แหล่งจ่ายในวงจร เป็ นแบบ balanced และ positive phase
sequence จงหา I aA , I bB , I cC และ the total complex power
supplied by the source
a
480
0o
A
Vrms
+
_
10 ohm
-j10 ohm
b
+
_
B
j5 ohm
+
_
c
ENE 104
C
AC Circuit Power Analysis
Week #12
Example: Final 2/47
Page 72
a
Van  480 0 V.
480
0o
A
Vrms
+
_
10 ohm
Vbn  480  120 V
-j10 ohm
b
+
_
B
j5 ohm
Vab  3 V p
30
Vbc  3  480  90
I
Vca  831.38  210 BC
+
_
Vcn  480  240 V.
c
C
Vab
I AB 
 83.1 30
10
831.38  90

 166.3
j5
831.38 150
I CA 
 83.1  120
 j10
ENE 104
AC Circuit Power Analysis
Week #12
Example: Final 2/47
Page 73
a
I AB  83.1 30
480
0o
A
Vrms
+
_
10 ohm
I BC  166.3
-j10 ohm
b
+
_
B
+
_
I CA  83.1  120
j5 ohm
c
C
I aA  I AB  I CA
I bB  I BC  I AB
I cC  I CA  I BC
Stotal  V I
*
AB AB
ENE 104
V I
*
BC BC
AC Circuit Power Analysis
V I
*
CA CA
Week #12
Example: Final 2/46
Page 74
เมื่อ Wattmeters ในวงจร อ่านค่าได้ W1  37,297.54 W.
และ W2  139,196.31 W. เมื่อค่า the magnitude ของ the
line voltage มีค่า 4160 V. จงหา Z
P2  P1
tan   3
P2  P1
ENE 104
AC Circuit Power Analysis
Week #12
Ex:
Page 75
A
a
480
0o
+
-
Vrms
Wm1
+
_
ZA= 60 -30˚
+
ZA
b
ZB= 24 30˚
B
+
_
ZC
o
480 120
+
_
ZB
480
-120
Wm2
o
c
+
-
+
-
ZC= 80 0˚
C
ให้หา VAB, VBC, IAB, IBC, IaA, IcC และค่าที่อ่านได้ที่
wattmeter ทั้งสอง
ENE 104
AC Circuit Power Analysis
Week #12
Ex:
Page 76
A
a
480
0
o
+
-
Vrms
Wm1
+
_
+
ZA
b
B
+
_
ZC
480 120
o
+
_
ZB
480
-120
Wm2
o
c
ENE 104
AC Circuit Power Analysis
+
-
+
-
C
Week #12
Hw:
ENE 104
Page 77
AC Circuit Power Analysis
Week #12
Reference:
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
ENE 104
Circuit Analysis and Electrical Engineering
Page 78