Three-Phase Delta Connected Balanced Load - 1 | Basic Electrical Technology - Electrical Engineering (EE) PDF Download

 Page 1


In the previous (first) lesson of this module, the two types of connections (star and delta), 
normally used for the three-phase balanced supply in source side, along with the line and 
phase voltages, are described. Then, for balanced star-connected load, the phase and line 
currents, along with the expression for total power, are obtained. In this lesson, the phase 
and line currents for balanced delta-connected load, along with the expression for total 
power, will be presented.  
Keywords: line and phase currents, star- and delta-connections, balanced load.    
After going through this lesson, the students will be able to answer the following 
questions: 
1. How to calculate the currents (line and phase), for the delta-connected balanced load 
fed from a three-phase balanced system? 
2. Also how to find the total power fed to the above balanced load, for the two types of 
load connections – star and delta? 
Currents for Circuits with Balanced Load (Delta-connected) 
 
 
 
V
BR
I
BR
V
RY
I
RY
V
YB
I
YB
120°
120°
F 
F 
F 
(b) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Fig. 19.1 (a) Balanced delta-connected load fed from a three-phase balanced 
supply 
 (b) Phasor diagram 
 
 
Version 2 EE IIT, Kharagpur 
Page 2


In the previous (first) lesson of this module, the two types of connections (star and delta), 
normally used for the three-phase balanced supply in source side, along with the line and 
phase voltages, are described. Then, for balanced star-connected load, the phase and line 
currents, along with the expression for total power, are obtained. In this lesson, the phase 
and line currents for balanced delta-connected load, along with the expression for total 
power, will be presented.  
Keywords: line and phase currents, star- and delta-connections, balanced load.    
After going through this lesson, the students will be able to answer the following 
questions: 
1. How to calculate the currents (line and phase), for the delta-connected balanced load 
fed from a three-phase balanced system? 
2. Also how to find the total power fed to the above balanced load, for the two types of 
load connections – star and delta? 
Currents for Circuits with Balanced Load (Delta-connected) 
 
 
 
V
BR
I
BR
V
RY
I
RY
V
YB
I
YB
120°
120°
F 
F 
F 
(b) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Fig. 19.1 (a) Balanced delta-connected load fed from a three-phase balanced 
supply 
 (b) Phasor diagram 
 
 
Version 2 EE IIT, Kharagpur 
A three-phase delta ( )-connected balanced load (Fig. 19.1a) is fed from a balanced 
three-phase supply. A balanced load means that, the magnitude of the impedance per 
phase, is same, i.e., 
?
BR YB RY p
Z Z Z Z = = = , and their angle is also same, as 
BR YB RY p
f f f f = = = . In other words, if the impedance per phase is given as, 
p p p p
X j R Z + = ? f , then 
BR YB RY p
R R R R = = = , and also 
BR Yb RY p
X X X X = = = . 
The magnitude and phase angle of the impedance per phase are: 
2 2
p p p
X R Z + = , and 
( )
p p p
R X / tan
1 -
= f .In this case, the magnitudes of the phase voltages 
p
V are same, as 
those of the line voltages 
BR YB RY L
V V V V = = = . The phase currents (Fig. 19.1b) are 
obtained as,  
p
RY
RY
p RY
RY
p RY
Z
V
Z
V
I f
f
f - ? =
?
° ?
= - ?
0
  
) 120 (
120
) 120 (
p
YB
YB
p YB
YN
p YB
Z
V
Z
V
I f
f
f + ° - ? =
?
° - ?
= + ° - ? 
) 120 (
120
) 120 (
p
BR
BR
p BR
BR
p BR
Z
V
Z
V
I f
f
f - ° ? =
?
° + ?
= - ° ? 
In this case, the phase voltage,  is taken as reference. This shows that the phase 
currents are equal in magnitude, i.e., (
RY
V
BR YB RY p
I I I I = = = ), as the magnitudes of the 
voltage and load impedance, per phase, are same, with their phase angles displaced from 
each other in sequence by . The magnitude of the phase currents, is expressed as ° 120
()
p p p
Z V I / = .  
The line currents (Fig. 19.1b) are given as 
) 30 ( 3 ) 120 ( ) (
p p p p p p BR RY R R
I I I I I I f f f ? + ° - ? = - ° ? - - ? = - = - ?  
) 30 (
p L
I f + ° - ? = 
) 150 ( 3 ) ( ) 120 (
p p p p p p RY YB Y Y
I I I I I I f f f ? + ° - ? = - ? - + ° - ? = - = - ?  
) 150 (
p L
I f + ° - ? = 
) 90 ( 3 ) 120 ( ) 120 (
p p p p p p YB BR B B
I I I I I I f f f ? - ° ? = + ° - ? - - ° ? = - = - ?  
) 90 (
p L
I f - ° ? = 
The line currents are balanced, as their magnitudes are same and 3 times the 
magnitudes of the phase currents (
p L
I I · = 3 ), with the phase angles displaced from 
each other in sequence by . Also to note that the line current, say , lags the 
corresponding phase current,  by .   
° 120
R
I
RY
I ° 30
If the phase current,  is taken as reference, the phase currents are  
RY
I
) 0 . 0 0 . 1 ( 0 j I I
p RY
+ = ° ? :    ) 866 . 0 5 . 0 ( 120 j I I
p YB
- - = ° - ? ; 
. ) 866 . 0 5 . 0 ( 120 j I I
p BR
+ - = ° + ?
The line currents are obtained as  
Version 2 EE IIT, Kharagpur 
Page 3


In the previous (first) lesson of this module, the two types of connections (star and delta), 
normally used for the three-phase balanced supply in source side, along with the line and 
phase voltages, are described. Then, for balanced star-connected load, the phase and line 
currents, along with the expression for total power, are obtained. In this lesson, the phase 
and line currents for balanced delta-connected load, along with the expression for total 
power, will be presented.  
Keywords: line and phase currents, star- and delta-connections, balanced load.    
After going through this lesson, the students will be able to answer the following 
questions: 
1. How to calculate the currents (line and phase), for the delta-connected balanced load 
fed from a three-phase balanced system? 
2. Also how to find the total power fed to the above balanced load, for the two types of 
load connections – star and delta? 
Currents for Circuits with Balanced Load (Delta-connected) 
 
 
 
V
BR
I
BR
V
RY
I
RY
V
YB
I
YB
120°
120°
F 
F 
F 
(b) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Fig. 19.1 (a) Balanced delta-connected load fed from a three-phase balanced 
supply 
 (b) Phasor diagram 
 
 
Version 2 EE IIT, Kharagpur 
A three-phase delta ( )-connected balanced load (Fig. 19.1a) is fed from a balanced 
three-phase supply. A balanced load means that, the magnitude of the impedance per 
phase, is same, i.e., 
?
BR YB RY p
Z Z Z Z = = = , and their angle is also same, as 
BR YB RY p
f f f f = = = . In other words, if the impedance per phase is given as, 
p p p p
X j R Z + = ? f , then 
BR YB RY p
R R R R = = = , and also 
BR Yb RY p
X X X X = = = . 
The magnitude and phase angle of the impedance per phase are: 
2 2
p p p
X R Z + = , and 
( )
p p p
R X / tan
1 -
= f .In this case, the magnitudes of the phase voltages 
p
V are same, as 
those of the line voltages 
BR YB RY L
V V V V = = = . The phase currents (Fig. 19.1b) are 
obtained as,  
p
RY
RY
p RY
RY
p RY
Z
V
Z
V
I f
f
f - ? =
?
° ?
= - ?
0
  
) 120 (
120
) 120 (
p
YB
YB
p YB
YN
p YB
Z
V
Z
V
I f
f
f + ° - ? =
?
° - ?
= + ° - ? 
) 120 (
120
) 120 (
p
BR
BR
p BR
BR
p BR
Z
V
Z
V
I f
f
f - ° ? =
?
° + ?
= - ° ? 
In this case, the phase voltage,  is taken as reference. This shows that the phase 
currents are equal in magnitude, i.e., (
RY
V
BR YB RY p
I I I I = = = ), as the magnitudes of the 
voltage and load impedance, per phase, are same, with their phase angles displaced from 
each other in sequence by . The magnitude of the phase currents, is expressed as ° 120
()
p p p
Z V I / = .  
The line currents (Fig. 19.1b) are given as 
) 30 ( 3 ) 120 ( ) (
p p p p p p BR RY R R
I I I I I I f f f ? + ° - ? = - ° ? - - ? = - = - ?  
) 30 (
p L
I f + ° - ? = 
) 150 ( 3 ) ( ) 120 (
p p p p p p RY YB Y Y
I I I I I I f f f ? + ° - ? = - ? - + ° - ? = - = - ?  
) 150 (
p L
I f + ° - ? = 
) 90 ( 3 ) 120 ( ) 120 (
p p p p p p YB BR B B
I I I I I I f f f ? - ° ? = + ° - ? - - ° ? = - = - ?  
) 90 (
p L
I f - ° ? = 
The line currents are balanced, as their magnitudes are same and 3 times the 
magnitudes of the phase currents (
p L
I I · = 3 ), with the phase angles displaced from 
each other in sequence by . Also to note that the line current, say , lags the 
corresponding phase current,  by .   
° 120
R
I
RY
I ° 30
If the phase current,  is taken as reference, the phase currents are  
RY
I
) 0 . 0 0 . 1 ( 0 j I I
p RY
+ = ° ? :    ) 866 . 0 5 . 0 ( 120 j I I
p YB
- - = ° - ? ; 
. ) 866 . 0 5 . 0 ( 120 j I I
p BR
+ - = ° + ?
The line currents are obtained as  
Version 2 EE IIT, Kharagpur 
 ) 866 . 0 5 . 1 ( )} 866 . 0 5 . 0 ( ) 0 . 0 0 . 1 {( 120 0 j I j j I I I I
p p BR RY R
- = + - - + = ° + ? - ° ? =   
° - ? = ° - ? = 30 30 3
L p
I I 
) 866 . 0 5 . 1 ( )} 0 . 0 0 . 1 ( ) 866 . 0 5 . 0 {( 0 120 j I j j I I I I
p p RY YB Y
+ - = + - - - = ° ? - ° - ? =   
° - ? = ° - ? = 150 150 3
L p
I I 
)} 866 . 0 5 . 0 ( ) 866 . 0 5 . 0 {( 120 120 j j I I I I
p YB BR B
- - - + - = ° - ? - ° + ? =   
° + ? = ° + ? = = 90 90 3 ) 732 . 1 (
L p p
I I j I 
Total Power Consumed in the Circuit (Delta-connected) 
In the last lesson (No. 18), the equation for the power consumed in a star-connected 
balanced circuit fed from a three-phase supply, was presented. The power consumed per 
phase, for the delta-connected balanced circuit, is given by  
( )
p p p p p p p p
I V I V I V W , cos cos · · = · · = f    
It has been shown earlier that the magnitudes of the phase and line voltages are same, i.e., 
L p
V V = . The magnitude of the phase current is ( 3 / 1 ) times the magnitude of the line 
current, i.e., ( ) 3 /
L p
I I = . Substituting the two expressions, the total power consumed 
is obtained as 
( )
p L L p L L
I V I V W f f cos 3 cos 3 / 3 · · = · · · =  
It may be observed that the phase angle, 
p
f is the angle between the phase voltage 
, and the phase current, . Also that the expression for the total power in a three-
phase balanced circuit is the same, whatever be the type of connection – star or delta.  
p
V
p
I
Example 19.1 
 The star-connected load having impedance of O - ) 16 12 ( j per phase is connected in 
parallel with the delta-connected load having impedance of O + ) 18 27 ( j per phase (Fig. 
19.2a), with both the loads being balanced, and fed from a three-phase, 230 V, balanced 
supply, with the phase sequence as R-Y-B. Find the line current, power factor, total 
power & reactive VA, and also total volt-amperes (VA).    
 
•
•
•
I
R
I
Y
I
N
Z
1
Z
1
Z
1
Z
2
 
B B
B 
Z
1
= (12-j16) O 
Z
2
= (27-j18) O
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
(a) 
R 
Y 
B 
Z
2
Page 4


In the previous (first) lesson of this module, the two types of connections (star and delta), 
normally used for the three-phase balanced supply in source side, along with the line and 
phase voltages, are described. Then, for balanced star-connected load, the phase and line 
currents, along with the expression for total power, are obtained. In this lesson, the phase 
and line currents for balanced delta-connected load, along with the expression for total 
power, will be presented.  
Keywords: line and phase currents, star- and delta-connections, balanced load.    
After going through this lesson, the students will be able to answer the following 
questions: 
1. How to calculate the currents (line and phase), for the delta-connected balanced load 
fed from a three-phase balanced system? 
2. Also how to find the total power fed to the above balanced load, for the two types of 
load connections – star and delta? 
Currents for Circuits with Balanced Load (Delta-connected) 
 
 
 
V
BR
I
BR
V
RY
I
RY
V
YB
I
YB
120°
120°
F 
F 
F 
(b) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Fig. 19.1 (a) Balanced delta-connected load fed from a three-phase balanced 
supply 
 (b) Phasor diagram 
 
 
Version 2 EE IIT, Kharagpur 
A three-phase delta ( )-connected balanced load (Fig. 19.1a) is fed from a balanced 
three-phase supply. A balanced load means that, the magnitude of the impedance per 
phase, is same, i.e., 
?
BR YB RY p
Z Z Z Z = = = , and their angle is also same, as 
BR YB RY p
f f f f = = = . In other words, if the impedance per phase is given as, 
p p p p
X j R Z + = ? f , then 
BR YB RY p
R R R R = = = , and also 
BR Yb RY p
X X X X = = = . 
The magnitude and phase angle of the impedance per phase are: 
2 2
p p p
X R Z + = , and 
( )
p p p
R X / tan
1 -
= f .In this case, the magnitudes of the phase voltages 
p
V are same, as 
those of the line voltages 
BR YB RY L
V V V V = = = . The phase currents (Fig. 19.1b) are 
obtained as,  
p
RY
RY
p RY
RY
p RY
Z
V
Z
V
I f
f
f - ? =
?
° ?
= - ?
0
  
) 120 (
120
) 120 (
p
YB
YB
p YB
YN
p YB
Z
V
Z
V
I f
f
f + ° - ? =
?
° - ?
= + ° - ? 
) 120 (
120
) 120 (
p
BR
BR
p BR
BR
p BR
Z
V
Z
V
I f
f
f - ° ? =
?
° + ?
= - ° ? 
In this case, the phase voltage,  is taken as reference. This shows that the phase 
currents are equal in magnitude, i.e., (
RY
V
BR YB RY p
I I I I = = = ), as the magnitudes of the 
voltage and load impedance, per phase, are same, with their phase angles displaced from 
each other in sequence by . The magnitude of the phase currents, is expressed as ° 120
()
p p p
Z V I / = .  
The line currents (Fig. 19.1b) are given as 
) 30 ( 3 ) 120 ( ) (
p p p p p p BR RY R R
I I I I I I f f f ? + ° - ? = - ° ? - - ? = - = - ?  
) 30 (
p L
I f + ° - ? = 
) 150 ( 3 ) ( ) 120 (
p p p p p p RY YB Y Y
I I I I I I f f f ? + ° - ? = - ? - + ° - ? = - = - ?  
) 150 (
p L
I f + ° - ? = 
) 90 ( 3 ) 120 ( ) 120 (
p p p p p p YB BR B B
I I I I I I f f f ? - ° ? = + ° - ? - - ° ? = - = - ?  
) 90 (
p L
I f - ° ? = 
The line currents are balanced, as their magnitudes are same and 3 times the 
magnitudes of the phase currents (
p L
I I · = 3 ), with the phase angles displaced from 
each other in sequence by . Also to note that the line current, say , lags the 
corresponding phase current,  by .   
° 120
R
I
RY
I ° 30
If the phase current,  is taken as reference, the phase currents are  
RY
I
) 0 . 0 0 . 1 ( 0 j I I
p RY
+ = ° ? :    ) 866 . 0 5 . 0 ( 120 j I I
p YB
- - = ° - ? ; 
. ) 866 . 0 5 . 0 ( 120 j I I
p BR
+ - = ° + ?
The line currents are obtained as  
Version 2 EE IIT, Kharagpur 
 ) 866 . 0 5 . 1 ( )} 866 . 0 5 . 0 ( ) 0 . 0 0 . 1 {( 120 0 j I j j I I I I
p p BR RY R
- = + - - + = ° + ? - ° ? =   
° - ? = ° - ? = 30 30 3
L p
I I 
) 866 . 0 5 . 1 ( )} 0 . 0 0 . 1 ( ) 866 . 0 5 . 0 {( 0 120 j I j j I I I I
p p RY YB Y
+ - = + - - - = ° ? - ° - ? =   
° - ? = ° - ? = 150 150 3
L p
I I 
)} 866 . 0 5 . 0 ( ) 866 . 0 5 . 0 {( 120 120 j j I I I I
p YB BR B
- - - + - = ° - ? - ° + ? =   
° + ? = ° + ? = = 90 90 3 ) 732 . 1 (
L p p
I I j I 
Total Power Consumed in the Circuit (Delta-connected) 
In the last lesson (No. 18), the equation for the power consumed in a star-connected 
balanced circuit fed from a three-phase supply, was presented. The power consumed per 
phase, for the delta-connected balanced circuit, is given by  
( )
p p p p p p p p
I V I V I V W , cos cos · · = · · = f    
It has been shown earlier that the magnitudes of the phase and line voltages are same, i.e., 
L p
V V = . The magnitude of the phase current is ( 3 / 1 ) times the magnitude of the line 
current, i.e., ( ) 3 /
L p
I I = . Substituting the two expressions, the total power consumed 
is obtained as 
( )
p L L p L L
I V I V W f f cos 3 cos 3 / 3 · · = · · · =  
It may be observed that the phase angle, 
p
f is the angle between the phase voltage 
, and the phase current, . Also that the expression for the total power in a three-
phase balanced circuit is the same, whatever be the type of connection – star or delta.  
p
V
p
I
Example 19.1 
 The star-connected load having impedance of O - ) 16 12 ( j per phase is connected in 
parallel with the delta-connected load having impedance of O + ) 18 27 ( j per phase (Fig. 
19.2a), with both the loads being balanced, and fed from a three-phase, 230 V, balanced 
supply, with the phase sequence as R-Y-B. Find the line current, power factor, total 
power & reactive VA, and also total volt-amperes (VA).    
 
•
•
•
I
R
I
Y
I
N
Z
1
Z
1
Z
1
Z
2
 
B B
B 
Z
1
= (12-j16) O 
Z
2
= (27-j18) O
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
(a) 
R 
Y 
B 
Z
2
 
•
•
•
R 
Y 
B 
I
R
I
Y
I
B
Z
2
I
RY
Z
2
Z
2
I
YB
I
BR
'
1
Z = 3.Z
1
(b) 
'
1
Z
'
1
Z
 
 
 
 
 
 
 
 
 
 
 
 
I
BN
(I
B
) 
V
BR
I
YN
(I
Y
) 
V
YB
I
RN
(I
R
) 
I
RY
I
YB
V
YN
I
BR
V
BN
V
RY
(c)
V
RN
Fig. 19.2 (a) Circuit diagram (Example 19.1) 
 (b) Equivalent circuit (delta-connected) 
 (c) Phasor diagram 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
Page 5


In the previous (first) lesson of this module, the two types of connections (star and delta), 
normally used for the three-phase balanced supply in source side, along with the line and 
phase voltages, are described. Then, for balanced star-connected load, the phase and line 
currents, along with the expression for total power, are obtained. In this lesson, the phase 
and line currents for balanced delta-connected load, along with the expression for total 
power, will be presented.  
Keywords: line and phase currents, star- and delta-connections, balanced load.    
After going through this lesson, the students will be able to answer the following 
questions: 
1. How to calculate the currents (line and phase), for the delta-connected balanced load 
fed from a three-phase balanced system? 
2. Also how to find the total power fed to the above balanced load, for the two types of 
load connections – star and delta? 
Currents for Circuits with Balanced Load (Delta-connected) 
 
 
 
V
BR
I
BR
V
RY
I
RY
V
YB
I
YB
120°
120°
F 
F 
F 
(b) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Fig. 19.1 (a) Balanced delta-connected load fed from a three-phase balanced 
supply 
 (b) Phasor diagram 
 
 
Version 2 EE IIT, Kharagpur 
A three-phase delta ( )-connected balanced load (Fig. 19.1a) is fed from a balanced 
three-phase supply. A balanced load means that, the magnitude of the impedance per 
phase, is same, i.e., 
?
BR YB RY p
Z Z Z Z = = = , and their angle is also same, as 
BR YB RY p
f f f f = = = . In other words, if the impedance per phase is given as, 
p p p p
X j R Z + = ? f , then 
BR YB RY p
R R R R = = = , and also 
BR Yb RY p
X X X X = = = . 
The magnitude and phase angle of the impedance per phase are: 
2 2
p p p
X R Z + = , and 
( )
p p p
R X / tan
1 -
= f .In this case, the magnitudes of the phase voltages 
p
V are same, as 
those of the line voltages 
BR YB RY L
V V V V = = = . The phase currents (Fig. 19.1b) are 
obtained as,  
p
RY
RY
p RY
RY
p RY
Z
V
Z
V
I f
f
f - ? =
?
° ?
= - ?
0
  
) 120 (
120
) 120 (
p
YB
YB
p YB
YN
p YB
Z
V
Z
V
I f
f
f + ° - ? =
?
° - ?
= + ° - ? 
) 120 (
120
) 120 (
p
BR
BR
p BR
BR
p BR
Z
V
Z
V
I f
f
f - ° ? =
?
° + ?
= - ° ? 
In this case, the phase voltage,  is taken as reference. This shows that the phase 
currents are equal in magnitude, i.e., (
RY
V
BR YB RY p
I I I I = = = ), as the magnitudes of the 
voltage and load impedance, per phase, are same, with their phase angles displaced from 
each other in sequence by . The magnitude of the phase currents, is expressed as ° 120
()
p p p
Z V I / = .  
The line currents (Fig. 19.1b) are given as 
) 30 ( 3 ) 120 ( ) (
p p p p p p BR RY R R
I I I I I I f f f ? + ° - ? = - ° ? - - ? = - = - ?  
) 30 (
p L
I f + ° - ? = 
) 150 ( 3 ) ( ) 120 (
p p p p p p RY YB Y Y
I I I I I I f f f ? + ° - ? = - ? - + ° - ? = - = - ?  
) 150 (
p L
I f + ° - ? = 
) 90 ( 3 ) 120 ( ) 120 (
p p p p p p YB BR B B
I I I I I I f f f ? - ° ? = + ° - ? - - ° ? = - = - ?  
) 90 (
p L
I f - ° ? = 
The line currents are balanced, as their magnitudes are same and 3 times the 
magnitudes of the phase currents (
p L
I I · = 3 ), with the phase angles displaced from 
each other in sequence by . Also to note that the line current, say , lags the 
corresponding phase current,  by .   
° 120
R
I
RY
I ° 30
If the phase current,  is taken as reference, the phase currents are  
RY
I
) 0 . 0 0 . 1 ( 0 j I I
p RY
+ = ° ? :    ) 866 . 0 5 . 0 ( 120 j I I
p YB
- - = ° - ? ; 
. ) 866 . 0 5 . 0 ( 120 j I I
p BR
+ - = ° + ?
The line currents are obtained as  
Version 2 EE IIT, Kharagpur 
 ) 866 . 0 5 . 1 ( )} 866 . 0 5 . 0 ( ) 0 . 0 0 . 1 {( 120 0 j I j j I I I I
p p BR RY R
- = + - - + = ° + ? - ° ? =   
° - ? = ° - ? = 30 30 3
L p
I I 
) 866 . 0 5 . 1 ( )} 0 . 0 0 . 1 ( ) 866 . 0 5 . 0 {( 0 120 j I j j I I I I
p p RY YB Y
+ - = + - - - = ° ? - ° - ? =   
° - ? = ° - ? = 150 150 3
L p
I I 
)} 866 . 0 5 . 0 ( ) 866 . 0 5 . 0 {( 120 120 j j I I I I
p YB BR B
- - - + - = ° - ? - ° + ? =   
° + ? = ° + ? = = 90 90 3 ) 732 . 1 (
L p p
I I j I 
Total Power Consumed in the Circuit (Delta-connected) 
In the last lesson (No. 18), the equation for the power consumed in a star-connected 
balanced circuit fed from a three-phase supply, was presented. The power consumed per 
phase, for the delta-connected balanced circuit, is given by  
( )
p p p p p p p p
I V I V I V W , cos cos · · = · · = f    
It has been shown earlier that the magnitudes of the phase and line voltages are same, i.e., 
L p
V V = . The magnitude of the phase current is ( 3 / 1 ) times the magnitude of the line 
current, i.e., ( ) 3 /
L p
I I = . Substituting the two expressions, the total power consumed 
is obtained as 
( )
p L L p L L
I V I V W f f cos 3 cos 3 / 3 · · = · · · =  
It may be observed that the phase angle, 
p
f is the angle between the phase voltage 
, and the phase current, . Also that the expression for the total power in a three-
phase balanced circuit is the same, whatever be the type of connection – star or delta.  
p
V
p
I
Example 19.1 
 The star-connected load having impedance of O - ) 16 12 ( j per phase is connected in 
parallel with the delta-connected load having impedance of O + ) 18 27 ( j per phase (Fig. 
19.2a), with both the loads being balanced, and fed from a three-phase, 230 V, balanced 
supply, with the phase sequence as R-Y-B. Find the line current, power factor, total 
power & reactive VA, and also total volt-amperes (VA).    
 
•
•
•
I
R
I
Y
I
N
Z
1
Z
1
Z
1
Z
2
 
B B
B 
Z
1
= (12-j16) O 
Z
2
= (27-j18) O
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
(a) 
R 
Y 
B 
Z
2
 
•
•
•
R 
Y 
B 
I
R
I
Y
I
B
Z
2
I
RY
Z
2
Z
2
I
YB
I
BR
'
1
Z = 3.Z
1
(b) 
'
1
Z
'
1
Z
 
 
 
 
 
 
 
 
 
 
 
 
I
BN
(I
B
) 
V
BR
I
YN
(I
Y
) 
V
YB
I
RN
(I
R
) 
I
RY
I
YB
V
YN
I
BR
V
BN
V
RY
(c)
V
RN
Fig. 19.2 (a) Circuit diagram (Example 19.1) 
 (b) Equivalent circuit (delta-connected) 
 (c) Phasor diagram 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
Solution  
For  the balanced star-connected load, the impedance per phase is, 
O ° - ? = - = 13 . 53 0 . 20 ) 16 12 (
1
j Z  
The above load is converted into its equivalent delta. The impedance per phase is,  
O ° - ? = - = - × = · = ' 13 . 53 0 . 60 ) 48 36 ( ) 16 12 ( 3 3
1 1
j j Z Z  
For  the balanced delta-connected load, the impedance per phase is, 
O ° + ? = + = 69 . 33 45 . 32 ) 18 27 (
2
j Z  
In the equivalent circuit for the load (Fig. 19.2b), the two impedances,  &   are 
in parallel. So, the total admittance per phase is, 
1
Z '
2
Z
° + ?
+
° - ?
= +
'
= + ' =
69 . 33 45 . 32
1
13 . 53 0 . 60
1 1 1
2 1
2 1
Z Z
Y Y Y
p
° - ? + ° + ? = 69 . 33 03082 . 0 13 . 53 0167 . 0
) 003761 . 0 03564 . 0 ( )] 017094 . 0 02564 . 0 ( ) 01333 . 0 01 . 0 [( j j j = - + + =
1
024 . 6 03584 . 0
-
O ° - ? = 
-
The total impedance per phase is,  
O + = ° + ? = ° - ? = = ) 928 . 2 748 . 27 ( 024 . 6 902 . 27 ) 024 . 6 03584 . 0 /( 1 / 1 j Y Z
p p
 
The phasor diagram is shown in Fig. 19.2c.  
Taking the line voltage,  as reference, 
RY
V V V
RY
° ? = 0 230 
The other two line voltages are,  
° + ? = ° - ? = 120 230 ; 120 230
BR YB
V V 
For the equivalent delta-connected load, the line and phase voltages are same. 
So, the phase current,   is,  
RY
I
A j
Z
V
I
p
RY
RY
) 8651 . 0 198 . 8 ( 024 . 6 243 . 8
024 . 6 902 . 27
0 0 . 230
- = ° - ? =
° + ?
° ?
= = 
The two other phase currents are,  
° + ? = ° - ? = 976 . 113 243 . 8 ; 024 . 126 243 . 8
BR YB
I I 
The magnitude of the line current is 3 times the magnitude of the phase current.  
So, the line current is A I I
p L
277 . 14 243 . 8 3 3 = × = · = 
The line current,  lags the corresponding phase current,  by . 
R
I
RY
I ° 30
So, the line current,  is 
R
I A I
R
° - ? = 024 . 36 277 . 14 
The other two line currents are,  
° + ? = ° - ? = 976 . 83 277 . 14 ; 024 . 156 277 . 14
B Y
I I 
Also, the phase angle of the total impedance is positive.  
So, the power factor is  lag
p
9945 . 0 024 . 6 cos cos = ° = f 
The total volt-amperes is kVA I V S
p p
688 . 5 243 . 8 230 3 3 = × × = · · = 
The total VA is also obtained as kVA I V S
L L
688 . 5 277 . 14 230 3 3 = × × = · · = 
The total power is W k I V P
p p p
657 . 5 9945 . 0 243 . 8 230 3 cos 3 = × × × = · · · = f 
The total reactive volt-amperes is, 
VAR I V Q
p p p
5 . 597 024 . 6 sin 243 . 8 230 3 sin 3 = ° × × × = · · · = f 
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