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 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 
Version 2 EE IIT, Kharagpur 
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FAQs on Three-Phase Delta Connected Balanced Load - 1 - Basic Electrical Technology - Electrical Engineering (EE)

1. What is a three-phase delta connected balanced load?
Ans. A three-phase delta connected balanced load refers to a type of electrical power system where three loads are connected in a delta configuration, and each load is balanced, meaning they draw equal current from each phase.
2. How does a three-phase delta connected balanced load differ from other types of load connections?
Ans. A three-phase delta connected balanced load differs from other types of load connections, such as star or wye connections, as it does not have a neutral connection. In a delta configuration, the loads are connected in a triangular formation, while other types of connections have a common neutral point.
3. What are the advantages of a three-phase delta connected balanced load?
Ans. Some advantages of a three-phase delta connected balanced load include a higher power factor, better voltage regulation, and improved efficiency compared to other load connections. Additionally, the absence of a neutral connection can result in cost savings in terms of wiring and equipment.
4. How do you determine the line and phase currents in a three-phase delta connected balanced load?
Ans. To determine the line and phase currents in a three-phase delta connected balanced load, you can use the formula: line current (IL) = phase current (Iph). In this configuration, the line current is equal to the phase current, making it easier to calculate and analyze the load.
5. Can a three-phase delta connected balanced load be converted to a star or wye connection?
Ans. Yes, a three-phase delta connected balanced load can be converted to a star or wye connection by introducing a neutral connection. This conversion can be achieved by adding a neutral wire and connecting it to the common point of the loads. However, it is important to ensure that the loads are balanced and the appropriate connections are made to maintain the desired electrical characteristics.
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