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 Page 1


Solution  
V V
L
400 =  O ° + ? = + = = 87 . 36 25 15 20 j Z Z
p RN
  
As the star-connected load is balanced, the magnitude of the phase voltage is,  
V V V
L p
231 3 400 3 = = =  
Taking the phase voltage,  as reference, the phase voltages are, 
RN
V
° + ? = ° - ? = ° ? = 120 231 ; 120 231 ; 0 231
BN YN RN
V V V 
The phasor diagram is shown in Fig. 20.6b. It has been shown that the line voltage, 
 leads the corresponding phase voltage,  by . So, the line voltages are, 
RY
V
RN
V ° 30
° + ? = ° - ? = ° + ? = 150 400 ; 90 400 ; 30 400
BR YB RY
V V V 
For a star-connected load, the phase and the line currents are same. 
The current in R-phase is, 
() ( ) A Z V I I
p RN R RN
° - ? = ° + ? ° ? = = = 87 . 36 24 . 9 87 . 36 0 . 25 0 0 . 231 /
A j ) 54 . 5 39 . 7 ( - = 
 Two other phase and line currents are,  
  13 . 83 9.24 ; 87 . 156 24 . 9 A I I A I I
B BN Y YN
° + ? = = ° - ? = =  
The power factor of the load is 8 . 0 87 . 36 cos cos = ° = f lagging, with ° + = 87 . 36 f , 
as the load is inductive. 
Total VA 403 . 6 24 . 9 0 . 231 3 3 = × × = · · =
p p
I V kVA 
This can be taken as 403 . 6 24 . 9 400 3 3 = × × = · ·
l L
I V kVA 
Total power 123 . 5 8 . 0 24 . 9 0 . 231 3 cos 3 = × × × = · · · = f
p p
I V kW 
Total reactive VA 842 . 3 87 . 36 sin 24 . 9 0 . 231 3 sin 3 = ° × × × = · · · = f
p p
I V kVAR 
The readings of the two wattmeters are, 
45 . 1 87 . 66 cos 24 . 9 400 ) 30 ( cos
1
= ° × × = + ° · · = f
R RY
I V W kW 
The phase angle between and  is  , obtained using two phasors. 
RY
V
R
I ° 87 . 66
67 . 3 87 . 6 cos 24 . 9 400 ) 30 ( cos
2
= ° × × = - ° · · = f
B BY
I V W kW 
The phase angle between and  is , obtained using two phasors, where 
. 
BY
V
B
I ° 87 . 6
° + ? = 90 400
BY
V
The sum of two readings is 12 . 5 ) 67 . 3 45 . 1 ( = + kW, which is same as the total power 
computed earlier  
Example 20.2 
 Calculate the readings of the wattmeter (W) connected as shown in Fig. 20.7a. The 
load is the same, as in Fig. 20.7a (Ex. 20.1), i.e., balanced star-connected one, with 
impedance of  per phase, fed from a three-phase, 400 V, balanced supply, 
with the phase sequence as R-Y-B.  
O + ) 15 20 ( j
 
 
 
Version 2 EE IIT, Kharagpur 
Page 2


Solution  
V V
L
400 =  O ° + ? = + = = 87 . 36 25 15 20 j Z Z
p RN
  
As the star-connected load is balanced, the magnitude of the phase voltage is,  
V V V
L p
231 3 400 3 = = =  
Taking the phase voltage,  as reference, the phase voltages are, 
RN
V
° + ? = ° - ? = ° ? = 120 231 ; 120 231 ; 0 231
BN YN RN
V V V 
The phasor diagram is shown in Fig. 20.6b. It has been shown that the line voltage, 
 leads the corresponding phase voltage,  by . So, the line voltages are, 
RY
V
RN
V ° 30
° + ? = ° - ? = ° + ? = 150 400 ; 90 400 ; 30 400
BR YB RY
V V V 
For a star-connected load, the phase and the line currents are same. 
The current in R-phase is, 
() ( ) A Z V I I
p RN R RN
° - ? = ° + ? ° ? = = = 87 . 36 24 . 9 87 . 36 0 . 25 0 0 . 231 /
A j ) 54 . 5 39 . 7 ( - = 
 Two other phase and line currents are,  
  13 . 83 9.24 ; 87 . 156 24 . 9 A I I A I I
B BN Y YN
° + ? = = ° - ? = =  
The power factor of the load is 8 . 0 87 . 36 cos cos = ° = f lagging, with ° + = 87 . 36 f , 
as the load is inductive. 
Total VA 403 . 6 24 . 9 0 . 231 3 3 = × × = · · =
p p
I V kVA 
This can be taken as 403 . 6 24 . 9 400 3 3 = × × = · ·
l L
I V kVA 
Total power 123 . 5 8 . 0 24 . 9 0 . 231 3 cos 3 = × × × = · · · = f
p p
I V kW 
Total reactive VA 842 . 3 87 . 36 sin 24 . 9 0 . 231 3 sin 3 = ° × × × = · · · = f
p p
I V kVAR 
The readings of the two wattmeters are, 
45 . 1 87 . 66 cos 24 . 9 400 ) 30 ( cos
1
= ° × × = + ° · · = f
R RY
I V W kW 
The phase angle between and  is  , obtained using two phasors. 
RY
V
R
I ° 87 . 66
67 . 3 87 . 6 cos 24 . 9 400 ) 30 ( cos
2
= ° × × = - ° · · = f
B BY
I V W kW 
The phase angle between and  is , obtained using two phasors, where 
. 
BY
V
B
I ° 87 . 6
° + ? = 90 400
BY
V
The sum of two readings is 12 . 5 ) 67 . 3 45 . 1 ( = + kW, which is same as the total power 
computed earlier  
Example 20.2 
 Calculate the readings of the wattmeter (W) connected as shown in Fig. 20.7a. The 
load is the same, as in Fig. 20.7a (Ex. 20.1), i.e., balanced star-connected one, with 
impedance of  per phase, fed from a three-phase, 400 V, balanced supply, 
with the phase sequence as R-Y-B.  
O + ) 15 20 ( j
 
 
 
Version 2 EE IIT, Kharagpur 
 
•
•
R 
Y 
B 
Balanced 
star-
connected 
load as in 
Fig. 20.6(a) 
W 
(a) 
I
R 
30° 
V
RY
V
RN
I
RN
V
YB
90°- F 
F
(b)
Fig. 20.7 (a) Circuit diagram (Example 20.2) 
 (b) Phasor diagram 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Solution  
 The steps are not repeated here, but taken from previous example (20.1).  
 The phasor diagram is shown in Fig. 20.7b.   
 The phase voltage,  is taken as reference as in Ex. 20.1. 
RN
V
 The phase current,  is 
RN
I A I
RN
° - ? = 87 . 36 24 .9 
 The phase angle, f of the load impedance is ° + = 87 . 36 f 
 The line voltage,  is 
RY
V V V
RY
° + ? = 30 400  
Version 2 EE IIT, Kharagpur 
Page 3


Solution  
V V
L
400 =  O ° + ? = + = = 87 . 36 25 15 20 j Z Z
p RN
  
As the star-connected load is balanced, the magnitude of the phase voltage is,  
V V V
L p
231 3 400 3 = = =  
Taking the phase voltage,  as reference, the phase voltages are, 
RN
V
° + ? = ° - ? = ° ? = 120 231 ; 120 231 ; 0 231
BN YN RN
V V V 
The phasor diagram is shown in Fig. 20.6b. It has been shown that the line voltage, 
 leads the corresponding phase voltage,  by . So, the line voltages are, 
RY
V
RN
V ° 30
° + ? = ° - ? = ° + ? = 150 400 ; 90 400 ; 30 400
BR YB RY
V V V 
For a star-connected load, the phase and the line currents are same. 
The current in R-phase is, 
() ( ) A Z V I I
p RN R RN
° - ? = ° + ? ° ? = = = 87 . 36 24 . 9 87 . 36 0 . 25 0 0 . 231 /
A j ) 54 . 5 39 . 7 ( - = 
 Two other phase and line currents are,  
  13 . 83 9.24 ; 87 . 156 24 . 9 A I I A I I
B BN Y YN
° + ? = = ° - ? = =  
The power factor of the load is 8 . 0 87 . 36 cos cos = ° = f lagging, with ° + = 87 . 36 f , 
as the load is inductive. 
Total VA 403 . 6 24 . 9 0 . 231 3 3 = × × = · · =
p p
I V kVA 
This can be taken as 403 . 6 24 . 9 400 3 3 = × × = · ·
l L
I V kVA 
Total power 123 . 5 8 . 0 24 . 9 0 . 231 3 cos 3 = × × × = · · · = f
p p
I V kW 
Total reactive VA 842 . 3 87 . 36 sin 24 . 9 0 . 231 3 sin 3 = ° × × × = · · · = f
p p
I V kVAR 
The readings of the two wattmeters are, 
45 . 1 87 . 66 cos 24 . 9 400 ) 30 ( cos
1
= ° × × = + ° · · = f
R RY
I V W kW 
The phase angle between and  is  , obtained using two phasors. 
RY
V
R
I ° 87 . 66
67 . 3 87 . 6 cos 24 . 9 400 ) 30 ( cos
2
= ° × × = - ° · · = f
B BY
I V W kW 
The phase angle between and  is , obtained using two phasors, where 
. 
BY
V
B
I ° 87 . 6
° + ? = 90 400
BY
V
The sum of two readings is 12 . 5 ) 67 . 3 45 . 1 ( = + kW, which is same as the total power 
computed earlier  
Example 20.2 
 Calculate the readings of the wattmeter (W) connected as shown in Fig. 20.7a. The 
load is the same, as in Fig. 20.7a (Ex. 20.1), i.e., balanced star-connected one, with 
impedance of  per phase, fed from a three-phase, 400 V, balanced supply, 
with the phase sequence as R-Y-B.  
O + ) 15 20 ( j
 
 
 
Version 2 EE IIT, Kharagpur 
 
•
•
R 
Y 
B 
Balanced 
star-
connected 
load as in 
Fig. 20.6(a) 
W 
(a) 
I
R 
30° 
V
RY
V
RN
I
RN
V
YB
90°- F 
F
(b)
Fig. 20.7 (a) Circuit diagram (Example 20.2) 
 (b) Phasor diagram 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Solution  
 The steps are not repeated here, but taken from previous example (20.1).  
 The phasor diagram is shown in Fig. 20.7b.   
 The phase voltage,  is taken as reference as in Ex. 20.1. 
RN
V
 The phase current,  is 
RN
I A I
RN
° - ? = 87 . 36 24 .9 
 The phase angle, f of the load impedance is ° + = 87 . 36 f 
 The line voltage,  is 
RY
V V V
RY
° + ? = 30 400  
Version 2 EE IIT, Kharagpur 
 The line voltage,  is 
YB
V V V
YB
° - ? = 90 400 
 The reading of the wattmeter (W) is, 
f ? sin 3 ) 90 ( cos ) , ( cos · · · - = + ° · · == · · =
p p p L RN YB RN YB
I V I V I V I V W
kW 218 . 2 87 . 36 sin 24 . 9 400 - = ° × × - = 
 The value is negative, as the load is inductive. 
The reading (W) is 577 . 0 3 / 1 = times the total reactive power as, 
f f sin 3 sin 3 · · · = · · ·
L L p p
I V I V ,  
or 3 times the reactive power per phase. 
Example 20.3 
 Calculate the readings of the two wattmeters ( & ) connected to measure the 
total power for a balanced delta-connected load shown in Fig. 20.8a, fed from a three-
phase, 200 V balanced supply with phase sequence as R-Y-B. The load impedance per 
phase is
1
W
2
W
O - ) 14 14 ( j . Also find the line and phase currents, power factor, total power, 
total reactive VA and total VA. 
  
X
C
= 14 O / R = 14 O
R 
Y 
B 
W
2 
•
•
I
B 
I
Y 
I
R 
(a) 
Z
1
I
RY
Z
1
Z
1
I
BR
I
YB
I
RY
Z
1
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
Page 4


Solution  
V V
L
400 =  O ° + ? = + = = 87 . 36 25 15 20 j Z Z
p RN
  
As the star-connected load is balanced, the magnitude of the phase voltage is,  
V V V
L p
231 3 400 3 = = =  
Taking the phase voltage,  as reference, the phase voltages are, 
RN
V
° + ? = ° - ? = ° ? = 120 231 ; 120 231 ; 0 231
BN YN RN
V V V 
The phasor diagram is shown in Fig. 20.6b. It has been shown that the line voltage, 
 leads the corresponding phase voltage,  by . So, the line voltages are, 
RY
V
RN
V ° 30
° + ? = ° - ? = ° + ? = 150 400 ; 90 400 ; 30 400
BR YB RY
V V V 
For a star-connected load, the phase and the line currents are same. 
The current in R-phase is, 
() ( ) A Z V I I
p RN R RN
° - ? = ° + ? ° ? = = = 87 . 36 24 . 9 87 . 36 0 . 25 0 0 . 231 /
A j ) 54 . 5 39 . 7 ( - = 
 Two other phase and line currents are,  
  13 . 83 9.24 ; 87 . 156 24 . 9 A I I A I I
B BN Y YN
° + ? = = ° - ? = =  
The power factor of the load is 8 . 0 87 . 36 cos cos = ° = f lagging, with ° + = 87 . 36 f , 
as the load is inductive. 
Total VA 403 . 6 24 . 9 0 . 231 3 3 = × × = · · =
p p
I V kVA 
This can be taken as 403 . 6 24 . 9 400 3 3 = × × = · ·
l L
I V kVA 
Total power 123 . 5 8 . 0 24 . 9 0 . 231 3 cos 3 = × × × = · · · = f
p p
I V kW 
Total reactive VA 842 . 3 87 . 36 sin 24 . 9 0 . 231 3 sin 3 = ° × × × = · · · = f
p p
I V kVAR 
The readings of the two wattmeters are, 
45 . 1 87 . 66 cos 24 . 9 400 ) 30 ( cos
1
= ° × × = + ° · · = f
R RY
I V W kW 
The phase angle between and  is  , obtained using two phasors. 
RY
V
R
I ° 87 . 66
67 . 3 87 . 6 cos 24 . 9 400 ) 30 ( cos
2
= ° × × = - ° · · = f
B BY
I V W kW 
The phase angle between and  is , obtained using two phasors, where 
. 
BY
V
B
I ° 87 . 6
° + ? = 90 400
BY
V
The sum of two readings is 12 . 5 ) 67 . 3 45 . 1 ( = + kW, which is same as the total power 
computed earlier  
Example 20.2 
 Calculate the readings of the wattmeter (W) connected as shown in Fig. 20.7a. The 
load is the same, as in Fig. 20.7a (Ex. 20.1), i.e., balanced star-connected one, with 
impedance of  per phase, fed from a three-phase, 400 V, balanced supply, 
with the phase sequence as R-Y-B.  
O + ) 15 20 ( j
 
 
 
Version 2 EE IIT, Kharagpur 
 
•
•
R 
Y 
B 
Balanced 
star-
connected 
load as in 
Fig. 20.6(a) 
W 
(a) 
I
R 
30° 
V
RY
V
RN
I
RN
V
YB
90°- F 
F
(b)
Fig. 20.7 (a) Circuit diagram (Example 20.2) 
 (b) Phasor diagram 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Solution  
 The steps are not repeated here, but taken from previous example (20.1).  
 The phasor diagram is shown in Fig. 20.7b.   
 The phase voltage,  is taken as reference as in Ex. 20.1. 
RN
V
 The phase current,  is 
RN
I A I
RN
° - ? = 87 . 36 24 .9 
 The phase angle, f of the load impedance is ° + = 87 . 36 f 
 The line voltage,  is 
RY
V V V
RY
° + ? = 30 400  
Version 2 EE IIT, Kharagpur 
 The line voltage,  is 
YB
V V V
YB
° - ? = 90 400 
 The reading of the wattmeter (W) is, 
f ? sin 3 ) 90 ( cos ) , ( cos · · · - = + ° · · == · · =
p p p L RN YB RN YB
I V I V I V I V W
kW 218 . 2 87 . 36 sin 24 . 9 400 - = ° × × - = 
 The value is negative, as the load is inductive. 
The reading (W) is 577 . 0 3 / 1 = times the total reactive power as, 
f f sin 3 sin 3 · · · = · · ·
L L p p
I V I V ,  
or 3 times the reactive power per phase. 
Example 20.3 
 Calculate the readings of the two wattmeters ( & ) connected to measure the 
total power for a balanced delta-connected load shown in Fig. 20.8a, fed from a three-
phase, 200 V balanced supply with phase sequence as R-Y-B. The load impedance per 
phase is
1
W
2
W
O - ) 14 14 ( j . Also find the line and phase currents, power factor, total power, 
total reactive VA and total VA. 
  
X
C
= 14 O / R = 14 O
R 
Y 
B 
W
2 
•
•
I
B 
I
Y 
I
R 
(a) 
Z
1
I
RY
Z
1
Z
1
I
BR
I
YB
I
RY
Z
1
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
 
I
R
V
RY
I
YB
V
YB
I
BR
V
BR
V
BY
I
Y
I
B
30°
30°
( F-30 °) 
( F+30 °)
60°
F 
F 
I
RY
(b) 
F 
Fig. 20.8 (a) Circuit diagram for a three-phase system with balanced delta-
connected load (Example 20.3) 
 (b) Phasor diagram 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Solution  
O ° - ? = - = = 45 8 . 19 14 14 j Z Z
p RY
  
For delta-connected load, V V V
p L
200 = =  
Taking the line or phase voltage  as reference, the line or phase voltages are,  
RY
V
° + ? = ° - ? = ° ? = 120 200 ; 120 200 ; 0 200
BR YB RY
V V V 
The phasor diagram is shown in Fig. 20.8b. The phase current,  is, 
RY
I
() ( ) A j Z V I
p RY RY
) 142 . 7 142 . 7 ( 45 1 . 10 45 9.8 1 / 0 0 . 200 / + = ° + ? = ° - ? ° ? = =  
The other two phase currents are, 
  165 .1 10 ; 75 1 . 10 A I A I
BR YN
° + ? = ° - ? =  
The power factor of the load is 707 . 0 45 cos cos = ° = f leading, with ° - = 45 f , as 
the load is capacitive. 
As the phase currents are balanced, the magnitude of the line current is 3 times the 
magnitude of the phase current, and the value is  
A I I
RY R
5 . 17 1 . 10 3 3 = × = · = 
It has been shown that the line current,  lags the corresponding phase current,  
by .  
R
I
RY
I
° 30
So, the line currents are, 
 15 5 . 17 165 1 . 10 45 1 . 10 A I I I
BR RY R
° + ? = ° + ? - ° + ? = - =  
 135 .5 17 ; 105 5 . 17 A I I I A I I I
YB BR B RY YB Y
° + ? = - = ° - ? = - = 
Version 2 EE IIT, Kharagpur 
Page 5


Solution  
V V
L
400 =  O ° + ? = + = = 87 . 36 25 15 20 j Z Z
p RN
  
As the star-connected load is balanced, the magnitude of the phase voltage is,  
V V V
L p
231 3 400 3 = = =  
Taking the phase voltage,  as reference, the phase voltages are, 
RN
V
° + ? = ° - ? = ° ? = 120 231 ; 120 231 ; 0 231
BN YN RN
V V V 
The phasor diagram is shown in Fig. 20.6b. It has been shown that the line voltage, 
 leads the corresponding phase voltage,  by . So, the line voltages are, 
RY
V
RN
V ° 30
° + ? = ° - ? = ° + ? = 150 400 ; 90 400 ; 30 400
BR YB RY
V V V 
For a star-connected load, the phase and the line currents are same. 
The current in R-phase is, 
() ( ) A Z V I I
p RN R RN
° - ? = ° + ? ° ? = = = 87 . 36 24 . 9 87 . 36 0 . 25 0 0 . 231 /
A j ) 54 . 5 39 . 7 ( - = 
 Two other phase and line currents are,  
  13 . 83 9.24 ; 87 . 156 24 . 9 A I I A I I
B BN Y YN
° + ? = = ° - ? = =  
The power factor of the load is 8 . 0 87 . 36 cos cos = ° = f lagging, with ° + = 87 . 36 f , 
as the load is inductive. 
Total VA 403 . 6 24 . 9 0 . 231 3 3 = × × = · · =
p p
I V kVA 
This can be taken as 403 . 6 24 . 9 400 3 3 = × × = · ·
l L
I V kVA 
Total power 123 . 5 8 . 0 24 . 9 0 . 231 3 cos 3 = × × × = · · · = f
p p
I V kW 
Total reactive VA 842 . 3 87 . 36 sin 24 . 9 0 . 231 3 sin 3 = ° × × × = · · · = f
p p
I V kVAR 
The readings of the two wattmeters are, 
45 . 1 87 . 66 cos 24 . 9 400 ) 30 ( cos
1
= ° × × = + ° · · = f
R RY
I V W kW 
The phase angle between and  is  , obtained using two phasors. 
RY
V
R
I ° 87 . 66
67 . 3 87 . 6 cos 24 . 9 400 ) 30 ( cos
2
= ° × × = - ° · · = f
B BY
I V W kW 
The phase angle between and  is , obtained using two phasors, where 
. 
BY
V
B
I ° 87 . 6
° + ? = 90 400
BY
V
The sum of two readings is 12 . 5 ) 67 . 3 45 . 1 ( = + kW, which is same as the total power 
computed earlier  
Example 20.2 
 Calculate the readings of the wattmeter (W) connected as shown in Fig. 20.7a. The 
load is the same, as in Fig. 20.7a (Ex. 20.1), i.e., balanced star-connected one, with 
impedance of  per phase, fed from a three-phase, 400 V, balanced supply, 
with the phase sequence as R-Y-B.  
O + ) 15 20 ( j
 
 
 
Version 2 EE IIT, Kharagpur 
 
•
•
R 
Y 
B 
Balanced 
star-
connected 
load as in 
Fig. 20.6(a) 
W 
(a) 
I
R 
30° 
V
RY
V
RN
I
RN
V
YB
90°- F 
F
(b)
Fig. 20.7 (a) Circuit diagram (Example 20.2) 
 (b) Phasor diagram 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Solution  
 The steps are not repeated here, but taken from previous example (20.1).  
 The phasor diagram is shown in Fig. 20.7b.   
 The phase voltage,  is taken as reference as in Ex. 20.1. 
RN
V
 The phase current,  is 
RN
I A I
RN
° - ? = 87 . 36 24 .9 
 The phase angle, f of the load impedance is ° + = 87 . 36 f 
 The line voltage,  is 
RY
V V V
RY
° + ? = 30 400  
Version 2 EE IIT, Kharagpur 
 The line voltage,  is 
YB
V V V
YB
° - ? = 90 400 
 The reading of the wattmeter (W) is, 
f ? sin 3 ) 90 ( cos ) , ( cos · · · - = + ° · · == · · =
p p p L RN YB RN YB
I V I V I V I V W
kW 218 . 2 87 . 36 sin 24 . 9 400 - = ° × × - = 
 The value is negative, as the load is inductive. 
The reading (W) is 577 . 0 3 / 1 = times the total reactive power as, 
f f sin 3 sin 3 · · · = · · ·
L L p p
I V I V ,  
or 3 times the reactive power per phase. 
Example 20.3 
 Calculate the readings of the two wattmeters ( & ) connected to measure the 
total power for a balanced delta-connected load shown in Fig. 20.8a, fed from a three-
phase, 200 V balanced supply with phase sequence as R-Y-B. The load impedance per 
phase is
1
W
2
W
O - ) 14 14 ( j . Also find the line and phase currents, power factor, total power, 
total reactive VA and total VA. 
  
X
C
= 14 O / R = 14 O
R 
Y 
B 
W
2 
•
•
I
B 
I
Y 
I
R 
(a) 
Z
1
I
RY
Z
1
Z
1
I
BR
I
YB
I
RY
Z
1
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Version 2 EE IIT, Kharagpur 
 
I
R
V
RY
I
YB
V
YB
I
BR
V
BR
V
BY
I
Y
I
B
30°
30°
( F-30 °) 
( F+30 °)
60°
F 
F 
I
RY
(b) 
F 
Fig. 20.8 (a) Circuit diagram for a three-phase system with balanced delta-
connected load (Example 20.3) 
 (b) Phasor diagram 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Solution  
O ° - ? = - = = 45 8 . 19 14 14 j Z Z
p RY
  
For delta-connected load, V V V
p L
200 = =  
Taking the line or phase voltage  as reference, the line or phase voltages are,  
RY
V
° + ? = ° - ? = ° ? = 120 200 ; 120 200 ; 0 200
BR YB RY
V V V 
The phasor diagram is shown in Fig. 20.8b. The phase current,  is, 
RY
I
() ( ) A j Z V I
p RY RY
) 142 . 7 142 . 7 ( 45 1 . 10 45 9.8 1 / 0 0 . 200 / + = ° + ? = ° - ? ° ? = =  
The other two phase currents are, 
  165 .1 10 ; 75 1 . 10 A I A I
BR YN
° + ? = ° - ? =  
The power factor of the load is 707 . 0 45 cos cos = ° = f leading, with ° - = 45 f , as 
the load is capacitive. 
As the phase currents are balanced, the magnitude of the line current is 3 times the 
magnitude of the phase current, and the value is  
A I I
RY R
5 . 17 1 . 10 3 3 = × = · = 
It has been shown that the line current,  lags the corresponding phase current,  
by .  
R
I
RY
I
° 30
So, the line currents are, 
 15 5 . 17 165 1 . 10 45 1 . 10 A I I I
BR RY R
° + ? = ° + ? - ° + ? = - =  
 135 .5 17 ; 105 5 . 17 A I I I A I I I
YB BR B RY YB Y
° + ? = - = ° - ? = - = 
Version 2 EE IIT, Kharagpur 
The procedure is only presented, with the steps given in brief.  
Total VA 06 . 6 1 . 10 0 . 200 3 3 = × × = · · =
p p
I V kVA 
This can be taken as 06 . 6 5 . 17 200 3 3 = × × = · ·
l L
I V kVA 
Total power 285 . 4 707 . 0 1 . 10 0 . 200 3 cos 3 = × × × = · · · = f
p p
I V kW 
Total reactive VA 285 . 4 . 45 sin 1 . 10 0 . 200 3 sin 3 = ° × × × = · · · = f
p p
I V kVAR 
The readings of the two wattmeters are, 
38 . 3 ) 15 cos( 5 . 17 200 ) 30 ( cos
1
= ° - × × = + ° · · = f
R RY
I V W kW 
The phase angle between and  is  , obtained using two phasors. 
RY
V
R
I ° 15
906 . 0 75 cos 5 . 17 200 ) 30 ( cos
2
= ° × × = - ° · · = f
B BY
I V W kW 
The phase angle between and is , obtained using two phasors, where 
. 
BY
V
B
I ° 75
° + ? = 60 200
BY
V
The sum of two readings is 286 . 4 ) 906 . 0 38 . 3 ( = + kW, which is same as the total 
power computed earlier  
Alternatively, the phase current,  can be taken as reference, with the corres-
ponding phase voltage,  leading the current by the angle of the load impedance, 
RY
I
RY
V
° = 45 f . So, the phase current and voltage are, 
; 0 1 . 10 A I
RY
° ? =     V Z I V
p RY RY
° ? = ° ? × = ? · ° ? = 45 200 45 ) 8 . 19 1 . 10 ( 0 f 
Two other phase currents and voltages are, 
° + ? = ° - ? = 120 1 . 10 ; 120 1 . 10
BR YB
I I      
° + ? = ° - ? = 165 200 ; 75 200
BR YB
V V 
The line current,  is,   
R
I
 30 5 . 17 120 1 . 10 0 1 . 10 A I I I
BR RY R
° + ? = ° + ? - ° ? = - =  
Two other line currents are,   
° + ? = ° - ? = 150 .5 17 ; 90 5 . 17
B Y
I I 
The other steps are not shown here. The readers are requested to study the previous 
lesson (No. 19) in this module. 
 
The measurement of power using two wattmeters for load (unbalanced or balanced), 
fed from a balanced three-phase supply, is discussed in this lesson. Also presented are the 
readings of the two wattmeters for balanced load, along with the determination of the 
load power factor from the two readings, and some comments on the way, the two 
readings vary with change in power factor of the load, with the magnitude of the load 
impedance remaining constant. Some examples are also described here. This is the last 
lesson in this module (No.5). In the next module (No.6) consisting of two lessons (no. 21-
22) only, the discussion on magnetic circuits will be taken up.  
 
20.1 Calculate the reading of the two wattmeter’s (W
1
, and W
2
) connected to measure 
the power for a balanced three-phase load as given in the following. The supply 
voltage is 200V and the phase. 
Version 2 EE IIT, Kharagpur

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FAQs on Three-Phase Delta Connected Balanced Load - 2 - 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 load where three identical loads are connected in a delta configuration. In this setup, each load is connected between two phase conductors, forming a closed loop resembling the Greek letter delta (∆). The loads in a delta configuration are balanced, meaning they have equal impedance and are evenly distributed across the three phases.

2. How does a three-phase delta connected balanced load differ from a star connected load?

Ans. A three-phase delta connected balanced load differs from a star connected load in terms of how the loads are connected. In a star connected load, each load is connected between one phase conductor and a common neutral point, forming a star configuration. On the other hand, in a delta connected load, each load is connected between two phase conductors, forming a closed loop. Additionally, the phase voltages and line voltages in a delta connected load differ in magnitude, while they are the same in a star connected load.

3. What are the advantages of using a three-phase delta connected balanced load?

Ans. There are several advantages of using a three-phase delta connected balanced load. Firstly, it allows for a higher power transfer capability compared to a single-phase load. Secondly, it provides a more balanced distribution of power across the three phases, reducing the likelihood of overload or voltage imbalance. Lastly, it offers better fault tolerance as a fault in one phase does not affect the other phases.

4. How can the power in a three-phase delta connected balanced load be calculated?

Ans. The power in a three-phase delta connected balanced load can be calculated using the formula P = √3 * V * I * cos(θ), where P is the power, V is the line voltage, I is the line current, and θ is the phase angle between the voltage and current. The factor of √3 accounts for the presence of three phases in the load.

5. Can a three-phase delta connected balanced load be easily converted to a star connected load?

Ans. Yes, a three-phase delta connected balanced load can be easily converted to a star connected load by connecting a neutral wire to the common connection point of the loads. However, it is important to ensure that the loads have the same impedance and are balanced to maintain the desired performance and distribution of power.

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