CHAPTER 20 THREE PHASE SYSTEMS
Exercise 112, Page 327
1. Three loads, each of resistance 50  are connected in star to a 400 V, 3-phase supply. Determine
(a) the phase voltage, (b) the phase current and (c) the line current.
400 V, 3-phase supply means that 400 V is the line voltage.
(a) For a star connection, VL  3 VP
Hence, phase voltage, VP 
(b) Phase current, I P 
VL 400

= 231 V
3
3
VP 231

= 4.62 A
R P 50
(c) For a star connection, IP  IL
Hence, line current, IL = 4.62 A
2. A star-connected load consists of three identical coils, each of inductance 159.2 mH and resistance
50 . If the supply frequency is 50 Hz and the line current is 3 A determine (a) the phase voltage
and (b) the line voltage.
Inductive reactance, XL = 2πfL = 2π(50)(159.2 × 10- 3) = 50 
Impedance of each phase, Zp =
For a star connection, IL = Ip =
R 2  X2L =
502  502 = 70.71 
Vp
Zp
Hence, phase voltage, Vp = Ip Zp = (3)(70.71) = 212 V
Line voltage,
VL =
3 Vp =
3 (212) = 367 V
3. Three identical capacitors are connected in star to a 400 V, 50 Hz, 3-phase supply. If the line
current is 12 A determine the capacitance of each of the capacitors.
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251
For a star connection, IL  12A  IP
VL  3 VP hence, VP 
XC 
VL 400
= 231 V

3
3
VP 231

 19.25 
IP
12
and capacitance, C =
thus,
1
1
 19.25
2 f C
2  50 19.25 
= 165.4 F
4. Three coils each having resistance 6  and inductance L H are connected in star to a 415 V, 50 Hz,
3-phase supply. If the line current is 30 A, find the value of L.
For a star connection, IL  30A  IP
VL  3 VP hence, VP 
ZL 
VL 415

= 239.6 V
3
3
VP 239.6

 7.987 
IP
30
X L  7.987 2  R 2  7.987 2  62 = 5.272 Ω
from which,
Hence,
and
R 2  XL 2  7.987
thus,
5.272 = 2π f L
inductance, L =
5.272
= 16.78 mH
2  50 
5. A 400 V, 3-phase, 4 wire, star-connected system supplies three resistive loads of 15 kW, 20 kW
and 25 kW in the red, yellow and blue phases respectively. Determine the current flowing in each
of the four conductors.
For a star connected system, VL  3 VP
from which, VP 
Power, P = VI for a resistive load, hence I 
Thus,
IR 
VL 400

= 230.94 V
3
3
P
V
PR 15000

= 64.95 A,
VR 230.94
IY 
PY
20000

= 86.60 A
VY 230.94
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252
and
IB 
PB 25000
= 108.25 A

VB 230.94
The phasor diagram of the three currents is shown in (i) below. Adding phasorially gives diagram
(ii) below, where I N is the neutral current.
(i)
(ii)
Total horizontal component = 64.95 cos 90 + 108.25 cos 210 + 86.60 cos 330 = - 18.75
Total vertical component = 64.95 sin 90 + 108.25 sin 210 + 86.60 sin 330 = - 32.475
Hence, magnitude of neutral current, I N = 18.752  32.4752 = 37.50 A
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253
Exercise 113, Page 329
1. Three loads, each of resistance 50  are connected in delta to a 400 V, 3-phase supply.
Determine (a) the phase voltage, (b) the phase current and (c) the line current.
(a) For a delta connection, VL  VP
Since VL = 400 V, then phase voltage, VP = 400 V
(b) Phase current, I P 
VP 400
=8A

RP
50
(c) For a delta connection, line current, IL  3 IP  3 8 = 13.86 A
2. Three inductive loads each of resistance 75  and inductance 318.4 mH are connected in delta to a
415 V, 50 Hz, 3-phase supply. Determine (a) the phase voltage, (b) the phase current, and (c) the
line current.
(a) For a delta connection, VL  VP
Since VL = 410 V, then phase voltage, VP = 415 V
(b) Phase impedance, ZP  R 2  X L 2 
=
Phase current, I P 
 75
2
  2 50  318.4 103 
2
752  1002  125 
VP 415

= 3.32 A
ZP 125
(c) For a delta connection, line current, IL  3 IP  3  3.32  = 5.75 A
3. Three identical capacitors are connected in delta to a 400 V, 50 Hz, 3-phase supply. If the line
current is 12 A determine the capacitance of each of the capacitors.
IL 12

 6.93 A
3
3
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For a delta connection, IL  3 IP hence IP 
254
VL  VP = 400 V
XC 
VP 400

 57.74 
I P 6.93
and capacitance, C =
thus,
1
2  50  57.74 
1
 57.74
2 f C
= 55.13 F
4. Three coils each having resistance 6  and inductance L H are connected in delta, to a 415 V,
50 Hz, 3-phase supply. If the line current is 30 A, find the value of L.
For a delta connection, IL  3 IP hence IP 
IL
30

 17.32 A
3
3
VL  VP = 415 V
ZP 
VP
415

 23.96 
I P 17.32
from which,
thus, 23.96 =
X L  23.962  62  23.197 
Hence, inductance, L =
R 2  XL2  62  XL2
i.e. 2π f L = 23.197 Ω
23.197
= 73.84 mH
2  50 
5. A 3-phase, star-connected alternator delivers a line current of 65 A to a balanced delta-connected
load at a line voltage of 380 V. Calculate (a) the phase voltage of the alternator, (b) the alternator
phase current and (c) the load phase current.
(a) In star, VL  3 VP
from which, phase voltage of alternator, VP 
VL 380

= 219.4 V
3
3
(b) In star, IP  IL hence, alternator phase current = 65 A
(c) In delta, IL  3 IP from which, load phase angle, IP 
IL
65

= 37.53 A
3
3
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6. Three 24 F capacitors are connected in star across a 400 V, 50 Hz, 3-phase supply. What value
of capacitance must be connected in delta in order to take the same line current?
In star, VL  3 VP hence, VP 
XC 
VL 400
= 230.94 V

3
3
1
1

= 132.63 
2 f C 2  50   24 106 
Hence,
IP 
VP 230.94
= 1.741 A = line current for star connection.

X C 132.63
In delta, if IL  1.741  3 IP hence, IP 
1.741
= 1.00517 A
3
VP VL
400
= 397.94 


IP
I P 1.00517
397.94 =
XC 
from which,
i.e.
capacitance in delta, C =
1
2 f C
1
= 8 F
2  50  397.94 
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256
Exercise 114, Page 331
1. Determine the total power dissipated by three 20  resistors when connected (a) in star and
(b) in delta to a 440 V, 3-phase supply.
(a) In star, VL  440V  3 VP hence, VP 
and
IP 
VP 254

 12.70 A  I L
RP
20
Hence in star, power, P =
or
440
 254 V
3
3 VL IL cos   3  440 12.70  cos 0 = 9.68 kW
P = 3I P 2 R P  3 12.70   20  = 9.68 kW
2
(b) In delta, VL  VP  440 V
and
IP 
VP 440

 22 A
RP
20
IL  3 IP  3  22  = 38.11 A
Hence in delta, power, P =
3 VL IL cos   3  440  38.11 cos 0 = 29.04 kW
P = 3I P 2 R P  3  22   20  = 1350 W = 29.04 kW
2
or
2. Determine the power dissipated in the circuit of Problem 2, Exercise 112, page 327.
X 
 50 
Circuit phase angle, ϕ = tan 1  L   tan 1    tan 1 1  45
 50 
 R 
Power, P =
or
3 VL IL cos   3  367  3 cos 45 = 1348 W = 1.35 kW
P = 3I P 2 R P  3  3  50  = 1.35 kW
2
3. A balanced delta-connected load has a line voltage of 400 V, a line current of 8 A and a lagging
power factor of 0.94. Draw a complete phasor diagram of the load. What is the total power
dissipated by the load ?
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If the power factor is 0.94, then 0.94 = cos ϕ
from which, ϕ = cos 1 0.94 = 19.95º
Line voltage = phase voltage = 400 V
Line current = 8 A
Phase current =
8
= 4.62 A
3
From the diagram above, IR  IRY  IBR , IY  IYB  IRY and IB  IBR  IYB
The complete phasor diagram is shown below, with the line current lagging the line voltage by
19.95º
Power, P =
3 VL IL cos   3  400 8 0.94 
since power factor = cos ϕ
= 5210 W = 5.21 kW
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4. Three inductive loads, each of resistance 4  and reactance 9  are connected in delta. When
connected to a 3-phase supply the loads consume 1.2 kW. Calculate (a) the power factor of the
load, (b) the phase current, (c) the line current and (d) the supply voltage.
(a) Phase impedance, ZP  42  92  9.849 
X 
9
and phase angle,   tan 1  L   tan 1    66.04
4
 R 
Hence, power factor of load = cos  = cos 66.04 = 0.406
i.e. 1.2 103  3IP 2  4
(b) Power, P = 3I P 2 R P
from which, phase current, I P 
1200
= 10 A
3(4)
(c) In delta, line current, IL  3 (10) = 17.32 A
(d) Power, P =
from which,
3 VL IL cos 
i.e.
1200 =
supply voltage, VL 
3 VL (17.32)(0.406)
1200
= 98.53 V
3 (17.32)(0.406)
5. The input voltage, current and power to a motor is measured as 415 V, 16.4 A and 6 kW
respectively, Determine the power factor of the system.
Power, P =
from which,
3 VL IL cos 
i.e.
6000 =
3 (415) (16.4) cos 
cos  = power factor of system =
6000
= 0.509
3 (415)(16.4)
6. A 440 V, 3-phase a.c. motor has a power output of 11.25 kW and operates at a power facror of
0.8 lagging and with an efficiency of 84%. If the motor is delta connected determine (a) the
power input, (b) the line current and (c) the phase current.
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259
(a) Efficiency =
from which,
(b) Power, P =
power output
power input
i.e.
power input =
0.84 =
11250
power input
11250
= 13393 W or 13.39 kW
0.84
3 VL IL cos  hence, line current, IL 
(c) In delta, IL  3 IP
from which, phase current, I P 
P
13393
= 21.97 A

3 VL cos 
3  440  0.80 
I L 21.97
= 12.68 A

3
3
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Exercise 115, Page 336
1. Two wattmeters are connected to measure the input power to a balanced three-phase load. If the
wattmeter readings are 9.3 kW and 5.4 kW determine (a) the total output power, and (b) the load
power factor
(a)Total input power, P = P1 + P2 = 9.3 + 5.4 = 14.7 kW
(b) tan  =
 P P 
3 1 2  =
 P1  P2 
 9.3  5.4 
3
 =
 9.3  5.4 
 3.9 
3
 = 0.459524
 14.7 
Hence,  = tan 1 (0.459524) = 24.68o
Power factor = cos  = cos 24.68o = 0.909
2. 8 kW is found by the two-wattmeter method to be the power input to a 3-phase motor.
Determine the reading of each wattmeter if the power factor of the system is 0.85
Working in kilowatts,
8 = P1  P2
If power factor = 0.85, then
cos  = 0.85
and
tan  = tan 31.79 = 0.6197
Hence,
tan  = 0.6197 =
from which,
and phase angle,  = cos 1 0.85 = 31.79
 P P 
 P P 
3 1 2   3 1 2 
 8 
 P1  P2 
P1  P2 
Adding equations (1) and (2) gives:
from which,
(1)
 0.6197 8
3
= 2.862
(2)
2 P1  10.862
P1 
10.862
= 5.431 kW
2
and from equation (1), P2  8  5.431 = 2.569 kW
Thus the readings of the two wattmeters are: 5.431 kW and 2.569 kW
© John Bird Published by Taylor and Francis
261
3. When the two-wattmeter method is used to measure the input power of a balanced load, the
readings on the wattmeters are 7.5 kW and 2.5 kW, the connections to one of the coils on the meter
reading 2.5 kW having to be reversed. Determine (a) the total input power, and (b) the load power
factor
Since the reversing switch on the wattmeter had to be operated the 2.5 kW reading is taken as
– 2.5 kW
(a) Total input power, P = P1 + P2 = 7.5 + (- 2.5) = 5 kW
(b) tan  =
 P P 
3 1 2  =
 P1  P2 
 7.5  ( 2.5) 
3
 =
 7.5  ( 2.5) 
 10 
3   2 3
 5
Angle  = tan 1 (2 3 ) = 73.90o
Power factor = cos  = cos 73.90o = 0.277
4. Three similar coils, each having a resistance of 4.0  and an inductive reactance of 3.46  are
connected (a) in star and (b) in delta across a 400 V, 3-phase supply. Calculate for each
connection the readings on each of two wattmeters connected to measure the power by the twowattmeter method.
(a) Star connection: VL  3 VP
Hence, VP 
IL  IP
and
VL 400

= 230.94 V
3
3
Phase impedance, ZP  4.02  3.462 = 5.289 
Phase current, I P 
VP 230.94

= 43.664 A
ZP
5.289
Total power, P = 3I P 2 R P  3  43.664   4.0  = 22.879 kW
2
If the wattmeter readings are P1 and P2 then:
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P1 + P2 = 22.879
(1)
X 
 3.46 
Phase angle,  = tan 1  L   tan 1 
  40.86
 4 
 R 
and
tan 40 86 =
 P P 
 P P 
3  1 2   3  1 2  from (1)
 22.879 
 P1  P2 
22.879 tan 40.86 
3
from which,
P1 - P2 =
i.e.
P1 - P2 = 11.426
(2)
2 P1 = 22.879 + 11.426 = 34.305
Adding equations (1) and (2) gives:
34.305
= 17.15 kW
2
and
P1 =
Substituting in (1) gives:
P2 = 22.879 – 17.15 = 5.73 kW
Hence, in star, the wattmeter readings are 17.15 kW and 5.73 kW
(b) Delta connection: VL  VP
Phase current, I P 
IL  3 IP
and
VP
400

= 75.629 A
ZP 5.289
Total power, P = 3I P 2 R P  3  75.629   4.0  = 68.637 kW
2
P1 + P2 = 68.637
Hence
tan 40 86 =
(3)
 P P 
 P P 
3  1 2   3  1 2  from (3)
 68.637 
 P1  P2 
68.637 tan 40.86 
3
from which,
P1 - P2 =
i.e.
P1 - P2 = 34.278
Adding equations (3) and (4) gives:
(4)
2 P1 = 68.637 + 34.278 = 102.915
102.915
= 51.46 kW
2
and
P1 =
Substituting in (3) gives:
P2 = 68.637 – 51.46 = 17.18 kW
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Hence, in delta, the wattmeter readings are 51.46 kW and 17.18 kW
5. A 3-phase, star-connected alternator supplies a delta connected load, each phase of which has a
resistance of 15  and inductive reactance 20 . If the line voltage is 400 V, calculate (a) the
current supplied by the alternator and (b) the output power and kVA rating of the alternator,
neglecting any losses in the line between the alternator and the load.
A circuit diagram of the alternator and load is shown below.
Vp
(a) Considering the load: Phase current, Ip =
Zp
Vp = VL for a delta connection, hence, Vp = 400 V
Phase impedance, Zp =
Hence, Ip =
Vp
Zp
=
R 2p  X 2L = 152  202 = 25 
400
= 16 A
25
For a delta-connection, line current, IL =
3 Ip =
3 (16) = 27.71 A
Hence, 27.71 A is the current supplied by the alternator
(b) Alternator output power is equal to the power dissipated by the load
i.e. P =
3 VL IL cos , where cos  =
Rp
Zp
=
15
= 0.6
25
Hence, P = 3 (400)(27.71)(0.6) = 11518.8 W = 11.52 kW
© John Bird Published by Taylor and Francis
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Alternator output kVA, S =
3 VL IL = 3 (400)(27.71) = 19198 VA = 19.20 kVA
6. Each phase of a delta-connected load comprises a resistance of 40  and a 40 F capacitor in
series. Determine, when connected to a 415 V, 50 Hz, 3-phase supply (a) the phase current,
(b) the line current, (c) the total power dissipated, and (d) the kVA rating of the load.
(a) Capacitive reactance, X C 
1
1

 79.58 
2 f C 2  50   40 106 
Phase impedance, ZP  R P 2  XC2  402  79.582  89.067 
Phase current, I P 
VP VL
415


= 4.66 A
ZP ZP 89.067
(since VP  VL in delta)
(b) Line current, IL  3 IP  3 (4.66) = 8.07 A
(c) From the impedance triangle, cos  =
Hence, total power dissipated, P =
(d) The kVA rating of the load, S =
RP
40
= 0.449

ZP 89.067
3 VL IL cos   3  4158.07  0.449  = 2.605 kW
3 VL IL  3  4158.07  = 5.80 kVA
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