Chapter 12, Solution 1.
(a) If Vab = 400 , then
400
Van = ∠ - 30° = 231∠ - 30° V
3
Vbn = 231∠ - 150° V
Vcn = 231∠ - 270° V
(b) For the acb sequence,
Vab = Van − Vbn = Vp ∠0° − Vp ∠120°
1 3
Vab = Vp 1 + − j = Vp 3∠ - 30°
2 2
i.e. in the acb sequence, Vab lags Van by 30°.
Hence, if Vab = 400 , then
400
Van = ∠30° = 231∠30° V
3
Vbn = 231∠150° V
Vcn = 231∠ - 90° V
Chapter 12, Solution 2.
Since phase c lags phase a by 120°, this is an acb sequence.
Vbn = 160∠(30° + 120°) = 160∠150° V
Chapter 12, Solution 3.
Since Vbn leads Vcn by 120°, this is an abc sequence.
Van = 208∠(130° + 120°) = 208∠ 250° V
Chapter 12, Solution 4.
Vbc = Vca ∠120° = 208∠140° V
Vab = Vbc ∠120° = 208∠260° V
Vab 208∠260°
Van = = = 120∠230° V
3 ∠30° 3 ∠30°
Vbn = Van ∠ - 120° = 120∠110° V
Chapter 12, Solution 5.
This is an abc phase sequence.
Vab = Van 3 ∠30°
Vab 420∠0°
or Van = = = 242.5∠ - 30° V
3 ∠30° 3 ∠30°
Vbn = Van ∠ - 120° = 242.5∠ - 150° V
Vcn = Van ∠120° = 242.5∠90° V
Chapter 12, Solution 6.
Z Y = 10 + j5 = 11.18∠26.56°
The line currents are
Van 220 ∠0°
Ia = = = 19.68∠ - 26.56° A
Z Y 11.18∠26.56°
I b = I a ∠ - 120° = 19.68∠ - 146.56° A
I c = I a ∠120° = 19.68∠93.44° A
The line voltages are
Vab = 200 3 ∠30° = 381∠30° V
Vbc = 381∠ - 90° V
Vca = 381∠ - 210° V
The load voltages are
VAN = I a Z Y = Van = 220∠0° V
VBN = Vbn = 220∠ - 120° V
VCN = Vcn = 220∠120° V
Chapter 12, Solution 7.
This is a balanced Y-Y system.
+
440∠0° V ZY = 6 − j8 Ω
−
Using the per-phase circuit shown above,
440∠0°
Ia = = 44∠53.13° A
6 − j8
I b = I a ∠ - 120° = 44∠ - 66.87° A
I c = I a ∠120° = 44∠173.13° A
Chapter 12, Solution 8.
VL = 220 V , Z Y = 16 + j9 Ω
Vp VL 220
I an = = = = 6.918∠ - 29.36°
ZY 3 ZY 3 (16 + j9)
I L = 6.918 A
Chapter 12, Solution 9.
Van 120 ∠0°
Ia = = = 4.8∠ - 36.87° A
Z L + Z Y 20 + j15
I b = I a ∠ - 120° = 4.8∠ - 156.87° A
I c = I a ∠120° = 4.8∠83.13° A
As a balanced system, I n = 0 A
Chapter 12, Solution 10.
Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
Van 220 ∠0°
Ia = = = 6.55∠36.53°
Z A + 2 27 − j20
For phase b,
Vbn 220 ∠ - 120°
Ib = = = 10 ∠ - 120°
ZB + 2 22
For phase c,
Vcn 220 ∠120°
Ic = = = 16.92 ∠97.38°
ZC + 2 12 + j5
The current in the neutral line is
I n = -(I a + I b + I c )
or - In = Ia + Ib + Ic
- I n = (5.263 + j3.9) + (-5 − j8.66) + (-2.173 + j16.78)
I n = 1.91 − j12.02 = 12.17 ∠ - 81° A
Chapter 12, Solution 11.
Vbc VBC 220∠10°
Van = = =
3 ∠ - 90° 3 ∠ - 90° 3 ∠ - 90°
Van = 127 ∠100° V
VAB = VBC ∠120° = 220∠130° V
VAC = VBC ∠ - 120° = 220∠ - 110° V
If I bB = 30 ∠60° , then
I aA = 30∠180° , I cC = 30 ∠ - 60°
I aA 30∠180°
I AB = = = 17.32∠210°
3 ∠ - 30° 3 ∠ - 30°
I BC = 17.32∠90° , I CA = 17.32 ∠ - 30°
I AC = -I CA = 17.32∠150° A
I BC Z = VBC
VBC 220 ∠0°
Z= = = 12.7 ∠ - 80° Ω
I BC 17.32 ∠90°
Chapter 12, Solution 12.
Convert the delta-load to a wye-load and apply per-phase analysis.
Ia
+
110∠0° V ZY
−
Z∆
ZY = = 20 ∠45° Ω
3
110∠0°
Ia = = 5.5∠ - 45° A
20∠45°
I b = I a ∠ - 120° = 5.5∠ - 165° A
I c = I a ∠120° = 5.5∠75° A
Chapter 12, Solution 13.
First we calculate the wye equivalent of the balanced load.
ZY = (1/3)Z∆ = 6+j5
Now we only need to calculate the line currents using the wye-wye circuits.
110
Ia = = 6.471∠ − 61.93° A
2 + j10 + 6 + j5
110∠ − 120°
Ib = = 6.471∠178.07° A
8 + j15
110∠120°
Ic = = 6.471∠58.07° A
8 + j15
Chapter 12, Solution 14.
We apply mesh analysis.
1 + j 2Ω A
a
+ ZL
100∠0 o V ZL
- I3
n I1 B C
- -
100∠120 o V 100∠120 o V Z L = 12 + j12Ω
+ - +
c I2 b 1 + j 2Ω
1 + j 2Ω
For mesh 1,
− 100 + 100∠120 o + I 1 (14 + j16) − (1 + j 2) I 2 − (12 + j12) I 3 = 0
or
(14 + j16) I 1 − (1 + j 2) I 2 − (12 + j12) I 3 = 100 + 50 − j86.6 = 150 − j86.6 (1)
For mesh 2,
100∠120 o − 100∠ − 120 o − I 1 (1 + j 2) − (12 + j12) I 3 + (14 + j16) I 2 = 0
or
− (1 + j 2) I 1 + (14 + j16) I 2 − (12 + j12) I 3 = −50 − j86.6 + 50 − j86.6 = − j173.2 (2)
For mesh 3,
− (12 + j12) I 1 − (12 + j12) I 2 + (36 + j 36) I 3 = 0 (3)
Solving (1) to (3) gives
I 1 = −3.161 − j19.3, I 2 = −10.098 − j16.749, I 3 = −4.4197 − j12.016
I aA = I 1 = 19.58∠ − 99.3 A o
I bB = I 2 − I 1 = 7.392∠159.8 o A
I cC = − I 2 = 19.56∠58.91o A
Chapter 12, Solution 15.
Convert the delta load, Z ∆ , to its equivalent wye load.
Z∆
Z Ye = = 8 − j10
3
(12 + j5)(8 − j10)
Z p = Z Y || Z Ye = = 8.076∠ - 14.68°
20 − j5
Z p = 7.812 − j2.047
Z T = Z p + Z L = 8.812 − j1.047
Z T = 8.874 ∠ - 6.78°
We now use the per-phase equivalent circuit.
Vp 210
Ia = , where Vp =
Zp + ZL 3
210
Ia = = 13.66 ∠6.78°
3 (8.874 ∠ - 6.78°)
I L = I a = 13.66 A
Chapter 12, Solution 16.
(a) I CA = - I AC = 10∠(-30° + 180°) = 10∠150°
This implies that
I AB = 10 ∠30°
I BC = 10∠ - 90°
I a = I AB 3 ∠ - 30° = 17.32∠0° A
I b = 17.32∠ - 120° A
I c = 17.32∠120° A
VAB 110 ∠0°
(b) Z∆ = = = 11∠ - 30° Ω
I AB 10 ∠30°
Chapter 12, Solution 17.
Convert the ∆-connected load to a Y-connected load and use per-phase analysis.
ZL Ia
+
Van ZY
−
Z∆
ZY = = 3 + j4
3
Van 120 ∠0°
Ia = = = 19.931∠ - 48.37°
Z Y + Z L (3 + j4) + (1 + j0.5)
But I a = I AB 3 ∠ - 30°
19.931∠ - 48.37°
I AB = = 11.51∠ - 18.37° A
3 ∠ - 30°
I BC = 11.51∠ - 138.4° A
I CA = 11.51∠101.6° A
VAB = I AB Z ∆ = (11.51∠ - 18.37°)(15∠53.13°)
VAB = 172.6∠34.76° V
VBC = 172.6∠ - 85.24° V
VCA = 172.6∠154.8° V
Chapter 12, Solution 18.
VAB = Van 3 ∠30° = (440 ∠60°)( 3 ∠30°) = 762.1∠90°
Z ∆ = 12 + j9 = 15∠36.87°
VAB 762.1∠90°
I AB = = = 50.81∠53.13° A
Z ∆ 15∠36.87°
I BC = I AB ∠ - 120° = 50.81∠ - 66.87° A
I CA = I AB ∠120° = 50.81∠173.13° A
Chapter 12, Solution 19.
Z ∆ = 30 + j10 = 31.62 ∠18.43°
The phase currents are
Vab 173∠0°
I AB = = = 5.47 ∠ - 18.43° A
Z ∆ 31.62 ∠18.43°
I BC = I AB ∠ - 120° = 5.47 ∠ - 138.43° A
I CA = I AB ∠120° = 5.47 ∠101.57° A
The line currents are
I a = I AB − I CA = I AB 3 ∠ - 30°
I a = 5.47 3 ∠ - 48.43° = 9.474∠ - 48.43° A
I b = I a ∠ - 120° = 9.474∠ - 168.43° A
I c = I a ∠120° = 9.474∠71.57° A
Chapter 12, Solution 20.
Z ∆ = 12 + j9 = 15∠36.87°
The phase currents are
210∠0°
I AB = = 14∠ - 36.87° A
15∠36.87°
I BC = I AB ∠ - 120° = 14∠ - 156.87° A
I CA = I AB ∠120° = 14∠83.13° A
The line currents are
I a = I AB 3 ∠ - 30° = 24.25∠ - 66.87° A
I b = I a ∠ - 120° = 24.25∠ - 186.87° A
I c = I a ∠120° = 24.25∠53.13° A
Chapter 12, Solution 21.
− 230∠120° − 230∠120°
(a) I AC = = = 17.96∠ − 98.66° A(rms)
10 + j8 12.806∠38.66°
230∠ − 120 230∠0°
I bB = I BC + I BA = I BC − I AB = −
10 + j8 10 + j8
(b) = 17.96∠ − 158.66° − 17.96∠ − 38.66°
= −16.729 − j6.536 − 14.024 + j11.220 = −30.75 + j4.684
= 31.10∠171.34° A
Chapter 12, Solution 22.
Convert the ∆-connected source to a Y-connected source.
Vp 208
Van = ∠ - 30° = ∠ - 30° = 120 ∠ - 30°
3 3
Convert the ∆-connected load to a Y-connected load.
Z (4 + j6)(4 − j5)
Z = Z Y || ∆ = (4 + j6) || (4 − j5) =
3 8+ j
Z = 5.723 − j0.2153
ZL Ia
+
Van Z
−
Van 120 ∠30°
Ia = = = 15.53∠ - 28.4° A
Z L + Z 7.723 − j0.2153
I b = I a ∠ - 120° = 15.53∠ - 148.4° A
I c = I a ∠120° = 15.53∠91.6° A
Chapter 12, Solution 23.
V AB 208
(a) I AB = =
Z∆ 25∠60 o
208 3∠ − 30 o
I a = I AB 3∠ − 30 = o
= 14.411∠ − 90 o
25∠60 o
I L =| I a |= 14.41 A
208 3
(b) P = P1 + P2 = 3VL I L cosθ = 3 (208)
25 cos 60 = 2.596 kW
o
Chapter 12, Solution 24.
Convert both the source and the load to their wye equivalents.
Z∆
ZY = = 20 ∠30° = 17.32 + j10
3
Vab
Van = ∠ - 30° = 240.2∠0°
3
We now use per-phase analysis.
1+jΩ Ia
+
Van 20∠30° Ω
−
Van 240.2
Ia = = = 11.24∠ - 31° A
(1 + j) + (17.32 + j10) 21.37 ∠31°
I b = I a ∠ - 120° = 11.24∠ - 151° A
I c = I a ∠120° = 11.24∠89° A
But I a = I AB 3 ∠ - 30°
11.24 ∠ - 31°
I AB = = 6.489∠ - 1° A
3 ∠ - 30°
I BC = I AB ∠ - 120° = 6.489∠ - 121° A
I CA = I AB ∠120° = 6.489∠119° A
Chapter 12, Solution 25.
Convert the delta-connected source to an equivalent wye-connected source and
consider the single-phase equivalent.
440 ∠(10° − 30°)
Ia =
3 ZY
where Z Y = 3 + j2 + 10 − j8 = 13 − j6 = 14.32 ∠ - 24°.78°
440 ∠ - 20°
Ia = = 17.74∠4.78° A
3 (14.32 ∠ - 24.78°)
I b = I a ∠ - 120° = 17.74∠ - 115.22° A
I c = I a ∠120° = 17.74∠124.78° A
Chapter 12, Solution 26.
Transform the source to its wye equivalent.
Vp
Van = ∠ - 30° = 72.17 ∠ - 30°
3
Now, use the per-phase equivalent circuit.
Van
I aA = , Z = 24 − j15 = 28.3∠ - 32°
Z
72.17 ∠ - 30°
I aA = = 2.55∠ 2° A
28.3∠ - 32°
I bB = I aA ∠ - 120° = 2.55∠ - 118° A
I cC = I aA ∠120° = 2.55∠122° A
Chapter 12, Solution 27.
Vab ∠ - 30° 220∠ - 10°
Ia = =
3 ZY 3 (20 + j15)
I a = 5.081∠ - 46.87° A
I b = I a ∠ - 120° = 5.081∠ - 166.87° A
I c = I a ∠120° = 5.081∠73.13° A
Chapter 12, Solution 28.
Let Vab = 400∠0°
Van ∠ - 30° 400∠ - 30°
Ia = = = 7.7 ∠30°
3 ZY 3 (30 ∠ - 60°)
I L = I a = 7.7 A
Van
VAN = I a Z Y = ∠ - 30° = 230.94∠ - 30°
3
Vp = VAN = 230.9 V
Chapter 12, Solution 29.
VL
P = 3Vp I p cos θ , Vp = , IL = Ip
3
P = 3 VL I L cos θ
P 5000
IL = = = 20.05 = I p
3 VL cos θ 240 3 (0.6)
Vp VL 240
ZY = = = = 6.911
Ip 3 IL 3 (20.05)
cos θ = 0.6
→ θ = 53.13°
Z Y = 6.911∠ - 53.13° (leading)
Z Y = 4.15 − j5.53 Ω
P 5000
S= = = 8333
pf 0.6
Q = S sin θ = 6667
S = 5000 − j6667 VA
Chapter 12, Solution 30.
Since this a balanced system, we can replace it by a per-phase equivalent, as
shown below.
+ ZL
Vp
-
3V 2 p VL
S = 3S p = , Vp =
Z*p 3
V 2L (208) 2
S= = = 1.4421∠45 o kVA
Z * p 30∠ − 45 o
P = S cosθ = 1.02 kW
Chapter 12, Solution 31.
PP
(a) Pp = 6,000, cosθ = 0.8, Sp = = 6 / 0.8 = 7.5 kVA
cos θ
Q p = S P sin θ = 4.5 kVAR
S = 3S p = 3(6 + j 4.5) = 18 + j13.5 kVA
For delta-connected load, Vp = VL= 240 (rms). But
3V 2 p 3V 2 p 3(240) 2
S=
→ Z*p = = , Z P = 6.144 + j 4.608Ω
Z*p S (18 + j13.5) x10 3
6000
(b) Pp = 3VL I L cosθ
→ IL = = 18.04 A
3 x 240 x0.8
(c ) We find C to bring the power factor to unity
Qc 4500
Qc = Q p = 4.5 kVA
→ C= = = 207.2 µF
ωV rms 2πx60 x 240 2
2
Chapter 12, Solution 32.
S = 3 VL I L ∠θ
S = S = 3 VL I L = 50 × 10 3
5000
IL = = 65.61 A
3 (440)
For a Y-connected load,
VL 440
I p = I L = 65.61 , Vp = = = 254.03
3 3
Vp 254.03
Z = = = 3.872
Ip 65.61
Z = Z ∠θ , θ = cos -1 (0.6) = 53.13°
Z = (3.872)(cos θ + j sin θ)
Z = (3.872)(0.6 + j0.8)
Z = 2.323 + j3.098 Ω
Chapter 12, Solution 33.
S = 3 VL I L ∠θ
S = S = 3 VL I L
For a Y-connected load,
IL = Ip , VL = 3 Vp
S = 3 Vp I p
S 4800
IL = Ip = = = 7.69 A
3 Vp (3)(208)
VL = 3 Vp = 3 × 208 = 360.3 V
Chapter 12, Solution 34.
VL 220
Vp = =
3 3
Vp 200
Ia = = = 6.73∠58°
ZY 3 (10 − j16)
I L = I p = 6.73 A
S = 3 VL I L ∠θ = 3 × 220 × 6.73∠ - 58°
S = 1359 − j2174.8 VA
Chapter 12, Solution 35.
(a) This is a balanced three-phase system and we can use per phase equivalent
circuit. The delta-connected load is converted to its wye-connected equivalent
1
Z '' y = Z ∆ = (60 + j 30) / 3 = 20 + j10
3
IL
+
Z’y Z’’y
230 V
-
Z y = Z ' y // Z '' y = (40 + j10) //( 20 + j10) = 13.5 + j 5.5
230
IL = = 14.61 − j 5.953 A
13.5 + j 5.5
(b) S = Vs I * L = 3.361 + j1.368 kVA
(c ) pf = P/S = 0.9261
Chapter 12, Solution 36.
(a) S = 1 [0.75 + sin(cos-10.75) ] =0.75 + 0.6614 MVA
S (0.75 + j 0.6614) x10 6
(b) S = 3V p I * p
→ I*p = = = 59.52 + j 52.49
3V p 3x 4200
PL =| I p | 2 Rl = (79.36) 2 (4) = 25.19 kW
(c) Vs = VL + I p (4 + j ) = 4.4381 − j 0.21 kV = 4.443∠ - 2.709 o kV
Chapter 12, Solution 37.
P 12
S= = = 20
pf 0.6
S = S∠θ = 20∠θ = 12 − j16 kVA
But S = 3 VL I L ∠θ
20 × 10 3
IL = = 55.51 A
3 × 208
2
S = 3 Ip Zp
For a Y-connected load, I L = I p .
S (12 − j16) × 10 3
Zp = 2 =
3 IL (3)(55.51) 2
Z p = 1.298 − j1.731 Ω
Chapter 12, Solution 38.
As a balanced three-phase system, we can use the per-phase equivalent shown
below.
110∠0° 110∠0°
Ia = =
(1 + j2) + (9 + j12) 10 + j14
1 2 1 (110) 2
Sp = I ZY = ⋅ ⋅ (9 + j12)
2 a 2 (10 2 + 14 2 )
The complex power is
3 (110) 2
S = 3S p = ⋅ ⋅ (9 + j12)
2 296
S = 551.86 + j735.81 VA
Chapter 12, Solution 39.
Consider the system shown below.
a 5Ω A
-j6 Ω 4Ω
100∠120° − + 100∠0° I1
+ − 8Ω I3 j3 Ω
5Ω
c − + C
b B
10 Ω
100∠-120° I2
5Ω
For mesh 1,
100 = (18 − j6) I 1 − 5 I 2 − (8 − j6) I 3 (1)
For mesh 2,
100 ∠ - 120° = 20 I 2 − 5 I 1 − 10 I 3
20∠ - 120° = - I 1 + 4 I 2 − 2 I 3 (2)
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