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V an 231 ∠ - 30 ° V
V bn = 231 ∠ - 150 ° V
V cn = 231 ∠ - 270 ° V
(b) For the acb sequence,
V ab = V an− V bn=Vp ∠ 0 °−Vp∠ 120 °
⎟⎟= ∠^ °
= + − V 3 - 30
j 2
V ab V (^) p (^1) p
i.e. in the acb sequence, V ab lags V an by 30°.
Hence, if V ab (^) = 400 , then
V an 231 ∠ 30 ° V
V bn = 231 ∠ 150 ° V
V cn = 231 ∠ - 90 ° V
Chapter 12, Problem 2.
What is the phase sequence of a balanced three-phase circuit for which V an = 160 ∠ 30 ° V
and V cn = 160 ∠ − 90 ° V? Find V bn.
Chapter 12, Solution 2.
Since phase c lags phase a by 120°, this is an acb sequence.
V bn = 160 ∠(30°+ 120 °) = 160 ∠ 150 ° V
Chapter 12, Problem 3.
Determine the phase sequence of a balanced three-phase circuit in which
V bn = 208 ∠ 130 ° V and V cn = 208 ∠ 10 ° V. Obtain V an.
Chapter 12, Solution 3.
Since V bn leads V cn by 120°, this is an abc sequence.
V an = 208 ∠(130°+ 120 °) = 208 ∠ 250 ° V
Chapter 12, Problem 4.
A three-phase system with abc sequence and V L = 200 V feeds a Y-connected load with
ZL = 40 ∠ 30 ° Ω. Find the line currents.
Chapter 12, Solution 4.
200 200 3 3
V L = = Vp ⎯⎯→ Vp =
2.887 30 A
o an o a (^) o Y
V
I
Z (^) x
120 2.887 150 A
o o I (^) b = Ia < − = < −
120 2.887 90 A
o o Ic = Ia < + = <
Chapter 12, Problem 5.
For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at
the terminals are:
vAN = 150 cos ( ω t + 32º) V
vBN = 150 cos ( ω t – 88º) V
vCN = 150 cos ( ω t + 152º) V
Write the time-domain expressions for the line-to-line voltages vAN , vBC , and vCA.
Chapter 12, Solution 5.
3 30 3 150 32 30 260 62
o o o o VAB = Vp < = x < + = <
Thus,
260 cos( 62 ) V
o
vAB = ω t +
Using abc sequence,
260 cos( 58 ) V
o
vBC = ω t −
Figure 12.
For Prob. 12.7.
Chapter 12, Solution 7.
This is a balanced Y-Y system.
Using the per-phase circuit shown above,
6 j 8
I (^) a 44 ∠ 53.13 ° A
I (^) b = I a∠- 120 ° = 44 ∠ - 66.87 ° A
I (^) c = I a∠ 120 ° = 44 ∠ 173. 13 ° A
440 ∠ 0 ° V
−
ZY = 6 − j8 Ω
Ω , while the per-phase
impedance of the load is 10 + j 14 Ω. Calculate the line currents and the load voltages.
Chapter 12, Solution 8.
Consider the per phase equivalent circuit shown below.
Z l
V an Z L
5.3958 35.1 A
o an o a L
V
I
Z Z j
120 5.3958 155.1 A
o o I (^) b = Ia < − = < −
120 5.3958 84.9 A
o o I (^) c = Ia < + = <
(4.4141 3.1033)(10 14) 92.83 19.35 V
o VLa = I Za L = − j + j = <
120 92.83 100.65 V
o o VLb = VLa < − = < −
120 92.83 139.35 V
o o VLc = VLa < + = <
_
5.396 ∠ –35.1˚ A
5.396 ∠ –155.1˚ A
5.396 ∠ 84.9˚ A
92.83 ∠ 19.35˚ A
92.83 ∠ –100.65˚ A
92.83 ∠ 139.35˚ A
Chapter 12, Problem 10.
For the circuit in Fig. 12.43, determine the current in the neutral line.
Figure 12.
For Prob. 12.10.
Chapter 12, Solution 10.
Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
27 j 10
A^2
an a Z
V
I
For phase b,
B^2
bn b Z
V
I
For phase c,
12 j 5
C^2
cn c Z
V
I
The current in the neutral line is
I (^) n = -( I (^) a+ I b+ I c)or - I (^) n= I a+ I b+ I c
- I n =( 7. 166 +j 2. 654 )+(- 5 −j 8. 667 )+(- 2. 173 +j 16. 783 )
I (^) n = 0. 007 −j 10. 77 = 10.77 ∠ 90°A
Chapter 12, Problem 11.
In the Y- ∆ system shown in Fig. 12.44, the source is a positive sequence with
V (^) an = 120 ∠^0 °^ V and phase impedance Z (^) p = 2 – j 3 Ω. Calculate the line voltage V (^) L and
the line current I (^) L.
Figure 12.
For Prob. 12.11.
Chapter 12, Solution 11.
o o VAB = Vab = Vp < = <
VL = | Vab | = 3 120 x =207.85 V
o AB p AB A
V V
I
Z j
o o p a AB
V (^) x I I j j j
I (^) L = | Ia | =99.846 A
