Download Chapter 12 Part 2-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity!
February 5, 2006
CHAPTER 12
P.P.12.1 For the abc sequence, V an leads V bn by 120° and V bn leads V cnby 120°.
Hence, V an = 110 ∠( 30 °+ 120 °)= 110 ∠ 150 ° V
V cn = 110 ∠( 30 °− 120 °) = 110 ∠ –90˚V
P.P.12.
(a) V ab = V an− V bn= 120 ∠ 30 °− 120 ∠- 90 °
V ab =( 103. 92 +j 60 )+j 120
V ab = 207.8 ∠ 60 ˚V
Alternatively, using the fact that leads by 30° and has a
magnitude of
V ab V an
3 times that of V an,
V ab= 3 ( 120 )∠( 30 °+ 30 °)= 207. 85 ∠ 60 °
Following the abc sequence,
V bc = 207.8 ∠ –60˚V
V ca = 207.8 ∠ 180 ˚V
(b) Z
V
I
an a =
Z =( 0. 4 +j 0. 3 )+( 24 +j 19 )+( 0. 6 +j 0. 7 )
Z = 25 +j 20 = 32 ∠38.66 °
I (^) a 3. 75 ∠ - 8.66 ° A
Following the abc sequence,
I (^) b = I a∠- 120 ° = 3. 75 ∠ - 128.66 ° A
I (^) c = I a∠- 240 ° = 3.75 ∠ 111.34˚A
P.P.12.
The phase currents are
AB^180 -^20
AB Z
V
I 9 ∠ - 60 ° A
I BC = I AB∠- 120 ° = 9 ∠ - 180 ° A
I CA = I AB∠ 120 ° = 9 ∠ 60 °
The line currents are
I (^) a = I AB 3 ∠- 30 °= 9 3 ∠- 90 ° = 15. 59 ∠ - 90 ° A
I (^) b = I a∠- 120 ° = 15.59 ∠ 150 ˚A
I (^) c = I a∠ 120 ° = 15. 59 ∠ 30 ° A
P.P.12.4 In a delta load, the phase current leads the line current by 30° and has a
magnitude 3
times that of the line current. Hence,
a AB
I
I 13 ∠ 65 ° A
Z Δ = 18 +j 12 = 21. 63 ∠33.69° Ω
V AB = I AB Z Δ=( 13 ∠ 65 °)( 21. 63 ∠33.69° )
V AB = 281. 2 ∠ 98.69 ° V
P.P.12.5 Z Y= 12 +j 15 = 19. 21 ∠ 51. 34 °
After converting the Δ-connected source to a Y-connected source,
V an
Y
an a Z
V
I 7. 21 ∠ - 66.34 ° A
I (^) b = I a∠- 120 ° = 7. 21 ∠ - 186.34 ° A
I (^) c = I a∠ 120 ° = 7. 21 ∠ 53.66 ° A
For load 2,
60 kVA
- 8
cos
P
S
2
2 2 = = θ
Q 2 =S 2 sinθ 2 =( 60 )( 0. 6 )= 36 kVAR
S 2 = 48 +j 36 kVA
S = S 1 + S 2 = 56. 47 + j 47. 29 kVA
S = 73. 65 ∠ 39. 94 ° kVA
with pf =cos( 39. 94 °)= 0. 7667
(b) Q (^) c =P(tanθold−tanθnew)
Q (^) c =( 56. 47 )(tan 39. 94 °−tan 0 °)= 47. 29 kVAR
For each capacitor, the rating is 15. 76 kVAR
(c) At unity pf, S =P = 56. 47 kVA
3 V
I
L
L
S
38. 81 A
P.P.12.
The phase currents are
10 j 5
AB
AB AB Z
V
I
- 5 - 120 -6.25 j10. 16
BC
BC BC = ∠ °= −
Z
V
I
8 j 6
CA
CA CA Z
V
I
The line currents are
I a = I AB− I CA=( 16 +j 8 )−( 2. 392 +j 19. 856 )
= 13. 608 −j 11. 856 = 18. 05 ∠ - 41.06 ° A
I b = I BC− I AB=(-6.25−j10.825)−( 16 +j 8 )
= - 22. 25 −j 18. 825 = 29. 15 ∠ 220.2 ° A
I c = I CA− I BC=( 2. 392 +j 19. 856 )−(- 6. 25 −j 10. 825 )
I (^) c= 8. 642 +j 30. 681 = 31. 87 ∠ 74.27 ° A
P.P.12.
The phase currents are
j 44
AB =
I =
j 10
I BC
I CA
The line currents are
I a = I AB− I CA=(j 44 )−(- 11 −j 19. 05 )
I (^) a= 11 +j 63. 05 = 64 ∠ 80.1 ° A
I b = I BC− I AB=( 19. 05 +j 11 )−(j 44 )
I (^) b= 19. 05 −j 33 = 38. 1 ∠ - 60 ° A
I c = I CA− I BC=(- 11 −j 19. 05 )−( 19. 05 +j 11 )
I (^) c= - 30. 05 −j 30. 05 = 42. 5 ∠ 225 ° A
The real power is absorbed by the resistive load
P = ( 10 )=( 22 ) ( 10 ) =
(^22) I (^) CA 4. 84 kW
P.P.12.11 The schematic is shown below. First, use the AC Sweep option of the
Analysis Setup. Choose a Linear sweep type with the following Sweep Parameters :
Total Pts = 1, Start Freq = 100, and End Freq = 100. Once the circuit is saved and
simulated, we obtain an output file whose contents include the following results.
FREQ IM(V_PRINT1) IP(V_PRINT1)
1.000E+02 8.547E+00 -9.127E+
FREQ VM(A,N) VP(A,N)
1.000E+02 1.009E+02 6.087E+
From this we obtain,
I (^) bB = 8. 547 ∠ - 91.27 ° A , V an = 100. 9 ∠ 60.87 ° V
Use the AC Sweep option of the Analysis Setup. Choose a Linear sweep type with the
following Sweep Parameters : Total Pts = 1, Start Freq = 0.1592, and End Freq =
0.1592. Once the circuit is saved and simulated, we obtain an output file whose contents
include the following results.
FREQ IM(V_PRINT1) IP(V_PRINT1)
1.592E-01 3.724E+01 8.379E+
FREQ IM(V_PRINT2) IP(V_PRINT2)
1.592E-01 1.555E+01 -7.501E+
FREQ IM(V_PRINT3) IP(V_PRINT3)
1.592E-01 2.468E+01 -9.000E+
From this we obtain,
I (^) ca = 24. 68 ∠ - 90 ° A I (^) cC= 37. 25 ∠ 83. 79 ° A I AB 15.55 ∠ –75.01˚A
P.P.12.
(a) If point o is connected to point B, P 2 = 0 W
P 1 =Re( V AB I *a )
P 1 = ( 200 )( 18. 05 )cos( 0 °+ 41. 06 °) = 2722 W
P Re( )
3 = V CB I c
where V CB = - V BC= 200 ∠(-120°+ 180 °)= 200 ∠ 60 °
P 3 = ( 200 )( 31. 87 )cos(60°− 74. 27 °) = 6177 W
(b) Total power is
P (^) T = P 1 +P 2 +P 3 = 2722 + 0 + 6177 = 8899 W
P.P.12.14 VL = 208 V, P 1 = -560W, P 2 = 800 W
(a) PT = P 1 +P 2 =-560+ 800 = 240 W
(b) (^) Q (^) T = 3 (P 2 −P 1 )= 3 ( 800 + 560 )= 2.356 kVAR
(c) θ^ = = =^9.^815 ⎯⎯→ θ=^84.^18 ° 240
P
Q
tan T
T
pf = cosθ= 0. 1014 (lagging / inductive)
It is inductive because P 2 >P 1
(d) For a Y-connected load,
I (^) p = I L, 120 V 3
V
V
L p = = =
6. 575 A
P (^) p =VpIpcosθ ⎯⎯→ Ip= =
I
V
Z
p
p p = = =
Z (^) p = Zp∠θ = 18. 25 ∠ 84. 18 °Ω
The impedance is inductive.
P.P.12.15 Z Δ = 30 −j 40 = 50 ∠-53.13°
The equivalent Y-connected load is
Δ
- 67 - 53. 3
Y
Z
Z
254 V
Vp = =
V 254
I
Y
p L =^ = = Z
P 1 =VLILcos(θ+ 30 ° )
P 1 = ( 440 )( 15. 24 )cos(-5 3. 13 °+ 30 °) = 6. 167 kW
P 2 =VLILcos(θ− 30 ° )
P 2 = ( 440 )( 15. 24 )cos(-53.13°− 30 °) = 0. 8021 kW
PT = P 1 +P 2 = 6. 969 kW
Q T = 3 (P 2 −P 1 )= 3 ( 802. 1 − 6167 )
Q (^) T = - 9.292kVAR
