P5 = ½Vm5.Im5.cos J5 = (½)(66)(0,846)cos(45Û + 36,15Û) = 4,295 W P = P0 + P1 + P3 + P5 = 0 + 2,451 + 384 + 4,295 = 390,746 W Assignment 4.2 A series RLC-circuit consists of a resistor of 18 ;, an inductor of 188 mH and a variable capacitor. The circuit resonates at the third harmonic. The voltage across the circuit is represented by the following equation: R(t) = 60 + 210 sin(314t + T9 ) + 159,3 sin(942t +
4.2.1 4.2.2
T
4
) + 105 sin(1570t +
T
6
) volts
Determine an expression for the current in the circuit. Calculate the overall power factor of the circuit.
Solution:
4.2.1
For
resonance: X L = X C
@
[.L =
1 [.C
1 (942)(C) = 5,994 QF
(942)(188 x 103) =
@
C
d.c.
3rd Harmonic 159,345r
5th Harmonic 10530r
Vm (V) R (;)
60
Fundamental 210 20r
18
18
18
18
XL (;) XC (;)
0 0
Z (;)
180r
59,032 531,317 472,62887,82r
177,036 177,036 180r
295,16 106,263 189,75384,56r
Im (A)
0
0,444107,82r
8,8545r
0,55354,56r
47
i (t)
= 0 + 0,444 sin (314t + 1,533) + 8,833 sin (942t+ 0,785) + 0,553 sin (1570t 0,952) A
4.2.2
V =
=
2 2 Vm5 V 2 Vm3 V02 m1 2
(60) 2
(210) 2 (159,3) 2 (105) 2 2
= 209,406 V I =
=
I 20
I 2m1 I 2m3 I 2m5 2
(0,444) 2 (8,85) 2 (0,553) 2 0 2
= 6,278 A P = I2.R = (6,278)2(18) = 709,439 W P = V.I.cos J 709,439 = (209,406)(6,278)(cos J)
@
cos J = 0,5396 leading
Assignment 4.3 A series RL-circuit consists of a 12- ; resistor and a 50-mH inductor. The radial velocity in the circuit is 470 rad/s and the voltage across the circuit is represented by the equation: R(t) = 120 + 90 sin [t + 60 sin 3 [t + 40 sin 5 [t volts
4.3.1 4.3.2 4.3.3 4.3.4
48
Determine an equation for the instantaneous value of the current in the circuit. Calculate the active power dissipated in the resistor. Calculate the overall power factor of the circuit. Calculate the active power contributed by the fundamental and each harmonic in the circuit. Verify that the sum of these powers equals the value calculated in Question 4.3.2.
Solution:
4.3.1 d.c.
Fundamental
3rd Harmonic
5th Harmonic
Vm (V) R (;) XL (;)
120
90
60
40
12
12
12
12
0
Z (;)
120r
23,5 26,38762,95r
70,5 71,51480,34r
117,5 118,11184,17r
Im (A)
100r
3,41162,95r
0,83980,34r
0,33984,17r
i (t)
= 10 + 3,411 sin ([t 62,95r) + 0,839 sin (3[t 80,34r) + 0,339 sin (5[t 84,17r) A
4.3.2
I =
=
I20
I2m1 I2m3 I2m5 2 2
(10)
(3,411) 2 (0,839) 2 (0,339) 2 2
= 10,307 A P = I2 .R = (10,307)2(12) = 1274,811 W 4.3.3
V =
=
V02
Vm21 Vm2 3 Vm2 5 2
(90)2 (60)2 ( 40)2 (120) 2 2
= 145,086 V P = V.I.ccos J 1274,811 = (145,086)(10,307)(cos J)
@
cos J = 0,8525 lagging
49
4.3.4
P0 = V0.I0 = (120)(10) = 1200 W P1 = ½Vm1.Im1.cos J1 = (½)(90)(3,411)(cos 62,95r) = 69,805 W P3 = ½Vm3.Im3.cos J3 = (½)(60)(0,839)(cos 80,34r) = 4,224 W P5 = ½Vm5.Im5.cos J5 = (½)(40)(0,339)(cos 84,17r) = 0,689 W P = P0 + P1 + P3 + P5 = 1200 + 69,805 + 4,224 + 0,689 = 1274,718 W
Assignment 4.4 Consider Figure
4.5 and determine the Fourier series expression for the current drawn from the supply. The expressions for the currents flowing through Z a and Zb respectively are given by: ia(t) = 8 + 24 sin 40t
15 sin (120t + 75 r) A
ib(t) = 12 + 18 sin (40t + 60r) + 12 sin (120t + 45 r) + 8 sin (200t + 30 r) A
is
Rs
ia
Za
ib
Zb
Figure 4.5 : Parallel a.c. circuit for Assignment 4.4 50
Solution:
d.c.
terms:
I0 = 8 + 12 = 20 A
[ = 40 rad/s:
I1 = 24 + 1860r = 36,49725,285r A
@
i1 = 36,497 sin (40t + 25,285 r) A
[ = 120 rad/s:
I3 = 1575r + 1245r = 26,09261,705r A
@
i3 = 26,092 sin (120t + 61,705r) A
[ = 200 rad/s:
i5 = 8 sin (200t + 30r) A
i(t) = ia + ib
@ i (t) =
20 + 36,497sin (40t + 25,285r) + 26,092 sin (120t + 61,705r) + 8 sin (200t + 30r) A
Assignment 5.1 Refer to the a.c. network of Figure 5.6. components in the network : E1 E2 Z1 Z2 Z3
= = = = =
The following data is applicable on the
22415 V 24835 V 12,432 ; 15,6 45 ; 14,829 ;
Calculate
the current flowing through the imped ance Z 3 using Kirchhoffs laws.
51
Z1
Z2 Z3 E2
E1
Figure 5.6: A.C. parallel network for Assignment 5.1 Solution:
Refer to Paragraph 5.1.1 for the solution procedure of Kirchhoffs laws : I1
a
I3
b
+
I2
Z1
+
Z2
+
+
E1
Z3
E2 +
c
f
e
d
Fig ure 5.7: A.C. network for Kirchhoff's laws with loops and polarities According to Kirchhoffs current law : I2 I 1 I 3 = 0 Or: Consider
I1 = I2 I3 loop abefa, using Kirchhoffs voltage law : I1.Z1 + I2.Z2 E1 E2 = 0
52
. (i)
Substitute
Equation (i) into this equation : (I2 I3)Z1 + I2.Z2 E1 E2 = 0
@
I2.Z1 I3.Z1 + I2.Z2 = E1 + E2 12,432I2 12,432I3 + 15,6 45I 2 = 22415 + 24835 22,003 11,69I2 12,432I3 = 464,84825,51
@
I2 = 21,127 37,2 + 0,56443,69I3
Consider
loop bcdeb, using Kirchhoffs voltage law : I3.Z3 + I2.Z2 E2 = 0
14,829I3 + 15,6 45I 2 = 24835
@
I2 = 15,89780 0,949 74I3
It is obvious that I 2 = I2, thus: 21,12737,2 + 0,564 43,69I 3 = 15,89780 0,949 74I3 1,46462,79I3 = 14,36168,42
@
I3 = 9,809105,63 A
Assignment 5.2 Refer to the a.c. network of Figure 5.8. components in the network : E1 E2 Z1 Z2 Z3 Z4 Z5
= = = = = = =
5.2.1 5.2.2 5.2.3
The following data is applicable on the
36036 V 44044 V 12,845 ; 15,4 36 ; 14,542 ; 18,260 ; 2452 ; Use Kirchhoffs laws and calculate the current through the impe dance Z 5. Calculate the voltage drop across the impedance Z5. Calculate the power factor of the impedance Z 5.
53
Z4
Z1
Z2 Z5
Z3 E1
E2
Figure 5.8: A.C. parallel network for Assignment 5.2 Solution:
5.2.1
Refer to Paragraph 5.1.1 for the solution procedure of Kirchhoffs laws : I1
a
b
I3
Z4
I4 +
d
I2 +
+
Z1
I5
c
Z2
Z3 + +
Z5 + +
E1
E2 h
g
f
e
Figure 5.9: A.C. network for Kirchhoff's laws with loops and polarities Refer to Figure 5.9 using Kirchhoffs current law : I 1 I 4 I3 = 0
@
I3 = I1 I4
. (ii)
I2 + I4 I5 = 0
@ Substitute
I2 = I5 I4 Equation (iii) into Equation (ii) : I3 = I1 I5 + I2
54
. (iii)
. (iv)
Consider
loop abgh a using Kirchhoffs voltage law :
I1.Z1 + I3.Z3 E1 = 0 Substitute
Equation (ii) into this equation : I1.Z1 + (I1 I4) Z3 = E1 I1.Z1 + I1.Z3 I4.Z3 = E1
12,845rI1 + 14,542rI1 14,542rI4 = 36036r
@
27,29143,41rI1 14,542rI4 = 36036r I1 = 13,1917,41 + 0,5311,41rI4
Consider
. (v)
loop bcfgb using Kirchhoffs voltage law : I2.Z2 + I3.Z3 I4.Z4 E2 = 0
Substitute
Equations (ii) and (iii) into this equation : (I 5 I4)Z2 + (I1 I4)Z3 I4.Z4 = E2
I5.Z2 I4.Z2 + I1.Z3 I4.Z3 I4.Z4 = E2 15,436rI5 15,436rI4 + 14,542rI1 14,542rI4 18,260rI4 = 44044r 15,436rI5 36,26126,91rI 4 + 14,542rI1
@
= 44044r
I1 = 30,3452r 1,06278rI5 + 2,50115,09rI4
Obviously I1 = I1, thus: 13,191 7,41r + 0,531 1,41rI4 = 30,3452r 1,06278rI5 + 2,50115,09rI4 1,989161,29I4 = 17,4659,09r 1,06278rI5
@
I4 = 8,781 152,2r 0,534120,71rI5
Consider
loop cdefc using Kirchhoffs voltage law : I2.Z2 + I5.Z5 E2 = 0
55
Substitute
Equation (iii) into this equation : (I5 I4)Z2 + I5.Z5 = E2 I5.Z2 I4.Z2 + I5.Z5 = E2
15,436rI5 15,436rI4 + 2452rI5 = 44044r 28,96519,9rI5 15,436rI4 = 44044r
@
I4 = 1,88155,9rI5 28,57180r
Again I 4 = I4, thus: 8,781152,2r 0,534120,71rI5 = 1,88155,9rI5 28,57180r
@
I5 = 11,19127,845r A
Assignment 5.3 Refer to the a.c. parallel network of Figure 5.10 and calculate the voltage drop across the impedance Z4 using Kirchhoffs Laws. Z4 = 18,260r ;
Z1 = 12,845r ;
E1 = 36036r V
Z2 = 15,4 36r ;
E2 = 38038r V
Z3 = 14,542r ;
E3 = 44044r V
Figure 5.10: Parallel impedance network for Assignment 5.3
56
Solution:
I1
a
Z4
I3
b
c
I2 Z1
Z2
E1
Z3
E2
f
e
d
Figure 5.11: A.C. network for Kirchhoff's laws with loops and polarities According to Kirchhoffs current law at p oint b: I1 = I2 + I3 Consider
. (vi)
loop abefa using Kirchhoffs voltage law : I1.Z1 + I2.Z2 E1 E2 = 0
Substituting
Equation (vi) into this eq uation :
(I2 + I3)(12,845r) + I2(15,436r) = 36036r + 38038r
@
I2 = 34,39837,03r I3(0,59545r)
Consider
loop bcdeb using Kirchhoffs voltage law : I3.Z4 + I3.Z3 E3 + E2 I2.Z2 = 0
I3 (18,260r + 14,542r) I2(15,4 36r) = 44044r 38038r
@ @ And:
I2 = I3 (2,098 88,03r) 4,786112,61r 34,39837,03r I3(0,59545r) = I3 (2,098 88,03r) 4,786112,61r I3 = 13,991 34,47 r A
57
Assignment 5.4 Refer to the a.c. network of Figure 5.12. The following data is applicable on the components in the network : E1 E2 Z1 Z2 Z3
= = = = =
22415 V 24835 V 12,432 ; 15,6 45 ; 14,829 ;
Calculate
the current through the imped ance Z3 by using the Superposition Theorem.
Z1
Z2 Z3
E1
E2
Figure 5.12: A.C. parallel network for Assignment 5.4 Solution: Consider
the voltage source E 1 and short-circuit the voltage source E 2: I1
Z1
I3(1)
Z2 Z3
E1
Fig ure 5.13: A.C. network for Assignment 5.4 with E 2 short-circuited
58
Consider
E1 in Figure 5.13: Z = Z1 +
Z 2 .Z 3 Z2 Z3
= 12,432r +
(15,6 45r)(14,829r) 15,6 45r 14 ,829r
= 20,68215,235r ; E1 = I1 .Z 22415 = (I1)(20,68215,235r)
@
I1 = 10,831 0,235r A I3(1) = I1 .
Z2 Z2 Z3
¨ ¸ 15,6 45r ¹¹ ª 15,6 45r 14,829r º
= (10,831 0,235r) ©© = 6,95836,1r A Consider
the voltage source E 2 and short-circuit the voltage source E 1: I3(2) I2 Z2
Z1
Z3 E2
Figure 5.14: A.C. network for Assignment 5.4 with E 1 short-circuited
59
Consider
E2 in Figure 5.14: Z = Z2 +
Z 1 .Z 3 Z1 Z 3
(12,4 32r)(14 ,829r) 12,432r 14 ,829r = 18,47124,27r ;
= 15,6 45r +
E2 = I2.Z 24835 = (I2)(18,471 24,27r)
@
I2 = 13,426559,27r A I3(2) = I2.
Z1 Z1 Z 3
¨ ¸ 12,432r ¹¹ ª 12,4 32r 14 ,829r º
= (13,426559,27r) ©© = 6,12360,9r A I3 = I3(2) I1(1)
= 6,12360,9r 6,95836,1r = 9,813105,63r A Assignment 5.5 Refer to the a.c. network of Figure 5.15. The following data is applicable on the components in the network : E1 E2 Z1 Z2 Z3 Z4 Z5
= = = = = = =
(285 + j165) V (180 + j160) V (14 j12) ; (18 + j24) ; (22 + j15) ; (18 j18) ; (16 + j12) ;
Use the Superposition Theorem and calculate the current drawn from e ach supply.
60
Z4
Z1
Z2 Z3
Z5
E1
E2
Figure 5.15: A.C. parallel network for Assignment 5.5 Solution: Consider
the voltage source E 1 and short-circuit the voltage source E 2: Z4 I1(1) I4 I2(1)
Z1
Z2 Z3
Z5
E1
Figure 5.16: A.C. parallel network for Assignment 5.5 with E 2 short-circuited Zs = =
Z 2 .Z 5 + Z4 Z2 Z5 (18 j24)(16 j12) + (18 j18) 18 j24 16 j12
= 35,6331,85r ;
Z = =
Z s .Z 3 + Z1 Zs Z3 (35,63 31,85r)(22 j15) + (14 j12) 35,63 31,85r (22 j15)
= 33,487 17,23r ; 61
E1 = I1(1).Z (285 + j165) = [I1(1)](33,487 17,23r)
@
I1(1) = 9,83447,3r A I4 = I1(1)
Z3 Z3 Zs
« » 22 j15 ¼ - (22 j15) 35,6331,85r ½ = 4,20748,695r A = (9,83447,3r) ¬
I2(1) = I4.
Z5 Z2 Z5
¨ ¸ 16 j12 ¹¹ ª 18 j24 16 j12 º
= ( 4,20748,695r) ©©
= 2,334113,88r A Consider
the voltage source E 2 and short-circuit the voltage source E 1: Z4
I1(2)
I4
I2(2)
Z1
Z2 Z3
Z5 E2
Figure 5.17: A.C. parallel network for Assignment 5.5 with E 1 short-circuited Zs = =
Z 1 .Z 3 + Z4 Z1 Z 3 (14 j12)(22 j15) + (18 j18) 14 j12 22 j15
= 37,509 33,33r ;
62
Z = =
Z s .Z 5 + Z2 Z s Z5 (37,509 33,33r)(16 j12) + (18+j24) 37,509 33,33r (16 j12)
= 43,21139,92r ; E2 = I2(2).Z (180 + j160) = [I2(2)](38,845 31,45r)
@
I2(2) = 5,5741,71 r A I1(2) = I2.
Z5 Z5 Zs
¨ ¸ 16 j12 ¹¹ ª (16 j12) 37,509 33,33r º
= ( 5,5741,71r) ©©
= 2,31748,89r A I1 = I1(1) + I1(2) = 9,83447,3r + 2,31748,89r = 12,1547,04r A I2 = I2(1) + I2(2) = 2,334 113,88r + 5,5741,71r = 5,02823,04r A Assignment 5.6 Refer to the a.c. parallel network of Figure 5.18 and calculate the voltage drop across the impedance Z4 using the Superposition theorem.
63
Z4 = 18,260r ;
Z1 = 12,845r ;
Z2 = 15,4 36r ;
E1 = 36036r V
E2 = 38038r V
Z3 = 14,542r ;
E3 = 44044r V
Figure 5.18: Parallel impedance network for Assignment 5.6 Solution: Consider
the voltage source E 1 and short-circuit the voltage sources E 2 and E3 and the impedance Z4: Z4 I1 I4(1)
Z1
Z2
Z3
E1
Figure 5.16: Parallel impedance network for Assignment 5.5 with E 2 and E3 shortcircuited Z = Z1 +
(Z 3 Z 4 )(Z 2 ) Z2 Z3 Z4
= 12,8 45r +
(14 ,542r 18,260r)(15,4 36r) 15,4 36r 14 ,542r 18,260r
= 23,431 16r ; E1 = I1.Z 36036r = (I1)(23,43116r)
@ 64
I1 = 15,364 20r A
I4(1) = I1 .
Z2 Z2 Z3 Z4
¨ ¸ (15,4 36r) ¹¹ ª 15,4 36r 14 ,542r 18,260r º
= (15,36420r) ©©
= 6,525 42,91 r A Z4
Z2
Z1
Z3
E2 I2 I4(2) Fig ure 5.17: Parallel impedance network for Assignment 5.5 with E 1 and E3 shortcircuited Z = Z2 +
(Z 3 Z 4 )(Z 1 ) Z1 Z 3 Z 4
= 15,436r +
(14 ,542r 18,260r)(12,845r) 12,845r 14 ,542r 18,260r
= 18,867 7,12 r ; E2 = I2 .Z 38038r = (I2)(18,8677,12r)
@
I2 = 20,141 45,12 r A I4(2) = I2 .
Z1 Z1 Z 3 Z 4
¨ ¸ (12,845r) ¹¹ ª 12,845r 14 ,542r 18,260r º
= (20,14145,12r) ©© = 5,725 44,09r A
65
Z4
Z2
Z1
Z3
E3
I4(3) Figure 5.18: Parallel impedance network for Assignment 5.5 with E 1 and E2 shortcircuited Z = Z3 + Z4 +
(Z 1 )( Z 2 ) Z1 Z 2
= 14,5 42r + 18,260r +
(12,845r)(15,4 36r) 12,845r 15,4 36r
= 39,5 42,92r ; E3 = I4(3).Z 44044r = [I4(3)](39,542,92r)
@
I4(3) = 11,139 1,08r A I4 = I4(1) + I 4(3) I4(2) = 6,525 42,91r + 11,1391,08r 5,72540,09r = 13,993 34,47 r A
Assignment 5.7 Refer to the a.c. network of Figure 5.20. The following data is applicable on the components in the network : E1 E2 Z1 Z2 Z4 66
= = = = =
22415 V 24835 V 12,432 ; 15,6 45 ; 14,829 ;
Calculate
the current through the imped ance Z4 by using Thevenins Theorem.
Z4
E2
E1
Figure 5.20: A.C. parallel network for Assignment 5.7 Solution:
Open-circuit Z4 and short-circuit the voltage sources E 1 and E2:
Z1
Z2
Figure 5.21: A.C. parallel network for Assignment 5.7 with Z 4 open-circuited and E1 and E2 short-circuited ZTH = =
Z1 Z 2 Z1 Z 2 (12,4 32r)(15,6 45r) 12,4 32r 15,6 45r
= 8,7911,31r ;
67
Consider
the voltage source E 1 and short-circuit the voltage source E 2 and open-circuit the impedance Z4:
Z1
Z2 VTH(1)
E1
Fig ure 5.22: A.C. parallel network for Assignment 5.7 with E 2 short-circuited and Z 4 open-circuited VTH(1) = E 1 =
Z2 Z1 Z 2
22415r¨©©
¸ 15,6 45r ¹¹ ª 12,432r 15,6 45r º
= 158,81218,31r V Consider
the voltage source E 2 and short-circuit the voltage source E 1 and open-circuit the impedance Z4:
Z1
Z2 VTH(2)
E2
Fig ure 5.23: A.C. parallel network for Assignment 5.7 with E 1 short-circuited and Z 4 open-circuited
68
VTH(2) = E 2
Z1 Z1 Z 2
¨ ¸ 12,432r ¹¹ ª 12,4 32r 15,6 45r º
= (24835) ©©
= 139,7678,69r V VTH = VTH(2) VTH(1) = 139,7678,69r 158,81218,31r = 223,973123,42r V The Thevenin equivalent circuit is shown in Figure 5.24. ZTH = 8,791 1,31r ; I4 VTH = 223,973123,42r V
Z4 = 14,829 ;
Figure 5.24: Thevenin equivalent circuit for Assignment 5.7 I4 = =
E TH Z TH Z 4 223,973123,42r 8,791 1,31r 14 ,829r
= 9,813105,63r A
69
Assignment 5.8 Refer to the a.c. network of Figure 5.25. The following data is applicable on the components in the network : E1 E2 Z1 Z2 Z3 Z4 Z5
= = = = = = =
36036 V 44044 V 12,845 ; 15,4 36 ; 14,542 ; 18,260 ; 2452 ;
5.8.1 5.8.2 5.8.3
Use Thevenins Theorem and calculate the current t hrough the impedance Z 5. Calculate the voltage drop across the impedance Z5. Calculate the power factor of the impedance Z 5. Z4
Z1
Z2 Z5
Z3 E1
E2
Figure 2.25: A.C. parallel network for Assignment 5.7
70
Solution:
5.8.1
Open-circuit the impedance Z5 and short-circuit the voltage sources E 1 and E2: Z4
Z1
Z2 Z3
Fig ure 5.26: A.C. parallel network for Assignment 5.7 with Z 5 open-circuited and E1 and E2 short-circuited Zp = =
Z 1 .Z 3 Z1 Z 3 (12,845r)(14 ,542r) 12,845r 14,542r
= 6,80143,59r ; ZTH = =
(Z p Z 4 )Z 2 Zp Z 4 Z 2 (6,80143,59r 18,260r)(15,4 36r) 6,80143,59r 18,260r 15,4 36r
= 13,2453,73r ;
71
Consider
the voltage source E 1, short-circuit the voltage source E 2 and opencircuit the impedance Z5: Z4
I1
I2 Z1
Z2 Z3
VTH(1)
E1
Figure 5.27: A.C. parallel network for Assignment 5.7 with Z 5 open-circuited and E2 short-circuited Z = =
(Z 2 Z 4 )Z 3 + Z1 Z2 Z4 Z3 (15,4 36r 18,260r)(14 ,542r) + 12,845 15,4 36r 18,260r 14,542r
= 21,7 39,78r ; E1 = I1.Z 36036r = (I1)(21,739,78r)
@
I1 = 16,593,78r A I2 = I1
Z3 Z3 Z4 Z2
¨ ¸ 14 ,542r ¹¹ ª 14 ,542r 18,260r 15,4 36r º
= (16,593,78r) ©©
= 6,63411,31r A VTH(1) = I2.Z2 = (6,63411,31 r)(15,436r) = 102,16424,69r V
72
Consider
the voltage source E 2, short-circuit the voltage source E 1 and opencircuit the impedance Z5: Z4 I2 Z1
Z2 Z3
VTH(2) E2
Fig ure 5.28: A.C. parallel network for Assignment 5.7 with Z 5 open-circuited and E1 short-circuited Z = =
Z 1 .Z 3 Z4 Z2 Z1 Z 3 (12,845r)(14 ,542r) + 18,260 + 15,4 36 12,845r 14 ,542r
= 28,83323,29r ; E2 = I2 .Z 44044r = (I2)(28,83323,29r)
@
I2 = 15,2620,71r A V2 = I2 .Z2 = (15,2615r)(15,436) = 235,00515,29r V VTH(2) = E2 V2 = 44044r 235,00415,29r = 378,42676,27r V VTH = VTH(1) + VTH(2) = 102,16424,69r + 378,42676,27 = 372,74360,66 V
73
The Thevenin equivalent circuit is shown in Figure 5.29. ZTH = 13,2453,73r ; I5 VTH = 395,75661,32r V
Z5 = 2452 ;
Figure 5.29: Thevenin equivalent circuit for Assignment 5.7 I5 = =
V TH Z TH Z 5 372,74360 ,66r 13,245 3,73r 24 52r
= 11,19127,845r A 5.7.2
V5 = I5.Z5 = (11,19127,845r)(2452r) = 268,5879,845r V
5.7.3
cos J5 = cos VI5 5
= cos (79,845r 27,845r) = 0,6157 lagging Assignment 5.8 Refer to the a.c. parallel network of Figure 5.30 and calculate the voltage drop across the impedance Z4 using Thevenins theorem.
74
Z4 = 18,260r ;
Z1 = 12,845r ;
Z2 = 15,4 36r ;
E1 = 36036r V
Z3 = 14,542r ;
E2 = 38038r V
E3 = 44044r V
Fig ure 5.30: Parallel impedance network for Assignment 5.8 Solution:
ZTH
Z1
Z2
Z3
Figure 5.31: Parallel impedance network for Assignment 5.8 to determine Z TH ZTH = Z3 +
Z 1 .Z 2 Z1 Z 2
= 14,542r +
(12,845r)(15,4 36r) 12,845r 15,4 36r
= 22,74 29,32r ;
75
Consider
the voltage source E 1 and short-circuit the voltage sources E 2 and E3 and open-circuit the impedance Z 4: VTH(1)
Z1
Z2
Z3
E1
Figure 5.32: Parallel impedance network for Assignment 5.8 with E 2 and E3 shortcircuited and Z4 open-circuited VTH(1) = E1 +
Z2 Z1 Z 2
= 36036r +
12,845r 12,845r 15,4 36r
= 257,743 0r V Consider
the voltage source E 2 and short-circuit the voltage sources E 1 and E3 and open-circuit the impedance Z 4: VTH(2)
Z1
Z2
Z3
E2
Figure 5.33: Parallel impedance network for Assignment 5.8 with E 1 and E3 shortcircuited and Z4 open-circuited
76
VTH(2) = E2 +
Z1 Z1 Z 2
= 38038r +
15,4 36r 12,845r 15,4 36r
= 226,129 83r V Consider
the voltage source E 3 and short-circuit the voltage sources E 1 and E2 and open-circuit the impedance Z 4: VTH(3)
Z1
Z2
Z3
E3
Fig ure 5.34: Parallel impedance network for Assignment 5.8 with E 1 and E2 shortcircuited and Z4 open-circuited VTH(3) = E3 = 440 44r V VTH = VTH(1) + VTH(3) VTH(2) = 257,743 + 44044r 226,12983r = 552,693 8,45 r V The Thevenin equivalent circuit is shown in Figure 5.35. I4 = =
V TH Z TH Z 4 552,6938,45r 22,74 29,32r 18,260r
= 13,992 34,47 r A
77
ZTH = 22,7429,32r ; I4 VTH = 552,6938,45r V
Z4 = 18,260 ;
Figure 5.35: Thevenin equivalent circuit for Assignment 5.8 Assignment 5.9 Refer to the a.c. network of Figure 5.37. The following data is applicable on the components in the network : E1 E2 Z1 Z2 Z3
= = = = =
22415 V 24835 V 12,432 ; 15,6 45 ; 14,829 ;
Calculate
the current through the impedance Z 3 by using Nortons Theorem.
Z1
Z2 Z3
E1
E2
Figure 5.37: A.C. parallel network for Assignment 5.9 Solution:
Open-circuit Z3 and short-circuit the voltage sources E 1 and E2:
78
Z1
Z2 ZN
Figure 5.38: A.C. parallel network for Assignment 5.9 with Z 3 open-circuited and E1 and E2 short-circuited ZN = =
Z1 Z 2 Z1 Z 2 (12,4 32r)(15,6 45r) 12,4 32r 15,6 45r
= 8,7911,31r ; Consider
the voltage source E 1 and short-circuit the voltage source E 2 and the impedance Z3:
Z1
Z2
E1 I1
IN(1)
Figure 5.39: A.C. parallel network for Assignment 5.9 with E 2 and Z3 short-circuited E1 = I1 .Z1 22415r = (I1)(12,432r)
@
I1 = 18,064517r A
79
Because no current will flow through Z 2 due to the short-circuit of Z 3: IN(1) = I1 Consider
the voltage source E 2 and short-circuit the voltage source E 1 and the impedance Z3: IN(2) I2 Z1
Z2
E2
Figure 5.40: A.C. parallel network for Assignment 5.9 with E 1 and Z3 short-circuited E2 = I2.Z2 24835r = (I2)(15,645r)
@
I2 = 15,89780r A
The same will happen as in the circuit of Figure 5.40 Thus: IN(2) = I2 IN = IN(2) IN(1) = 15,89780r 18,064517r = 25,476124,73r A
80
The Norton equivalent circuit is shown in Figure 5.41. I3
ZN
IN = 25,476124,73r A
Z3
Figure 5.41: Norton equivalent circuit for Assignment 5.9 I3 = I N
ZN ZN Z 3
¨ ¸ 8,791 1,31r ¹¹ ª 8,791 1,31r 14 ,829r º
= (25,476124,73r) ©©
= 9,812105,63r A Assignment 5.10 Refer to the a.c. network of Figure 5.42. The following data is applicable on the components in the network : E1 E2 Z1 Z2 Z3 Z4 Z5
= = = = = = =
36036 V 44044 V 12,845 ; 15,4 36 ; 14,542 ; 18,260 ; 2452 ;
5.10.1 Use Nortons Theorem and calculate the current through the load impedance. 5.10.2 Calculate the voltage drop across the l oad impedance. 5.10.3 Calculate the power factor of the load impedance.
81
Z4
Z1
Z2 Z5
Z3 E1
E2
Figure 5.42: A.C. parallel network for Assignment 5.10 Solution:
5.10.1 Open-circuit the impedance Z5 and short-circuit the voltages sources E 1 and E2: Z4
Z1
Z2 Z3
Figure 5.43: A.C. parallel network for Assignment 5.10 with Z 5 open-circuited and E1 and E2 short-circuited Zp = =
Z 1 .Z 3 Z1 Z 3 (12,845r)(14 ,542r) 12,845r 14,542r
= 6,80143,59r ;
82
ZN =
=
(Z p Z 4 )Z 2 Zp Z 4 Z 2 (6,80143,59r 18,260r)(15,4 36r) 6,80143,59r 18,260r 15,4 36r
= 13,2453,73r ; Consider
the voltage source E1 and short-circuit the voltage source E 2 and the impedance Z5: Z4 I1 IN(1)
Z1
Z2 Z3
E1
Figure 5.44: A.C. parallel network for Assignment 5.10 with E 2 and Z5 shortcircuited No
current will flow through Z 2 due to the short-circuit of Z 5, thus: Z = =
Z 3 .Z 4 Z3
Z4
+ Z1
(14 ,542r)(18,260r) + 12,845 14 ,542r 18,260r
= 20,9546,94 ; E1 = I1 .Z 36036r = (I1)(20,95146,94r)
@
I1 = 17,18310,94r A
83
IN(1) = I1
Z3 Z3 Z4
¨ ¸ 14 ,542r ¹¹ ª 14 ,542r 18,260r º
= (17,183 10,94r)©© = 7,71320,97r A Consider
the voltage source E 2 and short-circuit the voltage source E 1 and the impedance Z5: Z4
IN(2) I2
Z1
Z2 Z3 E2
Figure 5.45: A.C. parallel network for Assignment 5.10 with E 1 and Z5 shortcircuited E2 = I2.Z2 44044r = (I2)(15,436r)
@
I2 = 28,57180r A = IN(2) IN = IN(1) + IN(2) = 7,713 20,97r + 28,57180r = 28,14164,39r A
84
The Norton equivalent circuit is shown in Figure 5.46. I5
ZN
I N = 28,14164,39r A
Z5
Figure 5.46: Norton equivalent circuit for Assignment 5.10
I5 = I N
ZN ZN Z 5
¨ ¸ 13,245 3,73r ¹¹ ª 13,245 3,73r 24 52r º
= (28,14164 ,39r)©©
= 11,19127,845r A 5.10.2
V5 = I5 .Z5 = (11,19127,845r)(2452r) = 268,5879,845r V
5.10.3
cos J = cos(79,845r 27,845r) = 0,6157 lagging
Assignment 5.11 Refer to the a.c. parallel network of Figure 5.47 and calculate the voltage drop across the impedance Z4 using Nortons theorem.
85
Z4 = 18,260r ;
Z1 = 12,845r ;
Z2 = 15,4 36r ;
E1 = 36036r V
Z3 = 14,542r ;
E2 = 38038r V
E3 = 44044r V
Figure 5.47: Parallel impedance network for Assignment 5.11 Solution:
ZN
Z1
Z2
Z3
Figure 5.48: Parallel impedance network for Assignment 5.11 to determine Z N ZN = Z3 +
Z 1 .Z 2 Z1 Z 2
= 14,5 42r +
(12,845r)(15,4 36r) 12,845r 15,4 36r
= 22,74 29,32r ;
86
Consider
the voltage source E 1 and short-circuit the voltage sources E 2 and E3 and the impedance Z4: I1
Z1
IN(1)
Z2
Z3
E1
Figure 5.49: Parallel impedance network for Assignment 5.11 with E 2, E3 and Z4 shortcircuited Z = Z1 +
Z 2 .Z 3 Z2 Z3
= 12,845r +
(15,4 36r)(14,542r) 15,4 36r 14 ,542r
= 21,044 27,72 r ; E1 = I1 .Z 36036r = (I1)(21,04427,72r)
@
I1 = 17,107 8,28r A IN(1) = I1 .
Z2 Z2 Z3
¨ ¸ 15,4 36r ¹¹ ª 15,4 36r 14 ,542r º
= (17,1078,28r) ©©
= 11,334 29,32 r A
87
Consider
the voltage source E2 and short-circuit the voltage sources E 1, E3 and the impedance Z4:
Z1
Z2
Z3
E2 I2 IN(2) Figure 5.50: Parallel impedance network for Assignment 5.11 with E 1, E3 and Z4 shortcircuited Z = Z2 +
Z 1 .Z 3 Z1 Z 3
= 15,4 36r +
(12,845r)(14 ,542r) 12,845r 14 ,542r
= 17,923 14,09 r ; E2 = I2.Z 38038r = (I2)(17,92314,09r)
@
I2 = 21,201 52,09r A IN(2) = I2.
Z1 Z1 Z 3
¨ ¸ 12,845r ¹¹ ª 12,845r 14,542r º
= (21,20152,09r) ©© = 9,944 53,68r A
88
Consider
the voltage source E 3 and short-circuit the voltage sources E 1 and E2 and the impedance Z4: VTH(3)
Z1
Z2
Z3
E3
Figure 5.51: Parallel impedance network for Assignment 5.11 with E 1, E2 and Z4 shortcircuited Z = Z3 +
Z 1 .Z 2 Z1 Z 2
= 14,542r +
(12,845r)(15,4 36r) 12,845r 15,4 36r
= 22,74 29,32r ; E3 = IN(3) .Z 44044r = [IN(3)](22,7429,32r)
@
IN(3) = 19,349 14,68 r A
IN = IN(1) + IN(3) IN(2) = 11,33429,32r + 19,34914,68r 9,94453,68r = 24,304 20,87 r A
89
The Norton equivalent circuit is shown in Figure 5.52. I4
ZN
IN = 24,304 20,87r A
Z4
Figure 5.52: Norton equivalent circuit for Assignment 5.11 I4 = I N
ZN ZN Z 4
¨ ¸ 22,74 29,32r ¹¹ ª 22,74 29,32r 18,260r º
= (28,14164 ,39r)©©
= 13,99234,47r A Assignment 5.12 Consider Figure
5.12.1 5.12.2
5.54.
Calculate Calculate
the total impedance of the ne twork using delta-star conversion. the current drawn from the supply. (7,8 + j12,3) ;
a
(8,8 + j8,8) ;
(14,4 + j7,2) ;
(13,4 j9,3) ; E = 24012 V
b
c
(9,5 j16,2) ;
Figure 4.54: Circuit diagram for Assignment 5.12
90
Solution:
5.12.1
Convert
the delta to an equivalent star: Za = =
Z ab .Z ca Z ab Z bc Z ca (14,4 j7,2)(8,8 j8,8) 14 ,4 j7,2 9,5 j16,2 8,8 j8,8
= 6,127 71,92° ; Zb = =
Z ab .Z bc Z ab Z bc Z ca (9,5 j16,2)(14 ,4 j7,2) 14 ,4 j7,2 9,5 j16,2 8,8 j8,8
= 9,246 32,7° ; Zc = =
Z ab
Z bc .Z ca Z bc Z ca (8,8 j8,8)(9,5 j16,2)
14, 4 j7,2 9,5 j16,2 8,8 j8,8
= 7,147 14,26° ; The star-connection shown in Figure 5.55 now replaces the delta. (7,8 + j12,3) ;
a
6,12771,92r ; 7,14714,26r ;
9,24632,7r ;
c
b
(13,4 j9,3) ;
Figure 5.55: Equivalent star circuit diagram
91
Z1 = Za + (7,8 + j12,3) = 20,558 61,84° ; Z2 = Zc + (13,4 j9,3) = 23,141 28,55° ; Zp = =
Z 1 .Z 2 Z1 Z 2 (20 ,55861,84r)(23,141 28,55r) 20,55861,84r 23,141 28,55r
= 15,421 20,05° ; Z = Zb + Zp = 9,246 32,7° + 15,421 20,05° = 22,269 0,75° ; 5.12.2
E = I.Z 24012 = (I)(22,2690,75)
@
I = 10,77711,25 A
Assignment 5.13 Refer to the a.c. parallel network of Figure 5.56. The component values if the circuit are the following : E1 E2 Z1 Z2 Z3 Z4 Z5
= = = = = = =
36036r V 44044r V 12,845r ; 15,436r ; 14,542r ; 18,260r ; 2452r ;
Convert
the delta Z3, Z4 and Z5 to an equivalent star connection and draw this equivalent circuit.
92
Z4
Z1
Z2 Z5
Z3 +
+
E1
E2
Figure 5.56: Parallel network for Assignment 5.13 Solution:
5.13.1
Za =
Z 3 .Z 4 Z 3 Z 4 Z5
14 ,542r 18,260r 14,542r 18,260r 24 52r = 4,687 49,985 r ; =
Zb =
Z 4 .Z 5 Z 3 Z 4 Z5
18,260r 24 52r 14,542r 18,260r 24 52r = 7,758 59,985 r ;
=
Zc =
Z 3 .Z 5 Z3 Z4 Z5
14 ,542r 24 52r 14,542r 18,260r 24 52r = 6,181 41,985 r ;
=
93
Za
Zb
Z2
Z1
Zc
E1
E2
Figure 5.57: Equivalent circuit for Figure 5.56 Assignment 5.14 Consider
5.14.1 5.14.2
the parallel network in Figure 5.59.
Calculate Calculate
the load impedance for m aximum power transfer to the load. the value of the maximum power transferred. Z1 = 9,6 54 ;
E = 2400 V
Z2 = 8,245 ;
Figure 5.59: Parallel network for Assignment 5.14
94
Zload
Solution:
5.14.1 Open-circuit the impedance Zload and short-circuit the voltages source E : Z1 = 9,6 54 ;
Z2 = 8,245 ;
Figure 5.60: Parallel network for Assignment 5.14 with E short-circuited and Zload open-circuited ZTH = =
Z 1 .Z 2 Z1 Z 2 (9,6 54r)(8,245r) 9,6 54r 8,245r
= 6,7810,76 ;
@ 5.14.2
Zload = 6,781 0,76 ; Consider
E and open-circuit the impedance Z load: Z1 = 9,6 54 ;
E = 240 0 V
Z2 = 8,245 ;
VTH
Figure 5.61: Parallel network for Assignment 5.14 with Z load open-circuited
95
VTH = E.
Z2 Z1 Z 2
= (2400)
8,245r 8,245r 9,6 54r
= 169,52254,76r V Iload = =
V TH Z TH Z load 169,52254 ,76r 6,7810,76r 6,781 0,76r
= 12,50154,76r A Pout(max) =
2
I load load .R load load
= (12,501)2(6,78) = 1059,545 W Or:
ZTH = (6,78 j0,09) ;
@
RTH = 6,78 ;
Pout(max) =
=
2 V TH 4.R TH
(169,522) 2 (4)(6,78)
= 1059,65 W Assignment 5.15 Consider the parallel network in Figure 5.62. The components in the circuit have the following values:
E = 38016 V Zab = 6,448 ; Zbc = 7,236 ; Zca = 4,530 ; Zbd = 5,545 ;
96
5.15.1 5.15.2
Calculate Calculate
the load impedance for m aximum power transfer to the load. the value of the maximum power transferred. a
Zca
Zab Zbc
c
E
b
Zload
Zbd
d
Figure 5.62: Bridge circuit diagram for Assignment 5.15 5.15.1
Convert
the delta abc to star: Za = =
Z ab .Z ca Z ab Z bc Z ca (6,4 48r)(4 ,530r) 6,448r 7,236r 4 ,530r
= 1,60439,25r ; Zb = =
Z ab .Z bc Z ab Z bc Z ca (6,4 48r)(7,236r) 6,448r 7,236r 4 ,530r
= 2,56645,25r ; Zc = =
Z ab
Z bc .Z ca Z bc Z ca
(7,236r)(4 ,530r) 6,448r 7,236r 4 ,530r
= 1,80427,25r ; 97
Zb
Zbd
Zc
Zload
Za
E Figure 5.63: Equivalent delta-connected circuit for the circuit in Figure 5.62 Consider
E and open-circuit the impedance Z load: Zb
Zbd
Za Zc
Figure 5.64: Circuit diagram for Assignment 5.15 with E short-circuited and Zload open-circuited ZTH = =
Z a .(Z b Z bd ) + Zc Z a Z b Z bd (1,60439,25r)(2,56645,25r 5,545r) + 1,80427,25 1,60439,25r 2,56645,25r 5,545r
= 3,12332,77r ;
@
98
Zload = 3,12332,77r ;
5.15.2
Consider
E and open-circuit the impedance Z load: Zb
Zbd
Za Zc VTH
E Figure 5.65: Circuit diagram for Assignment 5.15 with Z load open-circuited
¨ Z b Z bd ¸ ¹¹ VTH = E. ©© ª Z a Z b Z bd º ¨ ¸ 2,56645,25r 5,545r ¹¹ ª 1,60439,25r 2,56645,25r 5,545r º
= (38016) ©©
= 317,19516,97r V ZTH = (2,626 + j1,69) ;
@
RTH = 2,626 ; P0(max) =
=
2 VTH
4R TH
(317,195) 2 (4)(2,626)
= 9578,51 W
99
Assignment 6.1 A balanced, three-phase, four-wire, star-connected load of (12,8 + j8,4) ohms per phase is connected to a symmetrical, three-phase, 380-V supply. Take Eab as reference and a positive phase sequence. 6.1.1 6.1.2 6.1.3
Calculate
the values of the phase and line currents. Calculate the sum of the line voltages. Calculate the value of the neutral current flowing in the neutral wire from the supply.
Solution:
No three-phase problem should be started without the phasor diagram!
Ecn
Eca 30
Eab 30 30 Ebn
Ean Ebc
Figure 6.15: Positive (abc) phase sequence wit h Eab as reference 6.1.1
Current
in each phase (line) : Ia = Ian = =
E an Z an 380 30r ( 3 )(12,8 j8,4)
= 14,3363,27° A
100
Ib = Ibn = =
E bn Z bn 380 150r ( 3 )(12,8 j8,4)
= 14,33176,73° A Ic = Icn = =
6.1.2
E cn Z cn
38090r ( 3 )(12,8 j8,4)
= 14,3356,73° A Eab + Ebc + Eca = 3800r + 380120r + 380120r = 0V
6.1.3
In = Ia + Ib + Ic = 14,33 63,27r + 14,33176,73r + 14,3356,73r = 0A The value of the neutral current is zero. Remember that this is only true for the special case of a balanced system.
Assignment 6.2 In a balanced, three-phase, 380-V, 50-Hz, star-delta system the 28,35 kW load is supplied at a power factor 0,875 lagging. Take Vbc as reference with a positive phase sequence. 6.2.1 6.2.2 6.2.3 6.2.4
Calculate
the power absorbed per phase. the value of the phase c urrents. Calculate the value of the line currents. Calculate the effective impedance of the load per phase. Calculate
101
Solution:
Three-phase supply
Three-phase load Ia
a
Iab Eca c
Eab
Vab
Vca Ica
b
Ib Ic
Ibc Ebc
Vbc
Figure 6.20: Line diagram of the star-delta system wi th the supply and load voltages and line and phase currents Vab
Vbc
Vca Figure 6.21: Positive (abc) phase sequence with Vbc as reference 6.2.1
Active power in each phase : Ptot = 3Pph 28,35 = (3)(Pph)
@
Pph = 9,45 kW = P ab = P bc = P ca
102
6.2.2
Current
in each phase : V Pbc = Vbc.I bc.cos I bc bc
9,45 x 10 3 = (380)(Ibc)(0,875)
@
Ibc = 28,421 28,96 r A Iab = 28,421( 28,96r + 120r) = 28,421 91,04 r A Ica = 28,421(28, 96r 120r) = 28,421 148,96 r A
6.2.3
Current
in each line : Ia = Iab Ica = 28,42191,04° 28,421 148,96° Ib
= 49,227 61,04° A = Ibc Iab = 28,42128,96° 28,42191,04° = 49,227 58,96° A
Ic = Ica Ibc = 28,421148,96° 28,42128,96° = 49,227 178,96° A For
a balanced load, once the phase currents are found, the line currents can be determined: IL = Where:
@
3 Iph(J ± 30)
+ | negative phase sequence | positive phase sequence Ib = ( 3 )(28,421( 28,96 30) = 49,227 58,96 A
103
@
Ia = ( 3 )(28,421(91,04 30) = 49,22761,04 A
@
Ic = ( 3 )(28,421( 148,96 30) = 49,227 178,96 A
6.2.4
Vab = Iab .Zab 380120r = (28,42191,04r)(Zab)
@
Zab = 13,3728,96° ; = Zbc = Zca
Assignment 6.3 A 360-V, symmetrical, three-phase supply is connected to a balanced, delta-connected, three-phase load of phase impedance (8,45 + j6,36) ;. Take Vac as reference phasor with a negative phase sequence. 6.3.1 6.3.2 6.3.3 6.3.4 6.3.5 6.3.6
Calculate
the value of the phase c urrents. the power absorbed per phase. Calculate the total power absorbed by the load. Calculate the phasor sum of the three line current s. Convert the load to an equivalent star-connected load. Calculate the total power absorbed by the star-connected load. Calculate
Solution:
Vbn Vbc 30
Vca
Vac 30 30
Vcn
Van
Vab Figure 6.22: Negative (cba) phase sequence with Vac as reference
104
6.3.1
Current
in each phase of the load : Vab = Iab .Zab
360 60r = (Iab)(8,45 + j6,36)
@
Iab = 34,039 96,97° A Vbc = Ibc .Zbc 36060r = (Ibc)(8,45 + j6,36)
@
Ibc = 34,039 23,03° A Vca = Ica .Zca 360180r = (Ica)(8,45 + j6,36)
@
Ica = 34,039 143,03° A *
6.3.2
S ab
*
= Vab. I ab = (36060°)(34,03996,97°) = 12254,0436,97° = (9790,372 + j7369,54) VA
@
Pab = 9790,372 W = P bc
Or:
Pab
= Pca 2 = I ab .Rab = (34,039)2(8,45) = 9790,622 W
6.3.3
P = 3.Pph = (3)(9790,622) = 29,372 kW
6.3.4
Current
in each line : Ia = Iab Ica = 34,03996,97° 34,039143,03° = 58,957 66,97° A
105