Electrical Circuit Theory Worked Examples

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 103) =

@

C

d.c.

3rd Harmonic 159,345r

5th Harmonic 10530r

 Vm (V) R (;)

60

Fundamental 210 20r

18

18

18

18

XL (;) XC (;)

0 0

Z (;)

180r

59,032 531,317 472,62887,82r

177,036 177,036 180r

295,16 106,263 189,75384,56r

Im (A)

0

0,444107,82r

8,8545r

0,55354,56r

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 (;)

120r

23,5 26,38762,95r

70,5 71,51480,34r

117,5 118,11184,17r

Im (A)

100r

3,41162,95r

0,83980,34r

0,33984,17r

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 + 1860r = 36,49725,285r A

@

i1 = 36,497 sin (40t + 25,285 r) A

[ = 120 rad/s:

I3 = 1575r + 1245r = 26,09261,705r 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

22415 V 24835 V 12,432 ; 15,6 45 ; 14,829 ;

Calculate

the current flowing through the imped ance Z 3 using Kirchhoffs 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 Kirchhoffs 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 Kirchhoffs current law : I2  I 1  I 3 = 0 Or: Consider

I1 = I2  I3 loop abefa, using Kirchhoffs 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 12,432I2  12,432I3 + 15,6 45I 2 = 22415 + 24835 22,003 11,69I2  12,432I3 = 464,84825,51

@

I2 = 21,127 37,2 + 0,56443,69I3

Consider

loop bcdeb, using Kirchhoffs voltage law : I3.Z3 + I2.Z2  E2 = 0

14,829I3 + 15,6 45I 2 = 24835

@

I2 = 15,89780  0,949 74I3

It is obvious that I 2 = I2, thus: 21,12737,2 + 0,564 43,69I 3 = 15,89780  0,949 74I3 1,46462,79I3 = 14,36168,42

@

I3 = 9,809105,63 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

36036 V 44044 V 12,845 ; 15,4 36 ; 14,542 ; 18,260 ; 2452 ; Use Kirchhoffs 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 Kirchhoffs 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 Kirchhoffs 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 Kirchhoffs 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

12,845rI1 + 14,542rI1  14,542rI4 = 36036r

@

27,29143,41rI1  14,542rI4 = 36036r I1 = 13,1917,41 + 0,5311,41rI4

Consider

. (v)

loop bcfgb using Kirchhoffs 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 15,436rI5  15,436rI4 + 14,542rI1  14,542rI4  18,260rI4 = 44044r 15,436rI5  36,26126,91rI 4 + 14,542rI1

@

= 44044r

I1 = 30,3452r  1,06278rI5 + 2,50115,09rI4

Obviously I1 = I1, thus: 13,191 7,41r + 0,531 1,41rI4 = 30,3452r 1,06278rI5 + 2,50115,09rI4 1,989161,29I4 = 17,4659,09r  1,06278rI5

@

I4 = 8,781 152,2r  0,534120,71rI5

Consider

loop cdefc using Kirchhoffs 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

15,436rI5  15,436rI4 + 2452rI5 = 44044r 28,96519,9rI5  15,436rI4 = 44044r

@

I4 = 1,88155,9rI5  28,57180r

 Again I 4 = I4, thus: 8,781152,2r  0,534120,71rI5 = 1,88155,9rI5  28,57180r

@

I5 = 11,19127,845r 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 Kirchhoffs Laws. Z4 = 18,260r ;

Z1 = 12,845r ;

E1 = 36036r V

Z2 = 15,4 36r ;

E2 = 38038r V

Z3 = 14,542r ;

E3 = 44044r 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 Kirchhoffs current law at p oint  b: I1 = I2 + I3 Consider

. (vi)

loop abefa using Kirchhoffs voltage law : I1.Z1 + I2.Z2  E1  E2 = 0

Substituting

Equation (vi) into this eq uation :

(I2 + I3)(12,845r) + I2(15,436r) = 36036r + 38038r

@

I2 = 34,39837,03r  I3(0,59545r)

Consider

loop bcdeb using Kirchhoffs voltage law : I3.Z4 + I3.Z3  E3 + E2  I2.Z2 = 0

I3 (18,260r + 14,542r)  I2(15,4 36r) = 44044r  38038r

@ @  And:

I2 = I3 (2,098 88,03r)  4,786112,61r 34,39837,03r  I3(0,59545r) = I3 (2,098 88,03r)  4,786112,61r 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

= = = = =

22415 V 24835 V 12,432 ; 15,6 45 ; 14,829 ;

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

= 12,432r +

(15,6  45r)(14,829r) 15,6  45r  14 ,829r

= 20,68215,235r ; E1 = I1 .Z 22415 = (I1)(20,68215,235r)

@

I1 = 10,831 0,235r A I3(1) = I1 .

Z2 Z2  Z3

¨  ¸ 15,6  45r ¹¹ ª 15,6  45r  14,829r º

= (10,831 0,235r) ©© = 6,95836,1r 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 ,829r) 12,432r  14 ,829r = 18,47124,27r ;

= 15,6 45r +

E2 = I2.Z 24835 = (I2)(18,471 24,27r)

@

I2 = 13,426559,27r A I3(2) = I2.

Z1 Z1  Z 3

¨  ¸ 12,432r ¹¹ ª 12,4 32r  14 ,829r º

= (13,426559,27r) ©© = 6,12360,9r A I3 = I3(2)  I1(1)

= 6,12360,9r  6,95836,1r = 9,813105,63r 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

= 35,6331,85r ;

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) = 9,83447,3r A I4 = I1(1)

Z3 Z3  Zs

« » 22   j15 ¼ - (22  j15)  35,6331,85r ½ = 4,20748,695r A = (9,83447,3r) ¬

I2(1) =  I4.

Z5 Z2  Z5

¨  ¸ 16   j12 ¹¹ ª 18  j24  16  j12 º

= ( 4,20748,695r) ©©

= 2,334113,88r 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)

= 43,21139,92r ; E2 = I2(2).Z (180 + j160) = [I2(2)](38,845 31,45r)

@

I2(2) = 5,5741,71 r A I1(2) =  I2.

Z5 Z5  Zs

¨  ¸ 16  j12 ¹¹ ª (16  j12)  37,509  33,33r º

= ( 5,5741,71r) ©©

= 2,31748,89r A I1 = I1(1) + I1(2) = 9,83447,3r + 2,31748,89r = 12,1547,04r A I2 = I2(1) + I2(2) = 2,334 113,88r + 5,5741,71r = 5,02823,04r 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 = 18,260r ;

Z1 = 12,845r ;

Z2 = 15,4 36r ;

E1 = 36036r V

E2 = 38038r V

Z3 = 14,542r ;

E3 = 44044r 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 ,542r  18,260r)(15,4  36r) 15,4   36r  14 ,542r  18,260r

= 23,431 16r ; E1 = I1.Z 36036r = (I1)(23,43116r)

@ 64

I1 = 15,364 20r A

I4(1) = I1 .

Z2 Z2  Z3  Z4

¨  ¸ (15,4   36r) ¹¹ ª 15,4   36r  14 ,542r  18,260r º

= (15,36420r) ©©

= 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

= 15,436r +

(14 ,542r  18,260r)(12,845r) 12,845r  14 ,542r  18,260r

= 18,867 7,12 r ; E2 = I2 .Z 38038r = (I2)(18,8677,12r)

@

I2 = 20,141 45,12 r A I4(2) = I2 .

Z1 Z1  Z 3  Z 4

¨  ¸ (12,845r) ¹¹ ª 12,845r  14 ,542r  18,260r º

= (20,14145,12r) ©© = 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 + 18,260r +

(12,845r)(15,4   36r) 12,845r  15,4  36r

= 39,5 42,92r ; E3 = I4(3).Z 44044r = [I4(3)](39,542,92r)

@

I4(3) = 11,139 1,08r A I4 = I4(1) + I 4(3)  I4(2) = 6,525 42,91r + 11,1391,08r  5,72540,09r = 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

= = = = =

22415 V 24835 V 12,432 ; 15,6 45 ; 14,829 ;

Calculate

the current through the imped ance Z4 by using Thevenins 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

= 8,7911,31r ;

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

22415r¨©©

 ¸ 15,6  45r ¹¹ ª 12,432r  15,6  45r º

= 158,81218,31r 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

¨  ¸ 12,432r ¹¹ ª 12,4 32r  15,6  45r º

= (24835) ©©

= 139,7678,69r V  VTH = VTH(2)  VTH(1) = 139,7678,69r  158,81218,31r = 223,973123,42r V The Thevenin equivalent circuit is shown in Figure 5.24. ZTH = 8,791 1,31r ; I4  VTH = 223,973123,42r V

Z4 = 14,829 ;

Figure 5.24: Thevenin equivalent circuit for Assignment 5.7 I4 = =

E TH Z TH  Z 4 223,973123,42r 8,791  1,31r  14 ,829r

= 9,813105,63r 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

= = = = = = =

36036 V 44044 V 12,845 ; 15,4 36 ; 14,542 ; 18,260 ; 2452 ;

5.8.1 5.8.2 5.8.3

Use Thevenins 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 (12,845r)(14 ,542r) 12,845r  14,542r

= 6,80143,59r ; ZTH = =

(Z p  Z 4 )Z 2 Zp  Z 4  Z 2 (6,80143,59r  18,260r)(15,4  36r) 6,80143,59r  18,260r  15,4  36r

= 13,2453,73r ;

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  18,260r)(14 ,542r) + 12,845 15,4   36r  18,260r  14,542r

= 21,7 39,78r ; E1 = I1.Z 36036r = (I1)(21,739,78r)

@

I1 = 16,593,78r A I2 = I1 

Z3 Z3  Z4  Z2

¨  ¸ 14 ,542r ¹¹ ª 14 ,542r  18,260r  15,4   36r º

= (16,593,78r) ©©

= 6,63411,31r A  VTH(1) = I2.Z2 = (6,63411,31 r)(15,436r) = 102,16424,69r 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 (12,845r)(14 ,542r) + 18,260 + 15,4 36 12,845r  14 ,542r

= 28,83323,29r ; E2 = I2 .Z 44044r = (I2)(28,83323,29r)

@

I2 = 15,2620,71r A  V2 = I2 .Z2 = (15,2615r)(15,436) = 235,00515,29r V  VTH(2) = E2  V2 = 44044r  235,00415,29r = 378,42676,27r V  VTH = VTH(1) + VTH(2) = 102,16424,69r + 378,42676,27 = 372,74360,66 V

73

The Thevenin equivalent circuit is shown in Figure 5.29. ZTH = 13,2453,73r ; I5  VTH = 395,75661,32r V

Z5 = 2452 ;

Figure 5.29: Thevenin equivalent circuit for Assignment 5.7 I5 = =

 V TH Z TH  Z 5 372,74360 ,66r 13,245  3,73r  24 52r

= 11,19127,845r A 5.7.2

V5 = I5.Z5 = (11,19127,845r)(2452r) = 268,5879,845r 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 Thevenins theorem.

74

Z4 = 18,260r ;

Z1 = 12,845r ;

Z2 = 15,4 36r ;

E1 = 36036r V

Z3 = 14,542r ;

E2 = 38038r V

E3 = 44044r 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

= 14,542r +

(12,845r)(15,4   36r) 12,845r  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

= 36036r +

12,845r 12,845r  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

= 38038r +

15,4  36r 12,845r  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 + 44044r  226,12983r = 552,693 8,45 r V The Thevenin equivalent circuit is shown in Figure 5.35. I4 = =

 V TH Z TH  Z 4 552,6938,45r 22,74 29,32r  18,260r

= 13,992 34,47 r A

77

ZTH = 22,7429,32r ; I4  VTH = 552,6938,45r V

Z4 = 18,260 ;

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

= = = = =

22415 V 24835 V 12,432 ; 15,6 45 ; 14,829 ;

Calculate

the current through the impedance Z 3 by using Nortons 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

= 8,7911,31r ; 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 22415r = (I1)(12,432r)

@

I1 = 18,064517r 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 24835r = (I2)(15,645r)

@

I2 = 15,89780r A

The same will happen as in the circuit of  Figure 5.40 Thus: IN(2) = I2 IN = IN(2)  IN(1) = 15,89780r  18,064517r = 25,476124,73r A

80

The Norton equivalent circuit is shown in Figure 5.41. I3

ZN

IN = 25,476124,73r 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 ,829r º

= (25,476124,73r) ©©

= 9,812105,63r 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

= = = = = = =

36036 V 44044 V 12,845 ; 15,4 36 ; 14,542 ; 18,260 ; 2452 ;

5.10.1 Use Nortons 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 (12,845r)(14 ,542r) 12,845r  14,542r

= 6,80143,59r ;

82

ZN =

=

(Z p  Z 4 )Z 2 Zp  Z 4  Z 2 (6,80143,59r  18,260r)(15,4  36r) 6,80143,59r  18,260r  15,4  36r

= 13,2453,73r ; 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 ,542r)(18,260r) + 12,845 14 ,542r  18,260r

= 20,9546,94 ; E1 = I1 .Z 36036r = (I1)(20,95146,94r)

@

I1 = 17,18310,94r A

83

IN(1) = I1 

Z3 Z3  Z4

¨  ¸ 14 ,542r ¹¹ ª 14 ,542r  18,260r º

= (17,183  10,94r)©© = 7,71320,97r 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 44044r = (I2)(15,436r)

@

I2 = 28,57180r A = IN(2) IN = IN(1) + IN(2) = 7,713 20,97r + 28,57180r = 28,14164,39r A

84

The Norton equivalent circuit is shown in Figure 5.46. I5

ZN

I N = 28,14164,39r 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

= (28,14164 ,39r)©©

= 11,19127,845r A 5.10.2

V5 = I5 .Z5 = (11,19127,845r)(2452r) = 268,5879,845r 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 Nortons theorem.

85

Z4 = 18,260r ;

Z1 = 12,845r ;

Z2 = 15,4 36r ;

E1 = 36036r V

Z3 = 14,542r ;

E2 = 38038r V

E3 = 44044r 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 +

(12,845r)(15,4  36r) 12,845r  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

= 12,845r +

(15,4   36r)(14,542r) 15,4   36r  14 ,542r

= 21,044 27,72 r ; E1 = I1 .Z 36036r = (I1)(21,04427,72r)

@

I1 = 17,107 8,28r A IN(1) = I1 .

Z2 Z2  Z3

¨  ¸ 15,4  36r ¹¹ ª 15,4  36r  14 ,542r º

= (17,1078,28r) ©©

= 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 +

(12,845r)(14 ,542r) 12,845r  14 ,542r

= 17,923 14,09 r ; E2 = I2.Z 38038r = (I2)(17,92314,09r)

@

I2 = 21,201 52,09r A IN(2) = I2.

Z1 Z1  Z 3

¨  ¸ 12,845r ¹¹ ª 12,845r  14,542r º

= (21,20152,09r) ©© = 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

= 14,542r +

(12,845r)(15,4   36r) 12,845r  15,4   36r

= 22,74 29,32r ; E3 = IN(3) .Z 44044r = [IN(3)](22,7429,32r)

@

IN(3) = 19,349 14,68 r A

IN = IN(1) + IN(3)  IN(2) = 11,33429,32r + 19,34914,68r  9,94453,68r = 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  18,260r º

= (28,14164 ,39r)©©

= 13,99234,47r 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 = 24012 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

6,12771,92r ; 7,14714,26r ;

9,24632,7r ;

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 ,55861,84r)(23,141  28,55r) 20,55861,84r  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 24012 = (I)(22,2690,75)

@

I = 10,77711,25 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

= = = = = = =

36036r V 44044r V 12,845r ; 15,436r ; 14,542r ; 18,260r ; 2452r ;

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 ,542r  18,260r 14,542r  18,260r  24 52r = 4,687 49,985 r ; =

Zb =

Z 4 .Z 5 Z 3  Z 4  Z5

18,260r  24 52r 14,542r  18,260r  24 52r = 7,758 59,985 r ;

=

Zc =

Z 3 .Z 5 Z3  Z4  Z5

14 ,542r  24 52r 14,542r  18,260r  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 = 2400 V

Z2 = 8,245 ;

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 = 8,245 ;

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)(8,245r) 9,6  54r  8,245r

= 6,7810,76 ;

@ 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 = 8,245 ;

 VTH

Figure 5.61: Parallel network for Assignment 5.14 with Z load open-circuited

95

 VTH = E.

Z2 Z1  Z 2

= (2400)

8,245r 8,245r  9,6  54r

= 169,52254,76r V Iload = =

 V TH Z TH  Z load 169,52254 ,76r 6,7810,76r  6,781  0,76r

= 12,50154,76r 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 = 38016 V Zab = 6,448 ; Zbc = 7,236 ; Zca = 4,530 ; Zbd = 5,545 ;

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 ,530r) 6,448r  7,236r  4 ,530r

= 1,60439,25r ; Zb = =

Z ab .Z bc Z ab  Z bc  Z ca (6,4 48r)(7,236r) 6,448r  7,236r  4 ,530r

= 2,56645,25r ; Zc = =

Z ab

Z bc .Z ca  Z bc  Z ca

(7,236r)(4 ,530r) 6,448r  7,236r  4 ,530r

= 1,80427,25r ; 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 (1,60439,25r)(2,56645,25r  5,545r) + 1,80427,25 1,60439,25r  2,56645,25r  5,545r

= 3,12332,77r ;

@

98

Zload = 3,12332,77r ;

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 ¨  ¸ 2,56645,25r  5,545r ¹¹ ª 1,60439,25r  2,56645,25r  5,545r º

= (38016) ©©

= 317,19516,97r 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)

= 14,3363,27° A

100

Ib = Ibn = =

E bn Z bn 380  150r ( 3 )(12,8  j8,4)

= 14,33176,73° A Ic = Icn = =

6.1.2

E cn Z cn

38090r ( 3 )(12,8  j8,4)

= 14,3356,73° A Eab + Ebc + Eca = 3800r + 380120r + 380120r = 0V

6.1.3

In = Ia + Ib + Ic = 14,33 63,27r + 14,33176,73r + 14,3356,73r = 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 = 28,42191,04°  28,421 148,96° Ib

= 49,227 61,04° A = Ibc  Iab = 28,42128,96°  28,42191,04° = 49,227 58,96° A

Ic = Ica  Ibc = 28,421148,96°  28,42128,96° = 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) = 49,22761,04 A

@

Ic = ( 3 )(28,421( 148,96  30) = 49,227 178,96 A

6.2.4

Vab = Iab .Zab 380120r = (28,42191,04r)(Zab)

@

Zab = 13,3728,96° ; = 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 36060r = (Ibc)(8,45 + j6,36)

@

Ibc = 34,039 23,03° A  Vca = Ica .Zca 360180r = (Ica)(8,45 + j6,36)

@

Ica = 34,039 143,03° A *

6.3.2

S ab

*

= Vab. I ab = (36060°)(34,03996,97°) = 12254,0436,97° = (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 = 34,03996,97°  34,039143,03° = 58,957 66,97° A

105