Electrical Circuit Theory Worked Examples

Assignment 1.1

The voltage measured across a coil is 135 V when a direct current of 5,625 A is passed through the coil. The voltage across the coil is 225 V with an alternating current of 5,625 A at 50 Hz. The coil is then connected to a 300-V, 60-Hz supply.

1.1.1 Calculate the current through the coil.1.1.2 Calculate the active and reactive components of this current.1.1.3 Calculate the power factor of the coil.

Solution:

1.1.1 The resistance of the coil is determined when it is connected to d.c.:

E = I.RL

135= (5,625)(RL)

And: RL= 24

The inductance of the coil can now be calculated when the coil is connected to a.c. at 50 Hz:

E = I.ZL

225= (5,625)(ZL)

And: Z= 40

ZL2 = RL

2 + X L

2

(40)2 = (24)2 + X L2

XL= 32

XL = 2...L

32 = (2)()(50)(L)

And: L = 101,859 mH

The new reactance of the coil and the current can now be calculated when the coil is connected to a.c. at 60 Hz:

XL = 2...L

= (2)()(60)(101,859 x 103)

= 38,4

1

E = IL.ZL

3000= (IL)(24 + j38,4)

And: IL= 6,62558 A

1.1.2 IL = (3,511 j5,618) A

Iactive = 3,511 A

And: Ireactive = 5,618 A

Iactive = I.cos

= (6,625)(cos 58)

= 3,511 A

Ireactive = I.sin

= (6,625)(sin 58)

= 5,618 A

1.1.3 cos L= cosI L

V L

= cos (0 + 58)

= 0,53 lagging

Assignment 1.2

A circuit draws 4,8 A from a 220-V, 50-Hz supply. The active power dissipated by the circuit is 219,65 W. A capacitor with a value of 110 F is connected in series with the circuit. This causes the magnitude of the input current to increase.

1.2.1 Calculate the values of the original circuit elements.1.2.2 Calculate the current after the capacitor is connected in the circuit.1.2.3 Calculate the power factor of the circuit.

Solution:

1.2.1 Since the input current increases when a capacitor is connected in series with the circuit, it means the original circuit is inductive.

P = V.I.cos

219,65= (220)(4,8)(cos )

cos = 0,208

2

And: = 78 E = I.Z

220= (4,8)(Z)

And: Z= 45,83378

Z = (9,529 + j44,831)

R = 9,529

XL = 2...L

44,831 = (2)()(50)(L)

And: L = 142,7015 mH

1.2.2 XC =

12. π . f .C

=

1

(2)( π )(50)(110 x 10−6 )= 28,937

Z = R + j(XL –XC)

= 9,529 + j(44,831 28,937)

= 18,53259,06 E = I.Z

2200= (I)(18,53259,06)

And: I= 11,87159,06 A

1.2.3 cos = cos ∠IE

= cos (0˚ + 59,06)

= 0,514 lagging

Assignment 1.3

A 12,96- resistor is connected in series with a 189-F capacitor. A coil with an unknown resistance and inductance is connected in series with this circuit. The circuit is connected across a supply that is represented by = 381,84 sin 345t volts and draws a current of 8,15 A at a power factor of 0,669 lagging from the supply.

1.3.1 Calculate the resistance and inductance of the coil.1.3.2 Calculate the potential difference across each component.

3

Solution:

1.3.1 XC =

1ω .C

=

1

(345 )(189 x 10−6 )

= 15,336

Em = √2 .E

E= 270 V

E = I.Z

2700= (8,1548)(Z)

And: Z = 33,12948

Z = R + ZL + XC

33,12948 = 12,96 + ZL + 15,33690

ZL= 41,00377,02

ZL = (9,208 + j39,956)

RL= 9,208

XL = ω.L

39,956 = (345)(L)

L = 115,8145 mH

1.3.2 VR = I.R

= (8,1548)(12,96)

= 105,62448 V

VL = I.ZL

= (8,1548)(41,00377,02)

= 334,174529,02 V

VC = I.XC

4

= (8,1548)(15,33690)

= 124,988138 V

Assignment 1.4

A series circuit consists of a resistor ab, a coil bc and a resistor cd. The circuit is connected across a 255-V, 50-Hz a.c. supply and a current of 6,5 A is drawn from the supply. The following voltage drops are measured in the circuit:

Vab = 52,65 VVbc = 107,25 VVac = 138,655 V

1.4.1 Calculate the resistance and inductance of the coil.1.4.2 Calculate the total impedance of the circuit.1.4.3 Calculate the total admittance of the circuit.1.4.4 Calculate the phasor value of the voltage Vbd.1.4.5 Calculate the power factor of the circuit.1.4.6 Draw a phasor diagram of the voltages and current in the circuit.

Solution:

Figure 1.5: Circuit diagram for Assignment 1.4

1.4.1 V bc2

= (I.RL)2 + (I.XL)2

(107,25)2= (I.RL)2 + (I.XL)2

(I.XL)2 = (107,25)2 (I.RL)2 .......... (i)

dcba

255 V, 50 Hz

6,5 A

138,655 V

107,25 V52,65 V

5

V ac2

= (Vab + I.RL)2 + (I.XL)2

= V ab2

+ 2Vab(I.RL) + (I.RL)2 + (107,25)2 (I.RL)2

(138,655)2= (52,65)2 + 684,45RL + (107,25)2

RL= 7,233

Substitute RL = 7,233 into equation (i):

[(6,5)(XL)]2 = (107,25)2 [(6,5)(7,233)]2

XL= 14,83

XL = 2...L

14,83= (2)()(50)(L)

L = 47,205 mH

1.4.2 VR1 = I.R1

52,65= (6,5)(R1) R1 = 8,1

E = I.Z

255 = (6,5)(Z)

Z= 39,231

Z2= (R1 + RL + R2)2 + X L2

(39,231)2= (8,1 + 7,233 + R2)2 + (14,83)2

R2 = 21

Z = (R1 + RL + R2) + jXL

= (8,1 + 7,233 + 21) + j14,83

= 39,24322,2

1.4.3 Y =

1Z

=

139 ,243∠22 ,2 °

= 0,025522,2 S

6

1.4.4 I = 6,522,2 A

Vcd = I.R2

= (6,522,2)(21)

= 136,522,2 V

Vbc = I.ZL

= (6,522,2)(7,233 + j14,83)

= 107,2541,8 V

Vbd = Vbc + Vcd

= 107,2541,8 + 136,522,2

= 207,2925,51 V

1.4.5 cos = ∠IE

cos (0 + 22,2)

= 0,9259 lagging1.4.6

Figure 1.6: Phasor diagram for Assignment 1.4

Assignment 1.5

A sinusoidal alternating current is represented by the equation:

i(t) = 12,35 sin 471,24t ampere

1.5.1 Calculate the time taken from t = 0 seconds for the current to reach a value of 7,85 A for the second time.

1.5.2 Calculate the energy dissipated when the current flows through a 22- resistor for one and a half hours.

1.5.3 Determine an expression for the instantaneous voltage drop across the resistor.

Solution:

I Vab

Vcd

Vbc

E22,2

41,8

7

1.5.1 i = 12,35 sin 471,24t1

7,85= (12,35){sin[(471,24)(t1)(180π )]}

And: t1 = 1,462 ms

Because 471,24t is in radians, it has to be multiplied by 180π to change

the factor to degrees in order to find the sine value. Refer to Figure 1.7. One full cycle equals one period. Thus, the second time that the current will reach 7,85 A, will be one period added to the value of the first time. Therefore:

ω = 2..

471,24= (2)()()

= 75 Hz

T =

1f

=

175

= 13,333 ms

t2 = T + t1

= 13,333 + 1,462

= 14,795 ms

Figure 1.7: Current wave for Assignment 1.5

Im = 12,35 A

i = 7,85 AIm = 12,35 A

Period, T (ms)

t2t1 t (ms)

i (A)

8

1.5.2 Im = √2 .I

12,35= (√2 )(I)

And: I= 8,733 A

E = I2.R.t

= (8,733)2(22)(1,5)(60)(60)

= 9,06 MJ

1.5.3 VRm = √2 .I.R

= (√2 )(8,733)(22)

= 271,707 V

R(t) = (271,707 sin 471,24t) volt

Assignment 1.6

Refer to the parallel network in Figure 1.12:

I1 = 5,96 A at a power factor of 0,906 I2 = 5,1 A at a power factor of 0,581 I = 10,95 A at a power factor of 0,829 lagging

1.6.1 Calculate the value of the components in each branch of the parallel circuit.

1.6.2 Calculate the total impedance of the circuit.1.6.3 Calculate the total admittance of the circuit.1.6.4 Draw a phasor diagram of the voltage and currents in the circuit.

9

Figure 1.12: Parallel network for Assignment 1.6

Solution:

1.6.1 Nothing else is mentioned which means that the supply voltage is the reference and at 0˚. Branch 1 is a series capacitive circuit, which means that the power factor is leading and the angle is therefore positive.

E = I1.Z1

2400˚ = (5,9625,04˚)(Z1)

Z1 = 40,268–25,04˚

= (36,484 – j17,044)

R1 = 36,484

XC =

12. π . f .C

17,044=

1(2)( π )(50)(C1)

C1 = 186,758 F

C1R1

L2R2

L3R3

I

I3

I2

I1

240 V/50 Hz

10

Branch 2 is a series inductive circuit, which means that the power factor is lagging and the angle is therefore negative, only because E is the reference at 0.

E = I2.Z2

2400˚ = (5,1–54,5˚)(Z2)

Z2 = 47,05954,5˚

= (27,327 + j38,311)

R2 = 27,327

XL = 2...L2

38,311= (2)()(50)(L2)

L2 = 121,95 mH

I = I1 + I2 + I3

10,95–34˚ = 5,9625,04˚ + 5,1–54,5˚ + I3

I3 = 4,551–80,94˚ A

E = I1.Z1

2400˚ = (4,551–80,94˚)(Z3)

Z3 = 52,73680,94˚

= (8,304 + j52,078) R3 = 8,304

XL = 2...L

52,078= (2)()(50)(L)

L3 = 165,769 mH

1.6.2

1Z =

1Z1 +

1Z2 +

1Z3

=

140 ,268∠−25 ,04 ° +

147 ,059∠54 ,5 ° +

152 ,736∠ 80 ,94 °

Z = 21,91734˚

Or: E = I.Z

11

2400˚ = (10,95–34˚)(Z)

Z= 21,91834˚

1.6.3 Y =

1Z

=

121 ,917∠34 °

= 0,0456–34˚ S

1.6.4

Figure 1.13: Phasor diagram for Assignment 1.6 (not to scale)

Assignment 1.7

A voltage that is represented by e(t) = 311,13 sin 375t volt supplies three parallel connected loads. The loads draw current from the supply as follow:

Load a: 6,4 A at a power factor of 0,8192 laggingLoad b: 9,6 A at unity power factorLoad c: 9,2 A at a power factor of 0,7431 leading

1.7.1 Calculate the total current flowing in the circuit.1.7.2 Calculate the active and reactive components of the total current.1.7.3 Calculate the value of the components of each load.1.7.4 Calculate the total impedance.1.7.5 Calculate the overall power factor of the circuit.1.7.6 If these loads were to be connected in series across the same supply,

what would be the current and overall power of the series circuit?

Solution:

1.7.1 I = I1 + I2 + I3

E

26,44˚

20,5˚

34˚

25,04˚

I3I2

I

I1

12

= 6,4–35˚ + 9,60˚ + 9,242˚

= 21,82156,54˚ A

1.7.2 I = (21,6795 + j2,485) A

Iactive = 21,6795 A

And: Ireactive = 2,485 A

1.7.3 Em = √2 .E

E =

311 ,13

√2

= 220 V

E = Ia.Za

2200˚ = (6,4–35˚)(Za)

Za = 34,37535˚

= (28,158 + j19,717) Ra = 28,158

XL = ω.La

19,717= (375)(La)

La = 52,579 mH

E = Ib.Zb

2200˚ = (9,6)(Zb)

Zb = 22,917

Rb = 22,917

E = Ic.Zc

2200˚ = (9,242˚)(Zc)

Zc = 23,913–42˚

= (17,771 – j16,001) Rc= 17,771

XC =

1ω .C

13

16,001=

1(375 )(Cc )

Cc = 166,656 F

1.7.4

1Z =

1Z1 +

1Z2 +

1Z3

=

134 ,375∠35 ° +

122 ,917 +

123 ,913∠−42 °

Z = 10,082–6,54˚


= cos (0 6,54˚)

= 0,9935 leading

1.7.6 Z = Z1 + Z2 + Z3

= 34,37535˚ + 22,917 + 23,913–42˚

= 68,9463,09˚

E = I.Z

2200˚ = (I)(68,9463,09˚)

I = 3,191–3,09˚ A

cos = cos ∠IE

= cos (0 3,09˚)

= 0,99855 lagging

Assignment 1.8

A coil is connected in parallel with an 18,75- resistor across a 50-Hz supply. The currents through the coil and resistor are 6,4 A and 4,8 A respectively. The current drawn from the supply by the circuit is 9,6 A.

14

1.8.1 Calculate the resistance and inductance of the coil.1.8.2 Calculate the power factor of the circuit.1.8.3 Draw a phasor diagram of the voltage and currents in the circuit.

Solution:

1.8.1

Figure 1.14: Phasor diagram of currents in Assignment 1.8 (not to scale)

Using the cosine rule:

IL2

= I2 + IR2

– 2.I.IR.cos∠I R

I

(6,4)2 = (9,6)2 + (4,8)2 – (2)(9,6)(4,8)(cos θ)

cos θ = 0,8056

And: θ = 36,34˚

The circuit is inductive and the overall power factor will therefore be lagging.

I = IL + IR

9,6–36,34˚ = IL + 4,80˚

IL = 6,4–62,73˚ A

E = I.Z

= (4,8)(18,75)

= 90 V

E = I.ZL

900˚= (6,4–62,73˚)(ZL)

ZL = 14,062562,73˚

= (6,443 + j12,5)

θ

IL I

IR

15

RL = 6,443

XL = 2...L

12,5= (2)()(50)(L)

L = 39,789 mH


= cos (0 + 36,34)

= 0,8055 lagging

1.8.3

Assignment 1.9

A single-phase a.c. parallel network consists of three branches with the current flowing in each branch represented by the equations:

i1(t) = 21,3 cos (ωt – π

1,5 ) A

i2(t) = 27,4 sin (ωt + π6 ) A

i3(t) = 18,6 sin (ωt + π3 ) A

The circuit is connected across a supply that is represented by the equation:

e(t) = 311,127 sin (315t + 0,085) V

1.9.1 Calculate the total current drawn from the supply expressing it in the same form as the branch currents.

1.9.2 Calculate the total admittance in the circuit.1.9.3 Calculate the overall power factor of the circuit.1.9.4 Draw the equivalent circuit for the parallel network.1.9.5 Calculate the total current drawn from the supply and the overall

power factor of the circuit if these impedances are connected in series across the same supply.

Solution:

26,39

36,34E

IL

I

IR

16

1.9.1 Im1 = 21,3(– 120˚ + 90˚) A

= 21,3–30˚ A

Im2 = 27,430˚ A

Im3 = 18,660˚ A

Im = I1 + I2 + I3

= 21,3–30˚ + 27,430˚ + 18,660˚

= 54,92520,41˚ A

i(t) = 54,925 sin (ωt + 0,356) A

1.9.2 Em = Im.Z

311,12715,3˚ = (54,92520,41˚)(Z)

Z = 5,665–5,11˚

Y =

1Z

=

15 ,665∠−5 ,11°

= 0,17655,11˚ S


= cos (15,3˚ – 20,41˚)

= 0,996 leading

1.9.4 Z = (5,6425 – j0,505)

XC =

1ω .C

0,505=

1(315 )(C )

C = 6286,343 F

17

Figure 1.16: Equivalent circuit for the parallel network of Assignment 1.9

1.9.5 Em = Im1.Z1

311,12715,3˚ = (21,3–30˚)(Z1)

Z1 = 14,60745,3˚

Em = Im2.Z2

311,12715,3˚ = (27,430˚)(Z2)

Z2 = 11,355–14,7˚

Em = Im3.Z3

311,12715,3˚ = (18,660˚)(Z3)

Z3 = 16,727–44,7˚

Z = Z1 + Z2 + Z3

= 14,60745,3˚ + 11,355–14,7˚ + 16,727–44,7˚

= 33,4205–7,33˚

Em = Im.Z

311,12715,3˚ = (√2 )(Im)(33,4205–7,33˚)

Im = 6,58322,63˚ A

cos = cos ∠IE

= cos (15,3˚ – 22,63˚)

= 0,9918 leading

Assignment 1.10

311,127 sin (315t + 0,805) V

6286,343 F5,6425

I3

I2

I1

bZs

Zx 12,3 290 F

c

25 mH(7,8 + j9)

(15 + j23,4) 96 F

a

18

Figure 1.17: Series-parallel a.c. network for Assignment 1.10

Refer to series-parallel a.c. circuit of Figure 1.17. The current drawn from the supply is 7,7 A at a power factor of 0,8033 lagging. The voltage drop Vcb

= 183,332,7 V.

1.10.1 Calculate the currents I1, I2 and I3.1.10.2 Calculate the values of the impedances Zx and Zs.

Solution:

1.10.1 E = 24020 V I = 7,716,55 A

XL1 = .L= (377)(0,025)

= 9,425

XC2 = (.C2)1

= [(377)(96 x 106)]1

= 27,63

XC3 =

1ω .C3

=

1

(377 )(290×10−6 )= 9,15

19

Vp = E Vcb

= 24020 183,332,7= 73,26313,37 V

Vp = I1.Z1

73,26313,37 = (I1)[7,8 + j(9 + 9,425)] I1 = 3,66280,425 A

Vp = I2.Z2

73,26313,37 = (I2)[15 + j(23,4 27,63)] I2 = 4,7012,38 A

I = I1 + I2 + I3 7,716,55= 3,66280,425 + 4,7012,38 + I3 I3 = 2,40830,5 A

1.10.2 Vp = I3.Z3

73,26313,37 = (2,40830,5)(Z3) Z3 = 30,42543,87

Z3 = Zx + XC3 + R 30,42543,87 = (Zx) + (12,3 j9,15)

Zx = 15,33951,09

Vs = Is.Zs

183,332,7 = (7,716,55)(Zs) Zs= 23,80549,25

Assignment 2.1

A coil of (7,2 + j9,6) is connected in series with a capacitor of 72 F. Another series circuit consists of a 15,5- resistor and a 72-mH inductor. The two circuits are then connected in parallel across a 240-V, 50-Hz supply.

2.1.1 Calculate the impedance of the circuit2.1.2 Calculate the admittance of the circuit.2.1.3 Calculate the current through the inductor.2.1.4 Calculate the current through the coil.2.1.5 Do the necessary calculations and draw the power triangle.2.1.6 Calculate the voltage drops across the inductor and the resistor.

20

2.1.7 Calculate the power factor of the circuit if a capacitor of 7,6 F is connected in parallel with the circuit to improve the power factor.

Solution:

2.1.1 XC =

12. π . f .C

=

1

(2)( π )(50)(72×10−6 )

= 44,21

XL = 2...L

= (2)()(50)(72 103)

= 22,6195

Z1 = RL + j(XL XC)

= 7,2 + j(9,6 44,21)

= 35,35178,25

Z2 = R + jXL

= 15,5 + j22,6195

= 27,42155,58

Z =

Z1⋅Z2

Z1+Z2

=

(35 ,351∠−78 ,25 °)(27 ,421∠55 ,58 ° )35 ,351∠−78 ,25 °+27 ,421∠55 ,58 °

= 37,7615,17

2.1.2 Y =

1Z

=

137 ,761∠5 ,17 °

21

= 0,02655,17 S

2.1.3 E = I2.Z2

2400˚ = (I2)(27,42155,58)

I2 = 8,75255,58 A

2.1.4 E = I1.Z1

2400˚ = (I1)(35,35178,25)

I1 = 6,78978,25 A

2.1.5 I = I1 + I2

= 6,78978,25 + 8,75255,58

= 6,3565,17 A

S* = E.I*

= (2400˚)(6,3565,17)

= 1525,445,17 VA

S* = (1519,234 + j 137,459) VA

P = 1519,234 W

And: Q = 137,459 VA’r

2.1.6 V L2 = I2.XL

= (8,75255,58)(22,619590)

= 197,96634,42 V VR = I2.R

= (8,75255,58)(15,5)

= 135,65655,58 V

2.1.7 XC =

12. π . f .C

=

1

(2)( π )(50)(7,6×10−6 )

= 418,829

22

QC =

V 2

XC

=

(240 )2

418 ,829= 137,526 VA’r

QC = P(tan 1 tan 2)

137,526 = (1519,234)(tan 5,17˚ – tan 2)

2 = 0

And: cos = 1

Assignment 2.2

A 12,8- resistor is connected in series with a 128-F capacitor. A coil is then connected in parallel to the series circuit. The parallel circuit is connected across an a.c. supply of 220-V, 50-Hz. A current of 7,24 A at a power factor of 0,724 is flowing through the coil.

2.2.1 Calculate the resistance and inductance of the coil.2.2.2 Calculate the current through the resistor.2.2.3 Calculate the total current.2.2.4 Calculate the power factor of the circuit.2.2.5 Calculate the total impedance.2.2.6 Calculate the complex, active and reactive power in the circuit.2.2.7 Calculate the value of three equal capacitors, connected in series, to

be connected in the circuit to raise the overall power factor to 0,978.

Solution:

2.2.1 E = IL.ZL

2200˚ = (7,2443,61)(ZL)

ZL = 30,38743,61

= (22 + j20,959)

RL = 22

And: XL = 2...L

20,959= (2)()(50)(L)

L = 66,715 mH

23

XC =

12. π . f .C

=

1

(2)( π )(50)(128×10−6 )

= 24,868

2.2.2 E = IS.ZS

2200˚ = (IS)(12,8 j24,868)

IS = 7,86662,76 A

2.2.3 I = IS + IL

= 7,2443,61 + 7,86662,76

= 9,06512,75 A


= (0 cos 12,75)

= 0,9753 leading

2.2.5 Z =

ZS⋅ZL

ZS+ZL

=

(12 ,8− j24 ,868 )(30 ,387∠43 ,61° )(12,8− j 24 ,868 )+30 ,387∠ 43 ,61°

= 24,26812,75 2.2.6 S* = E.I*

= (2200˚)(9,06512,75)

= 1994,312,75 VA

S* = (1945,125 j440,137) VA

P = 1945,125 W

And: Q = 440,137 VA’r

QC = P(tan 1 tan 2)

= (1945,125)(tan 12,75˚ – tan 12,04˚)

= 25,268 VA’r

24

QC =

V 2

XC

25,268=

(220 )2

XC

XC = 1915,466

XC =

12. π . f .C

1915,466 =

1(2)( π )(50)(C )

C = 1,662 F

C1 = C2 = C3

= 1,831 F

Assignment 2.3

Four elements are connected in parallel across a sinusoidal a.c. voltage supply that is represented by:

e(t) = 339,41 sin (314t π3 ) volt

The current drawn from the supply is represented by:

i(t) = 17,35 cos (314t 82,5˚) ampereThe instantaneous values of the current flowing through three of the elements are represented by:

i1(t) = 7,35 sin (314t +π

7,5 ) ampere

i2(t) = 8,1 sin (314t +π45 ) ampere

i3(t) = 7,1 sin (314t +π15 ) ampere

2.3.1 Determine the expression for the current of the fourth element.2.3.2 Calculate the component values of the fourth element.2.3.3 Calculate the impedance of the circuit.2.3.4 Calculate the power factor of the circuit.2.3.5 Draw a phasor diagram of the voltage and currents in the circuit.

25

2.3.6 If these impedances were connected in series, what would be the current flowing in the circuit?

2.3.7 Calculate the power factor of the series circuit.2.3.8 Calculate the value of a capacitor bank to be connected in parallel to

the parallel circuit to raise the overall power factor to 0,809 lagging.

Solution:

2.3.1 Im = Im1 + Im2 + Im3 + Im4

17,357,5 = 7,3524 + 8,14 + 7,112 + Im4

Im4 = 5,315148,64 A

i4(t) = 5,315 sin (314t 148,64) A

2.3.2 E = I4.Z4

339,41–60˚ = (5,315–148,64˚)(Z4)

Z4 = 63,85988,64

= (1,516 + j63,841)

R = 1,516

And: XL = ω.L

63,841= (314)(L)

L = 203,315 mH

2.3.3 E = I.Z 339,41–60˚ = (17,357,5˚)(Z)

Z = 19,562567,5


= cos (– 60˚ – 7,5˚)

= 0,3827 leading

5.3.5

I4

I2

I3I1

I

E

26

2.3.6 E = I1.Z1

339,41–60˚ = (7,3524˚)(Z1)

Z1= 46,17884

E = I2.Z2

339,41–60˚ = (8,14˚)(Z2)

Z2 = 41,902564

E = I3.Z3

339,41–60˚ = (7,112˚)(Z2)

Z3 = 47,80472

ZS = Z1 + Z2 + Z3 + Z4

= 46,17884 + 41,902564 + 47,80472 + 63,86388,64

= 76,22558,8

Em = Im.Z

339,41–60˚ = (Im)(76,225–58,8˚)

Im = 4,453–1,2 A

Im = √2 .I

4,453–1,2˚ = (√2 )(I)

I = 3,149–1,2˚ A


= cos [– 60˚ – (– 1,2˚)]

= 0,518 leading

2.3.8 P = E.I.cos

27

= (240)(12,268)(0,3827)

= 1126,791 W

QC = P(tan 1 tan 2)

= (1126,791)(tan 67,5˚ – tan 36˚)

= 1901,6525 VA’r

QC =

V C2

XC

1901,6525 =

(240 )2

XC

XC = 30,28945

XC =

1ω .C

30,28945 =

1(314 )(C )

C= 105,143 F

Assignment 2.4

A coil is connected in series with a resistor of 1,4 . These components are connected in parallel with a resistor of 19,5 and a capacitor of 1110 F. A coil with an impedance of (15,9 + j34,1) in series with a capacitor of 110 F is connected in series with the parallel circuit. The voltage drop measured across this coil is 28655,75˚ V. The whole is connected across a sinusoidal a.c. supply represented by the equation:

e(t) = 270 cos (310t – 72) volt

2.4.1 Calculate the voltage drop across the parallel part of the circuit.2.4.2 Calculate the resistance and inductance of the unknown coil.2.4.3 Calculate the power factor of the circuit.2.4.4 Calculate the value of each of three series capacitors connected in

parallel to the circuit to raise the overall power factor to 0,95106 lagging.

Solution:

28

2.4.1 VL = IL.ZL

28655,75˚ = (IL)(15,9 + j34,1)

IL= 7,601–9,25˚ A

= IC

XC =

1ωC

=

1

(310 )(110×10−6 )= 29,3255

VC = IC.XC

= (7,601–9,25˚)(29,3255–90˚)

= 222,903–99,25˚ V

VS = VL + VC

= 28655,75˚ + 222,903–99,25˚

= 126,2027,47˚ V

E= 190,91918˚ V

E = VP + VS

190,91918˚ = VP + 126,2027,47˚

VP = 70,70937,04˚ V

2.4.2 XC =

1ωC

=

1

(310 )(1110×10−6 )

= 2,906

VP = I2.Z2

70,70937,04˚ = (I2)(19,5 – j2,906)

I2 = 3,586545,52˚ A

I = I1 + I2

29

7,601–9,25˚ = I1 + 3,586545,52˚

I1 = 6,26–37,15˚ A

VP = I1.Z1

70,70937,04˚ = (6,26–37,15˚)(Z1)

Z1 = 11,29574,19˚

= (3,077 + j10,868)

RL = 3,077 – 1,4= 1,677

XL = ω.L

10,868 = (310)(L)

L = 35,058 mH


= cos(18˚ + 9,25˚)

= 0,889 lagging

2.4.4 P = E.I.cos

= (190,919)(7,601)(0,889)

= 1290,095 W

QC = P(tan 1 – tan 2)

= 1290,095(tan 27,25˚ – tan 18)

= 245,265 VAr

QC =

V C2

XC

245,265 =

(190 ,919)2

XC

XC = 148,615

30

XC =

1ω .C

148,615 =

1(310 )(C )

C = 21,706 F

1C =

1C1

+ 1C2

+ 1C3

121 ,706 =

3C1

C1 = 65,118 F

= C2

= C3

Assignment 2.5

A single-phase 39-kW, 380-V, 50-Hz induction motor operates on full load with an efficiency of 89% and at a power factor of 0,809 lagging. Three identical parallel-connected capacitors are used to raise the power factor to 0,96.

2.5.1 Calculate the kVAr-rating of the capacitor bank. 2.5.2 Calculate the capacitance of each capacitor.Solution:

2.5.1 =

Pout (motor )

Pin(motor )

0,89=

39P in(motor )

Pin(motor) = 43,82 kW

S1 =

P in(motor )

cos φ1

=

43 ,820 ,809

= 54,166 kVA

31

S2 =

P in(motor )

cos φ2

=

43 ,820 ,96

= 45,646 kVA

Q = S1.sin 1 S2.sin 2

= (54,166)sin(cos-1 0,809) (45,646)sin(cos-1 0,96)

= 19,058 kVAr

2.5.2 QC =

V C2

XC

19,058 x 103 =

(380 )2

XC

XC = 7,577

XC =

12. π . f .C

7,577=

1(2)( π )(50)(C )

C = 420,1 F

C = C1 + C2 + C3

= 3.C1

420,1= (3)(C1)

C1= 140 F

= C2

= C3

Assignment 2.6

A single-phase, 480-V, 50-Hz, induction motor develops 75 kW at a power factor of 0,76 lagging and an efficiency of 87%. A capacitor is connected in parallel across the supply to increase the power factor to 0,94 lagging. Calculate the capacitance of each capacitor.

Solution:

32

=

Pout (motor )

Pin(motor )

0,87=

75P in(motor )

Pin(motor) = 86,207 kW

QC = P(tan 1 tan 2)

= 86,201[tan (cos1 0,76) tan (cos1 0,94)]

= 42,429 kVAr ………. (i)

VC = IC.XC

IC = 2...C.VC

= (2)()(50)(C)(480)

= 150,79645(C) kA ………. (ii)

QC = VC.IC ………. (iii)

Substitute Equations (i) and (ii) into Equation (iii):

42,429= (480)(150,79645)(C)

And: C = 586,179 F

Assignment 3.1

A series circuit consists of a coil with an inductance of 640 mH and a resistance of 36 , a 11-F capacitor and a 16- resistor.

3.1.1 Calculate the value of the resonant frequency of the circuit.

The circuit is now connected to a 240-V a.c. supply at this resonant frequency:

3.1.2 Calculate the current drawn from the supply.3.1.3 Calculate the voltage drop across each component.

Solution:

33

3.1.1 r =

12. π √L .C

=

1

(2)( π )√(640×10−3 )(11×10−6 )

= 60 Hz

3.1.2 Z = R

Z = 52

E = I.Z

240= (I)(52)

I = 4,615 A

3.1.3 VR = I.R

= (4,615)(16)

= 73,84 V

XC =

12. π . f .C

=

1

(2)( π )(60)(11×10−6 )

= 241,144

= XL

VC = I.XC

= (4,615)(241,144–90˚)

= 1112,88–90˚ V

VL = I.ZL

= (4,615)(36 + j241,144)

= 1125,21381,51˚ V

Assignment 3.2

A coil is supplied from an a.c. source that is represented by the following equations for voltage and current:

34

eL(t) = 296,985 sin (315t + π

7,2 ) V

iL(t) = 5,303 sin (315t – π12 ) A

3.2.1 Calculate the resistance and inductance of the coil.3.2.2 Calculate the power factor of the circuit.3.2.3 If series capacitance is added to bring resonance to the circuit across

the same supply, determine the corresponding expression for the instantaneous value of the current.

Solution:

3.2.1 ELm = √2 .EL

296,98525˚ = (√2 )(EL)

= 21025˚ V

ILm = √2 .IL

5,303–15˚ = (√2 )(IL)

= 3,75–15˚ A

EL = IL.ZL

21025˚ = (3,75–15˚)(ZL)

ZL = 5640˚

= (42,8985 + j35,996)

RL = 42,8985

XL = ω.L

35,996= (315)(L)

L = 114,273 mH

3.2.2 cos = cos ∠IV

= cos [25˚ – (– 15˚)]

= 0,766 lagging

3.2.3 At resonance Z = R = 42,8985

35

E = I.Z

21025˚ = (I)(42,8985)

I = 4,89525˚ A

Im = √2 .I

= (√2 )(4,89525˚)

= 6,92325˚ A

i(t) = 6,923 sin (315t + π7,2 ) A

Assignment 3.3

A coil is connected in series with a capacitor across a variable-frequency source. The voltage is maintained constant at 230 V and the frequency is varied until a maximum current of 9,5 A flows through the circuit at 4,5 kHz. The Q-factor of the circuit under these conditions is 65.

3.3.1 Calculate the inductance of the coil.3.3.2 Calculate the value of the capacitance of the capacitor.3.3.3 Calculate voltage drop across each of the components.

Solution:

3.3.1 E = I.Z

230= (9,5)(Z)

Z = 24,2105

= R

Q =

2. π . f r . L

R

65 =

(2)( π )(4500 )(L)24 ,2105

L = 55,658 mH

36

3.3.2 XL = 2...L

= (2)()(4500)(55,658 x 10–3)

= 1573,693

= XC

XC =

12. π . f .C

1573,693 =

1(2)( π )(4500 )(C )

C= 22,474 nF

3.3.3 VC = I.XC

= (9,5)(1573,693–90˚)

= 14,95–90˚ kV

VL = I.ZL

= (9,5)(24,2105 + j1573,693)

= 14,9589,12˚ kV

Assignment 3.4

A 240-V, 50-Hz sinusoidal supply, with an overall power factor of unity, is connected across a parallel network consisting of the following three branches:

a: A coil with a resistance of 7,5 and an inductance of 31,831 mHb: A 10- resistor and a 425-F capacitor connected in seriesc: An unknown capacitor

3.4.1 Calculate the capacitance of the capacitor.3.4.2 Calculate the value of the current taken from the supply.

Solution:

37

3.4.1 XL = 2...L

= (2)()(50)(31,831 x 10–3)

= 10

XC =

12. π . f .C

=

1

(2)( π )(50)( 425×10−6 )

= 7,49

The parallel impedance of branches a and b is:

Zab =

Za .Zb

Za+Zb

=

(7,5+ j10 )(10− j7 ,49 )7,5+ j 10+10− j7 ,49

= 8,8348,135˚

= (8,745 + j1,25)

XL = 2...L

1,25= (2)()(50)(L)

L = 3,979 mH

A power factor of unity means that the circuit is in resonance. Thus:

r =

12. π √ 1

L .C−R2

L2

50 = [ 1(2)( π ) ]√ 1

(3 ,979×10−3 )(C )−

( 8 ,745 )2

(3 ,979×10−3)2

C = 51 F

3.4.2 XC =

12. π . f .C

38

=

1

(2)( π )(50)(51×10−6 )

= 62,414

Z =

Zab . XC

Zab+X C

=

(8 ,834∠8 ,135 ° )(62 ,414∠−90 ° )8 ,834∠8 ,135 °+62 ,414∠−90 °

= 8,9240˚

E = I.Z

2400˚ = (I)(8,9240˚)

I = 26,8940˚ A

Assignment 3.5

A coil with a resistance of 21,6 and an inductance of 216 mH is connected in parallel with a 108-F capacitor. The circuit is connected across a 180-V supply with a variable frequency set at 50 Hz.

3.5.1 Calculate the value of the current drawn from the supply.3.5.2 Calculate the value of the frequency at which the circuit will behave

as a pure resistor.3.5.3 Calculate the value of the impedance at resonance.3.5.4 Calculate the Q-factor of the circuit.3.5.5 Calculate the value of the current drawn from the supply at

resonance.

Solution:

3.5.1 XL = 2...L

= (2)()(50)(0,216)

= 67,858

XC =

12. π . f .C

=

1

(2)( π )(50)(108×10−6 )

39

= 29,473

Z =

Z L.ZC

Z L+ZC

=

(21 ,6+ j 67 ,858 )(− j29 ,473 )21 ,6+ j 67 ,858− j 29 ,473

= 47,652578,29

E = I.Z

180= (I)(47,652578,29)

I = 3,77778,29 A

3.5.2 r =

12. π √ 1

L .C−R2

L2

= [ 1(2)( π ) ]√ 1

(216×10−3 )(108×10−6 )−

(21 ,6 )2

(216×10−3 )2

= 28,854 Hz

3.5.2 Z =

LR .C

=

216×10−3

(21 ,6)(108×10−6 )

= 92,593

3.5.3 Q =

2. π . f r . L

R

=

(2)( π )(28 ,854 )(216×10−3 )21 ,6

= 1,813

Assignment 3.6

Refer to the parallel network in Figure 3.12. The final power factor of the circuit is unity.

3.6.1 Calculate the current in each branch of the circuit.

40

3.6.2 Calculate the capacitance of the unknown capacitor.3.6.3 Calculate the total current in the circuit.3.6.4 Calculate the total impedance of the circuit.


Solution:

3.6.1 XC =

12. π . f .C

=

1

(2)( π )(60)( 421×10−6 )

= 6,301

E = I1.Z1

3600˚ = (I1)(8,4 – j6,301)

I1 = 34,28436,87˚ A

RL = 7,2

C1 = 421 FR1 = 8,4

L = 25,465 mH

C3

I

I3

I2

I1

E = 360 V at 60 Hz

41

XL = 2...L

= (2)()(60)(25,465 x 10–3)

= 9,6

E = I2.Z2

3600˚ = (I2)(7,2 + j9,6)

I2 = 30–53,13˚ A

Ip = I1 + I2

= 34,28436,87˚ + 30–53,13˚

= 45,555–4,315˚ A

= (45,426 – j 3,4275) A

I3 = 3,427590˚ A

3.6.2 E = I3.XC

3600˚ = (3,427590˚)(XC)

XC = 105,033

XC =

12. π . f .C

105,033 =

1(2)( π )(60)(C3 )

C3 = 25,255 F3.6.3 I = Ip + I3

= 45,555–4,315˚ + 3,427590˚

= 45,4260˚ A

3.6.4 E = I.Z

3600˚ = (45,4260˚)(Z)

Z = 7,9250˚

Assignment 3.7

A sinusoidal a.c. supply is connected across a parallel network consisting of the following three branches:

a: A coil with a resistance of 7,5 and an inductance of 31,831 mHb: A 10- resistor and a 425-F capacitor connected in series

42

c: A coil with an impedance of (15,2 + j8,8) and a 150-F capacitor connected in series

The emf of the supply is represented by the equation:

e(t) = 311,127 sin (377t + π6 ) V

3.7.1 Calculate the current drawn from the supply.3.7.2 Calculate the frequency at which the circuit will be at resonance.

Solution:

3.7.1 Em = √2 .E

E = 22030 V

XL = .L

= (377)(31,831 x 10–3)

= 12

XCb =

1ω .C

=

1

(377 )(425×10−6 )

= 6,241

XCc =

1ω .C

=

1

(377 )(150×10−6 )

= 17,6835

1Z =

1Za +

1Zb +

1Zc

=

17,5+ j 12 +

110− j 6 ,241 +

115 ,2+ j 8,8− j17 ,6835

Z = 6,2874,92

E = I.Z

43

22030= (I)(6,2874,92)

I = 33,39934,92 A

3.7.2 C = 425 + 150

= 575 F

XLc = .Lc

8,8 = (377)(Lc)

Lc = 23,342 mH

L =

La .LcLa+Lc

=

(31 ,831)(23 ,342)31 ,831+23 ,342

= 13,467 mH

r =

12. π √ 1

L .C−R2

L2

=

1(2)( π ) √ 1

(0 ,013467 )(575×10−6 )−

(3 ,343 )2

(0 ,013467 )2

= 41,355 Hz

Assignment 4.1

A coil with a resistance of 12 and an inductance of 96 mH is connected in series with a variable capacitor. The circuit resonates at the third harmonic. The voltage across the circuit is represented by the following equation:

(t) = 72 + 121 sin 251t + 96 sin (753t + π3 ) + 66 sin (1255t +

π4 ) volts

4.1.1 Determine an expression for the current in the circuit.4.1.2 Calculate the overall power factor of the circuit.4.1.3 Calculate the active power contributed by each harmonic in the

circuit. Verify that the sum of these values is the same value calculated in Question 4.1.2.

Solution:

4.1.1 For resonance:

44

XL = XC

.L =

1ω .C

(753)(96 x 103) =

1(753 )(C )

C= 18,371 F

d.c. Fundamental 3rd Harmonic 5th Harmonic

Vm (V) 72 1210 9660 6645

R () 12 12 12 12

XL () 0 24,096 72,288 120,48

XC () 0 216,867 72,288 43,373

Im (A) 0 0,626586,44 860 0,846–36,15

For the d.c. part of the input Im = 0 since the capacitor acts as an open-circuit to d.c. when C has reached its final steady value.

i(t) = 0 + 0,6525 sin (251t + 1,509) + 8 sin (753t + 1,047)

+ 0,846 sin (1255t – 0,631) A

4.1.2 V= √V 02+

Vm12 +Vm3

2 +V m52

2

= √(72 )2+(121)2+(96 )2+(66)2

2

= 138,89 V

Although the d.c. term is present in the expression for the input voltage, the d.c. term for the current in the circuit is zero as a result of the capacitor in the circuit.

I = √ I 02+

I m12 + Im3

2 + Im52

2

45

= √0+(0,6525 )2+(8)2+(0,846 )2

2

= 5,706 A

P = I2.R

= (5,706)2(12)

= 390,701 W

P = V.I.cos

390,701 = (138,89)(5,706)(cos )

cos = 0,493 leading

4.1.3 P0 = V0.I0

= 0 W

P1 = ½Vm1.Im1.cos 1

= (½)(121)(0,6525)cos (0˚ – 86,44)

= 2,451 W


= (½)(96)(8)(1)

= 384 W


= (½)(66)(0,846)cos(45˚ + 36,15˚)

= 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:

46

(t) = 60 + 210 sin(314t +π9 ) + 159,3 sin(942t +

π4 ) + 105 sin(1570t +

π6 )

volts

4.2.1 Determine an expression for the current in the circuit.4.2.2 Calculate the overall power factor of the circuit.

Solution:

4.2.1 For resonance:

XL = XC

.L=

1ω .C

(942)(188 x 103) =

1(942 )(C )

C = 5,994 F


Vm (V) 60 21020 159,345 10530

R () 18 18 18 18

XL () 0 59,032 177,036 295,16

XC () 0 531,317 177,036 106,263

Z () 180 472,62887,82 180 189,75384,56

Im (A) 0 0,444107,82 8,8545 0,55354,56i(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 = √V 02+

Vm12 +Vm3

2 +V m52

2

= √(60 )2+(210 )2+(159,3 )2+(105)2

2

= 209,406 V

I =√ I 02+

I m12 + Im3

2 + Im52

2

47

= √0+(0,444 )2+(8,85 )2+( 0,553)2

2

= 6,278 A

P = I2.R

= (6,278)2(18)

= 709,439 W

P = V.I.cos

709,439 = (209,406)(6,278)(cos )

cos = 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:

(t) = 120 + 90 sin t + 60 sin 3t + 40 sin 5t volts

4.3.1 Determine an equation for the instantaneous value of the current in the circuit.4.3.2 Calculate the active power dissipated in the resistor.4.3.3 Calculate the overall power factor of the circuit.4.3.4 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


Vm (V) 120 90 60 40

R () 12 12 12 12

XL () 0 23,5 70,5 117,5

Z () 120 26,38762,95 71,51480,34 118,11184,17

Im (A) 100 3,41162,95 0,83980,34 0,33984,17

i(t) = 10 + 3,411 sin (t 62,95) + 0,839 sin (3t 80,34) + 0,339 sin (5t 84,17) A

48

4.3.2 I = √ I 02+

Im12 + Im3

2 + Im52

2

= √(10 )2+(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 = √V 02+

Vm12 +V m3

2 +Vm52

2

= √(120 )2+(90 )2+(60)2+(40 )2

2

= 145,086 V

P = V.I.ccos

1274,811 = (145,086)(10,307)(cos )

cos = 0,8525 lagging

4.3.4 P0 = V0.I0

= (120)(10)

= 1200 W


= (½)(90)(3,411)(cos 62,95)

= 69,805 W


= (½)(60)(0,839)(cos 80,34)

= 4,224 W

49


= (½)(40)(0,339)(cos 84,17)

= 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 Za and Zb respectively are given by:

ia(t) = 8 + 24 sin 40t 15 sin (120t + 75) A

ib(t) = 12 + 18 sin (40t + 60) + 12 sin (120t + 45) + 8 sin (200t + 30) A

Figure 4.5 : Parallel a.c. circuit for Assignment 4.4Solution:

d.c. terms: I0 = 8 + 12

= 20 A

= 40 rad/s: I1 = 24 + 1860

= 36,49725,285 A

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

= 120 rad/s: I3= 1575 + 1245

= 26,09261,705 A

i3 = 26,092 sin (120t + 61,705) A

ZbZa

ibiais

s

50

= 200 rad/s: i5 = 8 sin (200t + 30) A

i(t) = ia + ib

i(t) = 20 + 36,497sin (40t + 25,285) + 26,092 sin (120t + 61,705) + 8 sin (200t + 30) A

Assignment 5.1

Refer to the a.c. network of Figure 5.6. The following data is applicable on the components in the network:

E1 = 22415˚ VE2 = 24835˚ VZ1 = 12,432˚ Z2 = 15,6–45˚ Z3 = 14,829˚

Calculate the current flowing through the impedance Z3 using Kirchhoff’s laws.

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:

E2E1

Z3

Z2 Z1

51

Figure 5.7: A.C. network for Kirchhoff's laws with loops and polarities

According to Kirchhoff’s current law:

I2 – I1 – I3 = 0

Or: I1 = I2 – I3 ………. (i)

Consider loop abefa, using Kirchhoff’s voltage law:

I1.Z1 + I2.Z2 – E1 – E2= 0 Substitute Equation (i) into this equation:

(I2 – I3)Z1 + I2.Z2 – E1 – E2= 0

I2.Z1 – I3.Z1 + I2.Z2 = E1 + E2

12,432˚I2 – 12,432˚I3 + 15,6–45˚I2 = 22415˚ + 24835˚

22,003–11,69˚I2 – 12,432˚I3 = 464,84825,51˚

I2 = 21,12737,2˚ + 0,56443,69˚I3

Consider loop bcdeb, using Kirchhoff’s voltage law:

I3.Z3 + I2.Z2 – E2 = 0

14,829˚I3 + 15,6–45˚I2 = 24835˚

I2 = 15,89780˚ – 0,94974˚I3

It is obvious that I2 = I2, thus:

21,12737,2˚ + 0,56443,69˚I3 = 15,89780˚ – 0,94974˚I3

1,46462,79˚I3 = 14,36168,42˚

I2

I3I1

–

–

–

–

+

+

+

+

+

–

f e d

cba

E2E1

Z3

Z2 Z1

52

I3 = 9,809105,63˚ A

Assignment 5.2


E1 = 36036˚ VE2 = 44044˚ VZ1 = 12,845˚ Z2 = 15,4–36˚ Z3 = 14,542˚ Z4 = 18,260˚ Z5 = 2452˚ 5.2.1 Use Kirchhoff’s laws and calculate the current through the impedance

Z5.5.2.2 Calculate the voltage drop across the impedance Z5.5.2.3 Calculate the power factor of the impedance Z5.


Solution:

5.2.1 Refer to Paragraph 5.1.1 for the solution procedure of Kirchhoff’s laws:

Z4

E2

Z2

Z5

E1

Z3

Z1

+

–

–

–

++

–+

–

+

+

–

–

+

h g f e

dcba

I2I3

I5I4I1 Z4

E2

Z2

Z5

E1

Z3

Z1

53


Refer to Figure 5.9 using Kirchhoff’s current law:

I1 I4 I3 = 0

I3 = I1 I4 ………. (ii)

I2 + I4 I5 = 0

I2 = I5 I4 ………. (iii)

Substitute Equation (iii) into Equation (ii):

I3 = I1 I5 + I2 ………. (iv)Consider loop abgha 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

12,845I1 + 14,542I1 14,542I4 = 36036

27,29143,41I1 14,542I4 = 36036 I1 = 13,1917,41˚ + 0,5311,41I4 ……….

(v)

Consider loop bcfgb using Kirchhoff’s voltage law:

54

I2.Z2 + I3.Z3 I4.Z4 E2 = 0

Substitute Equations (ii) and (iii) into this equation:

(I5 I4)Z2 + (I1 I4)Z3 I4.Z4 = E2

I5.Z2 I4.Z2 + I1.Z3 I4.Z3 I4.Z4 = E2

15,436I5 15,436I4 + 14,542I1 14,542I4 18,260I4

= 44044

15,436I5 36,26126,91I4 + 14,542I1 = 44044

I1 = 30,3452 1,06278I5 + 2,50115,09I4

Obviously I1 = I1, thus:

13,1917,41 + 0,5311,41I4 = 30,3452 1,06278I5 + 2,50115,09I4

1,989161,29˚I4 = 17,4659,09 1,06278I5

I4 = 8,781152,2 0,534120,71I5

Consider loop cdefc using Kirchhoff’s voltage law:

I2.Z2 + I5.Z5 E2= 0

Substitute Equation (iii) into this equation:

(I5 I4)Z2 + I5.Z5 = E2

I5.Z2 I4.Z2 + I5.Z5 = E2

15,436I5 15,436I4 + 2452I5 = 44044

28,96519,9I5 15,436I4 = 44044

I4 = 1,88155,9I5 28,57180

Again I4 = I4, thus:

8,781152,2 0,534120,71I5 = 1,88155,9I5

28,57180

I5 = 11,19127,845 A

55

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.

Figure 5.10: Parallel impedance network for Assignment 5.3

Solution:

E2 = 38038 V E3 = 44044 V

Z2 = 15,436

E1 = 36036 V

Z4 = 18,260

Z3 = 14,542 Z1 = 12,845

Z3

I2

I3I1

f e d

cba

E2

Z2

E1

Z4

Z1

56


According to Kirchhoff’s current law at point b:

I1 = I2 + I3 ………. (vi)

Consider loop abefa using Kirchhoff’s voltage law:

I1.Z1 + I2.Z2 E1 E2 = 0

Substituting Equation (vi) into this equation:

(I2 + I3)(12,845) + I2(15,436) = 36036 + 38038

I2 = 34,39837,03 I3(0,59545)

Consider loop bcdeb using Kirchhoff’s voltage law:

I3.Z4 + I3.Z3 E3 + E2 I2.Z2 = 0

I3 (18,260 + 14,542) I2(15,436) = 44044 38038

I2 = I3(2,09888,03) 4,786112,61

34,39837,03 I3(0,59545) = I3(2,09888,03) 4,786112,61

And: I3 = 13,99134,47 A

Assignment 5.4


E1 = 22415˚ VE2 = 24835˚ VZ1 = 12,432˚ Z2 = 15,6–45˚ Z3 = 14,829˚

Calculate the current through the impedance Z3 by using the Superposition Theorem.

57


Solution:

Consider the voltage source E1 and short-circuit the voltage source E2:

Figure 5.13: A.C. network for Assignment 5.4 with E2 short-circuited

Consider E1 in Figure 5.13:

Z = Z1 +

Z2 .Z3

Z2+Z3

= 12,432 +

(15 ,6∠−45 °)(14 ,8∠29 °)15 ,6∠−45 °+14 ,8∠29 °

= 20,68215,235

E1 = I1.Z

22415˚ = (I1)(20,68215,235)

E2E1

Z3

Z2 Z1

I3(1)I1

E1

Z3

Z2Z1

58

I1 = 10,831–0,235 A

I3(1) = I1.

Z2

Z2+Z3

= (10,831–0,235)(15 ,6∠−45 °15 ,6∠−45 °+14 ,8∠29 ° )

= 6,95836,1 A


Figure 5.14: A.C. network for Assignment 5.4 with E1 short-circuited

Consider E2 in Figure 5.14:

Z = Z2 +

Z1 .Z3

Z1+Z3

= 15,645 +

(12 ,4∠32 °)(14 ,8∠29 °)12 ,4∠32°+14 ,8∠29 °

= 18,47124,27

E2 = I2.Z

24835˚ = (I2)(18,471–24,27)

I2 = 13,426559,27 A

I2

I3(2)

E2

Z3

Z2Z1

59

I3(2) = I2.

Z1

Z1+Z3

= (13,426559,27)(12 ,4∠32 °12 ,4∠32 °+14 ,8∠29 ° )

= 6,12360,9 A

I3 = I3(2) I1(1)

= 6,12360,9 6,95836,1

= 9,813105,63 A

Assignment 5.5


E1 = (285 + j165) VE2 = (180 + j160) VZ1 = (14 – j12) Z2 = (18 + j24) Z3 = (22 + j15) Z4 = (18 – j18) Z5 = (16 + j12)

Use the Superposition Theorem and calculate the current drawn from each supply.


Z4

E2

Z2

Z5

E1

Z3

Z1

60

Solution:


Figure 5.16: A.C. parallel network for Assignment 5.5 with E2 short-circuited

Zs =

Z2 .Z5

Z2+Z5 + Z4

=

(18+ j24 )(16+ j12 )18+ j24+16+ j12 + (18 – j18)

= 35,6331,85

Z =

Zs . Z3

Zs+Z3 + Z1

=

(35 ,63∠−31 ,85 °)(22+ j15 )35 ,63∠−31 ,85°+(22+ j15 ) + (14 – j12)

= 33,487–17,23 E1 = I1(1).Z

(285 + j165) = [I1(1)](33,487–17,23)

I1(1) = 9,83447,3 A

I4 = I1(1)

Z3

Z3+Zs

= (9,83447,3)[22+ j 15(22+ j 15)+35 ,63∠31 ,85 ° ]

= 4,20748,695 A

I2(1)I4I1(1)Z4

Z2

Z5

E1

Z3

Z1

61

I2(1) = – I4.

Z5

Z2+Z5

= (– 4,20748,695)(16+ j1218+ j24+16− j 12 )

= 2,334113,88 A


Figure 5.17: A.C. parallel network for Assignment 5.5 with E1 short-circuited

Zs =

Z1 .Z3

Z1+Z3 + Z4

=

(14− j 12)(22+ j15 )14− j 12+22+ j 15 + (18 – j18)

= 37,509–33,33

Z =

Zs . Z5

Zs+Z5 + Z2

=

(37 ,509∠−33 ,33 °)(16+ j12 )37 ,509∠−33 ,33°+(16+ j 12) + (18+j24)

= 43,21139,92

E2 = I2(2).Z

(180 + j160)= [I2(2)](38,845–31,45)

I2(2) = 5,5741,71 A

I4 I2(2)I1(2)Z4

E2

Z2

Z5 Z3

Z1

62

I1(2) = – I2.

Z5

Z5+Zs

= (– 5,5741,71)(16+ j12(16+ j12 )+37 ,509∠−33 ,33 ° )

= 2,31748,89 A

I1 = I1(1) + I1(2)

= 9,83447,3 + 2,31748,89

= 12,1547,04 A

I2 = I2(1) + I2(2)

= 2,334113,88 + 5,5741,71

= 5,02823,04 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.


E2 = 38038 V E3 = 44044 V

Z2 = 15,436

E1 = 36036 V

Z4 = 18,260

Z3 = 14,542 Z1 = 12,845

63

Solution:

Consider the voltage source E1 and short-circuit the voltage sources E2 and E3 and the impedance Z4:

Figure 5.16: Parallel impedance network for Assignment 5.5 with E2 and E3

short-circuited

Z = Z1 +

(Z3+Z 4)(Z2)Z2+Z3+Z4

= 12,845 +

(14 ,5∠42°+18 ,2∠60 ° )(15 , 4∠−36° )15 ,4∠−36°+14 ,5∠42 °+18 ,2∠ 60°

= 23,43116

E1 = I1.Z

36036= (I1)(23,43116)

I1 = 15,36420 A

I4(1) = I1.

Z2

Z2+Z3+Z4

= (15,36420)( (15 ,4∠−36 °)15 ,4∠−36°+14 ,5∠ 42°+18 ,2∠60 ° )

= 6,52542,91 A

Z4

I4(2)

I2

E2

Z2 Z1 Z3

Z4I4(1)I1

Z2

E1

Z1 Z3

64


short-circuited

Z = Z2 +

(Z3+Z 4)(Z1)Z1+Z3+Z4

= 15,436 +

(14 ,5∠42°+18 ,2∠60 ° )(12 ,8∠45 ° )12 ,8∠ 45°+14 ,5∠42°+18 ,2∠60 °

= 18,8677,12

E2 = I2.Z

38038= (I2)(18,8677,12)

I2 = 20,14145,12 A

I4(2) = I2.

Z1

Z1+Z3+Z4

= (20,14145,12)( (12 ,8∠45 ° )12 ,8∠45 °+14 ,5∠42 °+18 ,2∠60 ° )

= 5,72544,09 A


short-circuited

Z4

I4(3)

E3

Z2 Z1 Z3

65

Z = Z3 + Z4 +

(Z1)(Z2)Z1+Z2

= 14,542 + 18,260 +

(12 ,8∠ 45 ° )(15 ,4∠−36 ° )12 ,8∠ 45 °+15 ,4∠−36 °

= 39,542,92

E3 = I4(3).Z

44044= [I4(3)](39,542,92)

I4(3) = 11,1391,08 A

I4 = I4(1) + I4(3) I4(2)

= 6,52542,91 + 11,1391,08 5,72540,09

= 13,99334,47 A

Assignment 5.7


E1 = 22415˚ VE2 = 24835˚ VZ1 = 12,432˚ Z2 = 15,6–45˚ Z4 = 14,829˚ Calculate the current through the impedance Z4 by using Thevenin’s Theorem.


E2E1

Z4

66

Solution:

Open-circuit Z4 and short-circuit the voltage sources E1 and E2:

Figure 5.21: A.C. parallel network for Assignment 5.7 with Z4 open-circuited and E1 and E2 short-circuited

ZTH =

Z1⋅Z2

Z1+Z2

=

(12 ,4∠32 ° )(15 ,6∠−45 ° )12 ,4∠32 °+15 ,6∠−45 °

= 8,7911,31

Consider the voltage source E1 and short-circuit the voltage source E2 and open-circuit the impedance Z4:

Figure 5.22: A.C. parallel network for Assignment 5.7 with E2 short-circuited and Z4 open-circuited

Z2 Z1

VTH(1)

E1

Z2 Z1

67

VTH(1) = E1⋅

Z2

Z1+Z2

= (224∠15 ° )(15 ,6∠−45 °

12 ,4∠32 °+15 ,6∠−45 ° )= 158,81218,31 V

Consider the voltage source E2 and short-circuit the voltage source E1 and open-circuit the impedance Z4:

Figure 5.23: A.C. parallel network for Assignment 5.7 with E1 short-circuited and Z4 open-circuited

VTH(2) = E2⋅

Z1

Z1+Z2

= (24835˚)(12 ,4∠32 °12 ,4∠32 °+15 ,6∠−45 ° )

= 139,7678,69 V

VTH = VTH(2) VTH(1)

= 139,7678,69 – 158,81218,31

= 223,973123,42 V

The Thevenin equivalent circuit is shown in Figure 5.24.

VTH(2)

E2

Z2 Z1

I4

Z4 = 14,829˚

ZTH = 8,7911,31

VTH = 223,973123,42 V68

Figure 5.24: Thevenin equivalent circuit for Assignment 5.7

I4 =

ETH

ZTH+Z4

=

223 ,973∠123 ,42 °8 ,791∠−1 ,31 °+14 ,8∠29 °

= 9,813105,63 A

Assignment 5.8


E1 = 36036˚ VE2 = 44044˚ VZ1 = 12,845˚ Z2 = 15,4–36˚ Z3 = 14,542˚ Z4 = 18,260˚ Z5 = 2452˚ 5.8.1 Use Thevenin’s Theorem and calculate the current through the

impedance Z5.5.8.2 Calculate the voltage drop across the impedance Z5.5.8.3 Calculate the power factor of the impedance Z5.

69


Solution:

5.8.1 Open-circuit the impedance Z5 and short-circuit the voltage sources E1 and E2:

Z4

Z2

Z3

Z1

Z4

E2

Z2

Z5

E1

Z3

Z1

70


Zp =

Z1 .Z3

Z1+Z3

=

(12 ,8∠ 45 °)(14 ,5∠42 °)12 ,8∠ 45 °+14 ,5∠42 °

= 6,80143,59

ZTH =

(Z p+Z4 )Z2

Z p+Z4+Z2

=

(6 ,801∠43 ,59 °+18 ,2∠60 ° )(15 ,4∠−36 ° )6 ,801∠43 ,59 °+18 ,2∠60 °+15 ,4∠−36 °

= 13,2453,73

Consider the voltage source E1, short-circuit the voltage source E2 and open-circuit the impedance Z5:

Figure 5.27: A.C. parallel network for Assignment 5.7 with Z5 open-circuited and E2 short-circuited

I2

I1

VTH(1)

Z4

Z2

E1

Z3

Z1

71

Z =

(Z2+Z 4)Z3

Z2+Z4+Z3 + Z1

=

(15 ,4∠−36°+18 ,2∠ 60° )(14 ,5∠ 42° )15 ,4∠−36°+18 ,2∠ 60°+14 ,5∠ 42° + 12,845˚

= 21,739,78

E1 = I1.Z

36036= (I1)(21,739,78)

I1= 16,593,78 A

I2 = I 1⋅

Z3

Z3+Z4+Z2

= (16,593,78)(14 ,5∠ 42°14 ,5∠ 42°+18 ,2∠60°+ 15 ,4∠−36 ° )

= 6,63411,31 A

VTH(1) = I2.Z2

= (6,63411,31)(15,436)

= 102,16424,69 V

Consider the voltage source E2, short-circuit the voltage source E1 and open-circuit the impedance Z5:

Figure 5.28: A.C. parallel network for Assignment 5.7 with Z5 open-circuited and E1 short-circuited

I2

VTH(2)

Z4

E2

Z2

Z3

Z1

72

Z =

Z1 .Z3

Z1+Z3

+Z 4+Z2

=

(12 ,8∠ 45 °)(14 ,5∠42 °)12 ,8∠ 45°+14 ,5∠42 ° + 18,260˚ + 15,4–36˚

= 28,83323,29

E2 = I2.Z

44044= (I2)(28,83323,29)

I2= 15,2620,71 A

V2 = I2.Z2

= (15,2615)(15,436)

= 235,00515,29 V

VTH(2) = E2 V2

= 44044 235,00415,29

= 378,42676,27 V

VTH = VTH(1) + VTH(2)

= 102,16424,69 + 378,42676,27˚

= 372,74360,66˚ VThe Thevenin equivalent circuit is shown in Figure 5.29.


I5 =

V TH

ZTH+Z5

I5

Z5 = 2452˚

ZTH = 13,2453,73

VTH = 395,75661,32 V

73

=

372 ,743∠60 ,66 °13 ,245∠−3 ,73 °+24∠52 °

= 11,19127,845 A

5.7.2 V5 = I5.Z5

= (11,19127,845)(2452)

= 268,5879,845 V

5.7.3 cos 5 = cos∠I 5

V 5

= cos (79,845 – 27,845)

= 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.


Solution:

E2 = 38038 V E3 = 44044 V

Z2 = 15,436

E1 = 36036 V

Z4 = 18,260

Z3 = 14,542 Z1 = 12,845

ZTH

Z2 Z1 Z3 74

Figure 5.31: Parallel impedance network for Assignment 5.8 to determine ZTH

ZTH = Z3 +

Z1 .Z2

Z1+Z2

= 14,542 +

(12 ,8∠ 45 °)(15 ,4∠−36 °)12 ,8∠ 45 °+15 ,4∠−36 °

= 22,7429,32

Consider the voltage source E1 and short-circuit the voltage sources E2 and E3 and open-circuit the impedance Z4:


short-circuited and Z4 open-circuited

VTH(1)

Z2

E1

Z1 Z3

75

VTH(1) = E1 +

Z2

Z1+Z2

= 36036 +

12 ,8∠ 45 °12 ,8∠ 45 °+15 ,4∠−36 °

= 257,7430 V




VTH(2) = E2 +

Z1

Z1+Z2

= 38038 +

15 ,4∠−36 °12 ,8∠ 45 °+15 ,4∠−36 °

= 226,12983 V


VTH(2)

Z2

E2

Z1 Z3

VTH(3)

Z2

E3

Z1 Z3

76



VTH(3) = E3

= 44044 V

VTH = VTH(1) + VTH(3) VTH(2)

= 257,743 + 44044 226,12983

= 552,6938,45 V

The Thevenin equivalent circuit is shown in Figure 5.35.

I4 =

V TH

ZTH+Z4

=

552 ,693∠ 8 ,45 °22 ,74∠29 ,32 °+18 ,2∠ 60°

= 13,99234,47 A


Assignment 5.9


E1 = 22415˚ VE2 = 24835˚ V

I4

Z4 = 18,260˚

ZTH = 22,7429,32

VTH = 552,6938,45 V

77

Z1 = 12,432˚ Z2 = 15,6–45˚ Z3 = 14,829˚

Calculate the current through the impedance Z3 by using Norton’s Theorem.


Solution:

Open-circuit Z3 and short-circuit the voltage sources E1 and E2:


ZN =

Z1⋅Z2

Z1+Z2

=

(12 ,4∠32 °)(15 ,6∠−45 °)12 ,4∠32 °+15 ,6∠−45 °

= 8,7911,31

Consider the voltage source E1 and short-circuit the voltage source E2 and the impedance Z3:

E2E1

Z3

Z2 Z1

ZN

Z2 Z1

78

Figure 5.39: A.C. parallel network for Assignment 5.9 with E2 and Z3 short-circuited

E1 = I1.Z1

22415= (I1)(12,432)

I1 = 18,064517 A

Because no current will flow through Z2 due to the short-circuit of Z3:

IN(1) = I1



E2 = I2.Z2

I1 IN(1)

E1

Z2 Z1

I2

IN(2)

E2

Z2 Z1

79

24835= (I2)(15,645)

I2 = 15,89780 A

The same will happen as in the circuit of Figure 5.40 Thus:

IN(2) = I2

IN = IN(2) IN(1)

= 15,89780 – 18,064517

= 25,476124,73 A

The Norton equivalent circuit is shown in Figure 5.41.

Figure 5.41: Norton equivalent circuit for Assignment 5.9

I3 = IN⋅

ZN

ZN+Z3

= (25,476124,73)( 8 ,791∠−1 ,31 °

8 ,791∠−1 ,31 °+14 ,8∠29 ° )= 9,812105,63 A

Assignment 5.10

I3

Z3ZNIN = 25,476124,73 A

80


E1 = 36036˚ VE2 = 44044˚ VZ1 = 12,845˚ Z2 = 15,4–36˚ Z3 = 14,542˚ Z4 = 18,260˚ Z5 = 2452˚ 5.10.1 Use Norton’s Theorem and calculate the current through the load

impedance.5.10.2 Calculate the voltage drop across the load impedance.5.10.3 Calculate the power factor of the load impedance.


Solution:

5.10.1 Open-circuit the impedance Z5 and short-circuit the voltages sources E1 and E2:

Z4

E2

Z2

Z5

E1

Z3

Z1

Z4

Z2

Z3

Z1

81


Zp =

Z1 .Z3

Z1+Z3

=

(12 ,8∠ 45 ° )(14 ,5∠42 °)12 ,8∠ 45 °+14 ,5∠42 °

= 6,80143,59

ZN =

(Z p+Z4 )Z2

Z p+Z4+Z2

=

(6 ,801∠43 ,59 °+18 ,2∠60 °)(15 ,4∠−36 ° )6 ,801∠43 ,59 °+18 ,2∠60 °+15 ,4∠−36 °

= 13,2453,73


IN(1)I1 Z4

Z2

E1

Z3

Z1

82


No current will flow through Z2 due to the short-circuit of Z5, thus:

Z =

Z3 .Z4

Z3+Z4 + Z1

=

(14 ,5∠42 ° )(18 ,2∠ 60° )14 ,5∠42 °+18 ,2∠60 ° + 12,845˚

= 20,9546,94˚

E1 = I1.Z

36036= (I1)(20,95146,94)

I1 = 17,18310,94 A

IN(1) = I 1⋅

Z3

Z3+Z4

= (17 ,183∠−10 ,94 ° )(14 ,5∠42 °

14 ,5∠42 °+18 ,2∠60 ° )= 7,71320,97 A

Consider the voltage source E2 and short-circuit the voltage source E1

and the impedance Z5:

IN(2)

I2

Z4

E2

Z2

Z3

Z1

83


E2 = I2.Z2

44044= (I2)(15,436)

I2 = 28,57180 A

= IN(2)

IN = IN(1) + IN(2)

= 7,71320,97 + 28,57180

= 28,14164,39 A



I5 = IN⋅

ZN

ZN+Z5

= (28 ,141∠64 ,39 ° )(13 ,245∠−3 ,73°

13 ,245∠−3 ,73 °+24∠52° )= 11,19127,845 A

5.10.2 V5 = I5.Z5

I5

Z5ZN IN = 28,14164,39 A

84

= (11,19127,845)(2452)

= 268,5879,845 V

5.10.3 cos = cos(79,845 – 27,845)

= 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.


Solution:

E2 = 38038 V E3 = 44044 V

Z2 = 15,436

E1 = 36036 V

Z4 = 18,260

Z3 = 14,542 Z1 = 12,845

ZN

Z2 Z1 Z3

85

Figure 5.48: Parallel impedance network for Assignment 5.11 to determine ZN

ZN = Z3 +

Z1 .Z2

Z1+Z2

= 14,542 +

(12 ,8∠ 45 ° )(15 ,4∠−36 ° )12 ,8∠ 45 °+15 ,4∠−36 °

= 22,7429,32


Figure 5.49: Parallel impedance network for Assignment 5.11 with E2, E3

and Z4 short-circuited

Z = Z1 +

Z2 .Z3

Z2+Z3

= 12,845 +

(15 ,4∠−36 °)(14 ,5∠ 42° )15 ,4∠−36 °+14 ,5∠42 °

I1 IN(1)

Z2

E1

Z1 Z3

86

= 21,04427,72

E1 = I1.Z

36036= (I1)(21,04427,72)

I1 = 17,1078,28 A

IN(1) = I1.

Z2

Z2+Z3

= (17,1078,28)(15 ,4∠−36 °15 ,4∠−36°+14 ,5∠ 42° )

= 11,33429,32 A

Consider the voltage source E2 and short-circuit the voltage sources E1, E3

and the impedance Z4:



Z = Z2 +

Z1 .Z3

Z1+Z3

= 15,436 +

(12 ,8∠ 45 ° )(14 ,5∠42 °)12 ,8∠ 45 °+14 ,5∠42 °

I2

IN(2)

Z2

E2

Z1 Z3

87

= 17,92314,09 E2 = I2.Z

38038= (I2)(17,92314,09)

I2 = 21,20152,09 A

IN(2) = I2.

Z1

Z1+Z3

= (21,20152,09)(12 ,8∠45 °12 ,8∠45 °+14 ,5∠42 ° )

= 9,94453,68 A




Z = Z3 +

Z1 .Z2

Z1+Z2

= 14,542 +

(12 ,8∠ 45 ° )(15 ,4∠−36 ° )12 ,8∠ 45 °+15 ,4∠−36 °

VTH(3)

Z2

E3

Z1 Z3

88

= 22,7429,32

E3 = IN(3).Z

44044= [IN(3)](22,7429,32)

IN(3) = 19,34914,68 A

IN = IN(1) + IN(3) IN(2)

= 11,33429,32 + 19,34914,68 9,94453,68

= 24,30420,87 A



I4 = IN⋅

ZN

ZN+Z4

= (28 ,141∠64 ,39 ° )(22 ,74∠29 ,32 °

22 ,74∠29 ,32°+18 ,2∠60 ° )= 13,99234,47 A

Assignment 5.12

I4

Z4ZN IN = 24,30420,87 A

89

Consider Figure 5.54.

5.12.1 Calculate the total impedance of the network using delta-star conversion.

5.12.2 Calculate the current drawn from the supply.

Figure 4.54: Circuit diagram for Assignment 5.12

Solution:

5.12.1 Convert the delta to an equivalent star:

Za =

Zab .Zca

Zab+Zbc+Zca

=

(14 ,4+ j 7,2)(8,8+ j 8,8)14 , 4+ j7,2+9,5− j 16 ,2+8,8+ j 8,8

= 6,12771,92°

Zb =

Zab .Zbc

Zab+Zbc+Zca

=

(9,5− j 16 ,2 )(14 ,4+ j 7,2)14 , 4+ j7,2+9,5− j 16 ,2+8,8+ j 8,8

= 9,24632,7°

Zc =

Zbc . Zca

Zab+Zbc+Zca

E = 24012˚ V

cb

a

(14,4 + j7,2)

(9,5 j16,2)

(8,8 + j8,8)

(13,4 j9,3)

(7,8 + j12,3)

90

=

(8,8+ j8,8 )(9,5− j 16 ,2)14 , 4+ j7,2+9,5− j 16 ,2+8,8+ j 8,8

= 7,14714,26°

The star-connection shown in Figure 5.55 now replaces the delta.

Figure 5.55: Equivalent star circuit diagram

Z1 = Za + (7,8 + j12,3)

= 20,55861,84°

Z2 = Zc + (13,4 j9,3)

= 23,14128,55°

Zp =

Z1 .Z2

Z1+Z2

=

(20 ,558∠ 61,84 ° )(23 ,141∠−28 ,55 ° )20 ,558∠61 ,84 °+23 ,141∠−28 ,55 °

= 15,42120,05°

Z = Zb + Zp

= 9,24632,7° + 15,42120,05°

= 22,2690,75°

5.12.2 E = I.Z

24012˚ = (I)(22,2690,75˚)

I= 10,77711,25˚ A

c b

a

7,14714,26 9,24632,7

6,12771,92

(13,4 j9,3)

(7,8 + j12,3)

91

Assignment 5.13

Refer to the a.c. parallel network of Figure 5.56. The component values if the circuit are the following:

E1 = 36036 VE2 = 44044 VZ1 = 12,845 Z2 = 15,436 Z3 = 14,542 Z4 = 18,260 Z5 = 2452

Convert the delta Z3, Z4 and Z5 to an equivalent star connection and draw this equivalent circuit.


Solution:

5.13.1 Za =

Z3 .Z4

Z3+Z4+Z5

=

14 ,5∠42 °+18 ,2∠60 °14 ,5∠42 °+18 ,2∠60°+24∠52°

= 4,68749,985

++

E2 E1

Z5 Z3

Z2 Z1

Z4

92

Zb =

Z4 .Z5

Z3+Z4+Z5

=

18 ,2∠60 °+24∠52°14 ,5∠42 °+18 ,2∠60 °+24∠52°

= 7,75859,985

Zc =

Z3 .Z5

Z3+Z4+Z5

=

14 ,5∠42 °+24∠52 °14 ,5∠42 °+18 ,2∠60°+24∠52°

= 6,18141,985

Figure 5.57: Equivalent circuit for Figure 5.56

Assignment 5.14

Consider the parallel network in Figure 5.59.

5.14.1 Calculate the load impedance for maximum power transfer to the load.

Z1Z2

ZbZa

E2E1

Zc

93

5.14.2 Calculate the value of the maximum power transferred.


Solution:

5.14.1 Open-circuit the impedance Zload and short-circuit the voltages source E:

Figure 5.60: Parallel network for Assignment 5.14 with E short-circuited and Zload open-circuited

ZTH =

Z1 .Z2

Z1+Z2

ZloadZ2 = 8,245˚ E = 2400˚ V

Z1 = 9,6–54˚

Z2 = 8,245˚

Z1 = 9,6–54˚

94

=

(9,6∠−54 °)( 8,2∠ 45° )9,6∠−54°+8,2∠ 45°

= 6,7810,76˚

Zload = 6,7810,76˚

5.14.2 Consider E and open-circuit the impedance Zload:

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

VTH = E.

Z2

Z1+Z2

= (2400˚)

8,2∠ 45 °8,2∠ 45°+9,6∠−54 °

= 169,52254,76 V

Iload =

V TH

ZTH+Z load

=

169 ,522∠54 ,76 °6 ,781∠0 ,76 °+6 ,781∠−0 ,76 °

= 12,50154,76 A

Pout(max) = I load2 .R load

= (12,501)2(6,78)

= 1059,545 W

Or: ZTH = (6,78 – j0,09)

VTHZ2 = 8,245˚ E = 2400˚ V

Z1 = 9,6–54˚

95

RTH = 6,78

Pout(max) =

V TH2

4 . RTH

=

(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˚ VZab = 6,448˚ Zbc = 7,236˚ Zca = 4,530˚ Zbd = 5,545˚

5.15.1 Calculate the load impedance for maximum power transfer to the load.

5.15.2 Calculate the value of the maximum power transferred.

Figure 5.62: Bridge circuit diagram for Assignment 5.15

5.15.1 Convert the delta abc to star:

d

c b

a

Zbc

ZbdZload

ZabZca

E

96

Za =

Zab .Zca

Zab+Zbc+Zca

=

(6,4∠48 °)( 4,5∠ 30° )6,4∠48 °+7,2∠36 °+4,5∠30 °

= 1,60439,25

Zb =

Zab .Zbc

Zab+Zbc+Zca

=

(6,4∠48 ° )(7,2∠36 °)6,4∠48 °+7,2∠36 °+4,5∠30 °

= 2,56645,25

Zc =

Zbc . Zca

Zab+Zbc+Zca

=

(7,2∠36 ° )(4,5∠30 °)6,4∠48 °+7,2∠36 °+4,5∠30 °

= 1,80427,25

Figure 5.63: Equivalent delta-connected circuit for the circuit in Figure 5.62

Consider E and open-circuit the impedance Zload:

E

ZloadZc

Zbd

Za

Zb

Zc

Zbd

Za

Zb

97

Figure 5.64: Circuit diagram for Assignment 5.15 with E short-circuited and Zload open-circuited

ZTH =

Za .(Zb+Zbd )Za+Zb+Zbd + Zc

=

(1 ,604∠39 ,25 ° )(2,566∠45 ,25 °+5,5∠45 ° )1 ,604∠39 ,25°+2 ,566∠45 ,25 °+5,5∠45 ° +

1,80427,25˚

= 3,12332,77

Zload = 3,12332,77

5.15.2 Consider E and open-circuit the impedance Zload:

Figure 5.65: Circuit diagram for Assignment 5.15 with Zload open-circuited

VTH

E

Zc

Zbd

Za

Zb

98

VTH = E.( Zb+Zbd

Za+Zb+Zbd)

= (38016˚)( 2,566∠45 ,25 °+5,5∠45 °

1 ,604∠39 ,25 °+2 ,566∠45 ,25 °+5,5∠45 ° )= 317,19516,97 V

ZTH = (2,626 + j1,69)

RTH = 2,626

P0(max) =

V TH2

4 RTH

=

(317 ,195)2

(4 )(2 ,626 )

= 9578,51 W

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 Calculate the values of the phase and line currents.6.1.2 Calculate the sum of the line voltages.6.1.3 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!

Ebn Ean

EcnEca

Ebc

Eab

30˚

30˚

30˚

99

Figure 6.15: Positive (abc) phase sequence with Eab as reference

6.1.1 Current in each phase (line):

Ia = Ian

=

Ean

Zan

=

380∠−30 °(√3 )(12,8+ j8,4 )

= 14,3363,27° A

Ib = Ibn

=

Ebn

Zbn

=

380∠−150°(√3 )(12,8+ j8,4 )

= 14,33176,73° A

Ic = Icn

=

Ecn

Zcn

=

380∠90 °(√3 )(12,8+ j8,4 )

= 14,3356,73° A6.1.2 Eab + Ebc + Eca = 3800 + 380120 + 380120

= 0 V

100

6.1.3 In = Ia + Ib + Ic

= 14,3363,27 + 14,33176,73 + 14,3356,73

= 0 A

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 Calculate the power absorbed per phase.6.2.2 Calculate the value of the phase currents.6.2.3 Calculate the value of the line currents.6.2.4 Calculate the effective impedance of the load per phase.

Solution:

Figure 6.20: Line diagram of the star-delta system with the supply and load voltages and line and phase currents

Ica

Iab

Ibc

Three-phase loadThree-phase supply

Ic

Ib

Ia

VbcEbc

Eca VcaVabEab

c b

a

101

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

= Pab

= Pbc

= Pca

6.2.2 Current in each phase:

Pbc = Vbc.Ibc.cos ∠I bc

V bc

9,45 x 103 = (380)(Ibc)(0,875)

Ibc = 28,42128,96 A

Iab = 28,421( 28,96 + 120)

= 28,42191,04 A

Ica = 28,421(28, 96 120)

= 28,421148,96 A

6.2.3 Current in each line:

Ia = Iab Ica

= 28,42191,04° 28,421148,96°

= 49,22761,04° A Ib = Ibc Iab

= 28,42128,96° 28,42191,04°

Vab

Vca

Vbc

102

= 49,22758,96° A

Ic = Ica Ibc

= 28,421148,96° 28,42128,96°

= 49,227178,96° A

For a balanced load, once the phase currents are found, the line currents can be determined:

IL = √3 Iph( ± 30˚)

Where: + negative phase sequence – positive phase sequence

Ib = (√3 )(28,421(– 28,96˚ – 30˚)

= 49,227–58,96˚ A

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

380120= (28,42191,04)(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 Calculate the value of the phase currents.6.3.2 Calculate the power absorbed per phase.6.3.3 Calculate the total power absorbed by the load.6.3.4 Calculate the phasor sum of the three line currents.6.3.5 Convert the load to an equivalent star-connected load.

103

6.3.6 Calculate the total power absorbed by the star-connected load.

Solution:

Figure 6.22: Negative (cba) phase sequence with Vac as reference6.3.1 Current in each phase of the load:

Vab = Iab.Zab

36060 = (Iab)(8,45 + j6,36)

Iab = 34,03996,97° A

Vbc = Ibc.Zbc

36060= (Ibc)(8,45 + j6,36)

Ibc = 34,03923,03° A

Vca = Ica.Zca

360180= (Ica)(8,45 + j6,36)

Ica = 34,039143,03° A

6.3.2 Sab¿

= Vab.I ab¿

= (36060°)(34,03996,97°)

= 12254,0436,97°

= (9790,372 + j7369,54) VA

Pab = 9790,372 W

= Pbc

= Pca

Or: Pab = I ab2

.Rab

30˚

30˚

30˚VanVcn

Vbn

Vac

Vab

Vca

Vbc

104

= (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°

= 58,95766,97° A

Ib = Ibc Iab

= 34,03923,03° 34,03996,97°

= 58,95753,03° A

Ic = Ica Ibc

= 34,039143,03° 34,03923,03°

= 58,957173,03° A

Ia + Ib + Ic = 58,95766,97 + 58,95753,03 + 58,957173,03

= 0 A

As shown, the sum of the three line currents drawn by a balanced load is zero. This means that these currents are equal in magnitude and a phase difference of 120° exists between them.

6.3.5 Z = 3.ZY

(8,45 + j6,36) = (3)(ZY)

ZY = 3,52536,97˚

6.3.6 Van = Ian.Zan

360–30˚ = (√3 )(Ian)(3,52536,97˚)

Ian = 58,963–66,97˚ A

P = 3.Vph.Iph.cos ph

105

= (3)(360

√3 )(58,963)(cos 36,97˚)

= 29,374 kW

106

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