Solution manual fundamentals of electric circuits 3rd edition chapter12

If this is connected to a balanced delta-connected load, find the line and phase currents.. If the line impedance per phase is 0.4 + j1.2 Ω , find the total complex power delivered to th

3j2

11

What is the phase sequence of a balanced three-phase circuit for which Van = 160∠30° V

A three-phase system with abc sequence and V L = 200 V feeds a Y-connected load with

Z L = 40∠30°Ω Find the line currents

Y

V I

0220

Y

an a

Obtain the line currents in the three-phase circuit of Fig 12.42 on the next page

Figure 12.42

For Prob 12.7

Chapter 12, Solution 7

This is a balanced Y-Y system

Using the per-phase circuit shown above,

0440

Chapter 12, Problem 8

In a balanced three-phase Y-Y system, the source is an abc sequence of voltages and V an

= 100 ∠20° V rms The line impedance per phase is 0.6 + j1.2 Ω , while the per-phase impedance of the load is 10 + j14 Ω Calculate the line currents and the load voltages

a

L

V I

A balanced Y-Y four-wire system has phase voltages

120∠ °

=

cn

V

The load impedance per phase is 19 + j13 Ω , and the line impedance per phase is

1 + j2 Ω Solve for the line currents and neutral current

Chapter 12, Solution 9

=+

°

∠

=+

=

15j20

0120

Y L

an a

Z Z

22010

j27

02202

A

an a

Z

V I

22

120-2202

B

bn b

Z

V I

°

∠

=+

62.2213

1202205

j12

1202202

C

cn c

Z

V I

The current in the neutral line is

In the Y- ∆ system shown in Fig 12.44, the source is a positive sequence with

V an= 120 ∠0° V and phase impedance Z p= 2 – j3 Ω Calculate the line voltage V L and

the line current I L

AB

A

V V

Z Z

0110

In the balanced three-phase Y-∆ system in Fig 12.46, find the line current I L

and the average power delivered to the load

Figure 12.46

For Prob 12.13

– +

– +

– +

ZY A

N

Obtain the line currents in the three-phase circuit of Fig 12.47 on the next page

100 –120 °

Figure 12.47

For Prob 12.14

+ 2

1 j

For mesh,

0)1212()21()1614(120

1212()21()

120

100∠ o − ∠− o −I1 + j − + j I3 + + j I2 =

or

2.1736

.86506.8650)

1212()1614()

749.16098.10,

3.19161

2

o

The circuit in Fig 12.48 is excited by a balanced three-phase source with a line

voltage of 210 V If Z l = 1 + j1 Ω , Z∆ =24− j30Ω, and ZY = 12 + j5 Ω , determine the

magnitude of the line current of the combined loads

=

5j20

)10j8)(

5j12(

Z

047.1j812.8

L p

p a

V

Z Z I

Chapter 12, Problem 16

A balanced delta-connected load has a phase current I AC= 10 ∠ 30− ° A

(a) Determine the three line currents assuming that the circuit operates in the positive phase sequence

(b) Calculate the load impedance if the line voltage is V AB= 110 ∠0° V

If V an= 440 ∠60° V in the network of Fig 12.49, find the load phase currents I AB, I BC,

AB AB

ab AB

Refer to the ∆ - ∆ circuit in Fig 12.51 Find the line and phase currents Assume that the

load impedance is Z L = 12 + j9 Ω per phase

0210

Chapter 12, Problem 21

Three 230-V generators form a delta-connected source that is connected to a balanced

delta-connected load of ZL = 10 + j8 Ω per phase as shown in Fig 12.52

(a) Determine the value of IAC

(b) What is the value of Ib?

1202308

j10

120230

°

∠

−

= 17.96∠–98.66˚ A rms

(b)

A34.17110.31

684.4j75.30220.11j024.14536.6j729.16

66.3896.1766.15896

.17

8j10

02308

j10

120230

IIII

=

31.1∠171.34˚ A rms

Find the line currents Ia, Ib, and Ic in the three-phase network of Fig 12.53 below Take Z∆ =12− j15Ω, ZY = 4 + j6 Ω , and Z l = 2 Ω

3

)5j4)(

6j4()5j4(

||

)6j4(3

||

−+

=

−+

=

Z Z

2153.0j723

Chapter 12, Problem 23

A three-phase balanced system with a line voltage of 202 V rms feeds a delta-connected

load with Zp = 25 ∠60°Ω

(a) Find the line current

(b) Determine the total power supplied to the load using two wattmeters connected to the

6025

202Z

VI

o o

AB

6025

30320230

3I

3202)202(3cos

IV3P

=+

=

A balanced delta-connected source has phase voltage Vab = 416 ∠30° V and a positive phase sequence If this is connected to a balanced delta-connected load, find the line and phase currents Take the load impedance per phase as 60 ∠30°Ω and line impedance per phase as 1 + j1 Ω

Chapter 12, Solution 24

Convert both the source and the load to their wye equivalents

10j32.1730203

V V

We now use per-phase analysis

=

°

∠

=++

+

=

3137.21

2.240)

10j32.17()j1(

an a

31-24.11

Chapter 12, Problem 25

In the circuit of Fig 12.54, if Vab = 440 ∠10°, Vbc = 440 ∠250°, Vca = 440 ∠130°

V, find the line currents

3

)3010(440

20-440

For the balanced circuit in Fig 12.55, Vab = 125 ∠0° V Find the line currents IaA, IbB, and

30-17.72

Chapter 12, Problem 27

A ∆-connected source supplies power to a Y-connected load in a three-phase

balanced system Given that the line impedance is 2 + j1 Ω per phase while the load impedance is 6 + j4 Ω per phase, find the magnitude of the line voltage at the load

Assume the source phase voltage Vab = 208 ∠0° V rms

o a

Y

V I

The line-to-line voltages in a Y-load have a magnitude of 440 V and are in the positive

sequence at 60 Hz If the loads are balanced with Z 1 = Z 2 = Z 3 = 25∠30°, find all line currents and phase voltages

Chapter 12, Solution 28

P ab

A balanced three-phase Y-∆ system has Van = 120 ∠0° V rms and Z∆ =51 j+ 45Ω

If the line impedance per phase is 0.4 + j1.2 Ω , find the total complex power delivered to the load

Chapter 12, Solution 29

We can replace the delta load with a wye load, ZY = Z∆/3 = 17+j15Ω

The per-phase equivalent circuit is shown below

5.0475∠–42.96˚

Chapter 12, Problem 30

In Fig 12.56, the rms value of the line voltage is 208 V Find the average power

delivered to the load

3

p p

p p

V V Z

V S

kVA454421.14530

)208

*

2

o o

P

A balanced delta-connected load is supplied by a 60-Hz three-phase source with a line voltage of 240 V Each load phase draws 6 kW at a lagging power factor of 0.8 Find: (a) the load impedance per phase

(b) the line current

(c) the value of capacitance needed to be connected in parallel with each load phase to minimize the current from the source

Chapter 12, Solution 31

(a)

kVA5.78.0/6cos,

8.0cos,

000,

P S

P

kVAR5.4sin =

Q

kVA5.1318)5.46(3

For delta-connected load, Vp = VL= 240 (rms) But

Ω+

=+

)240(33

3

3

2 2

*

*

2

j Z

x j S

V Z

6000cos

=

x x I

I V

(c ) We find C to bring the power factor to unity

F2.207240

602

4500kVA

Q C Q

Q

rms

c p

c

Y

V I

A three-phase source delivers 4800 VA to a wye-connected load with a phase voltage of

208 V and a power factor of 0.9 lagging Calculate the source line current and the source line voltage

S= S =For a Y-connected load,

3(

4800V

3

SI

I

p p

Chapter 12, Solution 34

3

2203

02.127)

16j10(3

220V

Y

p a

∠

= 3VLIL 3 220 6.732 -58 2565 58

S

Three equal impedances, 60 + j30 Ω each, are delta-connected to a 230-V rms,

three-phase circuit Another three equal impedances, 40 + j10 Ω each, are

wye-connected across the same circuit at the same points Determine:

(a) the line current

(b) the total complex power supplied to the two loads

(c) the power factor of the two loads combined

Chapter 12, Solution 35

(a) This is a balanced three-phase system and we can use per phase equivalent circuit The delta-connected load is converted to its wye-connected equivalent

10203/)3060(3

)1040(//

Z

A953.561.145.5

A 4200-V, three-phase transmission line has an impedance of 4 + j10 Ω per phase If it supplies a load of 1 MVA at 0.75 power factor (lagging), find:

(a) the complex power

(b) the power loss in the line

(c) the voltage at the sending end

10)6614.075.0(33

V

S I

I

V

S

p p p

=

kW19.25)4()36.79(

The total power measured in a three-phase system feeding a balanced wye-connected load is 12 kW at a power factor of 0.6 leading If the line voltage is 208 V, calculate the

line current I L and the load impedance ZY

Chapter 12, Solution 37

206.0

12pf

P

kVA16j1220

1020I

3

p

2 p

L

p (3)(55.51)

10)16j12(I

0110

°

∠

=+++

)110(2

12

1

2 2

2 Y

2 a

)110(2

33

2

= S S

=

S 551 86+j 735 81 VA

Find the real power absorbed by the load in Fig 12.58

1 5 (8 j6))

6j18(

For mesh 2,

3 1

20120-

3 2

120-

For mesh 3,

3 2

1 10 (22 j3))

6j8(

To eliminate I2, start by multiplying (1) by 2,

3 2

1 10 (16 j12))

12j36(

Multiplying (2) by 5 4,

3 2

1 5 2.525

.1-120-

Adding (1) and (6),

3

1 (10.5 j6))

6j75.16(65.21j5

15j38-18j4465

.12j5.87

200

I I

25.26j5

192 −

=

∆ , ∆1 =900.25− j935.2, ∆3 =110.3−j1327.6

144.4j242.538.33-682.67.76-28.194

46.09-1.1298

85.25-2.1332

14

1120-

)347.3j7425.0()0359.1j3104.1()33.4j-2.5(

I

713.8j4471.0-

I

The average power absorbed by the 8-Ω resistor is

W89.164)8(551.2j756.3)8(

The average power absorbed by the 4-Ω resistor is

W1.188)4()8571.6()4(

)10(

∠

)8j7()5.0j1(

0100

a

I

For a wye-connected load,

567.8I

Ip = a = Ia =

)8j7()567.8)(

3(

A balanced delta-connected load draws 5 kW at a power factor of 0.8 lagging If the three-phase system has an effective line voltage of 400 V, find the line current

Chapter 12, Solution 41

kVA25.68.0

kW5pf

1025.6V3

SI

3

L

Chapter 12, Problem 42

A balanced three-phase generator delivers 7.2 kW to a wye-connected load with

impedance 30 – j40 Ω per phase Find the line current I L and the line voltage V L

Chapter 12, Solution 42

The load determines the power factor

°

=θ

(leading)6

.0cos

pf = θ=

kVA6.9j2.7)8.0(6.0

2.7j2

40j30)(

3(

10)6.9j2.7(3

A944

S= , Ip = for Y-connected loads IL

)047.2j812.7()66.13)(

A three-phase line has an impedance of 1 + j3Ω per phase The line feeds a balanced

delta-connected load, which absorbs a total complex power of 12 + j5 k VA If the line

voltage at the load end has a magnitude of 240 V, calculate the magnitude of the line voltage at the source end and the source power factor

S=

273.31)

240(3

10)512(V3

SI

3 2 2

L

At the source,

L L L

'

)3j1)(

273.31(0240

'

V

819.93j273.271

V 287 04 V

Also, at the source,

* L

' L ' 3V I

S =

)273.31)(

819.93j273.271(3

S

078.19273.271

819.93

=θ

= cos

A balanced wye-connected load is connected to the generator by a balanced transmission

line with an impedance of 0.5 + j2Ω per phase If the load is rated at 450 kW, 0.708 power factor lagging, 440-V line voltage, find the line voltage at the generator

V3

708.0

10450pf

3

)6.635(

V

)76062.2)(

45-834(440

V

°

∠+

L

V

7.885j1.1914

Chapter 12, Problem 46

A three-phase load consists of three 100-Ω resistors that can be wye- or delta-connected Determine which connection will absorb the most average power from a three-phase source with a line voltage of 110 V Assume zero line impedance

*

2 p

* p p

33

33

Z

V Z

V I

V

W121100

)110

For the delta-connected load,

*

2 p

* p p

33

3

Z

V Z

V I

V

W363100

)110)(

This shows that the delta-connected load will deliver three times more average

power than the wye-connected load This is also evident from

The following three parallel-connected phase loads are fed by a balanced phase source:

three-Load 1: 250 kVA, 0.8 pf lagging

Load 2: 300 kVA, 0.95 pf leading

Load 3: 450 kVA, unity pf

If the line voltage is 13.8 kV, calculate the line current and the power factor of the source Assume that the line impedance is zero

Chapter 12, Solution 47

°

=

=θ

.36250

.8-300

3 =

S

kVA45.37.93635.56j935

3 2 1

S

L L

107.936

=cos cos(3.45 )

Chapter 12, Problem 48

A balanced, positive-sequence wye-connected source has Van = 240 ∠0° V rms and

supplies an unbalanced delta-connected load via a transmission line with impedance

2 + j3Ω per phase

(a) Calculate the line currents if ZAB = 40 + j15 Ω , Z BC = 60 Ω , ZCA = 18 – j12 Ω

(b) Find the complex power supplied by the source

60Ω

923.1577.73

118

)1218)(

1540(

j j

j j

+

−+

=

105.7j52.203

j118

)

15j40(60

)1218(60

j j

(120240

or

85.207360

)105.1052.22()13.11

()(

120240

120

240∠ o − ∠− o −I1 Z B +Z l +I2 Z l +Z B +Z C =

or

69.415)

775.651.33()105.10

328.575

.10,

A64.1234

Chapter 12, Problem 49

Each phase load consists of a 20-Ω resistor and a 10- Ω inductive reactance With a line voltage of 220 V rms, calculate the average power taken by the load if:

(a) the three-phase loads are delta-connected

(b) the loads are wye-connected

Chapter 12, Solution 49

(a) For the delta-connected load, Z p =20+ j10Ω, V p =V L =220 (rms),

kVA56.26943.629045808

)1020(

2203

*

2

o p

p

j j

x Z

2203

*

2

o p

p

j

x Z

Chapter 12, Solution 50

kVA3kVA,

4.68.4)8.06.0(

p p

p

Z

V S

V V Z

=

Ω+

=

⎯→

⎯+

=

10)4.68.1(

240

3 2

2

*

x j S

V

Consider the ∆ - ∆ system shown in Fig 12.60 Take Zi = 8 + j6Ω , Z2 = 4.2 –

j2.2Ω , Z3 = 10 + j0Ω

(a) Find the phase current IAB, IBC, ICA

(b) Calculate line currents IaA, IbB, and IcC

120240Z

120240I

Z

120240

AN

an a

Z

V I

0120

BN

bn b

Z

V I

120-120

CN

cn c

Z

V I

Thus,

c b a n

-I =I +I +I

°

∠+

°

∠+

°

∠

=+

In the Y-Y system shown in Fig 12.61, loads connected to the source are unbalanced

(a) Calculate Ia, Ib, and Ic (b) Find the total power delivered to the load Take

– + + –

A balanced three-phase Y-source with VP = 210 V rms drives a Y-connected three-phase

load with phase impedance ZA = 80Ω , ZB = 60 + j90Ω , and ZC = j80Ω Calculate the line currents and total complex power delivered to the load Assume that the neutrals are connected

o a

°

∠

=

31.5617.108

21090

j60

0210

Chapter 12, Problem 55

A three-phase supply, with the line voltage 240 V rms positively phased, has an

unbalanced delta-connected load as shown in Fig 12.62 Find the phase currents and the total complex power

Refer to the unbalanced circuit of Fig 12.63 Calculate:

(a) the line currents

(b) the real power absorbed by the load

(c) the total complex power supplied by the source

10j0440120

)866.0j5.1)(

440(

For mesh 3,

05j)(

20)(

10

j I3−I1 + I3 −I2 − I3 =Substituting (1) and (2) into the equation for mesh 3 gives,

)(

440(

=

°

∠+

= 3 76.21 -60 114.315 j66 132 30

1 I I

From (2),

°

∠

=+

=

−

= 3 j38.1 76.21 j93.9 120.93 50.94

2 I I

2 3 2

S

kVA-j116.16-j5)

((152.42)-j5)

2 3

S

kVA08.58j04.29

CA BC

S

Real power absorbed = 29 04 kW

(c) Total complex supplied by the source is

=

S 29 04−j 58 08 kVA

Determine the line currents for the three-phase circuit of Fig 12.64 Let Va = 110∠0°,

3020()

80

100

53.190)

1080()

1.592 E–01 1.078 E+01 –8.997 E+01

i.e In = 10.78∠–89.97° A

The source in Fig 12.65 is balanced and exhibits a positive phase sequence If f = 60 Hz,

use PSpice to find V AN, VBN, and VCN

Figure 12.65

For Prob 12.59

Chapter 12, Solution 59

The schematic is shown below In the AC Sweep box, we set Total Pts = 1, Start Freq

= 60, and End Freq = 60 After simulation, we obtain an output file which includes

Use PSpice to determine I o in the single-phase, three-wire circuit of Fig 12.66 Let

Z1 = 15 – j10Ω , Z2 = 30 + j20Ω , and Z3 = 12 + j5Ω

Figure 12.66

For Prob 12.60

Chapter 12, Solution 60

The schematic is shown below IPRINT is inserted to give Io We select Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592 in the AC Sweep box Upon simulation, the output file includes

1.592 E–01 1.421 E+00 –1.355 E+02 from which, Io = 1.421∠–135.5° A

Given the circuit in Fig 12.67, use PSpice to determine currents I aA and voltage VBN

Figure 12.67

For Prob 12.61

Chapter 12, Solution 61

The schematic is shown below Pseudocomponents IPRINT and PRINT are inserted to measure IaA and VBN In the AC Sweep box, we set Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592 Once the circuit is simulated, we get an output file which

includes

from which

IaA = 11.15∠37° A, VBN = 230.8∠–133.4° V