lecture 16 phasor circuits, ac power, thevenin - courses · lecture 16 phasor circuits, ac power,...

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.

Lecture 16Phasor Circuits, AC Power,

Thevenin


Kirchhoff’s Laws in Phasor Form

0321 =−+ VVV

We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.

The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.

outin II =


Circuit Analysis Using Phasors and Impedances

1. Replace the time descriptions of the voltage and current sources with the corresponding phasors. (All of the sources must have the same frequency.)


2. Replace inductances by their compleximpedances ZL = jωL. Replace capacitances by their complex impedances ZC = 1/(jωC). Resistances have impedances equal to their resistances.

3. Analyze the circuit using any of the techniques studied earlier in Chapter 2, performing the calculations with complex arithmetic.


Example 5.3

)15500cos(707.0)(15707.0454.141

30100

454.141

10010050150100

50)10)(40)(500(

11150)3.0)(500(

30100)30500cos(100

6

ooo

o

o

oo

−=→−∠=∠∠

==

∠=

+=−+=++=

Ω−=−=−=

Ω===∠=→+=

−

ttiZ

jjjZZRZ

jjC

jZ

jjLjZtV

S

CLeq

C

L

SS

VI

V

ω

ω

Find i(t):


Example 5.3

oooo

oooo

o

o

o

1054.35)15707.0)(9050(9011751.106)15707.0)(90()90()(

)15707.0)(100(

15707.0

30100

−∠=−∠−∠=⎟⎠⎞

⎜⎝⎛ −∠=⎟

⎠⎞

⎜⎝⎛−=

∠=−∠∠=∠==

−∠==

−∠=

∠=

IIV

IIV

IV

I

V

CCj

LLjwL

R

C

L

R

S

S

ωω

ωω

Find the phasorvoltage across each element:


Example 5.4

50504571.704501414.0

01

01.001.01

1001

1001

111

1

100)1010)(1000(

11100)1.0)(1000(

9010)901000cos(10)1000sin(10)(

6

j

jjZR

Z

jx

jC

jZ

jjjwLZtttv

C

RC

C

L

ss

−=−∠=∠

∠=

+=

−+

=+

=

Ω−=−=−=

Ω===−∠=→−==

−ω

V

Find the voltage vc(t) in steady state:


Example 5.4

)1000cos(10)1801000cos(10)(

18010)9010)(901()9010(4571.704571.70

)9010(5050

4571.70)9010(100)5050(

4571.70

tttv

jjjZZZ

V

C

SLRC

RCC

−=−=

−∠=−∠−∠=−∠⎟⎟⎠

⎞⎜⎜⎝

⎛∠−∠

=

−∠⎟⎟⎠

⎞⎜⎜⎝

⎛+−∠

=−∠⎟⎟⎠

⎞⎜⎜⎝

⎛+−−∠

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

o

oooo

o

o

oo

oo

V


Example 5.4

ooo

oo

o

ooo

901.090100

1801010018010

1801.0100

18010

1351414.04571.709010

50509010

)5050(1009010

−∠=−∠

−∠=

−−∠

==

−∠=−∠

==

−∠=∠−∠

=+−∠

=−+

−∠=

+=

jZ

R

jjjZZ

C

CC

CR

RCL

S

VI

VI

VI

Find the phasor current through each element:


Example 5.5

5.11.02.005.1510

:2

22.0)2.01.0(902510

:1

21122

21211

=+−→∠=−−

+

−=−+→−∠=−−

+

VVVVV

VVVVV

jjjj

nodeatKCL

jjjj

nodeatKCL

o

o

Use the node voltage technique to find v1(t):


Example 5.5

)7.29100cos(1.16)(

74.2912.168142.01.0

2323)2.01.0(

:32.04.0

22.0)2.01.0(

5.11.02.022.0)2.01.0(

1

11

21

21

21

21

o

o

+=

∠=+=−−

=→−=−

=+−−=−+

=+−−=−+

ttv

jj

jjj

Addingjj

jjj

jjjjj

VV

VVVV

VVVV


Exercise 5.9

)135500cos(104.28)(

135104.28456.353

9010250250

:250)5.0)(500(

2509010)90500cos(10)500sin(10)(

3

3

o

oo

o

o

−=

−∠=∠−∠

=+

=+

=→=+

===Ω=

−∠=→−==

−

−

txti

xjZZ

ZZ

KVLjjLjZ

Ztttv

S

LR

SSLR

L

R

SS

VVIVII

V

ω

Find i(t):


Exercise 5.9

( )( )

9010

4507.7

135)250)(103.28(90)5.0)(500(

13507.7135)250)(103.28(

3

3

−∠=

−∠=

−∠∠==

−∠=−∠==

−

−

S

L

R

xLj

xR

V

IV

IV

o

o

oo

ω

Construct the a phasor diagram showing all three voltages and the current:


Find the phasor voltage and phasor current through each element:

o31.5647.5515.4677.3032

200

20032

1001

200501001

2001

5011

200)1)(200(

50)10100)(200(

11

1111

6

−∠=−=+

=

+=+−=++

−=

===

−=−=−=

++=

−

jj

Z

jjjjjZ

jjLjZ

jx

jC

jZ

ZZZZ

eff

eff

L

C

RLCeff

ωω

Exercise 5.10


Exercise 5.10

o

o

o

o

o

o

o

o

o

ooo

31.5677.20100

31.563.277

3.14639.190200

31.563.277

69.3355.59050

31.563.277

31.563.277)31.5647.55)(05(

−∠=∠−∠

==

−∠=∠−∠

==

∠=−∠−∠

==

−∠=−∠∠====

R

RR

L

LL

C

CC

effRLC

Z

Z

Z

Z

VI

VI

VI

IVVV

Find the phasor voltage and current through each element:


0)100100(100100)100100(

0)(

)(

21

21

1222

211

2

1

=−+−=−+

=−++

=−+

IjIVIIj

ZIIZIZI

VZIIZI

S

RLC

SRL

Exercise 5.11

Solve for the mesh currents:

100j

100j

-200j


Exercise 5.11

)451000cos(41.1)(

452009.1009.19881

1097.91097.90)100(0)141(

4514100

)0100)(0100()45141)(45141()0100)(0()45141)(0100(

451410100010045141

45141001000100

0)45141()0100(0100)0100()45141(

0)100100(100100)100100(

1

33

22

1

21

21

21

21

o−=

−∠=−=−

=∠−∠

−∠=

∠−∠−−−∠∠∠−−−∠∠

=

−∠∠−∠−∠−∠∠−∠

=

=−∠+∠−∠=∠−∠

=−+−=−+

tti

jxjx

I

IIII

IjIVIIj S


Power in AC Circuits

ZV

IwhereIZ

VZ

mmm

m =−∠=∠∠

== θθ

o0VI

For θ>0 the load is called “inductive” since Z=jωL for an inductor

For θ<0 the load is “capacitive” since Z=-j/ωC for a capacitor



Current, Voltage and Power for a Resistive Load

)(cos)(

)cos()()cos()(

0

2 tIVtp

tItitVtv

RZ

mm

m

m

ω

ωω

=

==

∠=



Current, Voltage and Power for an Inductive Load

)2sin(2

)sin()cos()(

)sin()90cos()(

9090900

0)cos()(

90

tIV

ttIVtp

tItIti

IL

VLV

VtVtv

LZ

mmmm

mm

mmm

mm

ωωω

ωωωω

ω

ω

==

=−=

−∠=−∠=∠

∠=∠=→=

∠=

o

ooo

o

IV

“Reactive” power

using cos(x)sin(x) = (1/2)sin(2x)


Power in AC Circuits• For a pure inductance half of the

time the power is positive and power flows to the inductor where it is stored as energy in the magnetic field

• Half of the time the power is negative and power flows from the inductor to the source

• Even though the average power is zero for a pure inductor, current flows between the source and the inductor and power is dissipated in the lines connecting the source to the inductor

Reactive Power



Current, Voltage and Power for a Capacitive Load

)2sin(2

)sin()cos()(

)sin()90cos()(

90901

00)cos()(

901

tIV

ttIVtp

tItIti

I

C

VVtVtv

CZ

mmmm

mm

mm

mm

ωωω

ωωω

ω

ω

−=−=

−=+=

∠=−∠

∠=→∠==

−∠=

o

o

o

oo

o

IV

Reactive power for a capacitor opposite the sign for an inductor



Inductor

CapacitorIf a load contains both inductance and capacitance with reactive powers of equal magnitude, the reactive powers cancel.


Power for a General Load

[ ]

)cos()cos(

)cos(22

)cos(2

)2sin()sin(2

)2cos(1)cos(2

)cos()cos()()cos()(

)cos()(

iirmsrms

imm

imm

av

imm

imm

imm

im

m

PFIV

IVIVP

tIV

tIV

ttIVtptItitVtv

θθ

θθ

ωθωθ

θωωθω

ω

==

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛==

++=

−=−=

=

For a general RLC circuit in which the phase can be any value between -90° to +90°

PF is the “Power Factor”


Power for a General Load

)cos()cos(

θθ

θθθ

==

−=

PFIVP rmsrms

currentvoltage

Power angle:

If the phase angle for the voltage is not zero, we define the power angle θ:


AC Power Calculations

( ) [ ]WIVP θcosrmsrms=

( ) [ ]VARIVQ θsinrmsrms=

Average Power:

Reactive Power:

Apparent Power: [ ]VAIV RMSRMS


( )2rmsrms22 IVQP =+

Power Triangles

Average power

Reactive power

Average power

Apparent power


Load Impedance in the Complex Plane

ZXZR

jXRZZ

=

=

+=∠=

)sin(

)cos(

θ

θ

θ


Additional Power Relations

RIP 2rms=

XIQ 2rms=

RVP R

2rms=

XVQ X

2rms=

Average power

Reactive power



Compute the power and reactive power from the source in this circuit:

AI

VV

rms

rms

s

ss

iv

1.02

1414.02

071.72

102

45)135(90

===

===

=−−−=−=

I

V

oooθθθ



VARx

IVQWx

IVP

rmss

rmss

rms

rms

5.0)45sin(1.0071.7

)sin(5.0)45cos(1.0071.7

)cos(

==

===

=

o

o

θ

θ



VARVARVARQQQ

VARXIQ

VARXIQ

CL

CCC

LrmsL

rms

5.05.00.1

5.0)100(21.0

0.1)100()1.0(2

2

22

=−=+=

−=−⎟⎟⎠

⎞⎜⎜⎝

⎛==

===

The reactive power delivered to the inductor and capacitor:



The power delivered to the resistor:

PWRRIP RRR rms

==⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛== 5.0100

21.0

222

2

I


Using Power Triangles

Find the power, reactive power and power factor for the source. Also find the phasor current I:

Load A: 10kVA apparent power, PF=0.5 leading (capacitive)

Load B: 5kW power, PF=0.7 lagging (inductive)



Load A: 10kVA apparent power, PF=0.5 leading (capacitive)

• Reactive power QA and the power angle θA are negative

Load B: 5kW power, PF=0.7 lagging (inductive)

• Reactive power QB and the power angle θB are positive



( ) ( ) ( )

kVARPQ

kVAR

PIVQ

kWIVP

BBB

B

AArmsA

AArmsA

rms

rms

101.5)57.45tan(5000)tan(

57.45)7.0arccos(

660.8

500010

5)5.0(10)cos(

22422

4

===

==

−=

−=−=

===

o

o

θ

θ

θ



( ) ( ) kVAxQPIV

PQ

kVARkVARkVARQQQkWkWkWPPP

rmsrms

BA

BA

61.1010559.310

59.1910

559.3arctanarctan

559.3101.5660.81055

232322 =−+=+=

−=⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛=

−=+−=+==+=+=

oθ



o

ooo

59.4915

59.49)59.19(30

15)61.10)(414.1(2

61.10161.10

1414.1

14142

∠=∠=

=−−=−=

===

===

===

i

vi

rms

rms

rmsrmsrms

rms

I

AAI

AkV

kVAV

IVI

kVV

θ

θθθ

I

I

V



o

o

59.4915

301414

∠=

∠=

I

V


Exercise 5.12o401.707 ∠=VVoltage source:

Delivers 5 kW to a load with a power factor of 100 percent. Find the reactive power and the phasor current:

o

o

4014.14

14.142

10500105

50041.11.707

50)sin(PowerReactive

01)cos(%100

3

∠=

==

==

==

===

==→=

I

AII

AVWxI

VVV

kWVIIVQ

PF

rms

rms

rms

rmsrms

rnsrms θθθ


Exercise 5.12If the PF=20% lagging, cos(θ) = 0.20 → θ = arcos(0.20) = 78.46°

θ

Q = 24.49kVAR

P = 5kW

o

ooo

o

46.3871.70

46.3846.7840

71.70)50)(41.1(2

50)cos(

49.24)46.78tan()5()tan()tan(

−∠=

−=−=−=

===

==

===→=

I

θθθ

θ

θθ

vi

rms

rmsrms

AAII

AV

PI

kVARkWPQPQ

VrmsIrms


Thevenin Equivalent Circuits

The Thevenin equivalent for an ac circuit consists of a phasor voltage source Vt in series with a complex impedance Zt


The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.

ocVV =t

We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.



The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.

scsc

oc

IV

IV t

tZ ==



Norton Equivalent Circuit

The Norton equivalent for an ac circuit consists of a phasor current source In in parallel with a complex impedance Zt

scII =n


Example 5.9


Example 5.9

50504571.704501414.0

0101.001.0

1)100/(1100/1

1 jjj

Zt −=−∠=∠

∠=

+=

−+= o

o

o

Circuit with the voltage and current source zeroed to find the Thevenin impedance


Example 5.9

Circuit with the output shorted to find the short circuit current.

ooo

o

oo

45414.1190101

901

01100

0100100

−∠=−=∠−∠=

∠=

∠=∠

==

−=

j

V

SC

S

SR

SRSC

I

I

I

III


Example 5.9

o

ooo

45414.1

90100)4571.70)(45414.1(

−∠==

−∠=−∠−∠==

SCn

tSCt Z

II

IV


The Thevenin equivalent of a two-terminal circuit delivering power to a load impedance.

Maximum Average Power Transfer


If the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.

If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.

Maximum Average Power Transfer


Maximum Average Power TransferFind the load for the maximum power transfer if the load can have any complex value and if the load must be a pure resistance:

Ω=+=

−=

+=

−=

71.705050

5050

5050

5050:

22

*

t

t

t

t

Z

jZloadReal

jZ

jZloadComplex


Exercise 5.14

50502

10010011

1100

1100

1001

1001

100111

jjjj

jjZ

jjRLjZ

eff

eff

+=+

=⎟⎟⎠

⎞⎜⎜⎝

⎛++

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=−

=

−=+

−=+=

ω

Find the Thevenin impedance, the Thevenin voltage and the Norton current.

25100255050502550 jjjjZZ efft +=−++=−+=

Zero the sourceZeff


Exercise 5.14

o

o

o

o

o

o

o

4571.704520100

4521000100

4521000100

100100100 2

−∠=∠∠

=∠∠

=∠∠

=+

= SSOC jVVV


Exercise 5.14

o

o

oo

04.59686.004.141.1034571.70

251004571.70

−∠=∠

−∠=

+−∠

===jZT

TSCN

VII

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