5.3 COMPLEX IMPEDANCES (複數阻抗 )

Inductance假設通過電感 L的 sinusoidal current 為

)sin()( tIti mL

則電感 L兩端的 voltage 為

dt

tdiLtv L

L

)()( )cos( tLIm ( 亦為 sinusoid)

InductanceVoltage & current 的 phasors 為

90 mL II

mmL VLIV

Current lags the voltage by 90o

)2

( tT

Im

LIm

Ohm’s Law

IRV

What is the relationship between phaor vltage and phasor current? Does it be similar to Ohm’s law?

90)90( mIL

90 mL II

mL LIV

LIL )90( LILj )90sin90(cos90 jLL ( )

90 LLjZ L 稱為 L的阻抗 (impedance)

Phasor voltage = impedance phasor current (similar to Ohm’s law)

LLL IZV

Note the impedance of an inductance is an imaginary number 虛數 (called reactance). Resistance is a real number.

Capacitance假設通過電容 C的 sinusoidal voltage 為

)cos()( tVtv mc

則電容 C兩端的 current 為

dt

tdvCti c

c

)()( )sin( tCVm

1809090 mmc CVCVI

mC VV

9090 mm ICV)180cos(1

Capacitance

CjCj

CCV

V

I

VZ

m

m

c

cc

11

901

90

( )

mC VV90 mc CVI

Current leads the voltage by 90o

Capacitance

Resistance

RR RIV

The current and voltage are in phase. ( 同相位 )

電容阻抗

電感阻抗

電阻

CjZc

1

LjZL

RZR

Current lags voltage by 90o

Current leads voltage by 90o

Current and voltage are in phase

Kirchhoff’s Laws in Phasor FormWe can apply KVL directly to

phasors. The sum of the phasor voltages equals zero for any closed path.KCL-The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.

5.4 Circuit Analysis with Phasors and Complex

Impedances

Circuit Analysis Using Phasors and Impedances1. 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 complex impedances ZL = jωL. Replace capacitances by their complex impedances ZC = 1/(jωC) or –j(1/ωC) . 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 5.4 Steady-state AC Analysis of a Series Circuit

I?VR?VL?VC?

1. Phasors

30100sV

2. Complex impedances

1503.0500 jjLjZ L

501040500

116

jjC

jZC CLeq ZZRZ 10010050150100 jjj

454.141 100

100arctan100100 22 ( )

3. Circuit Analysis

15707.0)4530(4.141

100

454.141

30100VI

eq

S

Z

)15500cos(707.0)( tti

3. Circuit Analysis

157.7015707.0100V IRR

751.106

15707.09015090V

ILILjL

1054.35

15707.090509011

V

IC

IC

jC Note

90,90 jj

Example 5.5 Series and Parallel Combinations of Complex Impedances.

VC?IL?IR?IC?

1. Phasors

9010 sV

2. Complex impedances 1001.01000 jjLjZ L

10010101000

116

jjC

jZC

5050)45sin45(cos71.704571.704501414.0

01

01.001.0

1

)100/(1100/1

1

/1/1

1

jj

jjZRZ

CRC

o

jj

j

jj

4571.7050

50arctan5050

505001.001.0

01.001.0

01.001.0

1

01.001.0

1

22

Or

3. Circuit Analysis

o

j

4571.70

5050

18010

4571.70

4571.709010

5050

4571.709010

5050100

4571.709010

j

jjZZ

ZVV

RCL

RCsC

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

o

j

4571.70

5050

1351414.04571.70

9010

5050

9010

5050100

9010

jjjZZ

VI

RCL

s

1801.0100

18010

R

VI CR

901.090100

18010

100

18010

jZ

VI

C

CC

Example 5.6 Steady-State AC Node-Voltage Analysis

v1(t)?

1. Phasors 902)100sin(2 t05.1)100cos(5.1 t

2. Complex impedances 101.0100 jjjwL

5102000100

116

jjwcj

3. Circuit Analysis

KCL 9025

VV

10

V 211

j

Node 1

KCL Node 2 05.1510

122 j

VV

j

V

22.0)2.01.0( 21 jVjVj

5.11.02.0 21 VjVj

2.05

1

5

1j

j

j

jj

∵

23)2.01.0( 1 jVj

7.291.1681405.0

4.07.0

2.01.0

2.01.0

2.01.0

23

2.01.0

23V1

jj

j

j

j

j

j

j

)7.29100cos(1.16)(1 ttv

5.5 Power in AC Circuits

阻抗有實數 ( 電阻 ) 或虛數 ( 電感 , 電容 ),功率有無實虛 ?

ZjXRZ

mm IZ

V

Z

VI Z

VI mm

1. If X=0 實阻抗 ( 純電阻 )

)cos()( tVtv m )cos()( tIti m

)(cos)()()( 2 tIVtitvtp mm

實功率 (real power)

( 參考Ch5.1)

0

,

RZ

0avgp

1. If X=0 實阻抗 ( 純電阻 )

RIR

VP rms

rmsavg

22

222

2222

2

mmmmrmsrmsavg

rmsrmsrmsrms

avg

IVIVIVP

IVRIR

VP

2m

rms

VV

2. If R=0, X>0 虛阻抗 ( 電感性 )

)cos()( tVtv m )sin()90cos()( tItIti mm

)2sin(2

)sin()cos()()()( tIV

ttIVtitvtp mmmm

虛功率 or 無效功率 (Reactive power)

90

,

jXZ

0avgp

3. If R=0, X<0 虛阻抗 ( 電容性 )

)cos()( tVtv m )sin()90cos()( tItIti mm

)2sin(2

)sin()cos()()()( tIV

ttIVtitvtp mmmm

虛功率 or 無效功率 (Reactive power)

90

,

jXZ

0avgp

)cos()( tVtv m

)cos()( tIti m

)2sin()sin(2

)]2cos(1)[cos(2

)sin()cos()sin()(cos)cos(

)cos()cos()(2

tIV

tIV

ttIVtIV

ttIVtp

mmmm

mmmm

mm

Power for a General Load一般狀況 R≠0 且 X≠0, RLC 電路 ( 有實與虛阻抗 ) 9090 , 0 jXRZ

)2sin(2

1)sin()cos( ),2cos(1(

2

1)(cos2 ttttt ( )

))sin()sin()cos()cos()cos(( ttt

)cos(2

)2sin()sin(2

)]2cos(1)[cos(2

10

mm

Tmmmm

IV

dtwtIV

wtIV

T

T

avg dttpT

P0

)(1

積分為 0 積分為 0

2,

2rmsrms

mm II

VV

)cos(rmsrmsavg IVP

)cos(rmsrmsavg IVPP

)cos( 為 power factor, PF ( 功率因子 )

iv 為 power angle ( 功率角 ) 代表電流 lags 電壓的角度

rmsrmsIV 為 apparent power ( 視出功律 )

)sin(rmsrmIVQ 為 reactive power ( 無效功律 or 虛功率 ), 單位為VAR (volt amperes reactive)

為 real power ( 有效功律 or 實功率 ) 單位為 W

單位為 VA(volt-amperes)

AC Power Calculations

cosrmsrmsIVP

cosPF

iv

sinrmsrmsIVQ

(W)

(VAR)

22 QPIV rmsrms (VA)

rmsrms22 IVQP

RIP 2rms

XIQ 2rms

(W)

(VAR)

(VA)

RIRI

IZ

RVIVIP rms

mrms

mrmsrmsrms

2

22cos

Z

Rcos

2m

rms

VV

RIP 2rms

Proof:

Z

VI

Z

VI m

m ,

XIQ rms2

Proof:

XIXI

IZ

XVIVIQ rms

mrms

mrmsrmsrms

2

22sin

0 1

90 0

pf

pf

L

L

Z

Z

若為純電阻性負載，則

若為純電抗性負載，則

RC pf

RL pf

在 負載中，電流領先電壓，負載有一領先的在 負載中，電流落後電壓，負載有一落後的

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