5.4 Circuit Analysis Using Phasors and Complex 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 complex impedances ZL = jωL. Replace capacitances by their complex impedances ZC = 1/(jωC). Resistances have complex impedances equal to their resistances3. Analyze the circuit using any of the techniques studied earlier in Chapter 2, performing the calculations with complex arithmetic.

•Find the steady-state current for the circuit as follows, and also find the phasor voltage across each element and construct a phasor diagram.

Answers:0.707 15 , 70.7 15

106.1 75 , 35.4 105

o oR

o oL C

I V

V V

0.707 15 , 70.7 15

106.1 75 , 35.4 105

o oR

o oL C

I V

V V

•Find the steady-state voltage uC(t) for the circuit as follows, and also find the phasor current through each element and construct a phasor diagram showing the currents and source voltage.

Answers:

0

10 90, 10 180 , 0.1414 135

0.1 180 , 0.1 90

o oS C

oR L

V V I

I I

)(5050100100

)100(100

jj

jZRC

0

10 90, 10 180 , 0.1414 135

0.1 180 , 0.1 90

o oS C

oR L

V V I

I I

•To find u1(t) in steady state.

Answers:

01 16.1 29.7V

•To find i(t) in steady state; construct a phasor diagram showing all three voltages and current; what is the phasor relationship between us(t) and i(t)?

Answers:0 0 0

0

0.0283 135 , 10 90 , 7.07 45

( ) ( ) 45

s L

s

I V V

i t lags u t by

•To find phasor voltage and the phasor current through each element in the circuit.

Answers:277 56.3, 5.55 33.7

1.39 146.3 , 2.77 56.3

oC

o oL R

V I

I I

j200Ω-j50Ω

•Steady-state Response of RLC in Series

R

Cj1

Lj

RU

LU

CU

U

I

XR

XXR

XXRZ

CL

CL

j

)j(

jj

Reactance

2222 )1

(||C

LRXRI

UZ

RC

L

R

X

1

arctanarctaniu

R

Cj1

Lj

RU

LU

CU

U

I

ZI

XIXIRI

UUUU CLR

CL jj

I RU

LU

CU

CU

XU

U

Voltage Triangle

RU

XUU

•Steady-state Response of RLC in Series

R

Cj1

Lj

RU

LU

CU

U

I

Impedances Triangle

IRU R

IX

UUU CLX

ZIU

R

X

Z

22|| XRZ

R

CLarctan

1

arctanU

UU

RC

L

•Steady-state Response of RLC in Series

cosrmsrmsIUP

sinrmsrmsIUQ

rmsrmsIUS

cosSP

sinSQ

S

Pcos

Power Factor功率因数

•Power and Power Factor

•Three Triangles

S

Q

P22 )( CL XXRZ

sin

cos

ZX

ZR

2)( CL2

R UUUU

sin

cos

UU

UU

X

R

22 QPS

sin

cos

SQ

SP

RU

U

CL UU R

CL XX

Z

•Example 5.6: Computer the power and reactive power taken from the source and each element in the circuit.

Answers:

CLSRS

RCL

RCL

SS

QQQPP

WPPP

QVarQVarQ

VarQWP

,

)(5.0,0,0

0),(5.0),(1

),(5.0),(5.0

Answers:

)(59.4915

94.0cos,59.19)arctan(

),(56.3

)(10

0

0

AI

P

Q

kVarQQQ

kWPPP

BA

BA

•Example 5.7:Find the power, reactive power and power factor for the source, find the phasor current i.

5.6 THÉVENIN EQUIVALENT CIRCUITS

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

ocVV t

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

5.6 THÉVENIN EQUIVALENT CIRCUITS

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

scsc

oc

I

V

I

V ttZ

scII n

5.6 THÉVENIN EQUIVALENT CIRCUITS

Answers:

•Example 5.9: Find Thevenin equivalent circuit for the circuit.

0 050 50( ), 100 90 ( ), 1.414 45 ( )t t nZ j V V I A

•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.

Answers:

)(71.20),(71.70250)(

)(25),(5050)(

WPwhenZb

WPjwhenZa

load

load

•Example 5.10: Determine the maximum power delivered to a load (a) the load can have any complex value; (b) the load must be a pure resistance.

5.7 BALANCED THREE-PHASE CIRCUITS

•Much of the power used by business and industry is supplied by three-phase distribution systems.

BALANCED THREE-PHASE CIRCUITS：

Three equal-amplitude ac voltages have phases that are 1200 apart.

Chapter 17 tells us how three-phase voltages are generated.

0120

Line a、 b、 c火线、相线

Neutral n零线、中线

Positive phase sequence： a → b → c

Wye (Y)-connected

0

0

0

, 0 ,

, 120 ,

, 120

an Y

bn Y

cn Y

phaseA V V

phaseB V V

phaseC V V

Phase voltage 相电压 VY

•Phase Sequence

•Three-phase sources can have either a positive or negative phase sequence.

•The direction of rotation of certain three-phase motors can be reversed by changing the phase sequence.

•Wye–Wye Connection

•Three-phase sources and loads can be connected either in a wye (Y) configuration or in a delta (Δ) configuration.

•The key to understanding the various three-phase configurations is a careful examination of the wye–wye (Y-Y) circuit.

sourcesBalanced loads对称负载

Line currents线电流 Neutral currents

中线电流Four-wire connectionBalanced loads:

All three load impedances are equal. •Under balanced three-phase sources and loads,

0 cCbBaANn IIII Therefore, we can omit the neutral wire.

we can eliminate the neutral wire. Then, compared with single phase circuit, only three wires are needed to connect the sources to the loads , it is less expensive.

In a balanced three-phase system, neutral current is zero.

•Wye–Wye Connection

cos3)( rmsrmsavg LY IVtpP

sin3sin2

3 rmsrms LYLY IVIV

Q

Compared with single-phase circuit, the total instantaneous power in a balanced three-phase system is constant rather than pulsating.

•Reactive power

•Wye–Wye Connection

A(相电压 )

B(相电压 )

(线电压 )

• A balanced positive-sequence wye-connected 60 Hz three-phase source has phase voltage UY=1000V. Each phase of the load consists of a 0.1-H inductance in series with a 50-Ω resistance.

• Find the line currents, the line voltages, the power and the reactive power delivered to the load. Draw a phasor diagram showing line voltages, phase voltages and the line currents. Assuming that the phase angle of Uan is zero.

0

0

37

3762.627.3750

jLjRZ

0

0 0

15.97 37

15.97 157 , 15.97 83

anaA

bB cC

VI

Z

I I

0

0

37

3762.627.3750

jLjRZ

0

0 0

15.97 37

15.97 157 , 15.97 83

anaA

bB cC

VI

Z

I I

0 0 0 03 30 1732 30 1732 90 , 1732 150ab an bc caV V V V

avg rms rms3 cos 19.13kWY LP V I

3 sin 14.42kVar2Y LV I

Q

•Delta (Δ)-connected Sources

According to KVL,

0ab bc caV V V

Thus, the current circulating in the delta is zero.

•Wye (Y) and Delta (Δ)-connected Loads

YZwhenZ 3 Two balanced loads are equal.

YZZ 3

•Wye (Y) and Delta (Δ)-connected Loads

•Delta - Delta (Δ- Δ) connection

0

0

0

303

,303

303

CAcC

BCbB

ABCAABaA

II

II

IIII

030abABAB

VVI I

Z Z

030ab LV V

Assuming line voltage

Then, we get phase current

Line current

IIL 3

LVIZ

A delta-connected source supplies power to a delta-connected load through wires having impedances of Zline=0.3+j0.4Ω, the load impedance are ZΔ=30+j6 Ω, the balanced source A voltage is

Find the line current, the line volatge at the load, the current in each phase of the load, the power delivered to the load, and dissipated in the line.

01000 30 ( )abV V

•Homework 5 P5.24 P5.34 P5.43 P5.53 P5.68 P5.72