1

Instructor:

Kai Sun

Fall 2015

ECE 325 – Electric Energy System Components2- Fundamentals of Electrical Circuits

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Content

•Fundamentals of electrical circuits (Ch. 2.0-2.15, 2.32-2.39)

•Active power, reactive power and apparent power

(Ch. 7)

•Three-phase AC systems (Ch. 8)

3

Notations: Current and Alternating Current

• Arbitrarily determine a positive direction, e.g. 12

– If a current of 2A flows from 1 to 2, I=+2A

– If a current of 2A flows from 2 to 1, I= -2A

G

2

1

I

I >0 I <0

4

Notations: Voltage

1. Double-subscript notation:

E21= +100V (the voltage between 2 and 1

is 100V, and 2 is positive w.r.t 1)

E12 = -100V

2. Sign notation:

Arbitrarily mark a terminal with (+), E=E21=+100V if that terminal is the actual

terminal (+); Otherwise, E=E12=-100V

100V

+

-

2

1

G

100V

2

1

+

E

Both the double-subscript notation and

sign notation apply to alternating voltage

5

Notations: Alternating Voltage

6

Notations: Sources and Loads

• Definition: given the instantaneous, actual polarity of voltage and direction of current

– Source: whenever current flows out of the terminal (+)

– Load: whenever current flows into the terminal (+)

• How about these?

– Resistor, battery cell, electric motor, capacitor and inductor

7

i(t)+

v(t)

Resistor

Inductor Capacitor

8

1-Phase AC System with Sinusoidal Voltage and Current

• e , i: instantaneous voltage (V) and current (A)

• Em , Im : peak values of the sinusoidal voltage (V) and current (A)

• = 2f (rad/s): angular frequency

• e , i : constant phase angles (rad or deg) of voltage and current

• Em/ 2, Im/ 2: RMS (root-mean-square, effective) values

( ) cos( )m ee t E t

( ) cos( )m ii t I t

Load

+

22 2[ ( )]

2 2

tm m

dc dct T

I RT II RT i t Rdt I

Equal heating effects

9

Phasor Representation

• E and I are called RMS phasors of e(t) and i(t);

E leads I by = e- i or in other words, I leads E by 2-

• Phasor:

– a complex number that carries the amplitude and phase angle

information of a sinusoidal signal of a common frequency () w.r.t. a chosen reference signal.

– a mapping from the time domain to complex number domain.

( ) cos( )

( ) cos( )

m e

m i

e t t

i

E

It t

2

2

me

mi

EE

II

2 cos( )| |

|2 c s( )| o

e

i

E

I

t

t

| | | |

| | | |

e

i

j

e

i

j

E E e

I I e

10

Impedance

• Impedance is a complex number () defined as

• Purely resistive:

• Purely inductive:

• Purely capacitive:

| | eE E

| | iI I +

def12

| || | ( )

| |

ee i

i

EE EZ Z R jX

I I I

| |Z Z R

| | 90 LZ Z jX j L

1

2

12E

1| | 90 CZ Z jX j

C

Z

Impedance of branch 1-2 =

(Voltage drop on 1-2)

(Current flow into 1)def

e i (Impedance angle)

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Example 2-5

• Draw the phasor diagram of the voltage and current at a frequency

of 60Hz. Calculate the time

interval t between the positive peaks of E and I

Solution:

=2f=377 (rad/s)=21600 (deg/s)

|E|=339/2=240 (V)

|I|=14.1/2=10 (A)

Choose an arbitrary reference to

draw phasors E and I

t=30/21600=0.00139 (s)

E240V

I10A

30o

( ) 339cos(21600 90 )

( ) 14.1cos(21600 120 )

e t t

i t t

,t

12

Kirchhoff’s Voltage Law (KVL)

• The algebraic sum of the voltages around a closed loop (CW/CCW) is zero

( voltage rises = voltage drops)

– Applied to both instantaneous voltages or voltage phasors

• Loop 24312 (BCDA, CW):

E24+E43+E31+E12=0 or

e24+e43+e31+e12=0

• Loop 2342 (ECF, CCW):

E23+E34+E42=0 or

e23+e34+e42=0

E12

E43

E31

E24

E23

42

1

3

E34

E42

13

Kirchhoff’s Current Law (KCL)

• The algebraic sum of the currents arriving at a node is equal to 0.

( currents in = currents out)

• Node A:

I1+I3+(-I2)+(-I4)+(-I5)=0

or i1+i3+(-i2)+(-i4)+(-i5)=0

I1+I3=I2+I4+I5

or i1+i3=i2+i4+i5

A

14

• KVL

– Loop 24312, CW:

E24 +E43 +E31 +E12 =0

I2Z2 -I3Z3 +Eb -I1Z1 =0

– Loop 2342, CCW:

E23 +E34 +E42 =0

I4Z4 +I3Z3 -I2Z2 =0

– Loop 242, CW:

E24 + E42 =0

Ea - I2Z2 =0

+

+

Kirchhoff’s Laws and AC Circuits

• KCL

– Node 2:

I5 -I2 -I4 -I1=0

– Node 3:

I4 +I1 – I3=0

Ea

Eb

15

Examples 2-14 ~ 2-16

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Impedance angle:

>0 for inductive load and

<0 for capacitive load

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

1 cos( ) cos(2 )

2

1 cos( ) cos[2( ) ( )]

2

1 cos cos[2( ) ]

2

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

2

m m e i

m m e i e i

m m e i e e i

m m e

m m e e

p t e t i t E I

E I

E I

E I

E I

| | / 2

| | / 2

m

m

E E

I I

e i

( ) cos [1 cos 2( )] sin sin 2( )

R X

e e

p p

p E I E I

Using trigonometric identity

1cos cos cos( - ) cos( )

2A B A B A B

Instantaneous Power

( ) cos( )m ee t E t

( ) cos( )m ii t I t

Load

Z

+

t

17

100cose 1.25 60 0.625 1.0825 ΩZ j

8060

2I

80cos( 60 )oi

Example:100

0 2

E

( ) cos [1 cos 2( )] sin sin 2( )

2000(1 cos 2 ) 3464sin 2

R X

XR

e e

p p

pp

p E I E I

2000

3464

ei

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Observations:

• The frequency of p, pR and pX is twice of e and i, which is 602=120Hz

• pR(t)

– changes between 0 and 4000 (=2|E||I|cos )

– always 0, and has an average value of 2000 (=|E||I|cos )

– is the power consumed by the load

• pX(t)

– changes between 3464 (= |E||I|sin)

– has an average value of 0

– is the power borrowed & returned by the load.

( ) ( )

( )( )

( ) cos [1 cos 2( )] sin sin 2( )

(1 cos 2 320 46 n 200 ) 4si

R X

XR

e e

p t p t

p tp t

p t E I E I

19

Apparent, Active (Real) and Reactive Powers

• Real or active power (average power)

• Unit ~ watt or W

• Power factor (PF): cos= cos(e-i)

– The PF is said to be lagging if =e-i >0 or leading if <0

def

| | | | cosP E I

def

| | | | sinQ E I

( ) ( )

[1 cos 2( )] sin 2( )

[1 cos 2( )]

cos

sin 2( )

sin

R X

e e

p t p t

e e

E I

QP

p E I

• Reactive power

• Unit ~ var (volt-ampere reactive). Some people use Var, VAR or VAr

• Q>0 if =e-i >0 (inductive) or Q<0 if <0 (capacitive)

P

Q

def

| | | | | |S E I

• Apparent power

• Unit ~ volt ampere or VA (kVA or MVA)

20

Complex Power and Power Triangle

• =tan-1(Q/P)

– If >0 (i <e, i.e. lagging PF), then Q>0, which means a reactive load (inductive)

– If <0 (i >e, i.e. leading PF), then Q<0, which

means a reactive source (capacitive)

def

= PS jQ cos i s nE I j E I

2 2| | I P QE S

R

XZ

S

• PF=cos=P/|S| (+/- sign of Q tells a lagging/leading PF)

P=|S| cos=|S| PF

• If the load impedance is Z=E/I, then

S = ZII*= Z|I|2= R|I|2+jX|I|2 = P + jQ

P=R|I|2 Q=X|I|2

is also called the load impedance angle

Power triangle

Q

p qE I Ij E Ip

Iq

*

e i e i EE I I IE

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Other Useful Formulas

• If Z is purely resistive

• If Z is purely reactive

2**

* *

* 2 | |

EEES EI

Z Z

II Z I Z

2

* 2| |

E SZ

S I

2E

PR

2 2 2

or 1/

E E EQ

X L C

22

Examples 7-2 & 7-3

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Saadat’s Example 2.2

1

2 3

1200 0 V, 60 0 ,

6 12 , 30 30

V Z j

Z j Z j

Find the power absorbed by each load

and the total complex power

1

1

2

2

3

3

1200 0 120020 0 A

60 0 60

1200 0 200 200(1 2)40 80 A

6 12 1 2 5

1200 0 40 40(1 )20 20 A

30 30 1 2

VI j

Z j

V jI j

Z j j

V jI j

Z j j

1 2 3 96,000 W+ 72,000 varS S S S j

*

1 1

*

2 2

*

3 3

1200 0 (20 0) 24,000 W+ 0 var

1200 0 (40 80) 48,000 W+ 96,000 var

1200 0 (20 20) 24,000 W 24,000 var

S VI j j

S VI j j

S VI j j

S1

S2

S3

S

24

• Other approaches

1 2 3

*

(20 0) (40 80) (20 20) 80 60 100 36.87 A

(1200 0 )(100 36.87 ) 120,000 36.87 VA 96,000 W 72,000 var

I I I I j j j j

S VI j

2 2

1 *

1

2 2

2 *

2

2 2

3 *

3

(1200)24,000 W 0 var

60

(1200)48,000 W 96,000 var

6 12

(1200)24,000 W 24,000 var

30 30

VS j

Z

VS j

Z j

VS j

Z j

/ / 1 2 3

1 2 3

2 2

*

/ /

1/ / / / 9.6 7.2

1/ 1/ 1/

(1200)96,000 W 72,000 var

9.6 7.2

Z Z Z Z jZ Z Z

VS j

Z j

or

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Three-Phase Systems

• Balanced 3-phase voltage:

• Sequence: a-b-c (CW).

• Phase voltages are displaced at

120o (a leads b, b leads c, and c leads a by 120o, respectively)

• Equal voltage magnitudes

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3-Phase, 4-Wire AC System

Ia+Ib+Ic

=Ea1/R+Eb2/R+Ec3/R

=(Ea1+Eb2+Ec3)/R

=0

27

3-Phase, 3-Wire AC System

• Line currents:

|Ia|=|Ib|=|Ic|= IL

• Phase voltages (phase-to-neutral or

line-to-neutral):

|Ean|=|Ebn|=|Ecn|= ELN

• From KVL, line voltages (line-to-line or phase-to-phase):

Eab=Ean+Enb=Ean-Ebn

Ebc=Ebn+Enc=Ebn-Ecn

Eca=Ecn+Ena=Ecn-Ean

|Eab|=|Ebc|=|Eca|= EL

• EL=3 ELN 1.73 ELN

28

Wye (Y) Connection

• Each line current (Ia, Ib and Ic) equals

the phase current.

IL = IZ

• Each line voltage is 3=1.73 times of

a phase voltage in magnitude

EL=3 ELN ( |Eab|= 3 |Ean| )

• Apparent power of each phase:

• Total 3-phase apparent, active and reactive power:

| |3

LZ LN Z L

ES E I I

22

3 3 3 | || |3

| | 3 L LL L

L L Y

Y

S EE E

II I ZZ

3 3 cosL LP E I

3 3 sinL LQ E I

| |Y YZ Z

a

b

c

a

b

c

ab

bc

ca

is the power factor angle

and load impedance angle

29

Delta () Connection

• From KCL:

Ia=Iab-Ica

Ib=Ibc-Iab

Ic=Ica-Ibc

Because three equations are symmetric, and Ia leads Ib, Ib leads Ic and Ic leads Ia all by

120o, we may easily conclude:

1. |Iab|=|Ibc|=|Ica|=IZ

2. Iab leads Ibc, Ibc leads Ica, and

Ica leads Iab by 120o, respectively

• Each line current is 3=1.73 times of a phase current in magnitude

IL=3 IZ ( |Ia|= 3 |Iab| )

• Each phase voltage equals the line voltage

ELN = EL

a

b

c

Iab

Ibc

Ica

30

Delta () Connection

• Apparent power of each phase:

• Total 3-phase apparent, active and reactive power:

• Given line voltage EL, line current IL and power factor cos, calculation of power is independent of the connection (Y / )

22

3

33 | |

| || 3

3| L L

L LL L

IS E II

EE Z

Z

|ZY|=|Z|/3

| |Z Z

a

b

c

Iab

Ibc

Ica

| |3

LZ L Z L

IS E I E

3 3 cosL LP E I

3 3 sinL LQ E I

is the power factor angle

and load impedance angle

31

Examples 8-8 & 8-11

32

Homework #1

• Read Ch. 2.0-2.15, 2.32-2.39, Ch. 7

• Answer Questions 2-1, 2-2, 2-10, 2-15, 2-26,

7-11, 7-12, 7-13, 7-14, 7-17, 7-20, 7-22

• Due date:

– hand in your solution in the class on 9/9 or

– to Denis at MK 205 or by email (dosipov@vols.utk.edu)

before the end of 9/11