Unit 2

Single-Phase Circuits

Lecture O

utlines

Phasor Relationships for Circuit Elements

Im

pedance of Circuit Elements

Components of Im

pedance

Admittance

Admittance

Im

pedance C

ombinations

Three common circuit elements will be studied.

Their characteristics when subjected to

sinusoidal input is to be analysed.

Circuit Elements

Resistors

Inductors

Capacitors

Tim

e D

omain

Phasor Domain θ∠

=mI

)co

s(θ

ω+

=t

I m

Phasor Relationship for Resistors

)co

s(φ

ω+t

Vm

φ∠

mV

The current that

passes through a

resistor is in

phasewith the

voltage across it.

Resistor Phase Diagram voltage across it.

Tim

e

V

I

dt

di

Lv=

IV

Ljω

=

)4

56

0co

s(1

2)

(°

+=

tt

vPhasor Relationship for Inductors

)4

56

0co

s(1

2)

(°

+=

tt

v

H 10.

L=

°−

∠=

°∠

°∠

=

⋅

°∠

==

45

290

6

45

12

1.0

60

45

12 j

LjωV

I

)45

60

cos(

2)

(°

−=

tti

dt

di

Lv=

IV

Ljω

=

)4

56

0co

s(1

2)

(°

+=

tt

vPhasor Relationship for Inductors

)4

56

0co

s(2

)(

°−

=t

ti Note that the current lags the voltage by 90o

Tim

e

VI

dt

iC

v

∫=

CjωI

V=

)3

01

00

cos(

6)

(°

−=

tt

vPhasor Relationship for Capacitors

µF

5

0=

C

°∠

=

°∠

°−

∠=

×⋅

°−

∠=

=−

60

03

.0

)90

005

.0

)(30

6(

)10

50

100

)(30

6(6

j

CjωV

I

)60

100

cos(

03

.0

)(

°+

=t

ti

dt

iC

v

∫=

CjωI

V=

)3

01

00

cos(

6)

(°

−=

tt

vPhasor Relationship for Capacitors

Tim

e

)60

100

cos(

03

.0

)(

°+

=t

ti Note that the current leads the voltage by 90o

VI

Inductor

Capacitor

Phase Diagrams for Inductors &

Capacitors

The current lagsthe

voltage by 90o

The current leads

the voltage by 90o

For the sake of consistency, always m

ake the voltage as the reference.

IV

R=

IV

Ljω

=C

jωI

V=

V“o

ppositio

n s

how

ed b

y

Impedance of Circuit Elements

Impedance

IVZ=

Resistor

R=

ZC

jω1=

ZL

jω=

Z

Inductor

Capacitor

“oppositio

n s

how

ed b

y

the c

ircuit t

o the f

low

of

a s

inusoid

al curr

ent”

Unit: ohm

s (

Ω)

Impedance

jXR+

=Z

resistance

reactance

Com

ponents of Impedance

resistance

reactance

If X > 0, im

pedance is inductive or lagging

If X < 0, im

pedance is capacitive or leading

If X = 0, im

pedance is resistive

Means the

current lags

the voltage

Ljω

=L

Z

0=

ω

Impedance of an Inductor

0=

ω

∞→

ω

Cjω1

=C

Z

Impedance of a Capacitor

0=

ω

∞→

ω

Admittance

VI

ZY

==

1“r

ecip

rocal of im

pedance”

Unit: sie

mens (

S)

Admittance U

nit: sie

mens (

S)

jBG+

=Y

conductance

susceptance

Find v(t)and i(t)in the circuit shown.

4=

ωV

010

s°

∠=

V

Cjω1

5+

=Z

1

Example of Impedance Calculations

5.25

010

j−

°∠

==ZV

Is

CjωI

IZV

C=

=

)4.

63

4co

s(47

.4

)(

°−

=t

tv

)6.

26

4co

s(79

.1

)(

°+

=t

ti

1.0

4

15

×+

=j

Ω 5.

25

j−

=

°∠

=6.

26

79

.1

°−

∠

°∠

=6.

26

59

.5

010

1.0

4

6.26

79

.1

×

°∠

=j

°−

∠=

°∠

°∠

=4.

63

47

.4

90

4.0

6.26

79

.1

Impedance in Series

N3

21

eqZ

ZZ

ZZ

++

++

=..

....

...

equivalent

impedance

Impedance in Parallel

N3

21

eqZ1

Z1

Z1

Z1

Z

1+

++

+=

....

....

.

N3

21

eqY

YY

YY

++

++

=.........

equivalent

impedance

equivalent

admittance

rad/s

50

=ω

Find the input im

pedanceZ

in

Z3

Z1

Z2andZ

3are parallel 10

8

1

23

11

11

++

=+

=j

-jZ

ZZ

Example 1

Z2

Ω−

=

××

=

=

−

10

10

250

11

3

j

j

Cjω

1Z

Ω−

=

××

×+

=−

)23(

10

10

50

13

3

j

j2Z

Ω+

=

×+

=

+=

)1

08(

2.0

50

8

8

jj

Lj ω

3Z

7.1

71

7.

46

0.3

66

0.

13

14

44

81

1

20

30

16

24

23

10

8

10

82

3

2

23

∠∠=

++=

−+

−

−+

+=

++

=+

=

jj

jj

j

jj

j-j

32Z

ZZ

07

.1

22

.3

3.18

39

.3

0.36

60

.13

7.17

17

.46

j−

=°

−∠

=∠∠

=23

Z

Ω−

=−

+−

=

+=

07

.11

22

.3

07

.1

22

.3

10

jj

j

23

1in

ZZ

Z

Ω∠

=−

=

73

.8-

11

.53

0

7.

11

22

.3

jinZ

Find the voltage across

the inductor vO

°−

∠=

⇒

°−

15

20

)15

4co

s(20

S

t

V

Example 2 (cont)

rad/s

4

=ω

Ω=

×=

⇒

20

54

H 5

jj

OZ

Ω−

=×

×=

⇒−

25

10

10

4

1m

F

01

3j

jCZ

A 0

.16

5-

0.0

47

0.

74

17

.0

0.5

96

2.

11

6

15

20

j=°

−∠

=

°∠

°−

∠=

=TOTAL

SSZ

VI

Z1

Z2

I O

I S

I COO

SO

SC

VZVI

II

I−

=−

=

Example 2 (cont)

Ω=

1

00

j2Z

100

20

1

25

11

11 2

j

j-j

−=

+=

+=

OC

ZZ

Z

OO

IV

Ljω

=

Ω°

∠=

+=

+=

0.

59

62

.1

16

10

06

0j

21

TOTAL

ZZ

Z

Z2

I C

20

165

.0

047

.0

jj

OV

−−

=

CC

OO

OZ

IZ

IV

==

20

165

.0

047

.0

25

jj

-j

OO

COC

VV

ZVI

−−

==

=

V 98

.15

16

.17

7.4

5.16

∠=

+=

jOV

)16

4co

s(2.

17

)(

°+

=t

tv