Impedance and Admittance

Objective of LectureDemonstrate how to apply Thévenin and

Norton transformations to simplify circuits that contain one or more ac sources, resistors, capacitors, and/or inductors.

Source TransformationA voltage source plus one impedance in

series is said to be equivalent to a current source plus one impedance in parallel when the current into the load and the voltage across the load are the same.

Equivalent Circuits

ThéveninVth = In Zn

NortonIn = Vth/Zth

Example 1

First, convert the current source to a cosine function and then to a phasor.

I1 = 5mA sin(400t+50o) = 5mA cos(400t+50o-90o)= 5mA cos(400t-40o)I1 = 5mA -40o

Example 1 (con’t)Determine the impedance of all of the

components when rad/s.In rectangular coordinates

kjFsradjCjZ

kRZ

jHsradjLjZ

jHsradjLjZ

kRZ

kjFsradjCjZ

C

R

L

L

R

C

57.37.0/400//

5

800)2)(/400(

120)3.0)(/400(

3

5.21/400//

2

2

2

1

1

1

2

2

2

1

1

1

Example 1 (con’t)Convert to phasor notation

oC

oR

oL

oL

oR

oC

kZ

kZ

Z

Z

kZ

kZ

9057.3

05

90800

90120

03

905.2

2

2

2

1

1

1

Example 1 (con’t)

o

oo

oo

V

kmA

kmA

1305.7

90405.2)5(

905.2405

1

1

11

th

th

C1th

V

V

ZIV

Example 1 (con’t)Find the equivalent impedance for ZC1 and ZR1

in series. This is best done by using rectangular coordinates for the impedances.

o

CReq

k

kkkk

kjkZZZ

8.3991.3

35.2tan5.23

5.23

122

1 11

1eq

1eq

Z

Z

Example 1 (con’t)Perform a

Norton transformation.

eq1n

n

n

n

eq1thn

ZZ

I

I

I

/ZVI

1

1

1

1

11

o

oo

oo

mA

kV

kV

2.9092.1

8.3913091.3/5.7

8.3991.31305.7

Example 1 (con’t)Since it is easier to combine admittances in

parallel than impedances, convert Zn1 to Yn1 and ZL1 to YL1. As Yeq2 is equal to YL1 + Yn1, the admittances should be written in rectangular coordinates, added together, and then the result should be converted to phasor notation.

Example 1 (con’t)

01

1122

12

112

11

01

11

11

1

6.88817.0

198.017.8tan17.8198.0

17.8198.0

33.8164.0198.0

33.8

9033.81

164.0198.0

8.39sin8.39cos256.0

8.39256.01

m

m

mjY

mjmjY

mjY

m

mjY

jmY

m

eq

eq

n

n

oon

o

eq2

eq2

L1L1

n1n1

Y

Y

ZY

ZY

Example 1 (con’t)Next, a Thévenin transformation will allow

Yeq2 to be combined with ZL2.

Example 1 (con’t)

o

o

o

V

m

mA

6.1235.0

6.8817.8

2.9092.1

/

1

2

2

212

th

th

thnth

V

V

YIV

Example 1 (con’t)

o

eq

eq

Ltheq

L

th

ooth

o

jZ

jjZ

ZZZ

jZ

jZ

jZ

8.89922

98.2922tan92298.2

92298.2

80012298.2

800

12298.2

6.88sin6.88cos122

6.88122/1

122

3

3

223

2

2

2

3

3

22

eq

eq

thth

Z

Z

YZ

Example 1 (con’t)Perform a Norton transformation after which

Zeq3 can be combined with ZR2.

Example 1 (con’t)

0

0

0

4.91255.0

8.89922

6.1235.0

mA

V

n2

n2

eq3th2n2

I

I

ZVI

Example 1 (con’t)

o

o

eq

ooeq

oo

m

mjY

mjmY

mm

4.79906

4.7910.1

08.1204.0

2.0)8.89sin()8.89cos(08.1

02.08.8908.1

/1/1

1

14

114

11

eq4

eq4

eq4

Req3eq4

Z

Y

Y

ZZY2

Example 1 (con’t)Use the equation for current division to find the current flowing through ZC2 and Zeq4.

o

o

o

o

o

eq

C

o

eq

A

mAm

m

mAmj

m

mAmjmj

m

mjY

mjY

m

3.740.86

4.91255.07.75826.0

90280.0

4.91255.08.0204.0

90280.0

4.91255.008.1204.0280.0

90280.0

08.1204.0

280.0

90280.0/1

0

0

0

4

4

2

2

2

2

2

22

2

2

2

C

C

C

C

CC

n2C

CC

I

I

I

I

ZY

IYY

YI

Example 1 (con’t)Then, use Ohm’s Law to find the voltage

across ZC2 and then the current through Zeq4.

o

o

o

o

oo

mA

V

V

kA

1.95341.0

4.79906

7.15309.0

7.15309.0

9057.33.740.86

4

4

4

4

42

2

222

eq

eq

eqeq

eqC

C

CCC

I

Z

VI

VV

V

ZIV

Example 1 (con’t)Note that the phase angles of In2, Ieq4, and IC2

are all different because of the imaginary components of Zeq4 and ZC2.The current through ZC2 leads the voltage,

which is as expected for a capacitor.The voltage through Zeq4 leads the current.

Since the phase angle of Zeq4 is positive, it has an inductive part to its impedance. Thus, it should be expected that the voltage would lead the current.

Electronic ResponseExplain why the circuit on the right is the

result of a Norton transformation of the circuit on the left. Also, calculate the natural frequency o of the RLC network.

SummaryCircuits containing resistors, inductors,

and/or capacitors can simplified by applying the Thévenin and Norton Theorems.Transformations can easily be performed using

currents, voltages, impedances, and admittances written in phasor notation.

Calculation of equivalent impedances and admittances requires the conversion of phasors into rectangular coordinates.

Use of the current and voltage division equations also requires the conversion of phasors into rectangular coordinates.