Chapter 5curt.nelson/engr228/lecture/chap5.pdf · The Superposition Theorem ... Engr228 - Chapter 5, Hayt 8E 12 Source Transformation and Equivalent Sources The sources are …

Engr228 - Chapter 5, Hayt 8E 1

Chapter 5

Engr228

Circuit Analysis

Dr Curtis Nelson

Chapter 5 Objectives

• State and apply the property of linearity

• State and apply the property of superposition

• Investigate source transformations

• Define and construct Thevenin and Norton equivalent

circuits

• Investigate maximum power transfer to a load


Linear Elements and Circuits

• A linear circuit element has a linear voltage-current

relationship:

– If i(t) produces v(t), then Ki(t) produces Kv(t)

– If i1(t) produces v1(t) and i2(t) produces v2(t), then i1(t) + i2(t) produces

v1(t) + v2(t),

• Resistors and sources are linear elements

– Dependent sources need linear control equations to be linear elements

• A linear circuit is one with only linear elements

The Superposition Concept

For the circuit shown below, the question is:

• How much of v1 is due to source ia, and how much is due to

source ib?

We will use the superposition principle to answer this question.


The Superposition Theorem

In a linear network, the voltage across or the current through

any element may be calculated by adding algebraically all the

individual voltages or currents caused by the separate

independent sources acting “alone”, i.e. with

– All other independent voltage sources replaced by short circuits (i.e. set

to a zero value) and

– All other independent current sources replaced by open circuits (also

set to a zero value).

Applying Superposition

• Leave one source ON and turn all other sources OFF:

– Voltage sources: set v=0

These become short circuits.

– Current sources: set i=0

These become open circuits.

• Then, find the response due to

that one source

• Add the responses from the other

sources to find the total response


Superposition Example (Part 1 of 4)

Use superposition to solve for the current ix


First, turn the current source off:

′ i x =3

6+9= 0.2



Then, turn the voltage source off:

ix′′ =

6

6 + 9(2) = 0.8


Finally, combine the results:

ix = ix′ + ix

′′ = 0.2 + 0.8 =1.0


Example: Power Ratings

Determine the maximum positive current to which the source Ix

can be set before any resistor exceeds its power rating.

Answer: Ix < 42.49 mA

Superposition with a Dependent Source

• When applying superposition to circuits with dependent

sources, these dependent sources are never “turned off.”


Superposition with a Dependent Source

Current source off

Voltage source off

ix = ix’+ix

’’=2 + (−0.6) = 1.4 A

Superposition Example

Find voltage Vx. First, solve by nodal analysis.

(42-Vx)/6 = (Vx-(-10))/4 + Vx/3

Vx = 54/9 = 6 volts


V

Vx V

333.9

42)7/12(6

)7/12(42

)4||3(6

)4||3()42(

=

×+

=×+

=

Example - Continued

V

Vx V

333.3

1042

210

4)3||6(

)3||6()10(

−=

×+

−=×+

−=

Example - Continued


V

VxVxVx VV

6333.3333.9

)10()42(

=−=

+=

Example - Continued

Practical Voltage Sources

• Ideal voltage sources: a first approximation model for a

battery.

• Why do real batteries have a current limit and experience

voltage drop as current increases?

• Two car battery models:


Practical Source: Effect of Connecting a Load

For the car battery example:

VL = 12 – 0.01 IL

This line represents

all possible RL

Practical Voltage Source

s

sLsc

soc

R

vi

vv

=

=

LssL iRvv −=

OR

s

L

s

sL

R

v

R

vi −=


Practical Current Source

sLsc

spLoc

ii

iRv

=

=

p

LsL

R

vii −=

Equivalent Practical Sources

p

LsL

R

vii −=

s

L

s

sL

R

v

R

vi −=

• These two practical sources are equivalent if

RS = RP S

SS

R

vi =


Source Transformation and Equivalent Sources

The sources are equivalent if

Rs=Rp and vs=isRs

Source Transformation

• The circuits (a) and (b) are

equivalent at the terminals.

• If given circuit (a), but circuit (b) is

more convenient, switch them.

• This process is called source

transformation.


Example: Source Transformation

We can find the current I in the circuit below using a source

transformation, as shown.

I = (45-3)/(5+4.7+3) = 3.307 mA

I = 3.307 mA

Example

Use source transformations to find Ix


Example - continued

Example - continued


AI X3

1

6

2

321

2==

++=

Example - continued

Textbook Problem 5.24 Hayt 7E

• Using source transformations, determine the power

dissipated by the 5.8 kΩ resistor.

42.97 mW



(a) Determine the individual contributions of each of the two

current sources to the nodal voltage v1

(b) Determine the power dissipated by the 2Ω resistor

v17A = 6.462V, v14A = -2.154V, v1tot = 4.31V, P2Ω= 3.41W


Determine the current labeled i after first transforming the

circuit such that it contains only resistors and voltage

sources.

i = -577µA



Using source transformations, reduce the circuit to a voltage

source in series with a resistor, both of which are in series

with the 6 MΩ resistor.

Vs = -6.284V Rs = 825.4 kΩ


P7v = 17.27W (generating)

Find the power generated by the 7V source.


Thévenin Equivalent Circuits

Thévenin’s theorem: a linear network can be replaced by its

Thévenin equivalent circuit, as shown below:

Thévenin Equivalent using Source Transformations

• We can repeatedly

apply source

transformations on

network A to find its

Thévenin equivalent

circuit.

• This method has

limitations - not all

circuits can be source

transformed.


Finding the Thévenin Equivalent

• Disconnect the load

• Find the open circuit voltage voc

• Find the equivalent resistance Req of the network with all

independent sources turned off

Then:

VTH = voc and

RTH = Req

Thévenin Example


Example

Find Thevenin’s equivalent circuit and the current

passing thru RL given that RL = 1Ω

0V

10V

0V 0V

6V 6V

VVTH 61032

3=×

+=

Find VTH

Example - continued


Short voltage source

RTH

Ω=

++

×+=

++=

2.13

232

3210

23||210THR

Example - continued

Find RTH

Thevenin’s equivalent circuit

The current thru RL = 1Ω is A423.012.13

6=

+

Example - continued


Example

Find Thevenin’s equivalent circuit

Find VTH

0V

5V

0V 0V

3V 3V

VVTH 331 =×=

Example - continued


Open circuit

current source

RTH

Ω=

++=

15

2310THR

Find RTH

Example - continued


Example - continued


Example: Bridge Circuit

Find Thevenin’s equivalent circuit as seen by RL

Find VTH

0V

10V

8V 2V

VTH = 8-2 = 6V

Example - continued


Find RTH

RTH

Example - continued

KKK

KKKKRTH

4.28.06.1

1||48||2

=+=

+=

Example - continued



Example - continued

Thevenin’s Equivalent Circuit - Recap


Norton Equivalent Circuits

Norton’s theorem: a linear network can be replaced by its Norton

equivalent circuit, as shown below:

Finding the Norton Equivalent

• Replace the load with a short circuit

• Find the short circuit current isc

• Find the equivalent resistance Req of the network with all

independent sources turned off

Then:

IN = isc and RN = Req


Source Transformation: Norton and Thévenin

• The Thévenin and Norton equivalents are source

transformations of each other!

RTH=RN =Req and vTH=iNReq

Example: Norton and Thévenin

Find the Thévenin and Norton equivalents for the network faced

by the 1-kΩ resistor.

The load

resistor

This is the circuit we will simplify


Example: Norton and Thévenin

NortonThévenin

Source

Transformation

One method to find the Thévenin equivalent of a circuit with a

dependent source: find VTH and IN and solve for RTH=VTH / IN

Example (Textbook p. 148):

Thévenin Example: Handling Dependent Sources



Finding the ratio VTH / IN fails when both quantities are zero!

Solution: apply a test source (Textbook p. 149)


Solve: vtest =0.6 V, and so RTH = 0.6 Ω

v test

2+

v test − (1.5i)

3=1

i = −1


Example

• Find Norton’s equivalent circuit and the current

through RL if RL = 1Ω

Find IN

Find R total:

Find I total:

Current divider:

Ω=+

×+=++ 4.4

123

1232)210(||32

AR

VI 27.2

4.4

10===

AI SC 45.027.2123

3=×

+=

Example - continued


Ω=

++

×+=

++=

2.13

232

3210

23||210THR

Find Rn

Example - continued

Norton’s Equivalent Circuit

The current thru RL = 1Ω is A418.045.012.13

2.13=×

+


Relationship Between Thevenin and Norton

I

VVOC

ISC

Slope = - 1/RTH

THNTH

NTH

RIV

RR

×=

=

Norton’s equivalent circuitThevenin’s equivalent circuit

Same R value

2.1345.06 ×=

THNTH

NTH

RIV

RR

×=

=

Recap: Thevenin and Norton


Thevenin’s equivalent circuitNorton’s equivalent circuit

0.2 x 15 = 3

Relationship Between Thevenin and Norton


Find the Thevenin equivalent of the circuit below.

RTH = 8.523 kΩ

VTH = 83.5 V


Equivalent Circuits with Dependent Sources

We cannot find RTH in circuits with dependent sources using

the total resistance method.

But we can use

SC

OCTH

I

VR =

Example

Find Thevenin and Norton’s equivalent circuits


Find Voc

I1 I2

12400014250

0)21(400012501

=−

=−++−

II

IIIKVL

loop1

024060801404000

)21(4000

010028022000)12(4000

=+−

−=

=−++−

II

IIVx

VxIIIIKVL

loop2

Example - continued

I1 I2

Solve equations

I1 = 3.697mA I2 = 3.678mA

V

mA

IIIVxIVOC

3.7

)678.3697.3(400000)678.3(80

)21(400000280100280

−=

−−=

−−=−=

Example - continued


Find Isc

I1 I2 I3

12400014250

0)21(400012501

=−

=−++−

II

IIIKVL

loop1

038024060801404000

)21(4000

0100)32(8022000)12(4000

=−+−

−=

=−−++−

III

IIVx

VxIIIIIKVL

loop2

KVL

loop3 038024000801400000

0100)23(80

=+−

=+−

III

VxII

Example - continued

Find Isc

I1 I2 I3

I1 = 0.632mA

I2 = 0.421mA

I3 = -1.052 A

Isc = I3 = -1.052 A

Example - continued


Ω=−

−== 94.6

052.1

28.7

SC

OCTH

I

VR

Thevenin’s equivalent circuit Norton’s equivalent circuit

Example - continued

Maximum Power Transfer

Thevenin’s or Norton’s equivalent circuit, which has an RTH

connected to it, delivers a maximum power to the load RL for

which RTH = RL


2

22

2

)( LTH

LTHL

LTH

TH

L

RR

RVR

RR

VP

RIP

+=⋅

+=

=LTH

TH

RR

VI

+=and

Plug it in

0)(

)(2)(4

222

=+

+⋅−+=

LTH

LTHLTHTHLTH

L RR

RRRVVRR

dR

dP

Maximum Power Theorem Proof

LTH

LLTH

LTHLTHTHLTH

RR

RRR

RRRVVRR

=

=+

+⋅=+

2)(

)(2)(222

0)(

)(2)(4

222

=+

+⋅−+=

LTH

LTHLTHTHLTH

L RR

RRRVVRR

dR

dP

For maximum power transfer

Maximum Power Theorem Proof - continued


Example

Evaluate RL for maximum power transfer and find the power.


RL should be set to 13.2Ω to get max. power transfer

Max. power is WR

V68.0

2.13

)2/6( 22

==

Example - continued


Chapter 5 Summary

• Stated and applied the property of linearity

• Stated and explored the property of superposition

• Investigated source transformations

• Defined and constructed the Thevenin and Norton

equivalent circuits

• Investigated maximum power transfer to a load

Documents

Chapter 5curt.nelson/engr228/lecture/chap5.pdf · The Superposition Theorem ... Engr228 - Chapter 5, Hayt 8E 12 Source Transformation and Equivalent Sources The sources are …