Chapter 2 Network Theorems

8/4/2019 Chapter 2 Network Theorems

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Network TheoremsNetwork Theorems

Chapter 2

Artemio P. MagaboArtemio P. MagaboProfessor of Electrical EngineeringProfessor of Electrical Engineering

Department of Electrical and Electronics EngineeringUniversity of the Philippines - Diliman

Department of Electrical and Electronics Engineering

Linear Element

A linear element is a passive element whosevoltage-current relationship is described by a linearequation; i.e. if the current through the element is

multiplied by a constant k, then the voltage acrossthe element is likewise multiplied by k.

Consider a resistor R with currenti=i1. From Ohms Law, we get +

R

vi -

11 Rivv ==

Suppose the current is increased by a factor k; i.e.

i2=ki1. The new voltage is

1122 kvRkiRiv === (R is a linear element)


L

+ vi -

Consider an inductor R with currenti=i1. The inductor voltage is

dt

diLvv 11 ==

Suppose the current is increased by a factor k; i.e.i2=ki1. The new voltage is

112

2 kvdt

)ki(dL

dt

diLv ===

Thus,L is a linear element.

Note: Following the same analysis, we can showthat a constant capacitor is a linear element.


Linear Dependent Source

A linear dependent source is a current or voltagesource whose output current or voltage isproportional only to the first power of somecurrent or voltage variable in the circuit, or to the

sum of such quantities.

Linear DependentVoltage Sources

+-

kvx+-

kiy

Linear DependentCurrent Sources

kvw kiz


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Linear Electric Circuit

An electric circuit is linear if it consists of

independent sources

linear dependent sources

linear elements

The response of a linear circuit is proportional to thesources; that is, if all independent sources aremultiplied by a constant k, all currents and voltages

will likewise increase by the same factor k.


Principle of Superposition

In an electric circuit containing N independentsources, the current (or voltage) in any branch isequal to the algebraic sum of N components, eachof which is due to one independent source actingalone.

1. For a voltage source, remove the source andreplace with a short circuit;

Note: Reducing an independent source to zero:

2. For a current source, remove the source andreplace with an open circuit.


Example: Find I1 andI2 using nodal analysis.

R1

REF

R2

+Vx

I1I2Vs

+

-

IsFrom KCL, we get

2

x

1

sxs

R

V

R

VVI +

=

or

s21

2

s21

21

x VRR

R

IRR

RR

V +++=

s

1

s

21

x VR

1I

R

1

R

1V +=

+

which can be simplified as


s21

2

s21

1 IRR

R

VRR

1

I ++=

Thus, we get

1

xs1

R

VVI

=

s21

1

s21

2 IRR

R

VRR

1

I +++=

and

2

x2

R

VI =


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Solution using Superposition:

s

21

a2a1 VRR

1II+

==

Assume Vs is acting alone.Replace Is by an open circuit.

R1

R2I1a

I2a

Vs

+

-

Next, assume Is is acting alone. Replace Vs with ashort circuit. We get

R1

R2I1b

I2b Is

s

21

2b1 I

RR

RI

+=

s

21

1b2 I

RR

RI

+=


Finally, we apply superposition to get I1 and I2.

b1a11 III +=

b2a22 III +=

Substitution gives

s

21

2s

21

1 IRR

RV

RR

1I

+

+=

s

21

1s

21

2 I

RR

RV

RR

1I

+

+

+

=


Example :Find all currents usingsuperposition.

6

81V

+

-

9A

12

6

6A

3

We have three independent sources.Solve for the currents with each source

acting alone.


A327

81Ia ==

A5)A9(1215

15Ib =

+=

Consider the 81-V sourceacting alone.

6

81V

+

-

12

6

3Ia

Consider the 9-A sourceacting alone. 6 12

6

39AIb Ic

A4I9I bc ==


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IdIe

Finally, consider the 6-Asource acting alone.

6A

6 12

6

3A2)A6(189

9

Id=

+=

A4I6I de ==

Apply superposition to getthe current in any resistor.

For example, the current in the 3 resistor is

=++=++= A11443IIII eca3


ax I2V =

which gives

A0.8Ia =

Example: Usesuperposition tofind the current I. I

34V

+ -

2A

+-Vx

+

-5Vx

12

Consider the 4-V source acting alone. From KVL,we get

Ia

34V

+ -

+-Vx

+

-

5Vx2

axa I25V3I4 ++=


Consider next the 2-Asource acting alone.

Ib

3

2A

+-Vx

+

-5Vx

12We get

)I-(22V bx =

Solving simultaneously, we find Ib=3.2 Amps.

Applying superposition, we get

A4III ba =+=

xbx V5I3V +=

and


Example: Find the voltageVx using the principle ofsuperposition. 20

16V

+

-

3A

10V

1.5A

+ -

+- Vx80

Consider the 16-V sourceacting alone. Using voltagedivision, we get 20

16V

+

-

+- Vx180)V16(

8020

20Vx1

+=

V3.2=


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Consider next the 3-Asource acting alone.

20

3A

+- Vx280Ix

From Ohms Law, we get

Next, consider the 10-Vsource acting alone. 20 10V

+ -

+- Vx3 80)V10(8020

20

Vx3 +=

V2=

A4.2=

)A3(8020

80Ix

+=


20

1.5A

+- Vx480

Finally, consider the 1.5-Asource acting alone.

Because of the shortcircuit, we get

0Vx4 =

Applying superposition, we get

x4x3x2x1x VVVVV +++=

V8.4602482.3 =+++=


Thevenins Theorem

Consider a circuit which can be represented by twonetworks: A which is linear and B, which may belinear or non-linear. Any dependent source innetwork A is controlled by a current or voltage in

network A. The same is true with network B.

LinearNetwork

Ay

x

Network

B

Network A can be replaced by a voltage source Vthwhich is connected in series with a resistor Rth.


Rth= the equivalent resistance from terminal xto terminal y, looking into network A, withall independent sources reduced to zero.

y

xRth

Vth

+

-

Network

B

The Thevenin equivalent of network A is shown.

Vth= open-circuit voltage from terminal x toterminal y, with network B removed

where


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Nortons Theorem

Network A can be replaced by a current source Inwhich is connected in parallel with a resistor Rn.

Consider a circuit which can be represented by twonetworks: A which is linear and B, which may belinear or non-linear. Any dependent source innetwork A is controlled by a current or voltage innetwork A. The same is true with network B.

LinearNetwork

Ay

x

Network

B


Rn=Rth=the equivalent resistance from terminal xto terminal y, looking into network A, withall independent sources reduced to zero.

y

x

In RnNetwork

B

The Norton equivalent of network A is shown.

In = short-circuit current from terminal x toterminal y, with network B removed

where


Thevenin EquivalentGeneral Procedure

1. Remove the load and find the voltage across theopen circuit terminals, VOC. This is the Theveninequivalent voltage.

+

VOC-

2. Determine the Thevenin equivalent resistanceRth at the open terminals with the load removed.

Rth

3. Connect the load to the Thevenin equivalentcircuit, consisting of VOC in series with Rth. Thedesired solution can now be obtained.

VOC+ Rload

Rth


General Procedure for DeterminingThevenin Resistance

If the circuit contains only independent sources,suppress all sources and compute the resistance atthe open terminals.

If the circuit contains only dependent sources,apply an independent voltage (current) source atthe open terminals and measure the correspondingcurrent (voltage). The Thevenin equivalentresistance is given by the voltage/current ratio.

If the circuit contains both independent anddependent sources, determine the short circuit

current ISC at the open terminals. The ratio VOC/ISCis the resistance Rth.


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Norton EquivalentGeneral Procedure

2. Determine the Norton equivalent resistance Rnat the open terminals with the load removed.Note that Rn is the same as Rth.

Rn

1. Remove the load and find ISC, the short-circuit

current. ISC is the Norton equivalent current.

ISC

3. Connect the load to the Norton equivalentcircuit, consisting of ISC in parallel with Rn. Thedesired solution can now be obtained.

ISC RloadRn


Example 1: Find

the voltage VO usingThevenins theorem.

I2

I1

300

9V +-

10mA300

1k 500+VO-

600

First, find VOC. Removethe 1k resistor and findthe open-circuit voltage.

300

9V +-

10mA300

500+VOC-

600

The current in mesh 2is known already.

I2 = 10mA

Circuits withIndependent Sources

Only


For mesh 1 we get

11 I600)01.0I(3009 +=

From KVL on the lower left loop

212OCI300)II(300V +=

Next, find Rth . Remove the 1k resistor, reduce allindependent sources to zero, and find the equiva-lent resistance from x to y.

300

300

500

600

xy=

++=

500

600300

)600(300300R th

I1 = -6.67mA

VOC = 8V


The Thevenin equivalent circuit is shown.

500x

y

8V

+

-

V33.585001000

1000VO =

+=

Finally, we return the 1k resistor and find thevoltage VO.

500 x

y

8V

+

-

1kVo+

-


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Example 2:Use Nortonstheorem to find VO.

300

9V +-

10mA300

1k 500

+VO-

600

First, find ISC. Replacethe 1k resistor witha short circuit andfind the current.

300

9V +-

10mA300

500ISC

600

ISC can be found usingsuperposition.


500300

10mA300

ISC1

600Consider the 10mA sourceacting alone.

ISC1 = 10mA

Consider next the 9Vsource acting alone.

9V +-

500300

300

ISC2

600

REF+Va

which gives

mA6300

VI a2SC ==Va=1.8 Volts and

9150600

150Va

+=

== 150300||300Req


Thus, we getISC = ISC1 + ISC2

= 16mA

We have previously found Rthto be 500.The

Norton equivalent circuit with the 1k

resistoris shown.

1k50016mA+

VO-

1000500

)1000(500016.0VO

+=

= 5.33V


Example 3:Find the Theveninresistance as seen

from xy.

Circuits with Dependent Sources Only

x

40 2020

y

20IX

IX

+ -

Attach a 1-A current source between terminalsxy and find the voltage across the source.

x

40 2020

y

20IX

IX

+ -+

VS-

IS=1A

REF

+Va +Vb


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20

V

40

V

20

V1 baa ++=KCL at supernode ab yields

Finally, the Thevenin resistance is

==== 5A1

V5

I

V

I

VRS

b

S

Sth

The voltage at the dependent source is

Va- Vb = 20Ix

But40

VI ax = so we get Va = 2Vb

Solving simultaneously, we have

Va=10 V Vb=5 V


Example 4: Find the Norton equivalent circuitbetween terminals xy.

We do not have an independent source so ISC=0.

2 x

4

3

2 4.5

y

0.5VX

+VX

To get the equivalent resistance Rn, attach a

1-V voltage source between terminals xy andfind the current passing through the source.


We apply source transformation on the dependentsource.

The resulting circuit is shown.

2 x

4

3

2 4.5

y

0.5VX

+VX

VS=1V+

-

Is

2 x4 3

+- 2 4.5

y

2VX

+VX

VS= 1V+

-Is

REF

+Va


From KCL

3

1V

2

V

24

2V0 aaa

++

+

=

We getVolts3

2

Va=

+= 5.43s III

5.4

1

3

V1 a +

=

We determine IS from KCL

A3

1Is =

Finally, the Nortonresistance is

S

Sn

I

VR =

== 3A3

1

V1


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Example 5:Find currentIx

usingNortonstheorem.

+vx-

100

400vx

400

300

+-

200 125

Ix

10V

Let us first find ISC. Remove the 125 resistorand find the short-circuit current.

ISC+vx-

100

400

vx

400

300

+-

200

10V

100300

10ISC

+

=

= 25 mA

Circuits with Independent and DependentSources


From mesh 1 we get

22x

1 I2

1

400

I200

400

VI ===

We find theopen-circuitvoltage VOC

+VOC-

+vx-

100

400

vx

400

300

+-

200

10V

I1 I2

212 I)200100()II(30010 ++=For mesh 2

Solving simultaneously, we have

I1=11.11 mA

I2= 22.22 mAAnd the open-circuit voltage

VOC = 200(22.22)(10-3) = 4.444 V


We get the equivalent resistance

=== 76.177A025.0

V444.4

I

VR

SC

OCth

12576.177

76.177025.0Ix

+=

= 14.68 mA

The Norton equivalent circuit with the 125

resistor is shown.

125177.7625mA Ix


The value of R that will dissipate maximum poweris equal to the Thevenin resistance as seen from

the location of R.

)1000(1500

8RIP

2

2

xR

==

mW44.28=

Maximum Power TransferTheorem

500 x

y

8V

+

-

R=1k

Ix

In Example 1:

But if R=Rth=500

mW32)500()500(2

8P

2

(max)R =

=


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In any passive linear bilateral network, if thesingle voltage source Vx in branch x produces acurrent Iy in branch y, then the removal of the

voltage source from branch x and its insertion inbranch y will produce a current response Iy inbranch x.

Reciprocity Theorem

In any passive linear bilateral network, if thesingle current source Ix between nodes x and xproduces a voltage response Vy between nodesy and y, then the removal of the current source

from nodes x and x and its insertion betweennodes y and y will produce a voltage responseVy between nodes x and x.


Example: Find the current Iy using nodal analysis.

Iy

6+V

36V

+

-3

4

REF

From KCL, we get

6

V

3

V

4

V36+=

which gives V=12 Volts. Thus

A26VIy ==


Let us transfer the source to the right side of thecircuit and this time find the current Iz.

Iz

6+V

36V

+

-

3

4

REF

From KCL, we get

4

V

3

V

6

V36+=

which gives V=8 Volts. Thus

A24

VIz ==

Note: A2II zy ==


Non-linear Resistance

Ohms Law, v=Ri, describes a linear resistance.

Physical resistors are generally linear within their

prescribed range of operation (limits on current orvoltage, temperature, frequency, etc.)

A resistor whose voltage-current relationship doesnot conform with Ohms Law is non-linear.

Note: Examples of non-linear devices include

diodes, fuses, surge arresters, etc.


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Ideal Diode

If v < 0, i = 0 Open Circuited(Reverse biased)

When i > 0, v = 0 Short Circuited(Forward biased)

i

v

Note: The ideal diode acts as a switch, which canbe opened or closed.

+ v

i

-


pn-Junction Diode

Note: The pn-Junction diode is a non-linearresistor whose current is a function of its voltage;i.e. a voltage-controlled non-linear resistor.

= 1

v

vexpIi

T

s

When conducting,the current is

i

vIs

+ v

i

-


Tunnel Diode

Note: For V1 < v < V2 , the resistance is negative.Used in amplifiers and oscillators.

For I2 < i < I1 , a given current i corresponds to

three values of voltage. Used in memory andswitching circuits.

voltage-controlled R

+ v

i

-

i

v

I1

V1 V2

I2


Glow Tube

current-controlled R

+ v

i

-

i

v

Note: The voltage is a single-valued function of

the current.


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Graphical Analysis

Note: Since v=E , simply locate E in the abscissa

and mark the corresponding current. From thegraph, we get i=ix.

Consider an ideal voltage source that is connectedacross a diode.

i

vE

ix+vi-

E+

-


i

vE

Note: This equation describes a straight line whoseintersection with the i vs. v curve gives the solution.

R

+vi-

E+

-

From KVL, we get

RiEv =

Consider the case when the source has a resistance.

R

E

ix


Example:Find the diodevoltage andcurrent.

+vi-

150

1.5V

+

-

+

-

6V

300

100

Remove the diode and find the Theveninequivalent circuit.

vOC-

+

150

1.5V

+

-

+

-

6V

300

100 iOC

V35.1i150v OCOC =+=

150300

5.16iOC

+

=

= 10 mA


The Thevenin resistance

Rth

150 300

100300150

)300(150100R th

++=

= 200

Finally, we attach thediode to the Theveninequivalent circuit.

200

3V

+

-

i +v

-

i

v

0.015

3

Chapter 2 Network Theorems

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