Chapter 1tl

8/8/2019 Chapter 1tl

1/98

Circuit Theorems

Chapter 1 :

Resistive circuit analysis techniques


2/98

Nodal analysis (revision)

Based on Kirchhoff Current Law (KCL).

Every point at the junction can be used as a

node. One of the nodes will be acting as a

reference node where this reference node

voltage will always equal to zero, V = 0.


3/98

Nodal analysis for

circuit with currentsource


4/98

Example

R1 R3

R2

IB2IB1


5/98

a) Label all the nodes including reference node

and all the currents flow in the circuit.

R2

R3R1

I1 I2 I5

I3 I4

IB2IB1

a b

c


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b) Write down KCL equation at node a and b.

KCL at node a: I1= I2 +I3 (1)

12

1

R

V

R

VVIaba

B

!


7/98

KCL at node b: I2 + I5 = I4 (2)

Equations (1) and (2) can be solved by usingsimultaneous equations or Cramer rule to

obtain the value of Va and Vb

32

2

R

VI

R

VV bba!


8/98

Nodal analysis for

circuit with current andvoltage source


9/98

Example

I1 I2

I3

I5

I4

R1 R3

R2

IBVB

a b

c


10/98

a) Nodal analysis equation for the above circuit.

Node a: Va = VB (1)

Node b: (2)32 R

VIR

VV bbba

!


11/98

a) Substitute equation (1) into equation (2).

IB = -VB(1/R2) + VB((1/R2) + (1/R3)) (3)

b) From equation (3), the value of Vb can be

obtained.

))/1()/1((

))/1((

22

2

RR

RVIV

BB

b

!


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Supernode


13/98

Occur when the existence of voltage

source between two nonreference nodesand any elements connected parallel with

it.

The first step is to write a KCL equation

for the nodes that being unite and thisequation is known as supernode equation.

Write down the equation that linked the

voltage for the unite node with the voltagesource. This equation is known as

supporting equation.


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supernode

I1

I2 I3

I4

IB2IB1

a

c

bVB


16/98

Node a and b.

a) Supporting equation,

Va Vb = VB

b) KCL equation at supernode,

Ientering supernode = Iout of supernode

I1 +I4 = I2 +I3

IB1 +IB2 = Va /R1 + Vb /R2


17/98

Nodal analysis for the

circuit with thedependent source


18/98

For the circuit that exist dependent

sources, the nodal analysis can be doneby the same method. The solution for the

circuit that deal with this dependent

sources are shown as in example.


19/98

Example

2 6

3

9V16V

Ia

6Ia


20/98

a) Label all the nodes (including reference node)

and the currents of each branch.

2 6

3

9V16V

Ia

6Ia

I1

I2

I3

b

a

Vb = 0


21/98

b) Write down KCL equation at node a.

((16 Va) /2) + ((6Ia - Va) /6) = ((Va +9) /3)

6Va 6Ia = 30 (1)

Where Ia = I1 = ((16 - Va) /2) (2)


22/98

c) Solve the equations (1) and (2).

Va = 8.67 V

Ia = ((16 - Va) /2) = ((16 8.67) /2) = 3.67 A


23/98

MESH

ANALYSIS(revision)


24/98

Step to Determine Mesh Currents :

1. Assign mesh currents i1, i2, i3.in to the

n meshes

2. Apply KVL to each of the n meshes. Use

Ohms law to express the voltages in

terms of the mesh currents.

3. Solve the resulting simultaneous

equations to get the mesh currents.


25/98

Mesh Analysis for circuit with voltage

source

Figure 1 : Circuit with 2 mesh and 1 voltage source

1

1

2


26/98

Solutions :-

a) Assign mesh currents i1 and i2 in to the 2 meshes

1

1

2

I1

I1 I2


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22211

22111

21

)(

)(

RIRRI

RIIRI

VVVB

!

!

!

b) Apply KVL to each of the n meshes. Use Ohms law

to express the voltages in terms of the mesh

currents.

mesh 1 :

mesh 2 :

0)(

0)(

0

32221

32221

34

!

!

!

RRIRI

RIRII

VV

c) Solve the resulting simultaneous equations to get the mesh

currents.


28/98

Figure 2 : Circuit with 3 mesh and 2 voltage source

R5

R4 2V

R1 R3

R2


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R5

R4 V2V1

R1 R3

R2

mesh 1 :

mesh 2 :

mesh 3 :

122211

122111

)(

)(

VRIRRI

VRIIRI

!

!

0)(

0)()(

43432221

43232212

!

!

RIRRRIRI

RIIRIRII

254342

253423

)(

)(

VRRIRI

VRIRII

!

!

c) Solve the resulting simultaneous equations to get the mesh

currents.

I1 I2 I3


30/98

Mesh Analysis for circuit with

Voltage and Current Source

Applying mesh analysis to circuits

containing current sources ( dependent or

independent ) , the presence of the

current sources reduces the numberof

equations.


31/98

Example :

Find current that flow in 3 resistor for the circuit below :

1 V 1

5

3

2


32/98

Solutions :

a) Assign mesh currents i1 and i2 in to the 2 meshes

18V 1A

5

3

2

I1 I2


33/98

b) Apply KVL to each of the n meshes. Use Ohms law to express the

voltages in terms of the mesh currents.

Mesh 1 :

Mesh 2 : amp

c) Solve the resulting simultaneous equations to get the mesh currents I1

I1 = 1.875 amp ; I3 = (I1-I2)= 2.875 amp

1

1838

2

21

!

!

I

II


34/98

Supermesh

A supermesh results when two meshes have a dependent or

independent current source between them.

We have to treat it as supermesh because mesh analysis

applies KVL, which wedo

no

t kno

w the vo

ltage acro

ss acurrent source in advance.


35/98

Figure below shows a current source ( IB ) exists between mesh 1

and mesh 2, therefore supermesh analysis is needed to solve the

circuit..

Mesh 1 & 2 :

Supermesh equation : R1I1 +R3I2 = VB1 + VB2

Supporting equation : I1 I2 = IB

R2

R1 R3

VB1 VB2

IB

I1 I2


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38/98

b) Write KVL equations for each mesh.

Mesh 1 : I1 = 2A (1)

Mesh 2 and 3 :

Supporting equation : ( I3 I2) = 5A (2)

Supermesh equation : -5I1 +4I2 +4I3 = 38 (3)

c) Solve equation (1), (2) dan (3) to get the value ofI2 and I3

I2 = 3.5 A ; I3 = 8.5 A

d) From circuit above :

V3 = 4 ( I2 I1 ) = 6V

Power absorb by the 4 resistor, P4 = = 9 W

@

@

4

2

4;V@


39/98

Mesh Analysis for circuit with

dependent source

To analyze circuit that have dependent source, mesh analysis

can be done as same as before.

Below is an example for circuit with dependent source

Example :

For circuit in figure below, get the value ofIa


40/98

a) Assign mesh currents I1 and I2 in to the 2 meshes

b) Write KVL equations for each mesh

Mesh 1 : 5I1 3I2 = 25 (1)

Mesh 2 : -3I1 +9I2 = -9 6Ia (2)

where Ia = I1 (3)

Substitude equation (3) into equation (2)

3I1 +9I2 = -9 (4)@

I1 I2

16V 6Ia

3

2 6

9V

Ia


41/98

c) From equation (1) and (4), using Cramers Rule to get the value ofI1.

I1 = 3.67 A Ia = I1 = 3.67 A

@ @

(End of revision)


42/98

SUPERPOSITION

THEOREM

The superposition principle statesthat the voltage across (or

current through) an element in a

linear circuit is the algebraic sumof the voltage across (or currentsthrough) that element due to eachindependent source acting alone.


43/98

Steps to Apply SuperpositionPrinciple :

Turn off all independent source exceptone source. Find the output (voltage orcurrent) due to that active source.

Repeat step 1 for each of the otherindependent source.

Find the total contribution by adding

algebraically all the contributions due tothe independent source.


44/98

Example 1:

Find the value of V.

Figure 1.1

Let, V = V1 + V2


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46/98

Current Division : i3 = 8 (3) = 2 A4 + 8

V2 = 4i3 = 8 V

The Value of V :

V = V1 + V2 = 2 + 8 = 10 V

Figure 1.1 (b)

Find V2 :


47/98

Example 2:

Find the value of io.

Figure 1.2

Let, i 0 = i0 + i0


48/98

Solution:

Figure 1.2 (a)

i1 = 4 A ....(1)

-3i1 +6 i2 - 1i3 - 5i0 = 0 ....(2)

-5 i1 - 1i2 + 10 i3 + 5i0 = 0 ......................................( 3)

At node 0 : i3= i1 - i0 = 4 - i0 . .(4)

Substitute (1) & (4) into (2) & (3) :3 i2 - 2i0 = 8 .(5)

i2 +5i0 = 20 ....(6)

i0 = 52 A ...(7)17


49/98

Figure 1.2 (b)

6i4 - i5 - 5i0 = 0..(8)

Loop 5 : - i4 + 10 i5 - 20 + 5i0 = 0 ... (9)

But , i5= - i0 substitute into (8) & (9) :

6i4 - i0 = 0 .(10)

i4 +5i0 = 0 ...(11)

i0 = - 60 A .....(12)

17The value ofi0 : i0= - 8 = - 0.4706 A

17


50/98

Example 3:

Figure 1.3

Find the value of i.

Let, i = i1 + i2 + i3


51/98

Solution:

i1 = 12 = 2 A6

Figure 1.3 (a)


52/98

Figure 1.3 (b)

RT = 8+ 4 + (12/7)= 13.714

I= 24/13.714 = 1.75A

i2 = -(4/7)(1.75) = -1 A

DC

I


53/98

Figure 1.3

(c)

I= (8/(5.714+8))x3 = 1.75 A

i3 = (4/7)x1.75 = 1 A

The value ofi: i = i1 + i2 + i3 = 2 - 1 + 1 = 2 A

I


54/98

THEVENINS

THEOREM

Thevenins theorem states that alinear two terminal circuit can bereplaced by an equivalent circuit

consisting of a voltage source VTh inseries with a resistor RTh ,where

VTh is the open circuit voltage atterminals a & b and RTh is the inputor equivalent resistance at theterminals when the independent

sources are turned off.


55/98


56/98

This complex circuit can be replacedby the simpler circuit known as

Thevenin equivalent circuit.


57/98

Thevenin Equivalent circuit


58/98

Assignment 1

Question

Prove that any complex linear electriccircuit can be replaced by itsThevenin

equivalent circuit consisting of a voltage

source in series with a resistor.

The deadline: 20th February 2008


59/98

Finding VTh and RTh

15 Volt

3 Ohm a

b

RL= 6 Ohm

I

Vab


60/98

Finding VTh and RTh

15 Volt

3 Ohm a

b

RL= 6 Ohm

I

Vab


61/98


62/98

Two cases when finding theThevenin resistance

Case 1 : Circuit with independent sources only.

RTh is the equivalent resistance of the

network looking between terminals a and bby turned off all the independent sources.

Voltage source Short circuit

Current source Open circuit


63/98

Case 2 :Circuit with independent and dependentsources.

Turn off all the independent sources asmentioned above but the dependent

sources are cannot be turned off. Apply the voltage source, V0 at terminals

a & b and determine the resultingcurrent i0. Then RTh = V0

i0

Fi di R h i i h


64/98

Finding RTh when circuit hasdependent source

Figure 1.6 (a) Figure 1.6 (b)

RTh = Voio


65/98

A circuit with a load

IL = VTh(RTh +RL)

VL = RL IL

= RL x VTh

(R

Th+

R

L)

Figure 1.7 (a) : Original circuit

Figure 1.7 (b) : Thevenin equivalent


66/98

Example 1:

Find the Thevenin equivalent circuit of the circuit shown in Fig.1.30

to the left of the terminals a-b. Then find the current through

RL = 6, 16 and 36.

Figure 1.8


67/98


68/98

Finding VTh :

Figure1.8 (b)

Loop i1 : - 32 + 4 i1 + 12 ( i1 - i2 ) = 0 .(1)

Loop i2 : i2 = - 2 A ...(2)

Substitute (2) into (1) :

i1 = 0.5 A

VTh = 12 ( i1 - i2) = 12 ( 0.5 + 2.0 ) = 30 V


69/98

IL = VTh = 30

RTh +RL 4+RL

When RL = 6, IL = 30 = 3A

10

When RL = 16, IL = 30 = 1.5A

20

When RL = 36, IL = 30 = 0.75A

40

Figure 1.8 (c)


70/98

Example 2 :

Find the Thevenin equivalent of the circuit.

Figure 1.9


71/98

Figure 1.9 (a)

Solution:

To find RTh, we set the independent sources equal to zero but the

dependent sources remains in the circuit. We may set Vo = 1V to ease

calculation.


72/98

Loop i1 : - 2 Vx + 2 ( i1 i2 ) = 0 or Vx = i1 i2

But, - 4i2 = Vx = i1 - i2

Hence, i1 = - 3i2

Loop i2& i3 : KVL : 4 i2 + 2 ( i2 - i1 ) + 6 ( i2 i3 ) = 0

6 ( i3 i2) + 2 i3 + 1 = 0

i3 = -1 A6

But, io = - i3 = 1 A6

Hence, RTh = 1V = 6io


73/98

Finding VTh :

Figure 1.9 (b)

Loop i1 : i1 = 5A

Loop i3 : - 2 Vx + 2 ( i3 i2 ) = 0

Vx = i3 i2

Loop i2 : 4 ( i2 i1) +2 ( i2 i3 ) +6 i2 = 0

Or 12i2 4i1 2i3 = 0


74/98

But, 4 ( i1 i2 ) = Vx

i2 = 10 A3

Hence, VTh = Voc = 6 i2 = 20V

Figure 1.9 (c) : Thevenin equivalent circuit


75/98

Example 3 :

Determine the Thevenin equivalent of the circuit.

Figure 1.10

Solution:

Since there is no dependent source in the circuit,

Vab = VTh = 0 V


76/98

Substitute (2) into (1) :

i0 = ix + V0 = - V0 + V0 = - V0

4 2 4 4

Or , V0 = - 4i0

Thus , RTh = V0 = - 4

i0

Nodal Analysis : i0 + ix = 2ix + V0 (1)4

But , ix = 0 - V0 = - V0 ........(2)2 2


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NORTONSTHEOREM

Nortons theorem states that alinear two terminal circuit can bereplaced by an equivalent circuit

consisting of a current source IN inparallel with a resistor RN where IN

is the short circuit current throughthe terminals and RN is the input orequivalent resistance at the

terminals when the independentsources are turned off.


78/98

Norton equivalent circuit

Figure 1.11 (a) : Original circuit Figure 1.11 (b) : The Norton equivalent circuit


79/98

To find the Norton current, IN, We determine the short-circuit-

current flowing from terminal a to b in both circuit.


80/98

Example 1:

Find the Norton equivalent circuit of the circuit

Figure 1.12

Solution:


81/98

RTh = RN

Solution:

RN = 5 // ( 8 +4+ 8 )= 5//20

= 20 5 = 4

25

Figure 1.12 (a)


82/98

Short circuit terminal a & b

Ignore 5 resistor

i1 = 2A

20 i2 4i1 12 = 0

i2 = 1A = isc = IN

Figure 1.12 (b)

Finding IN :


83/98


84/98

Example 2 :

Using the Nortons theorem,find RN and IN at terminals a-b

Figure 1.13


85/98

Fi di I


86/98

Short circuit terminals a & b and find the current isc

Note the 4 resistor, the 10V voltage source, the 5

resistor, and the dependent current source are all in parallel

Hence,

ix = 10 = 2.5 A

4 At node a, KCL : isc = 10 + 2ix

5

= 2 + 2 (2.5) = 7A

Thus, isc = IN = 7A

Figure 1.13 (b)

Finding IN :


87/98

MAXIMUM POWERTRANSFER

To determine the value of loadresistance which the maximum

power is transferred to the

load.


88/98

Maximum power transfer

RTh

VTh

P = I RL = ( VTh ) RL( RTh +RL )

Figure 1.14

I = ( VTh )

( RTh +RL )


89/98


RL

P



90/98


RL

P


The slope = dP

dRL

dRLdP


91/98


RL

P


The slope = dP

dRL

dRLdP = 0


92/98


dP = (RTH +RL)2.VTH 2 2VTH

2RL(RTH +RL)


The slope = dP

dRL= 0

dRL (RTH +RL)4 = 0

or

(RTH +RL)2.VTH 2 2VTH

2RL(RTH +RL) = 0


93/98


(RTH +RL)2.VTH 2 2VTH

2RL(RTH +RL) = 0

(RTH +RL)2.VTH 2 = 2VTH

2RL(RTH +RL)

(RTH +RL) = 2RL

RL-2RL = -2RTH

RL = RTH

When RL = RTH , RL absorbing the maximum power


94/98


RTh

VTh


Figure 1.14

For RL = RTh

P max = VTh

4RTh

I = ( VTh )

( RTh +RL )

E l 1


95/98

Example 1:

Find the value ofRL for maximum power transfer in the circuit. Find the

maximum power.

6 3 2

12 RL12V 2 A

Figure 1.15

S l ti


96/98

Solution:

Finding RTh :6 3 2

12

Figure 1.15 (a)

RTh = 2 + 3 + 6 //12

= 5 + 6 12

18= 9

RTh


97/98

Finding VTh :

Figure 1.15 (b)

Mesh analysis : - 12 +18 i1 12 i2 = 0,

i2= -2A

Get, i1 = - 2 A3

-12 + 6 i1 +3 i2 + 2 (0) + VTh = 0VTh = 22 V

RL = RTh = 9

Pmax = VTh = 22 = 13.44 watt

4RL (49)


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Chapter 1tl