d EG1108 Lecture Slides for Week 4 - Node Analysis to Maximum Power

EG1108 Electrical Engineering / Dr. Abdullah Al Mamun 1

Node Voltage Analysis

Systematic approach using KCL onlyStep 1: Define Unknowns# Choose a reference node (Ground)# All other nodes have voltage w.r.t.the reference node. Let these voltages beVa , Vb , …. (These are unknowns)# N-1 unknowns for a ckt with N nodes

Va Vb

Step 2:# Arbitrarily assign direction of current in all resistive branches# We can write current magnitude in terms of node voltages

I1

I2

I3

Step 3:# Write KCL equation for each of the N-1 nodes

Step 4:# Solve N-1 simultaneous equations to find N-1 unknowns# Once node voltages are known, you can find currents



Systematic approach using KCL only

21 RVV

RVI baa

a−

−−

33

22

111 ,,0

RVI

RVVI

RV

RVI bbaaa =

−==

−=

23 RVV

RVI bab

b−

+−

aba IVR

VRR

=−⎟⎟⎠

⎞⎜⎜⎝

⎛+⇒

221

111

bba IVRR

VR

=⎟⎟⎠

⎞⎜⎜⎝

⎛++−⇒

322

111

KCL to Node a:

KCL to Node b:

Equation at node 1: Gaa Va – Gab Vb = Ia

Gaa : Sum of conductance connected to node aGab : Conductance connected between node a and node bIa : Known current entering the node a



Generalizing the KCL equations

NNNNcNcbNbaNa

cNcNcccbcbaca

bNbNcbcbbbaba

aNaNcacbabaaa

IVGVGVGVG

IVGVGVGVGIVGVGVGVG

IVGVGVGVG

=+−−−−

=−−+−−=−−−+−

=−−−−

..........

..........

.....

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−−−−−−−

N

c

b

a

N

c

b

a

NNNcNbNa

cNcccbca

bNbcbbba

aNacabaa

I

III

V

VVV

GGGG

GGGGGGGGGGGG

.........

........................

Gxx : Sum of conductance connected to node xGxy = Gyx : Conductance connected between node x and node yIx : Sum of current entering node x

Using Matrix Notation:



Example: Matrix directly from Ckt

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡−

−23

2.12.02.07.0

b

a

VV

siemensRR

G

siemensR

GG

siemensRR

G

bb

baab

aa

2.10.12.011

2.01

7.02.05.011

32

2

21

=+=+=

===

=+=+=

22.12.032.07.0=+−

=−

ba

ba

VVVV

voltVvoltV

b

a

5.25

==

AIAIAI 5.215.2,5.0

55.25,5.2

25

321 ===−

===

R1 = 2Ω, R2 = 5Ω, R3 = 1ΩIa = 3 A, Ib = 2 A



Problem with voltage sourceCurrent through a voltage source depends on the circuit connected to it. Can not write KCL equation if a branch contains a voltage source

Method 1: If the voltage source has a series R,use source conversion

Alternatively, You may choose negative end of voltage source as ground.Then Vb =V is known. Current through R3 is (Va -V)/R3



Problem with voltage source

114)(3)(04)(833)(

3121

3121

−=×−+×−=×−−−−×−−

VVVVVVVV

4 S

3 S

1 S 5 S8 A 25 A

3 A

V1V2 V3

2 V

Current through 2V source is not known.KCL can not be written for nodes 2 & 3.

You can include node 2 & node 3 into aSupernode and write KCL

2254)(3)(0253514)(3)(

323121

323121

=−−×−+×−=−+×−×−×−+×−

VVVVVVVVVVVV

223 =−VV

KCL at Node 1:

KCL at Supernode:

We have 2 KCL equations for 3 unknowns. Third equation is:


Mesh Analysis

Systematic approach using KVL only R1 = 2 R3= 4

V1 = 6 V

V2 = 4 V

V3 = 2 V

a

b c

d

e

f

There are 3 loops in this circuit, but 2 mesh.‘bcefdab’ is a loop but not a mesh

Mesh is a closed loop that doesn’t include any other loop

Mesh →

Window

Step 1: Define Meshes and Unknowns# For each mesh, imagine a current circulating around it# Mesh currents are unknowns to be found (I1 , I2 , ….)# N unknowns for a ckt with N meshes

Step 2:# Assign voltage polarity according to direction of mesh currents# For a branch common between 2 meshes, you have do it twice

I1 I2

++

+

+


Mesh Analysis

Systematic approach using KVL only

( )46)(

046

22121

21211

−=−+=−−−−

IRIRRIIRIR

R1 = 2 R3= 4

V1 = 6 V

V2 = 4 V

V3 = 2 V

a

b c

d

e

f

For branches common between 2 meshes, Branch current is sum of mesh currents

Step 3:# Write KVL equation for each mesh# N equations for N meshes# Solve N equations to find N unknowns

I1 I2

++

+

+

( )24)(

024

23212

23122

+=++−=+−−−

IRRIRIRIIRMesh 1: Mesh 2:

Since KVL is used, it is better if all sources are voltage sourceIf there is a current source, use source transformation if possible.


Mesh Analysis

Systematic approach using KVL only

R1 = 2 R3= 4

6 V

4 V

2 V

a

b c

d

e

f

I1 I2

+

+

+--

-

+ -( ) 4622121 −=−+ IRIRRMesh 1 Mesh 2

( ) 2423212 +=++− IRRIR

Sum of all resistors in mesh i

Resistor common between mesh i and mesh j

Sum of all voltages in mesh i

We can generalize this for any circuit with N meshes


Mesh Analysis

If there are N mesh, you get N equations:

Using Matrix notation:

NNNNNNN

NN

NN

NN

VIRIRIRIR

VIRIRIRIRVIRIRIRIR

VIRIRIRIR

=+−−−−

=−−+−−=−−−+−

=−−−−

..........

.....

..........

332211

33333232131

22323222121

11313212111

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−

−−−−−−−−−

NNNNNNN

N

N

N

V

VVV

I

III

RRRR

RRRRRRRRRRRR

.........

........................

3

2

1

3

2

1

321

3333231

2232221

1131211

Rii : Sum of all resistors in mesh IRij = Rji : Resistor common between mesh I and mesh jVi : Sum of all known voltages in mesh i


Superposition Theorem

Superposition is a consequence of linearityThis theorem is applicable to linear circuits only

IRVV

RR=−⎥

⎦

⎤⎢⎣

⎡+

2

11

21

11

V I

R1

R2

V1

We want to determine V1[V, I, R1 and R2 are known]

V1 = aV + bI

IRVV

RRRR

+=+

11

21

21

IRR

RRVRR

RV21

21

21

21 +

++

=

Both coefficients (a & b) are constant for given values of R1 and R2

This is a linear circuit.Any circuit consisting of sources, R, L and C is linear



Linearity ⇒ Homogeneity & SuperpositionHomogeneity

Superposition

….

x1x2

xn

y ⇒ ….ax2

axn

ay

ax1

….

x1ax2a

xna

ya….x2b

xnb

yb

x1b

….x2a +x2b

xna +xnb

ya + yb

x1a +x1b

⇓



Specific Example of Superposition

V10 y1 V2

y2

0

0 + V2

y1 + y2

V1 + 0

⇓

We can consider both voltage source and current source as input to alinear systemVariables at the output (yi ) can be either current through a branch or voltage at a node



Any electrical quantity (voltage at a node or current through a branch) in a linear circuitwith multiple sources can be determined by summing that quantity obtained as response to each source acting alone

Method:Deactivate (nullify the effect of) all sources except one to find the desired voltage/current due to the source leftRepeat this for all sourcesThe desired quantity is sum of all voltages /currents found in previous steps



Deactivating/Nullifying the effect of a source

V = 0

I

V

+

-

I=0

⇒⇒

⇒⇒

Rest of the circuit is connected between these nodes. We want to

deactivate this source only.

Deactivating a voltage source⇒

replace it with a short circuit

Deactivating a current source⇒

replace it with an open circuit



Example:

Determine current IL .

Keep voltage source &deactivate current source:

.5.02416

201, AVIL =

Ω+Ω=

.2.11624

242, AIL −=⎟⎟

⎠

⎞⎜⎜⎝

⎛Ω+Ω

Ω−=

Keep current source &deactivate voltage source:

Superposition of two solutions:IL = IL,1 + IL,2

= (0.5 A – 1.2 A) = -0.7 A

Negative sign indicates that current throughRL is opposite to the assumed direction



A dependent source can not be nullifiedWhat is a dependent source?

There exists another category of sources whose output (voltage or current) is a function of some other voltage or current in the circuit. Such sources are called dependent sources.

A diamond shaped symbol is used to distinguish them from independent sources Voltage controlled voltage source: VS = kVX

Voltage controlled current source: IS = kVX

Current controlled voltage source: VS = kIy

Current controlled current source: IS = kIy

VX is voltage at node x in the circuit, Iy is current through branch y



Dependent sourceTransistor, amplifier etc can be modeled using dependent source

If VS is nullified, there will be no source in the loop on left ⇒

vd = 0

If dependent source is nullified, i.e., Avd = 0 ⇒

vd = 0⇓

Vs = 0

Nullifying a dependent source not only make that source 0 but also setsome other voltage/current in the circuit to 0

There will be no current through the speaker



Linearity ⇒ Superposition: You can use it to find voltage and current but not to find power

( ) WP ARx 3224 28, =×=

Examination: Semester I, 2008-09Find current through and power dissipated in RX

AAI ARx 488

8)8(8, =Ω+Ω

Ω×=

AAI VRx 83

2686

6, −=++

−=

AAI ARx 43

6106)2(2, −=

Ω+ΩΩ

×−=

AAIRx 823

43

834 =⎟

⎠⎞

⎜⎝⎛ −−=

WPRx 53.162823 2

=×⎟⎠⎞

⎜⎝⎛=

WP VRx 3292

83 2

6, =×⎟⎠⎞

⎜⎝⎛−=

WP ARx 892

43 2

2, =×⎟⎠⎞

⎜⎝⎛−=

ARxVRxARxRx PPPP 2,6,8, ++≠


Thevenin’s & Norton’s Theorem

0=−+ cbIaV

I

V+-

Problem: Find the simplest equivalent circuit model for A, such that the externalcircuit B would not feel any difference if A is replaced by that equivalent circuit. The solution is contained in two theorems – Thevenin’s & Norton’s

If we write down the Ohm’s law, KVL & KCL equations, we will have a number of equations with same number of unknowns.

If the equations are linear, we can get the following equation after appropriate eliminations and substitutions.

NN

TT

IRV

bcV

baIb

VIRacI

abVa

+−=+−

=⇒≠

+−=+−

=⇒≠

0

0 →

Thevenin’s

→

Norton’s



Circuit B(Externalapparatusor circuit)

Simplest modelfor circuit A

VT

RT I

V+

-

Circuit B(Externalapparatusor circuit)

Simplest modelfor circuit A

IN RN

I

V+

-

V = VT - IRTConsistent with equation on previous slide for a≠0Thevenin’s equivalent Model:

Voltage source plus a series resistor

I = IN – (V/RN )Consistent with equation on previous slide for b≠0Norton’s equivalent Model:

Current source plus a parallel resistor

These models are true even for an apparatus instead of circuit B.We can connect a voltmeter (to measure open loop voltage) or an ammeter(to measure short circuit current)

(Leon Charles Thevenin, French telegraph engineer)

(Edward Lawry Norton, Bell Labs & Hans Ferdinand Mayer, Siemens)



A linear 2-terminal circuit with sources and resistors can be replaced by an equivalent circuit consists of a single voltage source (VT) and a series resistor (RT).

A linear 2-terminal circuit with sources and resistors can be replaced by an equivalent circuit consisting of a single current source (IN) and a parallel resistor (RN).

VT

RT

RNIN

Thevenin’s Equivalent

Norton’s E

quivalent


Thevenin’s Theorem

Open circuit condition

Short circuit condition

Circuit withSources andResistances VT = VOC

T

TSC R

VI =

SC

OC

S

TT

T

TSC I

VIVR

RVI ==⇒=

Circuit withSources andResistances


Norton’s Theorem

Open circuit condition

Short circuit condition

IN RN

a

b

Circuit withSources andResistances VOC = IN RN

NSC II =

SC

OC

N

OCNNNOC I

VI

VRRIV ==⇒=

Circuit withSources andResistances

IN RN

a

b

ISC



Steps to determining the equivalent circuit:

Remove R from the branch you are interested in

Find open circuit voltage (VOC)

Replace R by a short circuitFind current through short circuit branch (ISC)

There are 3 methods for determining RT or RN.

Thevenin Equivalent:(1) Find VOC(2) Find RT

Norton Equivalent:(1) Find ISC(2) Find RN



Finding RT or RN:

Circuit contains only independent sources: all 3 methods Both independent & dependent sources: method 2 or 3Circuit with dependent sources only: method 3

Method 1: short circuit all voltage sources and open circuit all current sources (just like superposition).

Equivalent resistance of the remaining network is equal to RT or RN

Method 2: RT (or RN) = (VOC)/(ISC)



Finding RT or RN:Method 3: using test source

Deactivate independent sources but not dependent sources.Apply a test voltage (or current) at the open terminals and determine test current (or voltage). RT(or RN) = (VTest)/(ITest)

This method is suitable for any circuitThis method is useful if a circuit contains dependent sources only



Example 1: (independent sources only)

5.0520001050

12

21

+==+

IIII

Find VOC :VOC is voltage across 2 kΩ

Find ISC :ISC = I3 + 0.5 A.

I1

ISC

AVISC 5047.05.010505

=+=

VIVAIAI

OC 54.34720001737.03262.0

2

2

1

===−=

VOC

I3

I2



Example 1: (independent sources only)

Ω=×

=

=

5246.6883050

20001050)2000(||)1050()( NT RorR

Find RT or RN :

Method 2: VOC /ISC

( ) Ω=== 61.6885047.0

54.347

SC

OCNT I

VRorR

Method 1: Deactivate sources

Thevenin’sEquivalent

Norton’sEquivalent



Example 2: (Both dependent & independent sources)

b

b

IIII

101520001050

2

2

==+


Find ISC :ISC = 101Ib

ISC

AISC 48.01050

5101 =⎟⎠⎞

⎜⎝⎛=

( ) VIVmAI

bOC

b

05.51012000025.0

===

VOC

I2

Find RT (or RN ):

( ) Ω=== 5.1048.005.5

SC

OCNT I

VRorR



Example 3: (Using test voltage/current)

b

b

IIII

101520001050

2

2

==+


Find ISC :ISC = 101Ib

ISC

AISC 48.01050

5101 =⎟⎠⎞

⎜⎝⎛=

( ) VIVmAI

bOC

b

05.51012000025.0

===

VOC

I2

VT or IN is determined using similar steps as shown in the first 2 methods.Method of finding RT or RN using test voltage is shown in the next slide



Example 3: (Using test voltage/current)

( ) Ω===

=

+×

−=

34.1038.193

2000200038.193

20001050

1012000

Test

TestNT

TestTest

TestTestTest

IVRorR

IV

IVV( )Testb

TestbTest

IIIIV20001012000

1012000+×=+×=

( )

1050

501000

Testb

Testb

VI

VI

−=

−=+

I2

I2

Independent source must bedeactivated while using this methodfor finding RT (or RN )


Maximum Power Transfer Theorem

Defines the condition under which power transferred to a load is maximized

Consider a practical voltage source (VS and RS are known) providing power to a load of resistance RL

LS

SL RR

VI+

=

( )22

2

LS

LSLLL RR

RVRIP+

==VS

RS

RL

ILPL is always positive RL = 0 ⇒ PL = 0RL = ∞ ⇒ PL = 0

Value of RL that maximizes PL must satisfy

( )00 2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+⇒=

LS

LS

LL

L

RRRV

dRd

dRdP

RL

PL



Differentiating PL with respect to RL

( ) ( )

SL

LLS

LLSLSL

L

RR

RRR

RRRRRdRdP

=⇓

=−+⇓

=+−+⇒=

02

020 2

( )22

2

LS

LSLLL RR

RVRIP+

==

For a given source (VS & RS ), maximum power is transferred to load if RL = RS

( )( ) ( )[ ]

( )422

2

2 2

LS

LLSLSS

LS

LS

L RRRRRRRV

RRRV

dRd

++−+

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+

This condition is valid when VS & RS are known and you can choose RL

For a given RL , power transfer is maximum when RS =0



Any electrical system can be represented by a single source plus resistor

You can apply Maximum Power Transfer theorem for proper design

Abstraction of SpeakerResistance (RL )

Thevenin’sEquivalent

Documents

d EG1108 Lecture Slides for Week 4 - Node Analysis to Maximum Power