1

Chapter 3Resistive Network Analysis

Jaesung Jang

Node Voltage/Mesh Current MethodSuperposition

Thevenin and Norton Equivalent CircuitsVoltage and Current Source Transformation

Maximum Power Transfer

2

Network Analysis• The analysis of electrical network consists of determining each

of unknown branch currents (currents flowing through each branch) and node voltages (voltage at each node).

• Once the known (usually resistances and sources) and unknown (branch currents and node voltages) variables are identified, a set of equations relating these variables is constructed.

• If the number of unknown variables and equations are equal, the set of equations can be solved.

3

Node Voltage Method• Variables: voltage at each node (node voltages).• One node must be the reference point (usually ground)

for specifying the voltage at any other node.

• Solving procedure for a circuit containing n nodes– Step1: Select the reference node and define the remaining n-1 node

voltages as the independent or dependent variables. Each of m voltage sources is associated with dependent variables.

– Step2: Assume branch currents. (direction!)– Step3: Use KCL on the n-1-m nodes labeled as an independent

variables and express the branch currents in terms of node voltages.– Step4: Solve n-1-m linear equations where the number of unknowns

is n-1-m.

• The node-voltage analysis is especially useful when we have current sources.

4

Examples

0111

00

0 , nodeAt

111

0 , nodeAt

3222323

23

22121

21

=

++−→=

−−→=

−−

−→

=−

=−

+→=

−+

−→

=−+

bababbacb

SbaSbaca

S

vRR

vRR

vv

R

v

R

vv

R

vv

iib

ivR

vRR

iR

vv

R

vv

iiia

n=3, m=0, n-1-m=2

Ground node

5

Cramer’s rule

bcad

pcaq

dc

ba

qc

pa

ybcad

bqpd

dc

ba

dq

bp

x

q

p

y

x

dc

ba

−−==

−−==

=

,

AI

AI

I

I

B

A

B

A

129

2821

32

13

72

143

,529

742

32

13

37

114

7

14

32

13

−=−−=

−−

−−

=

−=−

+−=

−−

−−

=

−=

−−

This is to solve the linear equations using determi nants without matrix inversion.

6

Examples (cont.)

00

0 ,3 nodeAt

00 ,2 nodeAt

00 ,1 nodeAt

24

3

1

13241

22

12

3

2223

12

12

1

13121

=+−

−−

→=+−

=−−+→=−+

=+−

−−

−→=+−−

IR

v

R

vvIii

IR

vv

R

vIii

IR

vv

R

vvIii

00

0 , nodeAt

00 , nodeAt

4343

321321

=−−

+−

−→=−+−

=−++−−→=++−

Sccb

S

cbbbS

iR

v

R

vviiic

R

vv

R

v

R

vviiib

n=4, m=1, n-1-m=2n=4, m=0, n-1-m=3

i2

i3

i1

i4 i2

i3i1

i4

7

Mesh Current Method• Variables: current of meshes (mesh current).• A mesh is the simplest possible loop that does not contain

any other loops.

• Solving procedure for a circuit containing n meshes– Step1: Define the n mesh currents as the independent or dependent

variables. Each of m current sources is associated with dependent variables. The number of unknown variables is n-m.

– Step2: Define mesh current direction. Unknown mesh currents will always be defined in the clockwise direction except for known mesh currents (i.e. current sources) .

– Step3: Use KVL on each mesh containing unknown mesh currents and express each voltage in terms of the mesh currents.

– Step4: Solve n-m linear equations for the mesh currents.– Step5: Find the branch currents and voltages from the mesh currents.

8

Examplesn=2, m=0, n-m=2

Reference direction: pp 17 at Chap. 2.

Net current

9

Examples (cont.)

( )( )

( )( ) 32432221

212423322

24332

2122211

2212111

2211

0

0,2mesh at KVL

0

0 ,1mesh at KVL

VVRRRiRi

RiiRiVRiV

vvVvV

VVRiRRi

RiiVRiV

vVvV

−=+++−→=−++++−→=++++−

−=−+→=−+++−→

=+++−n=2, m=0, n-m=2

The voltage across the resistor 2 has different polarity depending on the meshes.

+

_

( ) currentsmesh :

2resistor at thecurrent Branch

12 ii −=

10

Examples (cont.)

( )( )

( )21

21

2211

2111

21

0

0 ,1mesh at KVL

RR

RIVi

RIVRRi

RIiRiV

vvV

SS

SS

SS

S

+−

=→

−=+→=+++−→

=++−

n=2, m=1, n-m=1

( ) ( )( )

( ) ( )( ) 1134332

1332343

134

233322

33222

32

0

0,3mesh at KVL

0

0 ,2mesh at KVL

IRRRRiRi

RIiRiiRi

vvv

IRVRiRRi

RiiRIiV

vvV

=+++−→=−+−+→

=++

+=−+→=−+−+−→

=++−

n=3, m=1, n-m=2

Mesh 2 is not considered because it has a current source (known).

+ + +

source)current (the1 Ii =+

+

+

++

+

+Three of the mesh current directions on resistors 1, 2, and 3 are different.

11

Node & Mesh Analysis w Controlled Sources

• Dependent source generates a voltage or current that depends on the value of another voltage or current in a circuit.

• Initially we treat them as ideal sources to set up the a set of equations and substitute constraint equation if you need.

• Constraint equation: an equation relating the dependent source with one of the circuit voltage or current.

( ) ( )Sb

SSCCb

Sb

SS

bS

Sb

b

CbC

b

bS

SS

bS

bSS

RR

iRRRiv

RR

iR

RR

iR

i

RivR

vi

RR

iviv

RR

R

v

R

vi

+−=−=

+=

+

=

−=→=−−

+

=→=

+→

=+−+−

ββ

ββ

2

22

11

11

&11

1

00

,2 nodeAt

11

11

00

,1 nodeAt

n=3, m=0, n-1-m=2

Constraint equation

12

Examples

( )( )

( )

( )( ) ( )

( )

( )( ) ( )

( ) ( ) 022

02

02,3mesh at KVL

033

03

02,2mesh at KVL

:Note

0

0 ,1mesh at KVL

54324212

22153423

53423

43243212

22143232

43232

221

122211

221111

111

=+++−→=−++−→

=++−

=−+++−→=−−−+→

=−−−+

−==−+→

=−++−→=++−

RRiRRiiR

RiiRiRii

vRiRii

RiRRRiiR

RiiRiiRi

vvRiiRi

Riiv

vRiRRi

RiiRiv

vRiv

n=3, m=0, n-m=3

++

+

+

( )( )

( ) ( )

=

++−−++−

−+

0

0

22

33

0 1

3

2

1

54242

42432

221 v

i

i

i

RRRRR

RRRRR

RRR

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Principle of Superposition• In a linear circuit containing more than one (voltage or current) source,

the current or voltage in any part of the circuit can be found by algebraically adding the effect of each (voltage or current) source separately.

• When we consider the effect of a single (voltage or current) source only in the circuit, the other (voltage or current) sources can be removed (zeroing) by

– shorting voltage source and opening current source.

+_ Shorting: replace the voltage source with a short circuit

Opening: replace the current source with a open circuit

14

Examples

only source voltage

only sourcecurrent

R

R

R

i

i

i

+

=

Determine the current flowing through the resistor R Current source only

Voltage source-> shortingVoltage source only Current source-> opening

B

GB

R i

RRR

Ri

++

=111

1

( )BG

G

BB

R RRR

V

RR

Ri

RR

Ri||11

1

11

1

1 +

+

=

+

=Note: iR can have a negative value depending on the polarity of a source.

i1 RG

RBi2

VG

RiR

+

15

One-Port Networks & Equivalent Circuits• The flow of energy from a source to a load can be described by showing

the connection of two “black boxes” labeled sources and load.

• It is always possible to view even a very complicated circuit in terms of much simpler equivalent source and load circuits.

16

Thevenin & Norton Theorem• Thevenin Theorem: When viewed from a load, any network composed of ideal

voltage and current sources, and of linear resistors can be represented by an equivalent circuit consisting of an ideal voltage source (VT) in series with an equivalent resistance (RT) .

• Norton Theorem: When viewed from a load, any network composed of ideal voltage and current sources, and of linear resistors can be represented by an equivalent circuit consisting of an ideal current source (IN) in parallel with an equivalent resistance (RN) .

Ideal current and voltage sources, Linear resistors, etc.

Note: RT= RN

17

Thevenin Equivalent Circuit• Determining Thevenin Resistance and Voltage

– Step1: Disconnect RL to find the voltage between open terminals A and B (->VT).

– Step2: Zero all the independent voltage and current sources (short the voltage source and open the current source) and calculating the circuit’s total resistance as seen from open terminals A and B. (-> RT). -> Thevenin equivalent circuit

– Step3: Reconnect RL with terminals A and B and find VL or IL.

network

A

B

+

-

RT

VT

A

B

+

-

network

A

B

+

-

Load RL

18

Example

Application of Thevenin’s theorem. (a) Actual circuit with terminals A and B across RL. (b) Disconnect RL to find that VAB is 24V (= VT). (c) Short-circuit V to find that RAB is 2Ω. (d) Thevenin equivalent circuit. (e) Reconnect RL at terminals A and B to find that VL is 12V.

Example

Procedure

Find VL

19

Example (Cont.)

Thevenizing a circuit with two voltage sources V1 and V2. (a) Original circuit with terminals A and B across the middle resistor R3. (b) Disconnect R3 to find that VAB is −33.6V (= VT). (c) Short-circuit V1

and V2 to find that RAB is 2.4 Ω(= RT). (d) Thevenin equivalent with RL reconnected to terminals A and B.

iL

20

Norton Equivalent Circuit• Determining Norton Current and Resistance

– Step1: Disconnect RL, short the terminal A and B, and find the currentbetween terminals A and B (-> IN).

– Step2: Zero all independent voltage and current sources (short the voltage source and open the current source) and calculating the circuit’s total resistance as seen from open terminals A and B. (-> RN=RTH). -> Norton equivalent circuit

– Step3: Reconnect RL with terminals A and B and find VL or IL.

network

A

B

+

-

RNIN

A

B

+

-

network

A

B

+

-

Load RL

21

Example

Same circuit as in Fig. 10-3, but solved by Norton’s theorem. (a) Original circuit. (b) Disconnect A and B and short circuit across terminals A and B. (c) The short-circuit current IN is 36/3 = 12 A. (d) Open terminals A and B but short-circuit V to find RAB is 2 Ω, the same as RN. (e) Norton equivalent circuit. (f) RL reconnected to terminals A and B to find that IL is 6A.

iL

22

Example (Thevenin & Norton Resistance)

Same circuit as in Fig. 10-3, but solved by Norton’s theorem. (a) Original circuit. (b) Disconnect A and B and short circuit across terminals A and B. (c) The short-circuit current IN is 36/3 = 12 A. (d) Open terminals A and B but short-circuit V to find RAB is 2 Ω, the same as RN. (e) Norton equivalent circuit. (f) RL reconnected to terminals A and B to find that IL is 6A.

( )[ ] 54321 |||| RRRRR

RR Tab

++==

( )[ ] 4321 |||| RRRR

RR Tab

+==

23

Source Transformations• Thevenin’s theorem says that any one-port network can be represented

by a voltage source and series resistance.• Norton’s theorem says that the same one-port network can be

represented by a current source and shunt (parallel) resistance.• Therefore, it is possible to convert directly from a Thevenin form to a

Norton form and vice versa.• Thevenin-Norton conversions are often useful.

RT=RN

VT=IN x RN

24

Example

I1 I2 I1+ I2

25

Maximum Power Transfer Theorem• How much power can be transferred to the load

from the source under the most ideal condition? -> Answer: Maximum power transfer theorem.

( )( ) ( )[ ]

( )( )( )

( )42

4

22

2

22

2

LT

LTLTT

LT

LTLLTT

L

L

LT

LTLLL

RR

RRRRv

RR

RRRRRv

R

P

RR

RvRiP

+−+=

++−+=

∂∂

+==

• The power curve peaks where RL = RT. At this point, the source transfers maximum power to the load.

• 50% of the power generated is dissipated in the source resistor and the other 50% is in the load resistance.