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Method of (Circuit)

Analysis

Chapter 8

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If two or more sources are present

in the circuit, what will you do?

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Methods for two or more souces

Branch-current Analysis

Mesh analysis and

Nodal analysis

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All the methods will be described

for linear bilateral networks only Liner indicates that the characteristics of

the of the network elements are

independent of voltage across or currentthrough them.

e. g. Resistor

A 10k resistor always shows 10kwhether the voltage across it is 10V or100V

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So resistor is a linear element

but diode is not

Bilateral: it refers to the fact that

there is no change in the behaviour

or characteristics of an element if

the current through or voltageacross the element is reversed

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II

I I

Bilateral ????

Bilateral ????

R R

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Current Source

It is the dual of the voltage source.

For a fixed dc voltage source, voltageacross its

terminal is constant but current may vary

For a fixed dc current source, current through it, is

constant but voltage may vary

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Current Source

IL

VL

+

+

E I

vL

IL

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Example 8.1. Find the Vs and I1 for the circuit of Fig 8.3

Soln: I1 = I = 10mAVs = V1 = I1R1 = 1010-320103 = 200 V

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Example: 8.2. Calculate V1, V2 and Vs for the circuit of Fig.

8.4.

Soln: V1 = I1 R1 = IR1 = 52 = 10 V (Ans)

V2 = I2 R2 = IR2 = 53 = 15 V (Ans)

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Example: 8.2. Continued

(Ans.)V251510

0

21

21

VVV

VVV

S

S

Applying KVL, we obtain

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Example: 8.3. Consider the series- parallel circuit of Fig. 8.5

Soln: Using current divider rule, we get

(Ans.)V422Law,sOhm'By

(Ans.)A421

26

(Ans.)A221

16

22333

23

23

23

32

RIRIV

RR

RII

RR

RII

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Example: 8.3. Continued

(Ans.)V1843640

0

directionCWin theKVLApplying

31

31

VIREV

VVEV

S

S

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Source Conversion

LS

SL

LS

S

LS

S

S

LS

S

S

L

LS

SS

LSLS

L

RR

RII

RRRI

RR

R

R

E

RR

E

R

R

I

RRERR

RRE

RREI

/)1(

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Source Conversion

LS

SL

LS

S

S

LS

SS

LSLS

L

RR

RII

RR

ERR

RR

ERR

RR

E

RR

EI

/)1(

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Example 8.4. Convert the voltage source of Fig. 8.9 to a

current source and calculate the current through the load for

each source.

(Ans.)A142

6

LS

LRR

EI

Soln:

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Example 8.4. Continued..

(Ans.)A142

23

LS

SL

RRRII

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Example 8.5. Convert the current source of Fig. 8.10 to a

voltage source and find the current through the load for each

source.

Soln:

(Ans.)mA3k6k3

k3109 3

LS

SL

RR

RII

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Example 8.5. Continued.

(Ans.)mA3k)63(

27

LS

LRR

EI

V27k3m9 SIRE

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Current Sources in Parallel

Two or more current sources can be connected in parallel

and they may be replaced (represented) by one current

source having the magnitude and direction of the resultant.

Sum if the currents enters a node and subtract if they haveopposite directions.

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Parallel Current Source

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Example 8.6. Reduce the left side Fig. 8.11 to a minimum

number of elements.

Soln.

(Ans.)A461012

III

Ohm263

63||

21

2121

RR

RRRRR

R

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Example 8.7. Reduce the network of Fig. 8.13 to a single

current source and calculate the current through RL.

Soln: Converting voltage source into a current source

A48

32

1

1

1

R

EI

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Here thevoltage

source is

converted

into a current

source

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A

A106421 IIIS

Ohm6248

248||

21

2121

RR

RRRRRS

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The reduced circuit can be represented as

(Ans.)A3146

6

10 LS

S

SL RR

R

II

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8.5 Current Sources in Series

Applying KCL at the node a, we get

I1= I2

So I1 and I2 must be same.

Current Sources of different ratings are not connected in series.

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8.6 Determinants

333

222

111

333

222

111

333

222

111

333

222

111

333

222

111

333

222

111

3333

2222

1111

cba

cba

cba

dba

dba

dba

z

cba

cba

cba

cda

cda

cda

y

cba

cba

cba

cbd

cbd

cbd

x

dzcybxa

dzcybxadzcybxa

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Determinant continued

1221

2112

22

11

22

11

22

11

22

11

222

111

baba

cbcbx

ba

ba

ca

ca

y

ba

ba

bc

bc

x

cybxa

cybxa

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8.7 Branch Current Method

Steps

Assign a distinct current of

arbitrary direction to each

branch of network

I1

I2

I3

Indicate the polarities for

each resistor as

determined by the

assumed current

direction.

Apply KVL around each

closed loop and KCL at

the node which includes

all the branch currents. Solve the linear

simultaneous equations.

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Branch Current Method Continueda

loop1

loop2

Applying KVL in loop 1, we get

1...........0

0

133211

33111

EIRIIR

IRIRE

3...........0321

321

III

III

Applying KVL in loop 2, we get

Applying KCL at the node a, we get

2...........0

0

233221

22332

EIRIRI

IRIRE

Solve equation 1, 2 and 3 to get I1,

I2 and I3.

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Example 8.15. Apply the branch-current method to the

network of Fig. 8.19

Soln.

Assign a distinct current of

arbitrary direction to each

branch of network

Indicate the polarities for

each resistor as

determined by the

assumed current

direction.

Apply KVL around each

closed loop and KCL at

the node which includes

all the branch currents. Solve the linear

simultaneous equations.

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Example 8.15 continued ..

)1.......(2402,

0,

0

321

133211

33111

IIIor

EIRIIRor

IRIRE

)2.......(6410,

0,

0

321

233221

22332

IIIor

EIRIRIor

IRIRELoop 1 Loop 2

Applying KVL around loop 1, we get

Applying KVL around loop 1, we get

Applying KCL at node a, we get

)3..(..........0-, 321

213

IIIor

III

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Example 8.15 continued ..

)3.(....................0

)2..(..........6410

)1..(..........2402

321

321

321

III

III

III

(Ans.)A114

14

14

2410

)10(40)41(2

)06(40)41(2

111

410

402

110

416

402

1

I

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A