Network Theorems. Mesh analysis Nodal analysis Superposition Thevenin’s Theorem Norton’s Theorem Delta-star transformation

Network Theorems

Mesh analysis Nodal analysis Superposition Thevenin’s Theorem Norton’s Theorem Delta-star transformation

Direct application in conjunction with Ohm’s law

Indirect application in conjunction with resistance

Simultaneous equations

VIRV 2765.422

162483 333443 IRIVRIV

AI 5.116

243

Since R3 and R4 are in parallel

Determine current and source e.m.f

IV 1

V 2

V 3

E

R 18

R 316

R 26

R 48

I4=3A

Therefore

By Kirchoff’s first law

VVVVE 87242736321

AIII 5.435.143 VIRV 3685.411 Also

By Kirchoff’s second law

VRIV 45153212

AR

VI 3

9

27

1

11

Determine I1, E, I3 and I

By Kirchoff’s second law

VVVVE 72452721

AIII 129331

AR

VI 9

8

72

33 Also

By Kirchoff’s first law

E

R 2

II3

V 3

V 2

V 1

R 3

R 1

I1

V

AII 312132

VV 1010110

Power dissipated in R3 is 20W. Calculate I3, R1,I1, I2 and E

By Kirchoff’s first law in node b

VVE 27225

112

22

1

11 I

VR

32

32

3 220 RRI By Kirchoff’s first law in node a

E

1AI3

I25A

V 1 R 3R 1

I1

31020 IWP

AI 210

203

By Kirchoff’s second law in loop 2

VV 2261061

AII 2355 21 53R

P.D across 1 is 5 X 1=5V

1 2 3

a b

cd

IV 1

V 2

V 3

E=87V

R 18

R 316

R 26

R 48

AR

VI

t

5.433.19

87

33.1933.56821 et RRRR

33.5816

816

43

43

RR

RRRe

First find the total effective resistance

AIRR

RI 35.4

816

16

43

34

Determine current I and I4

Then

Using current divisionV R t

I

VVAC 61266

6

81616

1616bR

61510

1510aR

Effective resistance for parallel resistor 10 // 15 and 16//16

VVBC 81248

8

Determine VAB

Then

Using voltage division

12V

10

A B

C

16 16

4

6

15

V AB

VVVV BCACAB 286

12VA B

C

8

4

6

6

V AB

V ACV BC

21 83128 III

121 38280 IIII

)(14340 211 III

Applying Kirchoff’s 2nd law for loop 1

21 831140 II

Calculate the current in each resistor

21 141740 II

3

I=5A

40V

A

B

C

D

12

3

3 I1

I

40V

A

B

C

D

12

3I1-I 2

I2

I1

I-I 1

I-I 1+I 2


AI 5But

Thus

---(b)

---(b)

---(a)

continue

AII 22421

21 56217980 II

12851140 I

Solving the 2 simultaneous equations

AIII 324521

Then (c) + (d)

21 5668160 II (a) X 4

(b) X 7

---(c)

---(d)

AII 1451

28124140 I AI 22 Substitute in (b)

AI 41

In 14 resistor

In 4 resistor

In 28 resistor

In 3 resistor

In 8 resistor

21 201820 II 212 18220 III

)(18110 211 III

21 180162180 II

Calculate the current in the network

21 181910 II

AI 85.21

(a)x10

(b)X 9

---(b)

---(d)

---(a)

AI 72.085.257.33

21 180190100 II


21834.5410 I


---(c)

213 III 20V10V

1

I1

I3

I2

18

1 2

(d)-(c) we get

Substitute I1 in(a)

Current in 18 resistor

AI 57.32

20V10V

1

I1

I3

I2

18

1 2

11020 1 I

31820 I

AI 1.111.1)10(2

Calculate the current in the network

AI 1.118

203


AI 101

Applying Kirchoff’s 2nd law for outside loop



AI 05.090

5.4

0905.4 I

The network shown is a 3 cells having an internal resistance of 30 . Calculate the current in the network

0305.1305.1305.1 III

Applying Kirchoff’s 2nd law

The voltage drop due to internal resistor is 0.05 x30=1.5V

Thus there is no potential different between two terminals

E=1.5V

E=1.5V

E=1.5V

R=30

R=30

R=30

A

BC

Create loop’s current rather than branch current

Use Kirchoff’s second (voltage ) law Ohm’s law to calculate p.d Branch is calculated by taking the

algebraic sum of the loop currents

50 60 10

40 30

20

20V50V100V

50 60 10

40 30

20

20V50V100V

I3

I1I2

3

21

Calculate the current in each branch

First create loop current ,i.e I1 , I2, I3 as shown

continue

321 401005070 III

40501040502050 312 III

305050306020100 321 III

321 9040300 III

321 305014080 III

---(b)

---(a)

40304020300 213 III

In loop 1

In loop 2

---(c)

In loop 350 60 10

40 30

20

20V50V100V

I3

I1I2

3

21

continue

AI 50.13 AI 65.11 AI 16.22 Solving these equations

AII 15.031

AII 51.012

AI 50.13

AI 16.22

AI 65.11 Current in 60

AII 66.032

In direction of I1

Current in 30 In direction of I1

Current in 50

Current in 40

Current in 10

Current in 20

In direction of I2

In direction of I2

In direction of I2

In direction of I3

Choose reference node where all nodes can refer

Assign currents going to/out the nodes Assign voltage at nodes as V1 , V2,

V3….which refer to reference node Apply Kirchoff’s current law at each node Relate the voltage , resistance andcurrent

using ohm’s law Solve the equations obtained

1A 5

node 1 node 2

reference

V 1 V 2

Simplified

321 III

351 211 VVV

73221 VVV

43 II

At node 1

133

1

5

1 21

VV ….(a)

Calculate V1 and V2

At node 2

5

node 1 node 2

reference

V 1 V 2

I1

I4

I2I3

1A

continue

07

1

3

1

3 21

V

V

VV3

101

Simplified

0377 21 VV

21 107 VV

VV3

7

3

10

10

72

1510

358 1

1 V

V

12 10

7VV

1558 21 VV

…..(b)

Solve for equations (a) and (b)

(b) X 21

(a) X 15

Substitute V2 we have

5

node 1 node 2

reference

V 1 V 2

I1

I4

I2I3

1A

Simplified

321 III

10155

4 2111 VVVV

24311 21 VV

At node 1

2121 332624 VVVV

….(a)

Calculate V1 and V2 and current in 8

Node 1

5

4V

referencenode

V 2

Node 2

6V15

V 1

Node 1

5

4V

referencenode

V 2

Node 2

6V15

V 1I1

I2I4

I5

I3

continueNode 1

5

4V

referencenode

V 2

Node 2

6V15

V 1I1

I2I4

I5

I3

Simplified

543 III

12

6

8102221

VVVV

603712 21 VV

VV 55.22

At node 2

6010151212 2221 VVVV

….(b)

Solving the simultaneous equations (a) and (b)

VV 88.21 AV

I 32.08

28

The superposition states that in any network containing more than one source , the current in , or the p.d. across in any branch can be found by considering each source separately and adding their effects: omitted sources of e.m.f are replaced by resistance equal to their internal resistances.

Separating the network into several circuit contenting only one source

I1

I1c

I1c+ I2c

I1b

I1+ I2

I2

I2c

I1b+ I2b

I2b

1

11

18

18 18

10V

10V

20V

20V

Original network

Separating into 2 networks

Network 1Total resistance

AR

VI

totalb 57.3

8.2

101

8.2182

1821

AII bb 36.021.357.321

AI b 21.357.3182

182

Also

Thus

and

I1b

I1b+ I2b

I2b

1 18

10V

Network 2

Total resistance

AR

VI

totalc 78.6

95.2

202

95.2181

1812

AII cc 36.042.678.612

AI c 42.678.6181

181

Also

Thus

and

I1c

I1c+ I2c

I2c

1 18

20V

combination

AIII cb 85.242.657.3111

AII 72.057.385.221

AIII cb 57.378.621.3222

Also

Thus

and

I1

I1+ I2

I2

1 18

10V 20V

Documents

Network Theorems. Mesh analysis Nodal analysis Superposition Thevenin’s Theorem Norton’s Theorem Delta-star transformation