Network Theorems

Network Theorems

Circuit analysis

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

An active network having two terminals A and B can be replaced by a constant-voltage source having an e.m.f Vth and an internal resistance Rth.

The value of Vth is equal to the open-circuited p.d between A and B.

The value of Rth is the resistance of the network measured between A and B with the load disconnected and the sources of e.m.f replaced by their internal resistances.

Networks to illustrate Thevenin theorem

VR 2

R

R 1

R 3

A

B

R 2

R th

R 1

R 3

A

B

VR 2

V th

R 1

R 3

A

B

V th

R

R th

A

B

(a)(b)

(c)(d)

313 RR

VIR

31

3

RR

VRVth

31

33 RR

VRVR

Since no current in R2, thus

Refer to network (b), in R2 there is not complete circuit, thus no current, thus current in R3

And p.d across R3 is

31

312 RR

RRRRth

RR

VI

th

th

Thus current in R (refer network (d))

Refer to network (c) the resistance at AB

R 3=10

R 1=2 R 2=3

E 1=6V E 2=4V

C

D

A B

R 1=2 R 2=3

E 1=6V E 2=4V

C

D

A BV

I1

ARR

I 4.032

246

311

VV 2.524.06

Calculate the current through R3

Solution

With R3 disconnected as in figure below

p.d across CD is E1-I1R1

continue

R 1=2 R 2=3

C

D

A B r

r=1.2 R 3=10

C

D

V=5.2V

I

2.132

32r

AI 46.0102.1

2.5

To determine the internal resistance we remove the e.m.f s

Replace the network with V=5.2V and r=1.2, then the at terminal CD, R3, thus the current

Determine the value and direction of the current in BD, using (a) Kirchoff’s law (b) Thevenin theorem

A

B

C

DE=2V

10

40

20 15

30 I1 I1-I 3

I3

I2I2+I 3

311 30102 III

31 30402 II

Solution

(a) Kirchoff’s lawUsing K.V.L in mesh ABC + the voltage E

31323 3015_400 IIIII

321 4020100 III Similarly to mesh ABDA

For mesh BDCB

321 8515300 III

…..(a)

……(b)

…..(c)

31 460900 II 31 111.5 II

Continue……

Multiplying (b) by 3 and (c) by 4and adding the two expressions, thus

321 12060300 III

mAAI 5.110115.03

Since the I3 is positive then the direction in the figure is correct.

321 340601200 III

Substitute I1 in (a)

continue

A

B

C

DE=2V

10

20 15

30 By Thevenin Theorem

VVAD 143.11520

202

VVBD 643.05.0143.1

VVAB 5.03010

102

P.D between A and B (voltage divider)

P.D between A and D (voltage divider)

P.D between B and D

continue

A

B

C

D

10

20 15

30

r

16.07

0.643V 10

57.81520

1520

07.1657.85.7r

For effective resistance,

5.73010

3010

AI 0115.01007.16

643.03

Substitute the voltage, resistance r and 10W as in figure below

DtoBfrom5.11 mA

10 parallel to 30

20 parallel to 15

Total

E

R S

R L

IL

sS R

EI

SLs

s

RRR

RE

LsL I

RR

R

RR

EI

s

Ls

s

Another of expressing the current IL

Where IS=E/RS is the current would flow in a short circuit across the source terminal( i.e when RL is replaced by short circuit)Then we can represent the voltage source as equivalent current source

E

R S

ISR S

1A

5

15

5

R s

V o

VVo 15151

20155 sR

Calculate the equivalent constant-voltage generator for the following constant current source

Vo

Current flowing in 15 is 1 A, therefore

Current source is opened thus the 5 W and 15 W are in series, therefore

Node 1

5

4V

referencenode

V 2

Node 2

6V15

V 1I1

I2I4

I5

I3

4V

5

50.8A

6V

0.5A

Analysis of circuit using constant current source

From circuit above we change all the voltage sources to current sources

AR

VI 5.0

12

6A

R

VI 8.0

5

4

continueNode 1

referencenode

V2

Node 2

15

V1

I2I4

I3

0.8A 0.5A12 5

I1 I5

101558.0 2111 VVVV

12

1

10

1

8

1

105.0 2

1 VV

1012151260 21 VV

1010

1

15

1

5

18.0 2

1

VV

21 332624 VV

101285.0 2122 VVVV

At node 1 At node 2

21 371260 VV 21 31124 VV …..(a) ……(b)

X 30 X 120

continue

65.3155.232411 1 V

2727.338.86 V

AV

I 32.08

55.2

82

4

21 273.3128.26 VV 11

12)( a

VV 55.22

………( c )

(c) + (b)

Hence the current in the 8 is

So the answers are same as before

VV 88.211

65.311

From (a)

Calculate the potential difference across the 2.0 resistor in the following circuit

10V 20V

2.0

8.0

8.0 4.0

10V 20V

8.0 4.0 AI 5.20.4

101

67.20.80.4

0.80.40.8//0.4sR

AIII s 55.25.221

20.820 I

10.410 I

………( c )

I2

First short-circuiting the branch containing 2.0 resistor

AI 5.20.8

202

I1 Is

continue

AI 06.151067.2

67.2

VV 1.20.206.1

Redraw for equivalent current constant circuit

Hence the voltage different in 8 isUsing current division method

5A

8.0

2.0

Is

I

V

Calculate the current in the 5.0 resistor in the following circuit

10A 8.0

2.0

4.0

6.0

10A 8.0

2.0

4.0

6.0

Is

AI s 0.8100.20.8

0.8

Short-circuiting the branch that containing the 5.0 resistor

Since the circuit is short-circuited across the 6.0 and 4.0 so they have not introduced any impedance. Thus using current divider method

continue

8.0

2.0

4.0

6.0

5.0 5.0 8.0A

0.50.40.60.80.2

0.40.60.80.2sR

AI 0.40.80.50.5

0.5

The equivalent resistance is a parallel (2.0+8.0)//(6.0+4.0)

Hence the current in the 5 is

Redraw the equivalent constant current circuit with the load 5.0

I

A

C BR 1

R a

R bR c

R 3R 2

BC

A

321

2131

RRR

RRRRRR ba

321

21

RRR

RRRc

baAB RRR

321

213

RRR

RRRRAB

321

13

RRR

RRRb

From delta cct , impedance sees from AB

Thus equating

Delta to star transformation

321

32

RRR

RRRa

Similarly from BC

321

3221

RRR

RRRRRR ca

321

3121

RRR

RRRRRR cb

321

2132

RRR

RRRRRR ca

From star cct , impedance sees from AB

and from AC

(a)

(b)

(c)

(b) – (c) (d)

By adding (a) and (d) ; (b) and (d) ;and (c) and (d) and then divided by two yield

(e) (f) (g)

1

3

R

R

R

R

c

a b

a

R

RRR 1

2 1

2

R

R

R

R

b

a Dividing (e) by (f)

Similarly

Delta to star transformation

c

baba R

RRRRR 3

Similarly, dividing (e) by (g)

a

cbcb R

RRRRR 1

c

a

R

RRR 1

3

b

acac R

RRRRR 2

therefore

We have

(i)

(j)

(j)

Substitude R2 and R3 into (e)

(k)

(l)

(m)

(n)Similarly

A

B

C D

R116

R3

6

R28

R412

R5

20

C

B

D

B '

R28

R412

R5

201

2

3

4

Rc

Ra Rb

Find the effective resistance at terminal between A and B of the network on the right side

Solution

R = R2 + R4 + R5 = 40 Ra = R2 x R5/R = 4 Rb = R4 x R5/R = 6 Rc = R2 x R4/R = 2.4

Substitute R2, R5 and R4 with Ra, Rb dan Rc:

R1+Ra20 R3+Rb12

A

B

R3 6R116

Rc 2.4

Ra

4

Rb

6

A

B

Rc 2.4

RAB = [(20x12)/(20+12)] + 2.4 = 9.9