4 Network Thoerem2

7/31/2019 4 Network Thoerem2

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Network Theorems


2/25

Circuit analysis

Mesh analysis

Nodal analysis

Superposition

Thevenins Theorem

Nortons Theorem Delta-star transformation


3/25

An active network having two terminals A and Bcan be replaced by a constant-voltage sourcehaving 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 loaddisconnected and the sources of e.m.f replacedby their internal resistances.


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Networks to illustrate Thevenin theorem

V

R2

R

R1

R3

A

B

R2

Rth

R1

R3

A

B

V

R2

Vth

R1

R3

A

B

Vth

R

Rth

A

B

(a)(b)

(c)(d)


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31

3RR

VIR

31

3

RR

VRV

th

31

3

3

RR

VRV

R

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

31

2 RR

RRRR

th

RR

VI

th

th

Thus current in R(refer network (d))

Refer to network (c) the resistance at AB


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R3=10

R1=2 R

2=3

E1=6V E

2=4V

C

D

A B

R1=2 R

2=3

E1=6V E

2=4V

C

D

A B

V

I1

ARR

I 4.032

246

31

1

VV 2.524.06

Calculate the current through R3

Solution

With R3 disconnected as in figure below

p.d across CD is E1-I1R1


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continue

R1=2 R

2=3

C

D

A B r

r=1.2

R3=10

C

D

V=5.2V

I

2.1

32

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


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

A

B

C

DE=2V

10

40

20 15

30I1 I

1

-I3

I3

I2

I2+I

3

311 30102III

3130402 II

Solution

(a) Kirchoffs law

Using K.V.L in mesh ABC + the voltage E

31323 3015_400 IIIII

3214020100 III

Similarly to mesh ABDA

For mesh BDCB

321 8515300III

..(a)

(b)

..(c)


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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.

321340601200 III

Substitute I1 in (a)


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continue

A

B

C

D

E=2V

10

2015

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


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continue

A

B

C

D

10

2015

30

r

16.07

0.643V10

57.8

1520

1520

07.1657.85.7r

For effective resistance,

5.7

3010

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


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E

RS

RL

IL

s

S

R

EI

S

Ls

s

R

RR

R

E

Ls

LI

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

RS

IS

RS


13/25

1A

5

15

5

Rs

Vo

VVo

15151

20155 s

R

Calculate the equivalent constant-voltage generator for thefollowing 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


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Node 1

5

4V

reference

node

V2

Node 2

6V15

V1I1

I2

I4

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

A

R

VI 5.0

12

6A

R

VI 8.0

5

4


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continueNode 1

reference

node

V2

Node 2

15

V1

I2

I4

I3

0.8A 0.5A12 5

I1 I

5

101558.0 2111

VVVV

12

1

10

1

8

1

10

5.02

1V

V

1012151260 21 VV

1010

1

15

1

5

1

8.02

1

V

V

21 332624 VV

101285.0 2122

VVVV

At node 1 At node 2

21371260 VV 21 31124 VV ..(a) (b)

X 30 X 120


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continue

65.3155.2324111 V

2727.338.86 V

AV

I 32.08

55.2

8

24

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)


17/25

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

10V 20V

2.0

8.0

8.04.0

10V 20V

8.04.0 AI 5.2

0.4

101

67.2

0.80.4

0.80.40.8//0.4

sR

AIIIs

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

i


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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


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Calculate the current in the 5.0 resistor in thefollowing circuit

10A 8.0

2.0

4.0

6.0

10A 8.0

2.0

4.0

6.0

Is

AIs

0.810

0.20.8

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


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continue

8.0

2.0

4.0

6.0

5.05.08.0A

0.5

0.40.60.80.2

0.40.60.80.2s

R

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


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A

C BR1

Ra

Rb

Rc

R3

R2

BC

A


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321

2131

RRR

RRRRRR

ba

321

21

RRR

RRR

c

baAB RRR

321

213

RRR

RRRR

AB

321

13

RRR

RRR

b

From delta cct , impedance sees from AB

Thus equating

Delta to star transformation

321

32

RRR

RRR

a

Similarly from BC

321

3221

RRR

RRRRRR

ca

321

3121

RRR

RRRR

RR cb

321

2132

RRR

RRRR

RR 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)


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1

3

R

R

R

R

c

a

b

a

R

RR

R1

2

1

2

R

R

R

R

b

a Dividing (e) by (f)

Similarly

Delta to star transformation

c

ba

ba

R

RRRRR

3

Similarly, dividing (e) by (g)

a

cb

cb

R

RRRRR

1

c

a

R

RRR 13

b

ac

ac

R

RRRRR 2

therefore

We have

(i)

(j)

(j)

Substitude R2 and R3 into (e)

(k)

(l)

(m)

(n)Similarly


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A

B

C D

R1

16

R3

6

R2

8

R4

12

R5

20

C

B

D

B'

R2

8

R4

12

R5

20

1

2

3

4

Rc

Ra Rb

Find the effective resistance at

terminal between A and B of thenetwork 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


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

R1+Ra 20 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

4 Network Thoerem2

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